b. gain of the tube? Q5. Identify the type of electron tube(s) that would be most suitable for the following applications.
a. Power amplifier b. Voltage amplifier with small signal inputs c. Low distortion amplifiers for use with large signal inputs
In the earlier discussion of conventional-electron tubes, you learned some of the limitations of tubes. One of these limitations was that the conventional tube was not able to operate (amplify) at extremely high frequencies such as those used in radar equipment. Even at frequencies lower than those used in radar equipment, problems occur. For example, at ultrahigh frequencies (300 MHz to 3000 MHz), transit time effects make the operation of a conventional-electron tube impossible. For this reason, the special ultrahigh frequency tubes were developed to operate within this frequency range.
Before we discuss the way in which special uhf tubes counter the effects of transit time, you should understand the manner in which transit time affects conventional tubes.
We will explain the limitation of transit time by using figure 2-9. In view A, the positive-going alternation of a uhf ac signal is applied to the grid of a conventional-triode tube. The first positive-going alternation reduces the negative bias on the grid, and electrons start to move toward the grid. Since the input is an ultrahigh frequency signal, the majority of the electrons cannot pass the grid before the input signal progresses to the negative alternation. The electrons that have not yet passed the grid are either stopped or repelled back toward the cathode. This is shown in view B. Before these electrons can move very far, the second positive alteration reaches the grid, and causes even more electrons to move from the cathode (view C). At the same time, the electrons that were repelled from the grid toward the cathode by the first negative alternation feel the effect of the positive-going grid. These electrons reverse direction and again move toward the grid. Because these electrons had to first reverse direction, they are now moving slower than the electrons that are attracted from the cathode by the second positive alteration. The result is that the electrons from the cathode catch up to the slower moving electrons and the two groups combine (view C). This action is called BUNCHING.
2-10
In effect, the area between the grid and cathode becomes highly negatively charged, as shown in view D. This negative charge is surrounded by an electrostatic field. The electrostatic field cuts the grid and repels electrons that are present in the grid. As electrons are forced from the grid, the grid tries to go positive. Unfortunately, this tendency toward a positive charge attracts electrons from the mass or bunched charge. Thus, as an electron is forced from the grid; it is replaced by another from the massed charge. Electrons forced from the grid represent grid current (Ig), as shown in view E. The grid current flows from the grid through Rg, to the cathode, from the cathode, to the massed charged, and back again to the grid, The movement of current in this manner is, in effect, a path for current flow from the cathode to the grid. Because current flows between the cathode and grid, the resistance (rgk) between these elements is lowered to the point of a short circuit. The grid, in effect, is short circuited to the cathode and ceases to function; and this, in turn, lowers tube efficiency dramatically. This is shown in view F of figure 2-9.
Transit time may be decreased by reducing the spacing between electrodes or by increasing the electrode voltages, which in turn increases electron velocity through the tube.
The problem with the last solution is that the tube does not present an infinite resistance to current flow. If the operating voltage is raised to an operating potential that is too high, arcing (arc over) occurs between the cathode and the plate and, most likely, will destroy the tube. For this reason, the effects of transit time are reduced in uhf tubes by placing the tube elements very close together.
2-11
Uhf tubes have very small electrodes placed close together and often are manufactured without socket bases. By reducing all the physical dimensions of the tube by the same scale, the interelectrode- capacitance and transit time effects are reduced, without reducing the amplification capability of the tube. A disadvantage to this type of tube construction is that the power-handling capability of these tubes is also reduced due to the close placement of the tube elements.
Uhf tubes are placed in three broad categories based on their shape and/or construction; ACORN, DOORKNOB, and PLANAR tubes.
ACORN TUBES, as shown in figure 2-10, are available for use as diodes, triodes, or pentodes. Acorns are very small tubes that have closely spaced electrodes and no bases. The tubes are connected to their circuits by short wire pins that are sealed in the glass or ceramic envelope. Because of their small size, acorn tubes are usually used in low-power uhf circuits.
The DOORKNOB TUBE is an enlarged version of the acorn tube. Because of its larger physical size, it can be operated at higher power than the acorn tube.
PLANAR TUBES are so named because of their construction. The ordinary (conventional) tube you studied earlier uses concentric construction. This means that each element (cathode, grid, and plate) is cylindrical in shape. The grid is placed over the cathode, and the plate, which is the largest cylinder, is
2-12
placed over the grid. The result is a tube composed of concentric cylinders like the one shown in figure 2-11. Thus, the name concentric tubes.
At ultrahigh frequencies, the problems of producing small tube elements while reducing the spacing between elements become very difficult. Not only are the elements hard to keep parallel with each other during the manufacturing process, but they also have a tendency to warp and sag under normal operating conditions. Since these elements are already as close together as possible, any reduction in element spacing can cause arcing. Therefore, a new type of tube was developed to prevent arcing or element sagging in conventional tubes. This tube is known as the planar tube.
Planar tubes are electron tubes in which the cathode, plate, and grids are mounted parallel to each other. Their physical construction greatly resembles a schematic diagram of a normal tube, as shown in figure 2-12.
2-13
A typical planar tube is depicted in figure 2-13. Notice that the tube elements are mounted close to each other and are parallel to one another. The oxide coating of the cathode is applied to the top surface only. Therefore, the emitting surface of the cathode is parallel to the plate and the grid.
The plate of the tube consists of a cylindrical stud. This stud-plate construction has two purposes. Its flat lower surface serves as a parallel plate, and its external upper end serves as the external-plate connection from the tube to the circuit.
Because of its construction, the planar tube cannot use the ladder-type grid, with which you are familiar. Instead, the grid, formed into a circle, is composed of a wire mesh similar to that of a common screen door.
The cathode structure is manufactured in two parts. Point A of figure 2-13 is the metallic shell of the tube and is used to couple (or connect) unwanted radio frequency signals from the cathode to ground. This connection is not, however, a direct coupling. The wafer at point C of figure 2-13 is composed of mica, which serves as a dielectric. The lower extension of the cathode serves as one plate of the capacitor, while the other plate is formed from the flattened upper portion of the cathode connector ring. The cathode has a direct connection to the tube pin through the connector labeled point B. You might think that this is a rather complicated method to connect the cathode to a circuit, but it serves a purpose. At high frequencies, the wiring of a circuit can pick up radio frequency signals and retransmit them. If the wiring involved happens to be the wiring used to supply dc voltages to the circuit, all the tubes in the circuit will receive the signal. The result will be massive distortion throughout the circuit.
The problem can be eliminated by isolating the dc and radio frequency circuits from each other. In planar tubes, this is fairly simple. The point A ring is grounded. Any rf signals that the cathode may pick up through tube conduction are grounded or shorted to ground through the capacitive coupling with the point A shell. In other words, the point A shell (capacitive ground) serves the same function as the bypass
2-14
capacitor in a cathode-biased circuit. Because the capacitor will not pass dc, bias voltages can be applied to the cathode through the tube pins.
Notice the external shape of the planar tube in figure 2-13. The tube is composed of five sections, or cylinders. As you go from the top to the bottom, each cylinder increases in diameter. Because of this piled cylinder construction, the tube resembles a lighthouse, and is therefore known as a LIGHTHOUSE TUBE.
Another type of planar tube is shown in figure 2-14. This type of tube, because of its external appearance, is called an OILCAN TUBE. The major difference between it and the lighthouse tube is the addition of cooling fins to allow it to handle more power than the lighthouse tube. Because of their planar construction, both types of tubes are capable of handling large amounts of power at uhf frequencies.
Q6. What effect does transit time have on a conventional triode operated at uhf frequencies? Q7. How do uhf tubes counter the effects of transit time? Q8. Why can acorn and doorknob tubes NOT handle large amounts of power? Q9. What type of uhf tube was developed to handle large amounts of power?
You know that great effort is made to produce a perfect vacuum within electron tubes. But, even the best vacuum pumps and getters cannot remove all of the air molecules. However, the chances of an electron hitting a molecule in a near-vacuum are very slim because of the great distance between the molecules, compared to the size of the electron. An electron can pass between two molecules of air inside the tube as easily as a pea could pass through a circle with a diameter equal to that of the earth!
In some tubes, the air is removed and replaced with an inert gas at a reduced pressure. The gases used include mercury vapor, neon, argon, and nitrogen. Gas-filled tubes, as they are called, have certain
2-15
electrical characteristics that are advantageous in some circuits. They are capable of carrying much more current than high-vacuum tubes, and they tend to maintain a constant IR drop across their terminals within a limited range of currents. The principle of operation of the gas-filled tube involves the process called ionization.
An operating gas-filled tube has molecules, ions, and free electrons present within the envelope. In a gas-filled diode, the electron stream from the hot cathode encounters gas molecules on its way to the plate. When an electron collides with a gas molecule, the energy transmitted by the collision may cause the molecule to release an electron. This second electron then may join the original stream of electrons and is capable of freeing other electrons. This process, which is cumulative, is a form of ionization. The free electrons, greatly increased in quantity by ionization, continue to the plate of the diode. The molecule which has lost an electron is called an ion and bears a positive charge. The positive ions drift toward the negative cathode and during their journey attract additional electrons from the cathode.
The velocity of the electrons traveling toward the plate varies directly with the plate voltage. If the plate voltage is very low, the gas-filled diode acts almost like an ordinary diode except that the electron stream is slowed to a certain extent by the gas molecules. These slower-moving electrons do not have enough energy to cause ionization when they hit the gas atoms. After the plate voltage is raised to the proper level of conduction, the electrons have enough energy to cause ionization when they hit the gas molecules. The plate potential at which ionization occurs is known as the IONIZATION POINT, or FIRING POTENTIAL, of a gas tube. If the plate voltage is reduced after ionization, it can be allowed to go several volts below the firing potential before ionization (and hence, high-plate current) win cease. The value of the plate voltage (Ep) at which ionization stops is called the DEIONIZATION POTENTIAL, or EXTINCTION POTENTIAL. The firing point is always at a higher plate potential than the deionization point.
The point at which the gas ionizes can be controlled more accurately by inserting a grid into the gas diode. A negative voltage on the grid can prevent electrons from going to the plate, even when the plate voltage is above the normal firing point. If the negative-grid voltage is reduced to a point where a few electrons are allowed through the grid, ionization takes place. The grid immediately loses control, because the positive ions gather about the grid wires and neutralize the grid’s negative charge. The gas triode then acts as a diode. If the grid is made much more negative in an effort to control the plate current, the only effect is that more ions collect about the grid wires—tube continues to conduct as a diode. Only by removing the plate potential or reducing it to the point where the electrons do not have enough energy to produce ionization will tube conduction and the production of positive ions stop. Only after the production of positive ions is stopped will the grid be able to regain control.
Such gas-filled triodes are known as THYRATRONS. Thyratrons are used in circuits where current flow in the thyratrons output circuit is possible only when a certain amount of voltage is present on the thyratrons grid. The flow of plate current persists even after the initiating grid voltage is no longer present at the grid, and it can be stopped only by removing or lowering the plate potential. The symbols for the gas-filled diode, the voltage regulator, and the thyratron are the same as those for high-vacuum tubes except that a dot is placed within the envelope circle to signify the presence of gas. Some examples of gas-filled tube schematic symbols are shown in figure 2-15.
2-16
Before leaving this section, you should be aware of one precaution associated with mercury-vapor tubes. The mercury vapor is not placed in the tube as a vapor; instead a small amount of liquid mercury is placed in the tube before it is sealed. When the liquid mercury comes in contact with the hot filament, the mercury vaporizes.
To ensure that the mercury has vaporized sufficiently, the filament voltage must be applied to mercury-vapor tubes for at least 30 seconds before the plate voltage is applied. If vaporization is incomplete, only partial ionization is possible. Under these conditions, the application of plate voltage results in a relatively high voltage drop across the tube (remember E = I × R), and the positive ions present are accelerated to a high velocity in the direction of the cathode. As the ions strike the cathode, they tear away particles of the emitting surface, usually causing permanent damage to the cathode and the tube. When the mercury is completely vaporized, the action of the gas is such that the voltage drop across the tube can never rise above the ionization potential (about 15 volts). At this low potential, positive-ion bombardment of the cathode does not result in damage to the emitting surface.
Generally, when gas-filled tubes are in the state of ionization, they are illuminated internally by a soft, blue glow. This glow is brightest in the space between the electrodes and of lesser intensity throughout the remainder of the tube envelope. This glow is normal and must not be confused with the glow present in high-vacuum tubes when gases are present. A high-vacuum tube with a bluish glow is gassy and should be replaced. The ionization of these gases will distort the output of the tube and may cause the tube to operate with much higher plate current than it can carry safely.
The cold-cathode, gas-filled tube differs from the other types of gas-filled tubes in that it lacks filaments. Thus, its name "COLD-CATHODE TUBE." In the tubes covered in this text thus far, thermionic emission was used to send electrons from the cathode to the plate. This conduction of electrons can be caused in another manner. If the potential between the plate and the cathode is raised to the point where tube resistance is overcome, current will flow from the cathode whether it is heated or
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not. In most applications in electronics, this method is not used because it is not as efficient as thermionic emission. There are two applications where cold-cathode emission is used. The first application you are already familiar with, although you may not be aware of it. Every time you look at a neon sign you are watching a cold-cathode tube in operation. Thus, the first application of cold-cathode tubes is for visual display. You are also familiar with the reason for this visual display. In the NEETS module on matter and energy, we explained that when energy is fed into an atom (neon in this case), electrons are moved, or promoted, to higher orbits. When they fall back, they release the energy that originally lifted them to their higher orbits. The energy is in the form of light. Cold-cathode tubes are also used as VOLTAGE REGULATORS. Because voltage regulators will be dealt with extensively in the next chapter, we will not cover their operation now. At this point, you only need to understand that a cold-cathode tube has the ability to maintain a constant voltage drop across the tube despite changes of current flow through the tube. The tube does this by changing resistance as current flow varies.
Examine figure 2-16. Here you see a cold-cathode tube connected to a variable voltage source. The variable resistor rkp does not exist as a physical component, but is used to represent the resistance between the cathode and the plate. Most cold-cathode tubes have a firing point (ionizing voltage) at about 115 volts. Thus, the tube in view A of the figure is below the firing point. Because the tube lacks thermionic emission capabilities, no current will flow and the tube will have a resistance (rkp) near infinity. The potential difference between the plate and ground under these conditions will be equal to the source (Ebb) voltage, as shown on the voltmeter.
In view B, the source voltage has been raised to the firing point of 115 volts. This causes the gas to ionize and 5 milliamperes of current will flow through the tube. Because the tube represents a resistance (rkp), voltage will be dropped across the tube; in this case, 105 volts. The plate-load resistor (RL) will drop the remaining 10 volts. The resistance of the tube at this time will be equal to:
2-18
In view C, the source voltage has been raised to 200 volts. This will cause more gas in the tube to ionize and 40 milliamperes of current to flow through the tube. The increased ionization will lower the resistance of the tube (rkp). Thus, the tube will still drop 105 volts. The tube’s resistance (rkp) at this time will be equal to:
As you can see, increasing the current flow will cause more ionization in the tube and a corresponding decrease in the tube’s resistance. Because of this, the tube will always have a constant voltage drop between its plate and cathode throughout its operating range.
Q10. What are two advantages that gas-filled tubes have over conventional electron tubes? Q11. Once ionization has occurred in a thyratron, what control does the control grid have over the
tube’s operation?
Q12. What precautions should be exercised when using mercury-vapor thyratrons? Q13. Cold-cathode tubes can be used as voltage regulators because of what characteristic?
in all probability, the one tube with which you are most familiar. Before you started your study of electronics, you probably referred to cathode-ray tubes as picture tubes. The cathode-ray tube (CRT) and the picture tube of a television set are one and the same.
2-19
Cathode-ray tubes are used in more applications than just television. They can be considered as the heart of the many types of information.
Cathode-ray tubes have one function that cannot be duplicated by any other tube or transistor; namely, they have the ability to convert electronic signals to visual displays, such as pictures, radar sweeps, or electronic wave forms.
All CRT’s have three main elements: an electron gun, a deflection system, and a screen. The electron gun provides an electron beam, which is a highly concentrated stream of electrons. The deflection system positions the electron beam on the screen, and the screen displays a small spot of light at the point where the electron beam strikes it.
The ELECTRON GUN is roughly equivalent to the cathodes of conventional tubes. The cathode of the electron gun in the CRT is required not only to emit electrons, but also to concentrate emitted electrons into a tight beam. In the electron tubes that you have studied, the cathode was cylindrical and emitted electrons in all directions along its entire length. This type of cathode is not suitable for producing a highly concentrated electron-beam. The cathode of the CRT consists of a small diameter nickel cap. The closed end of the cap is coated with emitting material. This is shown in figure 2-18. Because of this type of construction, electrons can only be emitted in one direction. Notice that the emitted electrons shown in figure 2-18 are leaving the cathode at different angles. If these electrons were allowed to strike the screen, the whole screen would glow. Since the object of the electron gun is to concentrate the electrons into a tight beam, a special grid must be used. This special grid is in the form of a solid metal cap with a small hole in the center. The grid is placed over the emitting surface of the cathode and charged negatively in relation to the cathode. The dotted lines represent the direction of cathode emitted electron repulsion, as shown in figure 2-19. Since all emitted electrons leave the cathode (point C), their paths can be identified. An electron attempting to travel from point C to point B (downward) will instead follow the path from point C to point E to point P. Consider an electron leaving from C in the direction of point A (upward). Its path will be curved from point C to point P by electrostatic repulsion. These curving electron paths are due to the negative potential of the grid coupled with the high positive potential of the anode. The potential of the anode attracts electrons out of the cathode-grid area past point P toward the screen. The grid potential may be varied to control the number of electrons allowed to go through the control-grid opening. Since the brightness or intensity of the display depends on the number of electrons that strike the screen, the control grid is used to control the brightness of the CRT.
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The proper name, BRIGHTNESS CONTROL, is given to the potentiometer used to vary the potential applied to the control grid. The control grid actually serves as an electron lens. It is this electronic lens that you adjust when you turn up the brightness control on your TV set. Notice that the effect of the grid is to focus the electron beam at point P in figure 2-19.
After passing point P, the electrons start to spread out, or diverge, again. Therefore, it becomes necessary to provide some additional focusing to force the electrons into a tight beam again. This is done by two additional positively-charged electrodes as shown in figure 2-20. The first electrode is commonly called the FOCUSING ANODE. Generally, the focusing anode is charged a few hundred volts positive with respect to the cathode. Electrons emitted by the cathode are attracted to the focusing anode. This is the reason that they travel through the small hole in the grid. The second electrode, called the ACCELERATING ANODE, is charged several thousand volts positive in relation to the cathode. Any electrons approaching the focusing anode will feel the larger electrostatic pull of the accelerating anode and will be bent through the opening in the focusing anode and will travel into the area labeled D. You might think that once an electron is in this region, it is simply attracted to the accelerating anode and that is the end of it. This does not happen. Because the accelerating anode is cylindrical in shape, the electrostatic field radiating from it is equal in all directions. Thus, an electron is pulled in all directions at once, forcing the electron to travel down the center of the tube. Then, the electron is accelerated into the accelerating anode. Once it passes the mid-point (point E), it feels the electrostatic attraction from the front wall of the accelerating anode, which causes it to move faster toward the front. Once the electron reaches point F, equal electrostatic attraction on either side of the opening squeezes it through the small opening in the front of the anode. From there, it is joined by millions of other electrons and travels in a tight beam until it strikes the screen (point S).
2-21
The inside of the large end of a CRT is coated with a fluorescent material that gives off light when struck by electrons. This coating is necessary because the electron beam itself is invisible. The material used to convert the electrons’ energy into visible light is a PHOSPHOR. Many different types of phosphor materials are used to provide different colored displays and displays that have different lengths of PERSISTENCE (duration of display).
In one way, the CRT screen is similar to a tetrode vacuum tube. Both suffer from the effects of secondary emission. In order to reach the screen, electrons from the cathode are accelerated to relatively high velocities. When these electrons strike the screen, they dislodge other electrons from the material of the screen. If these secondary emission electrons are allowed to accumulate, they will form a negatively-charged barrier between the screen and the electron beam, causing a distorted image on the CRT screen. The method used to control secondary emission, which you are already familiar with, i.e., a suppressor grid, is not practical in CRT’s. Instead, a special coating called an AQUADAG COATING is applied to the inside of the tube as shown in figure 2-21. This coating is composed of a conductive material, such as graphite, and has the same high-positive potential applied to it that is applied to the accelerating anode. This allows the aquadag to perform two functions. First, since the aquadag coating is positive, it attracts the secondary emitted electrons and removes them. Second, because the aquadag is operated at a high-positive potential and is mounted in front of the accelerating anode, it aids in the acceleration of electrons toward the screen.
Before going on, let’s review what you have already learned about CRT operation.
1. Electrons are emitted from a specially constructed cathode and move toward the front of the CRT.
2. The number of electrons that leave the area of the cathode is determined by the cap-shaped grid. In addition, the grid concentrates the emitted electrons into a beam.
3. The electron beam is focused and accelerated toward the screen by two electrodes: the focusing anode and the acceleration anode.
4. The electron beam strikes the screen and causes a bright spot to appear at the point of impact.
5. Any electrons released by secondary emission are removed from the tube by the aquadag coating.
2-22
At this point, you have a bright spot in the center of the CRT screen as shown in figure 2-22. Having watched TV, you know that a TV picture consists of more than just a bright spot in the center of the picture tube. Obviously, something is necessary to produce the picture. That something is called DEFLECTION. For the CRT to work properly, the spot must be moved to various positions on the screen. In your TV set for example, the spot is moved horizontally across the CRT face to form a series of tightly packed lines. As each line is displayed, or traced, the electron beam is moved vertically to trace the next line as shown in figure 2-23. This process starts at the top of the tube and ends when the last line is traced at the bottom of the CRT screen. Because the beam is swept very quickly across the CRT and the phosphor continues to glow for a short time after the beam has moved on, you do not see a series of lines, but a continuous picture.
2-23
These same principles also apply to the CRT used in your use of your major tool: the OSCILLOSCOPE. Remember, the unique function of a CRT is to convert electronic (and electrical) signals to a visual display. This function of a CRT is used in the oscilloscope to show the waveform of an electronic signal. To help you understand better how an oscilloscope works, we will discuss the type of deflection used in oscilloscopes. Bear in mind that the following discussion is only about deflection; we will cover the actual operation of an oscilloscope in a later NEETS module that deals specifically with test equipment.
As you should know, there are two ways to move an electron (and thus an electron beam): either with a magnetic or with an electrostatic field. Because of this, there are three possible ways to move or deflect an electron beam in a CRT: magnetically, electromagnetically, and electrostatically. All three ways are used in electronics. In general, though, electrostatic and electromagnetic deflection are used most often. Your TV set, for example, uses electromagnetic deflection, while much of the test equipment in the Navy uses electrostatic deflection.
ELECTROSTATIC DEFLECTION uses principles you are already familiar with. Namely, opposites attract, and likes repel. Look at figure 2-24, view A. Here you see an electron traveling between two charged plates, H1 and H2. As you can see, before the electron reaches the charged plates, called DEFLECTION PLATES, its flight path is toward the center of the screen. In view B, the electron has reached the area of the deflection plates and is attracted toward the positive plate, H2, while being repelled from the negative plate, H1. As a result, the electron is deflected to the right on the inside of the screen. You, the viewer, will see the spot of light on the left side of the CRT face (remember, you are on the opposite side of the CRT screen). This is shown in view C.
A spot of light on the left-hand side of the CRT screen, however, is no more useful than a spot of light in the center of the screen.  We will explain the manner in which this is done by using
2-24
figure 2-25. In view A, five electrons are emitted in sequence, 1 through 5, by the electron gun. The right deflection plate, H
2, has a large positive potential on it while the left plate, H1 has a large negative potential on it. Thus, when electron 1 reaches the area of the deflection plates, it is attracted to the right plate while being repelled from the left plate. In view B, electron 2 has reached the area of the deflection plates. However, before it arrives, R1 and R2 are adjusted to make the right plate less positive and the left plate less negative. Electron 2 will still be deflected to the right but not as much as electron 1. In view C, electron 3 has reached the area of the deflection plates. Before it gets there, R1 and R2 are adjusted to the mid-point. As a result, both plates have 0 volts applied to them. Electron 3 is not deflected and simply travels to the center of the CRT screen. In view D, electron 4 has reached the area of the deflection plates. Notice that R1 and R2 have been adjusted to make the right plate negative and the left plate positive. As a result, electron 4 will be deflected to the left. Finally, in view E, the left plate is at its maximum positive value. Electron 5 will be deflected to the extreme left. What you see when you are facing the CRT is a bright luminous line, as shown in view E. While this description dealt with only five electrons, in reality the horizontal line across a CRT face is composed of millions of electrons. Instead of seeing five bright spots in a line, you will see only a solid bright line.
2-25
In summary, the horizontal line displayed on a CRT or on the face of a television tube is made by sweeping a stream of electrons rapidly across the face of the CRT. This sweeping action, or scanning, is performed by rapidly varying the voltage potential on the deflection plates as the electron stream passes.
As we mentioned earlier, a CRT can be used to graphically and visually plot an electronic signal, such as a sine wave. This is done by using a second set of deflection plates called VERTICAL- DEFLECTION PLATES. Examine figure 2-26. You are looking at the front view (facing the screen) of a CRT, back into the tube at the deflection plates. In normal usage, the horizontal plates sweep a straight line of electrons across the screen from left to right while the signal to be displayed is applied to the vertical deflection plates. A circuit of this type is shown in figure 2-27. We will use this figure to explain how a sine wave is displayed. First, however, you need to understand what is happening in view A. The box on the left of the CRT labeled HORIZONTAL-DEFLECTION CIRCUITS is an electronic circuit that will duplicate the actions of R1 and R2 used earlier in making up a horizontal line. How it works will be discussed in a later NEETS module. Notice T1; the output of this transformer is applied to the vertical- deflection plates. The signals applied to the vertical plates are 180º out of phase with each other. Thus, when one plate is attracting the electron beam, the other will be repelling the electron beam. Because you are only concerned with what happens inside the CRT, this circuitry will be eliminated and only the CRT and its deflection plates will be shown, as in view B.
2-26
Now look at view C. While this illustration looks complicated, don’t let it worry you. You have already analyzed more complicated diagrams. The sine wave in the center of the screen is the signal that will be displayed as a result of the two 180º out-of-phase sine waves applied to the vertical-deflection plates. The five spots on the center sine wave represent the five electrons used to explain horizontal deflection. Only now these electrons will be deflected both vertically and horizontally. Time lines T1 through T5 represent the time when each like-numbered electron reaches the area of the deflection plates. Because you already know how the electron beam is swept or deflected horizontally, we will not discuss horizontal deflection. Just remember that from T1 to T5, the electron beam will be continuously moved from your left to your right. Now that you know where everything is on the illustration, you are ready to discover how a sine wave is displayed on a CRT.
At time 1 (T1), the sine waves applied to both vertical-deflection plates are at their null points, or zero volts. As a result, electron 1 is not vertically deflected and strikes the CRT at its vertical center. At time 2 (T2), the sine wave applied to the top plate is at its maximum negative value. This repels electron 2 toward the bottom of the CRT. At the same time, the sine wave applied to the bottom plate is at the most positive value, causing electron 2 to be attracted even further toward the bottom of the CRT. Remember, the beam is also being moved to the left. As a result, electron 2 strikes the CRT face to the right of and below electron 1. At time 3 (T3), both sine waves applied to the vertical-deflection plates are again at the null point, or zero volts. Therefore, there is no vertical deflection and electron 3 strikes the CRT face in
2-27
the center of the vertical axis. Because the electron beam is still moving horizontally, electron 3 will appear to the right of and above electron 2. At time 4 (T4), the sine wave applied to the top vertical- deflection plate is at its maximum positive value. This attracts electron 4 toward the top deflection plate. The upward deflection of electron 4 is increased by the negative-going sine wave (at time 4) applied to the bottom deflection plate. This negative voltage repels electron 4 upward. Thus, electron 4 strikes the CRT face to the right of and above electron 3. Finally, at time 5 (T5) both input sine waves are again at zero volts. As a result, electron 5 is not deflected vertically, only horizontally. (Remember, the beam is continually moving from right to left.)
While this discussion is only concerned with five electrons, vertical scanning, or deflection, involves millions of electrons in a continuous electron beam. Instead of seeing five spots on the CRT screen, you will actually see a visual presentation of the sine wave input. This was, as you remember, described earlier as the unique feature of the CRT. You may have wondered why so much space in this chapter was taken up with the discussion of the CRT. There are two reasons for this. First, the field of electronics is in a constant state of evolution. Transistors replaced most vacuum tubes. Transistors are being replaced by integrated circuits (ICs). As you progress in your career in electronics, you will find that the equipment you work on will follow this evolution, from transistors to IC chips. Of all the tubes discussed in this text, the CRT is the least likely to be replaced in the near future. Thus, in all probability, whether your career in electronics lasts for only the time you spend with this text or 20 years, the CRT will be your constant companion and co-worker.
The second reason for this rather extensive coverage of the CRT is that, while the CRT has a unique ability, it operates exactly like all the tubes previously discussed.
This summary will not only review the CRT, but will also point out the similarities between the CRT and other tubes.
Look at figure 2-28.  Each element is identified by a circuit number. We will review briefly the function of each element in a CRT and its similarity to elements in conventional tubes. This summary will help you tie together everything you have learned about the CRT and electron tubes in general.
2-28
1. The Heater—serves as the source of heat for the cathode in both the CRT and indirectly heated tubes.
2. The Cathode—serves as the source of thermionically emitted electrons in both the CRT and conventional tubes. The major difference is that in the CRT, the cathode is circular in shape and the outer surface is coated to ensure that electron emission is roughly unidirectional.
3. The Control Grid-in both the CRT and conventional vacuum tubes, the control grid controls the number of electrons that will be fired across "the gap." The major difference is in the physical construction. Conventional tubes use a wire-mesh ladder-type grid, while the CRT uses a cap- like grid.
4. The Focusing Anode—in the CRT, this anode serves a dual purpose of attracting electrons from the area of the control grid and focusing the electrons into a beam. Its function of attracting electrons from the area of the grid is similar to the action of the plate in a conventional tube. The focusing action of the anode is similar to that performed by beam-forming plates in the beam- forming tetrode. Bear in mind, though, that beam-forming plates are negatively charged and repel electrons into electron sheets, while the focusing anode is positively charged and attracts electrons into beam.
5. The Accelerating Anode—in the CRT, this anode is used to accelerate the electrons toward the front of the tube. Its action is similar to the screen grid of tetrodes and pentodes. But remember, while the screen grid in conventional tubes accelerates electrons toward the plate, its primary purpose is to reduce interelectrode capacitance, NOT accelerate electrons.
6. The Vertical-Deflection Plates—in the CRT, these plates move the electron beam up and down the screen. The input signal is usually applied to these plates. While no equivalent element is
2-29
found in conventional tubes, the principle employed (electrostatic attraction and repulsion) forms the heart of all vacuum tube operation.
7. The Horizontal-Deflection Plates—in the CRT, these plates move the electron beam by electrostatic attraction and repulsion, horizontally across the CRT screen. In most equipment using the CRT, including television sets, electronic signals are supplied to these plates to trace or paint a horizontal line.
8. The Aquadag Coating—in the CRT, this coating performs the same function as the suppressor grid in conventional tubes; namely, eliminating the effects of secondary emission. In conventional tubes, the suppressor grid is negatively charged and repels secondary emission electrons back to the plate. In the CRT, the aquadag is positively charged and attracts secondary emission away from the screen.
9. The Screen—also called the face, is a unique element of the CRT. When struck by electrons, the phosphor coating becomes luminous, or glows, thus enabling the tube to visually present electronic signals.
From this comparison of the CRT and other types of electron tubes, one fact should be clear. Almost all tubes, no matter what their function, operate on two principles: electrostatic attraction and repulsion, and thermionic emission. By keeping these two principles in mind, you should be able to analyze any type of tube operation.
Q14. What is the unique ability of the CRT? Q15. What are the three main parts of CRT? Q16. What term is used for the ability of a spot on a CRT screen to continue to glow after the electron
bean has struck it and moved away?
Q17. The electron beam in a CRT is made to sweep from left to right across the screen. What tube element causes this sweeping motion?
Q18. In applications where electronic waveforms are displayed on a CRT screen, the input signal is normally applied to what CRT element?
There are certain safety precautions you should follow when you work with or handle the special tubes covered in this chapter. We will examine these tubes and their associated precautions in the following sections.
The average electron tube is a rugged device capable of withstanding the shocks and knocks of everyday usage and handling. However, they are not indestructible. You should remember that most electron tubes contain a near vacuum enclosed by a glass envelope. Because of this, the glass is under constant stress from atmospheric pressure. Any undue stress, such as striking the envelope against a hard surface, may cause the envelope to shatter, resulting in an IMPLOSION.
An implosion is just the opposite of an explosion. When the glass envelope of an electron tube shatters, the outside atmosphere rushes into the tube to fill the vacuum. As the air rushes into the tube, it
2-30
carries glass fragments of the envelope with it. Once these fragments reach the center of the tube, they continue outward with considerable velocity. The result is similar to an explosion, in that the immediate area surrounding the electron tube is filled with fast-moving glass fragments. You, as a nearby object, may find yourself the target for many of these glass fragments. For this reason you should handle all electron tubes with care.
small. There are two exceptions to this: CRT’s and radioactive tubes.
The glass envelope of a CRT encloses a high vacuum. Because of its large volume and surface area, the force exerted on a CRT by atmospheric pressure is considerable. The total force on a 10-inch CRT may exceed 4,000 pounds. Over 1000 pounds is exerted on the CRT face alone.
When a CRT is broken, a large implosion usually occurs. Almost two tons of force hurl glass fragments toward the center of the tube. At the same time, the electron gun is normally thrown forward inside the tube. The face, because of its size, tends to move very slowly toward the center of the tube. This presents one of the main hazards of a broken CRT. The electron gun passes through the center of the tube with considerable force. It continues until it strikes the CRT face. The impact from the electron gun normally breaks the CRT face into many small fragments, which are hurled outward. The face is coated with a chemical coating that is extremely toxic. If you are unfortunate enough to experience an accidental implosion of a CRT and are nicked by one of these fragments, seek immediate medical aid. As you can see, improper handling of a CRT can be very hazardous to your health.
The CRT is, in essence, a tiny fragmentation bomb. The major difference between a CRT and a bomb is that a bomb is designed to explode; a CRT is not. As long as you handle a CRT properly, it represents no danger to you. Only when you mishandle it do you risk the danger of being pelted with an electron gun and toxic glass fragments. When handling a CRT, you should take the following precautions:
1. Avoid scratching or striking the surface of the CRT.
2. Do not use excessive force when you remove or replace a CRT’s deflection yoke or socket.
3. Do not try to remove an electromagnetic-type CRT from its yoke until you have discharged the high voltage from the CRT’s anode connector (hole).
4. Never hold the CRT by its neck.
5. Always set the CRT with its face down on a thick piece of felt, rubber, or smooth cloth.
6. Always handle the CRT gently. Rough handling or a sharp blow on the service bench can displace the electrodes within the tube, causing faulty operation.
7. Wear safety glasses and protective gloves.
One additional handling procedure you should be aware of is how to dispose of a CRT properly. When you replace a CRT, you cannot simply throw the old CRT over the side of the ship, or place it in the nearest dumpster. When thrown over the side of a ship, a CRT will float; if it washes ashore, it is dangerous to persons who may come in contact with it. A CRT thrown in a dumpster represents a hidden booby trap. Therefore, always render the CRT harmless before you dispose of it. This is a fairly simple procedure, as outlined below.
Note: Be sure to wear safety goggles.
2-31
• Place the CRT face down in an empty carton and cover its side and back with protective material.
• Carefully break off the plastic locating pin from the base (fig. 2-29) by crushing the locating pin with a pair of pliers.
• Brush the broken plastic from the pin off the CRT base.
• Look into the hole in the base where the locator pin was. You will see the glass extension of the CRT called the vacuum seal. Grasp the vacuum seal near the end with the pliers and crush it.
This may sound a little risky but it isn’t. The vacuum seal can be crushed without shattering the tube. Once the seal has been crushed, air will rush into the tube and eliminate the vacuum.
Another type of tube that can prove hazardous to you, if you handle it improperly, is the radioactive tube.
These tubes contain radioactive material and are used as voltage-regulator, gas-switching, and cold- cathode, gas-rectifier tubes. Some of these tubes have dangerous radioactive intensity levels. Radioactive tubes are marked according to military specifications.
Radioactive material is added to a tube to aid in ionization. The radioactive material emits relatively slowly moving particles. This should not worry you because the glass envelope is thick enough to keep these particles inside the tube. Therefore, proper handling is nothing more than ensuring that the envelope remains unbroken. If these tubes are broken and the radioactive material is exposed, or escapes from the confines of the electron tube, the radioactive material becomes a potential hazard.
The concentration of radioactivity in an average collection of electron tubes in a maintenance shop does not approach a dangerous level, and the hazards of injury from exposure are slight. However, at major supply points, the storage of large quantities of radioactive electron tubes in a relatively small area may create a hazard. For this reason, personnel working with equipment using electron tubes containing radioactive material, or in areas where a large quantity of radioactive tubes are stored, should read and become thoroughly familiar with the safety practices contained in Radiation, Health, and Protection Manual, NAVMED P-5055. Strict compliance with the prescribed safety precautions and procedures of this manual will help to prevent accidents, and to maintain a safe working environment which is conducive to good health.
2-32
The clean-up procedures listed below are based on NAVMED P-5055. Your ship or station may have additional procedures that you should follow. Be sure you are aware of your command’s policy concerning decontamination procedures before you begin working on equipment containing radioactive tubes. Some important instructions and precautions for the proper handling of radioactive tubes are listed below:
1. Do not remove radioactive tubes from their carton until you are ready to install them.
2. When you remove a tube containing a radioactive material from equipment, place it immediately in an appropriate carton to prevent possible breakage.
3. Never carry a radioactive tube in a manner that may cause it to break.
4. If a radioactive tube that you are handling or removing breaks, notify the proper authority and obtain the services of qualified radiological personnel immediately.
5. Isolate the immediate area of exposure to protect other personnel from possible contamination and exposure.
6. Follow the established procedures set forth in NAVMED P-5055.
7. Do not permit contaminated material to come in contact with any part of your body.
8. Take care to avoid breathing any vapor or dust that may be released by tube breakage.
9. Wear rubber or plastic gloves at all times during cleanup and decontamination procedures.
10. Use forceps to remove large fragments of a broken radioactive tube. Remove the remaining small particles with a designated vacuum cleaner, using an approved disposal collection bag. If a vacuum cleaner is not designated, use a wet cloth to wipe the affected area. In this case, be sure to make one stroke at a time. DO NOT use a back-and-forth motion. After each stroke, fold the cloth in half, always holding one clean side and using the other for the new stroke. Dispose of the cloth in the manner stated in instruction 14 below.
11. Do not bring food or drink into the contaminated area or near any radioactive material.
12. Immediately after leaving a contaminated area, if you handled radioactive material in any way, remove all of your clothing. Also wash your hands and arms thoroughly with soap and water, and rinse with clean water.
13. Notify a medical officer immediately if you sustain a wound from a sharp radioactive object. If a medical officer cannot reach the scene immediately, stimulated mild bleeding by applying pressure about the wound and by using suction bulbs. DO NOT USE YOUR MOUTH if the wound is a puncture-type wound. If the opening is small, make an incision to promote free bleeding, and to make the wound easier to clean and flush.
14. When you clean a contaminated area, seal all debris, cleaning cloths, and collection bags in a container such as a plastic bag, heavy wax paper, or glass jar, and place them in a steel can until they can be disposed of according to existing instructions.
15. Use soap and water to decontaminate all tools and implements you used to remove the radioactive substance. Monitor the tools and implements for radiation with an authorized radiac set to ensure that they are not contaminated.
2-33
As you can see, the cleanup that results from breaking a radioactive tube is a long and complicated procedure. You can avoid this by simply ensuring that you don’t break the tube.
While conventional tubes present few safety problems, beyond broken glass and the possibility of cutting yourself, there is one precaution you must know. Namely, electron tubes are hot. The filaments of some tubes may operate at several thousand degrees. As a result, the envelopes can become very hot. When you work on electron tube equipment, always deenergize the equipment and allow the tubes sufficient time to cool before you remove them. If this is impossible, use special tube pullers which the Navy stocks for this purpose. Never attempt to remove a hot tube from its socket with your bare fingers.
The following summary covers the main points of this chapter. Study it to be sure you understand the material before you begin the next chapter.
MULTI-UNIT TUBES were developed to reduce the size of vacuum tube circuits. Incorporating more than one tube in the same envelope allowed the size of a vacuum tube circuit to be reduced considerably. While a single envelope may contain two or more tubes, these tubes are independent of each other.
MULTI-ELECTRODE TUBES were developed to extend the capability of conventional tubes. In some cases, a multi-element tube may contain up to seven grids. These types of tubes are normally classified by the number of grids they contain.
POWER PENTODES are used as current or power amplifiers. Power pentodes use in-line grid arrangements. In this manner, more electrons can reach the plate from the cathode. In effect, this lowers plate resistance and allows maximum conduction through the tube.
2-34
BEAM-POWER TUBES are also used as power amplifiers. In addition to the in-line grid arrangement, beam-power tubes use a set of negatively charged beam-forming plates. The beam-forming plates force electrons that would normally be deflected from the plate back into the electron steam and, thus, add to the number of electrons the tube can use for power amplification.
VARIABLE-MU (µ) TUBES or REMOTE-CUTOFF TUBES were developed to extend the amplification range of electron tubes by avoiding the possibility of driving the tube into cutoff. This is done by uneven spacing of the grid wires.
2-35
UHF TUBES are special-purpose tubes designed to operate at ultrahigh frequencies between 300 MHz and 3000 MHz with minimum effect from transit time limitations. Among these are acorn tubes, and doorknob tubes, lighthouse tubes, and oilcan tubes.
2-36
PLANAR TUBES have their plates and grids mounted parallel to each other. Because of their planar construction, they can handle large amounts of power at uhf frequencies.
GAS-FILLED TUBES contain a small amount of gas that ionizes and reduces the internal resistance of the tubes. Because of this, gas-filled tubes can handle relatively large amounts of power while maintaining a constant voltage drop across the tube.
2-37
COLD-CATHODE TUBES lack heaters or filaments and, therefore, do not use thermionic emission. Instead, a voltage potential applied across the tube causes the internal gas to ionize. Once ionization has occurred, the voltage drop across the tube remains constant, regardless of increased conduction.
The CRT is a special-purpose tube that has the unique ability to visually display electronic signals. The CRT uses the principles of electrostatic attraction, repulsion, and fluorescence. Because of its unique ability, the CRT makes up the heart of many types of test equipment that you will become familiar with during your career in electronics.
2-38
ANSWERS TO QUESTIONS Q1. THROUGH Q18. A1. Conventional pentodes have a staggered grid arrangement, while power pentodes have a shielded
grid arrangement.
A2. Beam-forming plates. A3. By increasing the number of electrons that reach the plate, plate current is increased. A4. A large negative voltage causes conduction to occur only at the center of the grid A5. Decreases gain.
a. Power pentode or beam-forming tetrode. b. Conventional tube.
c. Variable-mu tube.
A6. It causes the control grid to short to the cathode.
A7. By reducing the spacing between tube elements.
A8. The close spacing of tube elements allows for the ready formation of arcs or short circuits. A9. Planar
A10.
a. They can carry more current.
b. They maintain a constant IR drop across the tube.
A11. None.
A12. The filament’s voltage should be applied to the tube at least 30 seconds before attempting to operate the tube.
A13. They have the ability to maintain a constant voltage drop across the tube despite changes in current flow.
A14. To visually display electronic signals.
2-39
A15.
a. Electron gun.
b. Deflection system c. Screen.
A16. Persistence.
A17. The horizontal-deflection plate. A18. The vertical-deflection plate.
3-1
Upon completion of this chapter you will be able to:
1. Identify the various sections of a power supply.
2. State the purpose of each section of a power supply.
3. Describe the operation of the power supply from both the whole unit standpoint and from the
subunit standpoint.
4. Describe the purpose of the various types of rectifier circuits used in power supplies.
5. Describe the purpose of the various types of filter circuits used in power supplies.
6. Describe the operation of the various voltage and current regulators in a power supply.
7. Trace the flow of ac and dc in a power supply, from the ac input to the dc output on a schematic
diagram.
8. Identify faulty components through visual checks.
9. Identify problems within specific areas of a power supply by using a logical isolation method of
troubleshooting.
10. Apply safety precautions when working with electronic power supplies.
powered by batteries. While the use of batteries allowed the equipment to be portable (to some degree), it
also placed several limitations on how the equipment could be used. Because of their general inefficiency,
batteries had to be either replaced frequently or, if they were rechargeable, kept near a battery charger.
the batteries frequently.
effective. Since batteries failed to satisfy these requirements, the "electronic power supply" was
developed.
supply. Therefore, this chapter is of extreme importance to you. We will discuss the sections and
individual components of the power supply and their purposes within the power supply. We will also
discuss troubleshooting of each section and its components.
3-2
Figure 3-1 shows the block diagram of the basic power supply.
As you can see, the first section is the TRANSFORMER. The transformer serves two primary purposes: (1) to step up or step down the input line voltage to the desired level and (2) to couple this
voltage to the rectifier section. The RECTIFIER section converts the ac signal to a pulsating dc voltage. However, you will see later in this chapter that the pulsating dc voltage is not desirable. For this reason, a
FILTER section is used to convert the pulsating dc voltage to filtered dc voltage. The final section, the REGULATOR, does just what the name implies. It maintains the output of the power supply at a constant level in spite of large changes in load current or in input line voltage. Depending upon the design
of the equipment, the output of the regulator will maintain a constant dc voltage within certain limits.
what changes are made to the input signal. In figure 3-2, the input signal of 120 volts ac is applied to the
primary of the transformer, which has a turns ratio of 1:3. We can calculate the output by multiplying the
input voltage by the ratio of turns in the secondary winding to turns in the primary winding. Therefore,
the output voltage of our example is: 120 volts ac × 3, or 360 volts ac. Depending on the type of rectifier
used (full-wave or half-wave), the output from the rectifier will be a portion of the input. Figure 3-2
shows the ripple waveform associated with a full-wave rectifier. The filter section contains a network of
remains at a more constant dc level. You will see this more clearly in the discussion of the actual filter
circuits. You can see that the output of the filter is at a 180-volt dc level with an ac RIPPLE voltage riding on it. (Ripple voltage is a small ac voltage riding at some dc voltage level. Normally, ripple voltage
is an unwanted ac voltage created by the filter section of a power supply.) This signal now goes to the
regulator where it will be maintained at approximately 180 volts dc to the load.
Q1. What are the four basic sections to a power supply?
Q2. What is the purpose of the regulator?
3-3
down the ac voltage to the desired level. Additionally, most input transformers have separate step-down
(load). Such a transformer is shown in figure 3-3. Because the input transformer is located in the power
POWER TRANSFORMER. Notice that the transformer has the ability to deliver both 6.3 and 5 volts ac filament voltages to the electron tubes. The High-voltage winding is a 1:3 step-up winding and delivers
360 volts ac to the rectifier. This transformer also has what is called a center tap. This center tap provides the capability of developing two high-voltage outputs from one transformer.
Q3. What are the purposes of the transformer in a power supply?
Q4. For what are the low voltage windings in a transformer used?
Q5. For what is the center tap on a transformer used?
dc voltage. Now let's discuss the process of rectification.
alternating current to direct current. If an ac voltage is applied to a diode, the diode will conduct ONLY DURING THE POSITIVE ALTERNATION OF VOLTAGE when the plate of the diode is made positive with respect to the cathode.
Figure 3-4 shows a diode connected across the 120-volt ac line. During the positive alternation of the
source voltage, the sine wave applied to the tube makes the plate positive with respect to the cathode. At
milliammeter, through the tube, and to the positive supply lead. This is indicated by the shaded area of the
output waveform. This current exists during the entire period of time that the plate is positive with respect
to the cathode (for the first 180 degrees of the input sine wave).
3-4
and the tube cannot conduct. When conditions prevent the tube from conducting, the tube is said to be in
CUTOFF. This is indicated by the dotted waveform. The tube will be in cutoff and no current will flow for the entire negative alternation.
degrees. The circuit current therefore has the appearance of a series of positive pulses, as shown by the
shaded areas. Notice that although the current is in the form of pulses, the current always flows through
the circuit in THE SAME DIRECTION. Current that flows in pulses in the same direction is called PULSATING DC. The diode has thus RECTIFIED the input voltage. Although the principle of rectification applies to all rectifier circuits, some rectifiers are more efficient than others. For this reason,
full-wave, and bridge.
Figure 3-5 is a diagram of a complete half-wave rectifier circuit. For the diode to be used as a
current flows. Because Navy electronic equipment requires various input voltages, it is necessary to have
a rectified voltage that is greater (or smaller in some cases) than the source voltage. The rectifier plate
circuit is supplied power from a step-up (or step-down) transformer. Notice that the transformer has the
two secondary windings mentioned earlier. The lower winding supplies high voltage to the plate and
cathode of the diode, and the upper winding supplies a low ac voltage to the filaments of the diode.
the load resistor (RL). Any current flowing through the tube also flows through the load resistor, causing a
voltage to be developed across it. The magnitude of the voltage developed across the load resistor is
directly proportional to the amount of current flowing through it (Ohm's law: E = IR).
3-5
of a simplified series circuit. As you can see in figure 3-6, the diode (V1) and load resistor (RL) are in
series with the secondary winding of the transformer. During the positive alternation of the input, as the
increases. Since the diode tube and the load resistor form a series circuit, the same current flows through
both the tube and the resistor. This current produces a voltage drop across the tube and the load resistor,
which have polarities as shown. Since the plate resistance of the tube is only about 500 ohms and the
dropped across the load resistor (425 × .95 = 404 V) and 5 percent (425 × .05 = 21 V) across the tube.
flows in the circuit. Since there is no current flow through RL, the load voltage remains at zero volts
throughout the negative alternation. During this time the entire negative alternation is felt across the tube.
The reason for this is derived from Kirchhoff's law, which states:
The sum of the load voltage and diode voltage equals the applied voltage.
3-6
pulses is the same as the frequency of the input sine wave. The output pulse frequency is called RIPPLE FREQUENCY. If the rectifier circuit is supplied power from a 60-hertz ac line voltage, 60 pulses of load current will occur each second.  This average dc
power is determined by the amplitude of the pulses and the time delay between pulses. The higher the
supplied to the load. To determine average dc voltage (Eavg), it is necessary to know the average value of
the pulses and the peak value of load voltage. This is illustrated in figure 3-7.
we can develop a conversion factor. The formula for average value was discussed earlier in NEETS, module 2. By now you should know that the average value for a full sine wave is .637 times its peak or
maximum value. Therefore, if you want the average value of a half-wave rectifier output, you should
multiply half the value of .637 (.318) times the peak or maximum voltage, as expressed in the following
equation:
Eavg (the average load voltage) = .318 × Emax
Where:
can assume Emax in our equation to be the same as the peak value of the input sine wave.
that it applies to the load current. Thus,
Iavg (the average load current) = .318 × Imax
Where:
If a line is drawn through the rectified waveform at a point that is 0.318 of the distance from zero to
maximum, the waveform will be divided so that area A is equal to area B (fig. 3-7). Therefore, current or
3-7
voltage pulses with a value of .318 of the peak value have the same effect on the load as a steady voltage
or current.
The half-wave rectifier uses the transformer during only one-half of the cycle. Therefore, for any
cycle. In other words, to obtain large amounts of power, the half-wave transformer must be relatively
large in comparison to what it would have to be if both halves of the cycle were used. This disadvantage
limits the use of the half-wave rectifier to applications that require a very small current drain. The half-
inexpensive voltage supplies will suffice. As you can see from your study on half-wave rectifiers, this
type of circuit placed many limitations on electronic equipment. For this reason another type of rectifier
circuit had to be developed. One of the factors that had to be considered was how to use the full output
from the transformer to obtain the highest average voltage and current. Thus, the FULL-WAVE rectifier was developed.
Q6. Does a rectifier tube conduct on the positive or negative alternation of the input signal?
Q7. What term is used to describe the period when the diode is not conducting?
Q8. Current that flows in pulses in the same direction is called _____.
Q9. For a diode to act as a rectifier, should it be connected in series or parallel with the load?
Q10. What is the Ripple frequency of a half-wave rectifier if the input frequency is 60 Hz?
Q11. What is the equation for determining average voltage in a half-wave rectifier?
same direction during each half cycle of the ac supply.
A schematic diagram of a simple full-wave rectifier is shown in figure 3-8. The transformer supplies
the source voltage for two rectifier tubes (V1 and V2). This power transformer has a CENTER-TAPPED high-voltage secondary winding that is divided into two equal parts (W1 and W2). W1 provides the
source voltage for V1 and the other winding (W2) provides the source voltage for V2. The connections to
the diodes are arranged so that the diodes conduct on alternate half cycles.
3-8
During one alternation of the secondary voltage, the polarities will be as shown in figure 3-8. The
source for diode V2 is the voltage induced into the lower half of the transformer secondary winding (W2).
conduct.
Throughout the period of time during which the plate of V2 is negative, the plate of V1 is positive.
This is illustrated by the polarity signs across W1, which indicate the source for V1. Since the plate of V1
arrow.
Figure 3-9 shows the next half cycle of secondary voltage. As you can see, the polarities across W1
and W2 are reversed. During this alternation the plate of V1 is driven negative and V1 cannot conduct.
For the period of time that V1 is negative, the plate of V2 is positive, permitting V2 to conduct.
current of V1. In this circuit arrangement, a pulse of load current flows during each alternation of the
input cycle. Since both alternations of the input voltage cycle are used, the circuit is called a
FULL-WAVE RECTIFIER.
practical full-wave rectifier and its waveforms.
A practical full-wave rectifier circuit is shown in figure 3-10. It uses two diodes (V1 and V2) and a
center-tapped transformer (T1). When the center tap is grounded, the voltages at the opposite ends of the
secondary windings are 180 degrees out of phase with each other. Thus, when the voltage at point A is
positive with respect to ground, the voltage at point B is negative with respect to ground. Let's examine
the operation of the circuit during one complete cycle.
3-9
ground and the plate of V2 is negative. As shown, current flows from ground (center tap), up through the
load resistor (RL), through diode V1 to point A. In the transformer, current flows from point A, through
the upper winding and back to ground (center tap). When V1 conducts, it acts like a closed switch so that
the positive half-cycle is felt across the load.
plate of V2 is positive with respect to ground and the plate of V1 is negative. Only V2 can conduct.
V2 to point B of T1. In the transformer, current flows from point B up through the lower windings and
back to ground (center tap). Notice that the current flows across the load resistor (R L) in the SAME DIRECTION for both halves of the input cycles.
each cycle of input voltage. The ripple frequency at the output of the full-wave rectifier is therefore
TWICE THE LINE FREQUENCY.
the higher pulse frequency, the output is closely approximate to pure dc. This higher frequency also
makes filtering much easier than the output of the half-wave rectifier.
rectifier is twice as great as the average current or voltage at the output of the half-wave rectifier. The
relationship between peak and average values is illustrated in figure 3-11.
3-10
average current or voltage is 63.7 percent (or .637) of the peak current or voltage.
As an equation:
Eavg (the average load voltage) = .637 × Emax
Iavg (the average load current) = .637 × Imax
Where:
wave rectifier is 600 volts. Find the average load voltage. (Ignore the drop across the rectifier tube.)
or 300 volts. As the secondary voltage is an rms value, the peak load voltage is:
Emax = 1.414 × Es
Emax = 1.414 × 300
The average load voltage is:
Eavg = .637 × Emax
Eavg = .637 × 424
NOTE: If you have problems with this equation, review NEETS, module 2, pertaining to this area.
every circuit. The full-wave rectifier is no exception. In studying the full-wave rectifier, you have found
much easier to filter because of the high-ripple frequency. The only disadvantage is that the peak voltage
in a full-wave rectifier is only half the peak voltage in a half-wave rectifier. This is because the secondary
goes to each diode.
same ripple frequency as a full-wave rectifier. This circuit, called the BRIDGE RECTIFIER, will be the chapter of our next discussion.
Q12. What is the ripple frequency of a full-wave rectifier with an input frequency of 60 Hz?
3-11
Q13. What is the average voltage (Eavg) output of a full-wave rectifier that has an output of 100 volts peak?
When four diodes are connected as shown in figure 3-12, the circuit is called a BRIDGE RECTIFIER. The input to the circuit is applied to the diagonally opposite corners of the network, and the output is taken from the remaining two corners.
amount of voltage induced into the secondary of the transformer. During this time, the voltage between
points A and B may be considered to be impressed across V1, the load resistor RL, and V3, in series. The
from point B through tube V1 in an upward direction across the load resistor, through tube V3, to point A.
This path is indicated by the solid arrows. The waveform is shown as numbers (1) and (2).
negative with respect to their cathodes. At the same time, the plates of V2 and V4 become positive with
respect to their cathodes, and current flows in the direction indicated by the dashed arrows. The current
through RL is always in the same direction. This current, in flowing through RL, develops a voltage
corresponding to that shown in waveform (5) of the figure. The bridge rectifier is a full-wave rectifier
since current flows through the load during both half cycles of the applied alternating voltage.
full-wave circuit. We can show this by assigning values to some of the components as shown in figure
3-13, views (A) and (B). Assume that the same transformer is used in both circuits. The peak voltage
developed between points X and Y is 1,000 volts in both circuits. In the conventional full-wave circuit,
3-12
view (A), the peak voltage from the center tap to either X or Y is 500 volts. Since only one diode can
conduct at any instant, the maximum voltage that can be rectified at any instant is 500 volts. Therefore,
of the small voltage drop across the tube). In the bridge rectifier of view (B), the maximum voltage that
can be rectified is the full secondary voltage, 1,000 volts. Therefore, the peak output voltage across the
load resistor is nearly 1,000 volts. Thus, with both circuits using the same transformer, the full-wave
bridge circuit produces a higher output voltage than the conventional full-wave rectifier.
voltage. For this reason bridge rectifiers that use vacuum tubes are widely used in high-voltage power
supply applications.
required. This is due to the different potentials existing at the filaments of the diodes. The filaments of V2
either V2 or V4. The three filament transformers must be well insulated from each other, and from
ground, because of the high potentials to which they are subjected. The use of indirectly heated diodes
heater would be likely to result in arcing.
3-13
Q14. What is the main disadvantage of the conventional full-wave rectifier?
Q15. What main advantage does a bridge rectifier have over a conventional full-wave rectifier?
dc for proper operation. This type of output may be provided by placing single or multisection filter
circuits between the output of the rectifier and the load.
There are four basic types of filter circuits:
We will cover the function of each of these filters in detail later in this chapter.
Filtering is done by using various combinations of capacitors, inductors, and resistors. Inductors are
used as series impedances to oppose the change in flow of alternating (pulsating dc) current. Capacitors
are used as shunt elements to bypass the alternating components of the signal around the load (to ground).
Resistors are used in place of inductors in low current applications.
Let's briefly review the properties of a capacitor. First, a capacitor opposes any change in voltage.
The opposition to a change in voltage is called capacitive reactance (XC) and is measured in ohms. The
capacitor.
3-14
From the formula, you can see that if frequency or capacitance is increased, the XC will decrease.
high XC. This is done by providing a better shunting effect of the ac around the load, as shown in figure
3-15.
applied voltage. Once charged, the capacitor must retain the charge as long as possible. The capacitor
must have a short charge time constant (view A) and a long discharge time constant (view B). This can be
resistance of the load as large as possible (slow discharge time).
defined as the time it takes a capacitor to charge to 63.2 percent of the applied voltage or to discharge to
36.8 percent of its total charge. This can be expressed by the following equation:
You should also recall that a capacitor is considered fully charged after five RC time constants.
charge rapidly and discharge as slowly as possible.
3-15
In filter circuits the capacitor is the common element to both the charge and discharge paths.
Therefore, to obtain the longest possible discharge time, you want the capacitor to be as large as possible.
the larger the value of the capacitor (C), the smaller the opposition (XC) or resistance to ac.
Now let's look at inductors and their application in filter circuits. Remember, AN INDUCTOR OPPOSES ANY CHANGE IN CURRENT. In case you have forgotten, a change in current through an inductor produces a changing electromagnetic field. The changing field, in turn, cuts the windings of the
wire in the inductor and thereby produces a counterelectromotive force (cemf). It is the cemf that opposes
the change in circuit current. Opposition to a change in current at a given frequency is called inductive
reactance (XL) and is measured in ohms. The inductive reactance (XL) of an inductor is determined by the
applied frequency and the inductance of the inductor. Mathematically,
XL = 2!fL
will increase. Since inductors are placed in series with the load (fig. 3-17), the larger the XL, the larger the
ac voltage developed across the inductor and the smaller the ac voltage developed across the load.
3-16
Now back to our circuit. As illustrated in figure 3-18, when the current starts to flow through the
coil, an expanding magnetic field builds up around the inductor. This magnetic field around the coil
develops the cemf that opposes the change in current. When the rectifier current decreases as shown in
into the coil. This additional current adds to the rectifier current and attempts to keep it at its original
level.
rectifier, let's discuss the different types of filter circuits in use today.
3-17
Q16. If you increase the value of the capacitor will the XC increase or decrease? Why?
The simple capacitor filter is the most basic type of power supply filter. The use of this filter is very
limited. It is sometimes used on extremely high-voltage, low-current power supplies for cathode-ray and
similar electron tubes that require very little load current from the supply. This filter is also used in
circuits where the power-supply ripple frequency is not critical and can be relatively high.
The simple capacitor filter shown in figure 3-20 consists of a single-filter element. This capacitor
(C1) is connected across the output of the rectifier in parallel with the load. The RC charge time of the
using this filter. In other words, the capacitor must charge up fast with preferably no discharge at all.
filter than the half-wave rectifier because of its higher frequency.
in figure 3-21 to those with filters in figure 3-22. The output waveforms in figure 3-21 represent the
unfiltered outputs of the half-wave and full-wave rectifier circuits. Current pulses flow through the load
resistance (RL) each time a diode conducts. The dashed line indicates the average value of output voltage.
For the half-wave rectifier, Eavg is less than half the peak output voltage (or approximately 0.318 of the
peak output voltage). For the full-wave rectifier, Eavg is approximately 0.637. This value is still much less
than the applied voltage. With no capacitor connected across the output of the rectifier circuit, the
waveform has a large pulsating component (ripple) compared with the average or dc component. Now
refer to figure 3-22. When a capacitor is connected across the output (in parallel with RL), the average
value of output voltage (Eavg) is increased due to the filtering action of capacitor C1.
3-18
reactance to the pulsating current and stores a substantial charge. The rate of charge for the capacitor is
limited only by the relatively low resistance of the conducting diode. The RC charge time of the circuit is,
therefore, relatively short. As a result, when the pulsating voltage is first applied to the circuit, the
cycles. The capacitor attempts to charge to the peak value of the rectified voltage anytime a diode is
conducting, and tends to retain its charge when the rectifier output falls to zero. (The capacitor cannot
discharge immediately). The capacitor slowly discharges through the load resistance (RL) during the time
the rectifier is nonconducting.
load resistance. If the capacitance and load resistance values are large, the RC discharge time for the
circuit is relatively long.
ripple component (Er) present across the load resistance.
and a load resistor (RL).
pulsating rectified current. The resistance of RL is assumed to be much greater than the reactance of C1 at
the input frequency.
3-19
through the circuit allowing C1 to charge. C1 will charge to approximately the peak value of the input
voltage. The charge is less than the peak value because of the voltage drop across diode V1. The charge
on C1 is indicated by the heavy, solid line on the waveform.
plate of V1 is negative in respect to the cathode. During this interval, C1 discharges through load
resistance RL. The discharge of C1 produces the downward slope indicated by the solid line on the
waveform in the figure.
of rectifier operation. Keep in mind that for good filtering, the filter capacitor should charge as fast as
possible and discharge as little as possible.
a substantial charge remains on the capacitor at the time of the next half cycle of operation. As a result, no
voltage of the charge remaining on C1. The charge on C1 is the cathode potential of the diode. When the
and C1 commences to charge to approximately the peak value of the applied voltage.
After the capacitor has charged to its peak value, it begins to discharge. Since the fall of the ac input
quickly becomes more positive than the plate, and the diode ceases to conduct.
3-20
operation we discussed for the half-wave rectifier. Notice in figure 3-25 that because one of the diodes is
always conducting on either alternation, the filter capacitor charges and discharges during each half cycle.
than the charge across the capacitor.)
the ease of filtering their output voltages. You can now see the reason for this. The ripple frequency is
doubled; therefore, the time period the capacitor is allowed to discharge is cut in half. This means that the
capacitor discharges less. Thus, ripple amplitude is less, and a smoother output voltage occurs.
and the average output voltage (Eavg) is the dc component of the output. Since the filter capacitor offers a
relatively low impedance to ac, the majority of the ac component flows through the filter capacitor. The
Eavg) flows through the load resistance. To clarify this statement, let's take a look at the formula for XC in
a half-wave and full-wave rectifier. First, you must establish some values for the circuit.
3-21
impedance of the capacitor is reduced by one-half. Therefore, when the simple capacitor filter is used in
3-22
ripple frequency decreases the capacitive reactance of the filter capacitor. This allows the ac component
to be passed through the capacitor more easily. Therefore, the output of a full-wave rectifier is much
easier to filter than that of a half-wave rectifier.
better the filtering action. By using the largest possible capacitor, we achieve the best filtering. The load
resistance is also an important consideration. If load resistance is made small, the load current increases,
and the average value of output voltage (Eavg) decreases. The RC discharge time constant is a direct
function of the current through the load. The greater the load current, the more rapid the discharge of the
capacitor, and the lower the average value of output voltage. For this reason, the simple capacitor filter is
seldom used with rectifier circuits that must supply a relatively large load current.
Q17. What is the most basic type of filter?
Q18. In a capacitor filter, is the capacitor in series or parallel with the load?
Q19. Is better filtering achieved at a high frequency or at a low frequency at the input of the filter?
Q20. Does a filter circuit increase or decrease the average output voltage?
Q21. What determines the rate of discharge of the capacitor in a filter circuit?
Q22. Does low ripple voltage indicate good or bad filtering?
Q23. Is a full-wave rectifier output easier to filter than that of a half-wave rectifier?
filter capacitor can be disconnected from the circuit.
REMEMBER-AN UNDISCHARGED CAPACITOR RETAINS ITS POLARITY AND HOLDS ITS CHARGE FOR LONG PERIODS OF TIME. TO BE SAFE, USE A PROPER SAFETY SHORTING PROBE TO DISCHARGE THE CAPACITOR TO BE TESTED WITH THE POWER OFF BEFORE CONNECTING TEST EQUIPMENT OR DISCONNECTING THE CAPACITOR.
and leakage resistance. During these checks it is very important that you observe correct polarity if the
capacitor is an electrolytic. A decrease in capacitance or losses within the capacitor can cause the output
to be below normal and also cause excessive ripple amplitude.
using an ohmmeter. You can make resistance measurements across the terminals of the capacitor to
determine whether it is shorted, leaky, or open. When you test electrolytic capacitors, set the ohmmeter to
the high range, and connect the test probes across the capacitor. Be careful to observe polarity. This is
than in the other. If you do not observe the correct polarity, you will get an incorrect reading and you may
damage the meter. When you first connect the test probes, a large deflection of the meter should take
charges. For a good capacitor with a rated working voltage of 450 volts dc, the final reading on the
3-23
ohmmeter should be over 500,000 ohms. (A rough rule of thumb for high-voltage capacitors is at least
1000 ohms per volt.) Low-voltage electrolytic capacitors (below 100 volts rating) should indicate
approximately 100,000 ohms.
found an open-circuit capacitor.
shorted.
minimum acceptable value. (Be certain this reading is not caused by an in-circuit shunting part.) To be
in which it operates.
mica and ceramic capacitors (for low-capacitance values). In this case, polarity is of no importance unless
the capacitor terminals are marked plus or minus. It is, however, good maintenance practice to use the
output polarity of the circuit as a guide, connecting positive to positive, and negative to negative. Thus,
or to test equipment is eliminated.
important and where the output current is relatively high and subject to varying load conditions. This
power supplies.
output filter capacitor (C1).
Inductor L1 is placed at the input to the filter and is in series with the output of the rectifier circuit.
constant current flowing to the load throughout the complete cycle of the applied voltage. As a result, the
approximates the average value of the rectified input to the filter, as shown in figure 3-27.
3-24
output voltage by an appreciable amount. (The dc resistance of the inductor is just a few ohms.)
from the rectifier circuit to the load. In addition, the reactance of the shunt capacitor (XC) presents a low
component around the load. The capacitor attempts to hold the output voltage relatively constant at the
average value of the voltage.
pulsating current and to store a substantial charge. The rate of the charge for the capacitor is limited by
developed by the coil. Therefore, the RC charge time constant (fig. 3-28) is short compared to its
discharge time.
3-25
filter choke (L1) produces a counter emf that opposes the constantly increasing input voltage. The net
result is to effectively prevent the rapid charging of the filter capacitor (C1). Thus, instead of reaching the
peak value of the input voltage, C1 only charges to the average value of the input voltage. After the input
load resistance (RL). C1 will attempt to discharge as indicated in figure 3-29. Because of its relatively
long discharge time constant, C1 can only partially discharge.
The larger the value of the filter capacitor, the better the filtering action. However, due to the
physical size, there is a practical limitation to the maximum value of the capacitor.
capacitor.
of the applied input voltage. This is shown in figure 3-30. As far as the ripple component is concerned,
the inductor offers a high impedance (Z) and the capacitor offers a low impedance. As a result, the ripple
component (Er) appearing across the load resistance is greatly attenuated (reduced). Since the inductance
compared with the value of ripple produced across the coil.
3-26
circuit is used to supply the input voltage to the filter. The rectifier voltage is developed across capacitor
C1. The ripple voltage in the output of the filter is the alternating component of the input voltage reduced
in amplitude by the filter section.
charges. Conduction occurs twice during each cycle for a full-wave rectifier. For a 60-hertz supply, this
produces a ripple frequency of 120 hertz. Although the diodes alternate (one conducts while the other is
nonconducting), the filter input voltage is not steady. As the plate voltage of the conducting diode
the counter emf developed by the choke. During the nonconducting interval, (when the plate voltage
drops below the capacitor charge voltage), C1 discharges through the load resistance RL. The components
in the discharge path cause a long time constant; thus C1 discharges slower than it charges.
inductive reactance to the 120-hertz ripple component produced by the rectifier. Therefore, the effect that
L1 has on the charging of the capacitor (C1) must be considered. Since L1 is connected in series with the
occurs. The greater the impedance of the choke, the less the ripple voltage that appears across C1 and the
output. The dc output voltage is fixed mainly by the dc resistance of the choke.
component values. For simplicity, the input frequency at the primary of the transformer will be 117 volts
60 hertz. We will use both half-wave and full-wave rectifier circuits to provide the input to the filter.
(1.414) = 165 V]. You should recall that the ripple frequency of this half-wave rectifier is 60 hertz.
Therefore, the capacitive reactance of C1 is:
3-27
This means that the capacitor (C1) offers 265 ohms of opposition to the ripple current. Note,
however, that the capacitor offers an infinite impedance to direct current. The inductive reactance of L1
is:
XL = 2!fL
XL = (2) (3.14) (60) (10)
XL = 3.8 Kilohms
This shows that L1 offers a relatively high opposition (3.8 kilohms) to the ripple in comparison to
the opposition offered by C1 (265 ohms). Thus, more ripple voltage will be dropped across L1 than across
C1. In addition, the impedance of C1 (265 ohms) is relatively low in respect to the resistance of the load
(10 kilohms). Therefore, more ripple current flows through C1 than the load. In other words, C1 shunts
most of the ac component around the load.
Let's go a step further and redraw the filter circuit so that you can see the voltage divider action.
(Refer to figure 3-33.) Remember, the 165 volts peak-to-peak 60 hertz provided by the rectifier consist of
both an ac and a dc component. The first discussion will be about the ac component. Looking at figure
therefore, the greatest amount of ac will flow through C1. (The heavy line indicates current flow through
the capacitor.) Thus the capacitor bypasses, or shunts, most of the ac around the load.
equivalent impedance of 265 ohms.
3-28
You now have a voltage divider as illustrated in figure 3-34. You should see that because of the
amount is dropped across C1 and RL. You can further increase the ripple voltage across L1 by increasing
the inductance:
XL = 2!fL
Now let's discuss the dc component of the applied voltage. Remember, a capacitor offers an infinite
(") impedance to the flow of direct current. The dc component, therefore, must flow through RL and L1.
As far as the dc is concerned, the capacitor does not exist. The coil and the load are, therefore, in series
with each other. The dc resistance of a filter choke is very low (50 ohms average). Therefore, most of the
the coil, as shown in figure 3-35.
3-29
is frequency sensitive, it provides a large resistance to ac and a small resistance to dc. In other words, the
coil opposes any change in current. This property makes the coil a highly desirable filter component. Note
with a full-wave rectifier as shown in figure 3-36. This is due to the decrease in the XC of the filter
capacitor and the increase in the XL of the choke. Remember, the ripple frequency of a full-wave rectifier
is twice that of a half-wave rectifier. For a 60-hertz input, the ripple will be 120 Hertz. Let's briefly
calculate the X C of C1 and the XL of L1:
3-30
the flow of ac. The greater the ac flow through the capacitor, the lower the flow through the load.
choke; consequently, less ripple is developed across the load. This condition provides better filtering.
Q24. In an LC choke-input filter, what prevents the rapid charging of the capacitor?
Q25. What is the value usually chosen for a filter choke?
Q26. If the inductance of a choke-input filter is increased, will the output ripple voltage amplitude (Er) increase or decrease?
An LC choke-input filter is subject to several problems that can cause it to fail. The filter capacitors
are subject to open circuits, short circuits, and excessive leakage. The series inductor is subject to open
windings and, occasionally, shorted turns or a short circuit to the core.
of the protection offered by the inductor; however, the capacitor can become open, leaky, or shorted.
Shorted turns in the choke may reduce the value of inductance below the critical value. This will
ripple amplitude, and poor voltage regulation. A choke winding that is open, or a choke winding that is
shorted to the core will result in a no-output condition. A choke winding that is shorted to the core may
cause overheating of the rectifier element(s), blown fuses, and so forth.
To check the capacitor, first remove the supply voltage from the input to the filter circuit. Then
disconnect one terminal of the capacitor from the circuit. Check the capacitor with a capacitance analyzer
to determine its capacitance and leakage resistance. When the capacitor is electrolytic, be sure to use the
correct polarity at all times. A decrease in capacitance or losses within the capacitor can decrease the
efficiency of the filter and produce excessive ripple amplitude. If a suitable capacitance analyzer is not
available, you can use an ohmmeter to check for leakage resistance. The test procedure is the same as that
described for the input capacitor filter.
and capacitive filter circuits. For the two remaining types of filters, we will discuss only the differences
between them and the other basic filters.
3-31
The RC capacitor-input filter is limited to applications in which the load current is small. This type
necessary. For example, RC filters are used in high-voltage power supplies for cathode-ray tubes and as
part of decoupling networks for multistage amplifiers.
Figure 3-37 shows an RC capacitor-input filter and its associated waveforms. Both half-wave and
full-wave rectifiers are used to provide the inputs.
output filter capacitor (C2). Although not part of the RC filter, RL is shown to help explain the circuit.
This filter is sometimes referred to as an RC pi-section filter because its schematic symbol resembles the Greek letter !.
capacitor must have an extremely large value. While electrolytic capacitors are available in sizes up to
10,000 µF or greater, the larger sizes are quite expensive. A more practical approach is to use a more
sophisticated filter that can do the same job but that has lower capacitor values, such as the RC filter.
The waveforms shown in the figure represent the unfiltered output from a typical rectifier circuit.
rectifier, is less than half the amplitude of the voltage peaks (approximately 0.318). The average value of
output voltage (Eavg) for the full-wave rectifier is greater than half (approximately 0.637), but is still much
less than, the peak amplitude of the rectifier-output waveform. With no filter circuit connected across the
as compared to the average (or dc) component.
3-32
An RC filter, such as a pi-section filter, does a much better job than a single capacitor filter.
Figure 3-37 illustrates an RC filter connected across the output of a rectifier. C1 performs the same
function that it did in the single capacitor filter. It is used to reduce the percentage of ripple to a relatively
low value. Thus, the voltage across C1 might consist of an average dc value of +100 volts with a ripple
voltage of 10 volts. This voltage is passed on to the R1-C2 network, which reduces the ripple even further
(view C).
C2 offers an infinite impedance (resistance) to the dc component of the output voltage. Thus, the dc
voltage is passed to the load, but reduced in value by the amount of the voltage drop across R1. However,
R1 is generally small compared to the load resistance. Therefore, the drop in the dc voltage by R1 is not a
drawback.
at the ripple frequency. C2 offers a very low impedance to the ac ripple frequency. Thus, the ac ripple
senses a voltage divider consisting of R1 and C2 between the output of the rectifier and ground.
Therefore, most of the ripple voltage is dropped across R1. Only a trace of the ripple voltage can be seen
across C2 and the load.
filtering can be added. In practice, the second stage is rarely required. The RC filter is extremely popular
because smaller capacitors can be used with good results.
The RC filter has some disadvantages, however. First, the voltage drop across R1 takes voltage away
from the load. Second, power is wasted in R1 and is dissipated in the form of unwanted heat.
 Even so, the
advantages of the RC filter overshadow these disadvantages in many cases.
Q27. Is an RC filter used when a large current or a small current demand is required?
Q28. Why is the use of large value capacitors in filter circuits discouraged?
Q29. When is a second RC filter stage used?
The resistor-capacitor (RC) filter is also subject to problems that can cause it to fail. The shunt
capacitors (C1 and C2) are subject to an open circuit, a short circuit, or excessive leakage. The series filter
resistor (R1) is subject to changes in value and occasionally opens. Any of these troubles can be easily
detected.
voltage surges. As a result, it is frequently subject to voltage breakdown and shorting. The remaining
by the series filter resistor (R1). However, a shunt capacitor can become open, leaky, or shorted.
A shorted capacitor or an open filter resistor results in a no-output indication. An open filter resistor
results in an abnormally high dc voltage at the input to the filter and no voltage at the output of the filter.
in value, result in an excessive ripple amplitude in the output of the supply.
3-33
The LC input filter is one of the most commonly used filters. This type of filter is used primarily in
current is low and the load current is relatively constant.
Figure 3-38 shows an LC capacitor-input filter and its associated waveforms. Both half-wave and
full-wave rectifier circuits are used to provide the inputs.
The waveforms shown in the figure represent the unfiltered output from a typical rectifier circuit.
amplitude of the voltage peaks. This is indicated by the dashed line. The average value of output voltage
rectifier-output waveform. With no filter circuit connected across the output of the rectifier circuit
(or dc) component.
A common type of LC filter is shown in figure 3-38. C1 performs the same functions as discussed
earlier by reducing the ripple to a relatively low level. L1 and C2 form the LC filter, which reduces the
ripple even further (view C).
L1 is a large value iron-core inductor (choke.) It has a high value of inductance and, therefore, a high
value of XL, which offers a high reactance to the ripple frequency. At the same time, C2 offers a very low
reactance to the ac ripple. L1 and C2 form an ac voltage divider and, because the reactance of LI is much
higher than that of C2, most of the ripple voltage is dropped across L1. Only a slight trace of the ripple
appears across C2 and the load.
While the L1-C2 network greatly reduces the ac ripple, it has little effect on the dc. You should
recall that an inductor offers no reactance to dc. The only opposition to current flow is the resistance of
3-34
the wire in the choke. Generally, this resistance is very low and the dc voltage drop across the coil is
minimal. Thus, the LC filter overcomes the disadvantages of the RC filter.
Aside from the voltage divider effect, the inductor improves filtering in another way. You should
recall that an inductor resists changes in the magnitude of the current flowing through it. Consequently,
when the inductor is placed in series with the load, the inductor tends to hold the current steady. This, in
turn, helps to hold the voltage across the load constant.
The LC filter provides good filtering action over a wide range of currents. The capacitor filters best
when the load is drawing little current. Thus, the capacitor discharges very slowly and the output voltage
remains almost constant. On the other hand, the inductor filters best when the current is highest. The
of components is a factor.
The LC filter has two disadvantages. The first is cost. The LC filter is more expensive than the RC
filter because its iron-core choke costs more than the resistor of the RC filter. The second disadvantage is
size, since the iron-core choke is bulky and heavy. Thus, the LC filter may be unsuitable for some
applications but is still one of the most widely used.
Q30. What is the most commonly used filter in use today?
Q31. What are the two main disadvantages of an LC capacitor filter?
Several problems may cause the LC capacitor filter to fail. Shunt capacitors are subject to open
occasionally shorted turns or a short circuit to the core.
voltage surges, and has a generally higher average voltage applied. As a result, the input capacitor is
frequently subject to voltage breakdown and shorting. The output capacitor (C2) is not as susceptible to
become open, leaky, or shorted.
no-output indication. A shorted capacitor, depending on the magnitude of the short, may cause a shorted
rectifier, transformer, or filter choke and result in a blown fuse in the primary of the transformer. An open
of the filter. A leaky or open capacitor in the filter circuit results in a low dc output voltage. This
condition is generally accompanied by an excessive ripple amplitude. Shorted turns in the winding of a
filter choke reduce the effective inductance of the choke and decrease its filtering efficiency. As a result,
the ripple amplitude increases.
Ideally, the output of most power supplies should be a constant voltage. Unfortunately, this is
difficult to achieve. There are two factors that can cause the output voltage to change. First, the ac line
voltage is not constant. The so-called 115 volts ac can vary from about 105 volts ac to 125 volts ac. This means that the peak ac voltage to which the rectifier responds can vary from about 148 volts to 177 volts.
The ac line voltage alone can be responsible for nearly a 20 percent change in the dc output voltage.
The second factor that can change the dc output voltage is a change in the load resistance. In
complex electronic equipment, the load can change as circuits are switched in and out. In a television
3-35
settings, or even the channel selected.
has a fixed internal resistance. If the load resistance decreases, the internal resistance of the power supply
drops more voltage. This causes a decrease in the voltage across the load.
Many circuits are designed to operate with a particular supply voltage. When the supply voltage
changes, the operation of the circuit may be adversely affected. Consequently, some types of equipment
resistance or changes in the ac line voltage. This constant supply of power may be achieved by adding a
circuit called the VOLTAGE REGULATOR at the output of the filter.
A commonly used FIGURE OF MERIT for a power supply is its PERCENT OF REGULATION. The figure of merit gives us an indication of how much the output voltage changes over a range of load resistance values. The percent of regulation aids us in determining of the type of load
regulation needed. Percent of regulation is determined by the equation:
produced at full loading. For example, assume that a power supply produces 12 volts when the load
current is zero. If the output voltage drops to 10 volts when full load current flows, then the percent of
regulation is:
Ideally, the output voltage should not change over the full range of operation. That is, a 12-volt
power supply should produce 12 volts at no load, at full load, and at all points in between. In this case, the
percent of regulation would be:
3-36
Thus, zero-percent load regulation is the ideal situation. It means that the output voltage is constant
under all load conditions. While you should strive for zero percent load regulation, in practical circuits
you must settle for something less. Even so, by using a voltage regulator, you can hold the percent of
regulation to a very low value.
If you are interested in reading more on this subject, refer to the Electronic Installation and Maintenance Book (EIMB) series or other similar books from your technical library.
Q32. What two factors can cause output dc voltage to change?
Q33. What is the commonly used figure of merit for a power supply?
Q34. If a power supply produces 20 volts with no load and 15 volts under full load, what is the percent of regulation?
Q35. What percent of regulation would be ideal?
current requirements. Because many military electronic equipments require operating voltages and
currents that must remain constant, some form of regulation is necessary. The circuits that maintain power
supply voltage or current outputs within specified limits, or tolerances, are called regulators. They are
designated as dc voltage or dc current regulators, depending on their specific application.
and filter sections. The purpose of the voltage regulator is to provide an output voltage with little or no
variation. Regulator circuits sense changes in output voltages and compensate for the changes. Regulators
that maintain voltages within plus or minus (±) 0.1 percent are quite common. The diagram in figure 3-39
clearly illustrates the purpose of the voltage regulator.
3-37
There are two basic types of voltage regulators, series and shunt. Whether a voltage regulator is
the circuit load resistance.
Figure 3-40 illustrates the two basic types of voltage regulators. In actual practice the circuitry of
regulating devices may be quite complex. We use the simplified drawings in the figure to emphasize that
there are two basic types of voltage regulators. Broken lines highlight the differences between the series
and shunt regulators.
The schematic in view (A) is that of a shunt-type regulator. It is called a shunt-type regulator because
the regulating device is connected in parallel with the load resistance. This is a characteristic of all shunt-
type regulators. The schematic in view (B) is that of a series regulator. It is called a series regulator
because the regulating device is connected in series with the load resistance.
Figure 3-41 illustrates the principle of series voltage regulation. As you study the figure, notice that
the regulator is in series with the load resistance and that all current passes through the regulator. In this
example, variable resistor Rv is used for regulation. Examine the circuit to determine how the regulator
functions. When the input voltage increases, the output voltage also increases. However, since the voltage
greater voltage drop through Rv. This causes the output voltage to decrease to normal or, for all practical
purposes, to remain constant.
3-38
decreases almost simultaneously, thereby compensating for the voltage drop. Since there is a smaller
voltage drop across Rv, the output voltage remains almost constant. Voltage fluctuations within the circuit
occur in microseconds.
The diagram in figure 3-42 represents a shunter voltage regulator. Notice that variable resistor Rv is
in parallel with the load resistance RL and that fixed resistor RS is in series with the load resistance. You
or decrease) in the current through it.
resistor (RS). For now, assume that the circuit in figure 3-42 is operating under normal conditions, that the
input is 120 volts dc, and that the desired regulated output is 100 volts dc. For a 100-volt output to be
maintained, 20 volts must be dropped across the series resistor (RS). If you assume that the value of RS is
2 ohms, then you must have 10 amperes of current through Rv and RL. (Remember: E = IR.) If the values of the resistance of Rv and RL are equal, then 5 amperes of current will flow through each resistance (Rv
and RL).
Now, if the load resistance (RL) increases, the current through RL will decrease. For example, assume
that the current through RL is now 4 amperes and that the total current through RS is 9 amperes. With this
increased to 102 volts. At this time, the regulating device (Rv) decreases in resistance, and 6 amperes of
current flows through this resistance (Rv). Thus, the total current through RS is once again 10 amperes (6
output to decrease back to 100 volts.
3-39
decreases its resistance to compensate for the change. If RL decreases, the opposite effect will occur and
Rv will increase. Now take a look at the circuit when a decrease in load resistance takes place.
When RL decreases, the current through RL subsequently increases to 6 amperes. This action causes a
total of 11 amperes to flow through RS which now drops 22 volts. As a result, the output is now 98 volts.
amperes) flows through Rv. The total current again becomes 10 amperes, and the output is again 100
volts.
output voltage by sensing the current change that occurs in the parallel resistance of the circuit.
compensate for changes in input voltages. You know, of course, that the input voltage may vary and that
any variation must be compensated for by the regulating device. Consider an increase in input voltage.
between Rv and RS. You should see, therefore, that the regulator operates in the opposite way to
compensate for a decrease in input voltage.
type of regulation has limitations. Obviously, the variable resistor cannot be adjusted rapidly enough to
compensate for frequent fluctuations in voltage. Since input voltages fluctuate frequently and rapidly, the
variable resistor is not a practical method for voltage regulation. A voltage regulator that operates
continuously and automatically to regulate the output voltage without external manipulation is required.
Q36. The purpose of a voltage regulator is to provide an output voltage with little or no ____.
Q37. The two basic types of voltage regulators are _______ and _______.
Q38. When a series voltage regulator is used to control output voltages, any increase in the input voltage results in an increase/a decrease in the resistance of the regulating device.
Q39. A shunt type voltage regulator is connected in series/parallel with the load resistance.
now need to cover the capabilities and limitations of the VR tube itself.
Figure 3-43 shows a basic VR tube regulating circuit. The voltage produced by the source is 150
volts. The VR 90 will provide a constant 90 volts across the load resistance (RL) if the tube is operated in
the normal glow discharge region. This means that 60 volts is dropped across RS, which is the series
limiting resistance used to limit the current through the VR tube.
3-40
normal point of operation. The normal point of operation, which allows maximum variation in current,
must be midway between the current limits of the tube. This median current is called Imean. and can be
calculated by the use of the following equation:
values:
To calculate the value of series dropping resistance R S, we use the following equation:
series dropping resistance in the following manner:
3-41
current of 100 milliamperes flows through RL. The internal resistance of the VR tube can be calculated in
a similar manner. With 22.5 milliamperes flowing and 90 volts dropped across the VR tube, its resistance
is 4 kilohms.
of 150 volts and a variable load resistance. If the value of RL were to decrease to 857 ohms, the load
current would increase to approximately 105 milliamperes to maintain 90 volts across the load resistance.
RS must drop 60 volts. To do so requires a current of 122.5 milliamperes flowing through the series
resistance. Since 105 milliamperes is now flowing through the load, the current through the VR tube must
decrease from 22.5 milliamperes to 17.5 milliamperes. We will discuss the sequence of events in more
detail to help you better understand how the tube current is made to vary.
The original load resistance was 900 ohms. Changes in this resistance will not occur instantaneously,
but will require some time to vary from 900 ohms to a new value. As resistance of the load begins to
decrease, load current begins to increase. The minute increase in load current will flow through the series
resistance RS causing a slight increase in ERS. This slight increase in voltage across RS will result in the
VR tube voltage dropping slightly. This slight drop in tube voltage will cause a decrease in the ionization
of the tube gas, which in turn increases the resistance of the tube. As a result, less current flows through
the tube.
Note that tube current can decrease only to a value of 5 milliamperes before deionization occurs.
Therefore, the load current cannot exceed 117.5 milliamperes, for beyond this value, tube current
becomes less than 5 milliamperes and regulation ceases.
If load resistance were to increase, load current would decrease. This would result in the VR tube
current increasing to maintain a current of 122.5 milliamperes. The VR tube current can only increase to
40 milliamperes. Beyond this value of current, the tube enters the abnormal glow region and tube voltage
increases.
The upper limit of the VR tube current will occur when load current decreases to a value of 82.5
milliamperes. When load current drops below this value, the VR tube ceases to regulate the load voltage.
reached when current in the load exceeds 117.5 milliamperes or drops below 82.5 milliamperes.
The VR tube regulator can also compensate for changes in power supply voltage. Under these
conditions, the load resistance will remain constant while the power supply voltage will be variable. Refer
to figure 3-44 for the following discussion.
volts. As this voltage increases, current through RS increases from its original value of 122.5
milliamperes. Initially, this additional current is drawn from the load, causing a slight increase in load
voltage. This increase in load voltage is felt across the VR tube and causes an increase in tube ionization.
This decreases the internal resistance of the VR tube with a resultant increase in tube current. When
source voltage reaches 155 volts, current through RS is approximately 133 milliamperes (RS = 490 ohms).
Most of the additional current through RS flows through the VR tube. As a result, approximately 33
milliamperes flows through the VR tube, maintaining the load voltage at 90 volts.
limit of 5 milliamperes at some point. When source voltage drops below 141.4 volts, tube current will be
less than 5 milliamperes and regulation will cease. The upper and lower limits of the supply voltage
variations that can be allowed and still provide regulation in the circuit are 158.6 volts and 141.4 volts,
3-42
respectively. Remember that tube voltage varies slightly through its operating range, but this voltage
change is less than that which would exist without the use of a VR tube.
As the source voltage increases, the current through the VR tube increases. Since the upper limit of
tube current is 40 milliamperes, there is a limit in the ability of the tube to regulate increasing voltage.
When the supply voltage exceeds 158.6 volts, tube current will be greater than 40 milliamperes and
regulation will cease.
490-ohm series resistance (RS) to maintain the load voltage at 90 volts. Current through R S for a 55 volt
drop is 112 milliamperes. Since load current is 100 milliamperes, the remaining 12 milliamperes must
flow through the VR tube. This represents a decrease in the ionization level of the VR tube, with a
resultant increase in tube resistance. Under these conditions, 90 volts will be maintained across the load
resistance.
two or more tubes may be placed in series as shown in figure 3-45.
In the figure, a VR75-30 and a VR105-40 are shown connected in series. The source voltage is 250
volts, and 82.5 milliamperes flows through the load resistance. Since current through the two VR tubes is
common, the limits of regulation are determined by the tube having the smaller current limitations. (In
this case, the VR 75-30). In computing Imean for this circuit, Imax and Imin will be 30 milliamperes and 5
milliamperes, respectively. Therefore, the mean current will be 17.5 milliamperes.
current through R S (load current + Imean). Using these values, RS = 700 ohms. Note that the regulated
voltage to the load is 180 volts. This provides a regulated voltage greater than would be possible using
either VR tube by itself.
Another advantage of using VR tubes in series is illustrated in figure 3-46. In this circuit, several
values of regulated voltages are obtained from a single power supply.
3-43
current through V1. The current through V3, on the other hand, is the sum of the currents through V2 and
R2. Since V3 has more current flowing through it than any of the other VR tubes, it places or determines
the limit on the maximum current in the VR tube circuit. Since the maximum rating of V3 is 40
V3 will be exceeded and regulation will cease.
single power supply. The primary disadvantage is in the current limitations. Since it is impossible to have
wide ranges of variations in load resistance or source voltage.
current handling capacity of the network. Although this is true for some gas-filled tubes, it is not true for
VR tubes. In figure 3-47, two VR tubes are constructed in exactly the same way. The only difference will
be a slight variation in their ionization potential. For the purpose of this discussion, VR tube VR1 will
have a lower ionization potential than VR2. The potential that must be reached before a VR tube ionizes
is considerably higher than its normal operating voltage.
3-44
begins to conduct and the potential across it decreases to its operating voltage. The potential across VR2
never becomes high enough to cause it to ionize. Therefore, placing the VR tubes in parallel accomplishes
no useful purpose. When greater current handling capacity and better regulation are desired, electronic
(vacuum tube) regulator circuits are used.
Several conditions may either indicate or cause problems with a VR tube regulator. Initially, you can
determine the presence of the characteristic glow from the ionized gas within the tube. When current
through the tube is near its maximum rating, the tube is highly ionized. When the current is near its
tube is an indication of tube conduction. If the tube is not ionized, however, this does not necessarily
mean that the tube is defective. The same indication (lack of characteristic glow) may also result from the
normal, the load current is below normal, or the load current is excessive. You therefore need to make dc
whether the problem is inside the regulator circuit or outside of it.
whether any change in resistance has occurred. If the maximum current rating of the regulator tube is
characteristics; therefore, you can suspect the regulator tube itself as a possible source of trouble.
greater degree of regulation than a VR tube can provide. For these circuits, an electron tube voltage
regulator is used.
An electron tube may be considered a variable resistance. When the tube is passing a direct current,
the dc plate resistance (Rp). For a given plate voltage, the value of Rp depends upon the tube current, and
the tube current depends upon the grid bias.
Refer to figure 3-48, view (A). The resistance of V1 is established initially by the bias on the tube.
Assume that the voltage across the load is at the desired value. Then the cathode is positive with respect
to ground by some voltage (EL). The grid can be made positive relative to ground by a voltage (E2) that is
less than E1. The potentiometer R2 is adjusted until the bias (grid-to-cathode voltage), which is E2 - E1,
is sufficient to allow V1 to pass a current equal to the desired load current. With this bias, the resistance
of V1 is established at the proper value to reduce the rectifier output voltage to the desired load voltage.
3-45
If the rectifier output voltage increases, the voltage at the cathode of V1 tends to increase. As E1
increases, the negative bias on the tube increases and the plate resistance of the tube becomes greater.
Consequently, the voltage drop across V1 increases with the rise in input voltage. If the circuit is designed
input. Thus the load voltage remains essentially constant.
The resistor (R1) is used to limit the grid current. This is necessary in this particular circuit because
the battery is not disconnected when the power is turned off. However, the battery can be eliminated from
the grid of the tube. The action of the circuit in view (B) is the same as the action of the circuit in view
(A). The output voltage of the simple voltage regulators shown in the figure cannot remain absolutely
constant. As the rectifier output voltage increases, the voltages on the cathode of V1 must rise slightly if
the regulator is to function.
from the rectifier, but also for changes in the load. For example, in view (B) if the load resistance
decreases, the load current will increase. The load voltage will tend to fall because of the increased drop
across V1. The decrease in load voltage is accompanied by a decrease in bias voltage on V1. The bias
voltage on V1 is equal to E1 - E2. Thus the effective resistance of V1 is reduced at the same time the load
current is increased. The IR drop across V1 increases only a slight amount because R decreases about as
much as I increases. Therefore, the tendency for the load voltage to drop when the load is increased is
checked by the decrease in resistance of the series triode.
Q40. In an electron tube regulator, the electron tube replaces what component?
3-46
regulation. In most power supplies, current is not regulated directly. Fuses and other circuit protection
devices are used to set an upper limit on the amount of current that can flow in a power supply. Once this
limit is exceeded, the fuse simply opens and the power supply is deenergized. Beyond this, current is
it needs. Decreases and increases in the power supply voltage caused by the variations in load current are
usually controlled by the voltage regulator.
There are some cases in which current must be regulated or kept at a relatively constant value. The
designed to supply filament power to many tubes. You can see this in view (A) of figure 3-49, which is a
portion of a power supply designed to supply 50 vacuum tubes with both plate and filament voltages.
Under normal conditions, circuit current will not exceed 2.5 amperes. For this reason, the power supply
has been fused at 3 amperes. Because you are only interested in current regulation at this time, only the
and four of the 50 parallel connected vacuum-tube filaments. At operating temperatures, the resistance of
each filament is 1 kilohm. Because the filaments are connected in parallel, the total filament resistance at
operating temperature is 20 ohms. Ohm's law,
3-47
tells you that the filaments draw .315 amperes. You should know from your previous study that as
conductors are heated, their resistance increases. Therefore, the cold resistance of the filaments is
considerably lower than the hot resistance. In this case, assume 100 ohms per filament. The total
filaments draw 3.15 amperes of current. If the current for the rest of the power supply is added to the
energized. Unfortunately, the power supply is fused at 3 amperes. Under these conditions, it would be
temperature.
There are three possible solutions to this problem. The first is simply to fuse the power supply at 5
amperes, but this could allow excessive current to flow in the power supply. Another solution is to use a
slow-blow fuse. Unfortunately, the duration of the current surge may exceed the time limit that a slow-
blow fuse can handle. Therefore, current regulation is the best solution to this problem.
Because of its quick-heating ability, the amperite tube is ideal as a current regulator. The amperite
regulator is nothing more than an iron wire enclosed in a hydrogen-filled envelope. Because of its
construction, the iron filament will heat quickly when current is applied to it.
load. When the power supply is first energized, the iron wire of the amperite gets hot quickly and presents
a large resistance connected in series with the 2 ohms of filament resistance. As a result, most of the
voltage is dropped across the amperite. Because of the large resistance of the amperite regulator, current
in the circuit is held to an acceptable level in accordance with Ohm's law:
As the filaments warm up, their resistance increases, which causes circuit current to decrease. The
decreasing circuit current allows the iron wire of the amperite to cool. As it cools, its resistance decreases
until it reaches the approximate resistance of the circuit wiring. You might think that decreasing the
resistance of the amperite would allow circuit current to increase again, but this does not happen. As the
increases. Throughout the entire heating cycle of the filaments, the total resistance of the series circuit,
consisting of the amperite and tube filaments, remains fairly constant. When power is first applied, most
of the resistance is in the amperite. Therefore, most of the voltage is dropped across the resistance of the
amperite. Halfway through the cycle, the resistance of the amperite and the resistance of the filaments are
approximately equal, and the voltage drops across the two series elements are equal. Finally, when the
filaments have reached their operating temperature, most of the resistance is in the filaments of the tube.
Therefore, most of the voltage is dropped across the tube filaments.
throughout the heating cycle. As the cycle progresses, the resistance of the amperite decreases as the
resistance of the tube filaments increases. Because resistance and voltage (6.3 volts) remain constant,
necessary to heat up the iron wire of the amperite.
many variables that affect good regulation.
because your electronic equipment depends upon good regulation to operate property.
3-48
Up to this point we have discussed only the individual sections of the electron tube power supply. In
power supply.
Q41. What is the purpose of the amperite regulator?
Q42. As the tube filaments in the load heat up, will the circuit current increase or decrease?
Whenever you work with electricity, you must follow all the appropriate safety precautions. In the front of all electronic technical manuals, you will always find a section on safety precautions. You should
also find posted on each piece of equipment a sign listing the specific precautions for that equipment. One