AM RECEIVERS WITH TUBES [Radio Service Training Manual (1966)]


Fig. 2-1 shows a block diagram of a typical AM broadcast receiver. The order of the blocks in the signal path must be kept clearly in mind when studying the isolation procedures for symptoms in the following sections. In this section the operation of the circuits will be described briefly, with emphasis on those points of circuit theory which are important to the servicing technician.


The power supply converts the 120-volt AC line voltage to the required DC voltage for B+ in the radio. The following are the types of power supplies used in radio receivers.

1. The half-wave rectifier. Used in a transformerless circuit, it utilizes either a diode tube, such as the 35W 4, or a selenium or silicon rectifier.

2. The full-wave, transformer-powered type. It uses a dual diode, such as the 5U4, 5Y3, 5V3, 6X4, or 6X5.

3. The half-wave voltage doubler.

4. The full-wave voltage doubler.

5. The full-wave bridge.

Fig. 2-1. Block diagram of a 5-tube AC/DC receiver.

Half-Wave Transformerless Rectifier

Fig. 2-2. Half-wave transformerless rectifiers. (A) Selenium type. (B) Vacuum-tube type.

The first type listed above is shown in Fig. 2-2. Rectification occurs only on that half of the input cycle when the polarity on the anode (arrowhead) of the diode is positive, and the cathode is negative. The currents shown in solid lines flow during this half of the cycle, charging filter capacitors C1 and C2. On the other half of the cycle, the currents shown by dotted lines will flow. It can be seen that the electrons stored on the negative plates of the capacitors are released to flow through the load during the time when the input voltage puts a negative polarity on the plate of the diode. In this manner, a continuous current is available for the tubes in the load, provided the capacities of the electrolytics are large enough to support the total load cur rent for one-half cycle.

If no load current is drawn, the electrolytic capacitors will be charged by the solid-line current to approximately 1.41 times the 120-volt input voltage. But, in practice, the output voltage of the supply will be less than this value because of the voltage drops across R1, R2, and the rectifier when current is drawn through the load. With a moderate load, the normal output voltage will be from 110 to 130 volts DC with a 120-volt rms input.

R1 is called a surge resistor. Its purpose is to reduce the first surge of current into the completely discharged electrolytic capacitors when the switch is turned on. This current might be high enough to exceed the rating of the rectifier because the uncharged capacitors represent a short circuit.

R2 serves to slow down the discharge of C1 during the off duty half-cycle of the input. The result is a filling of the deep valley between the peaks of voltage across C1. This reduces the 60-cycle ripple on the DC output voltage and maintains it at a higher value. R2 is sometimes replaced by a choke.

R3 and C3 isolate the B- from the chassis. Since B- is connected directly to one side of the 120-volt lines, a dangerous condition would result if this 120 volts were also connected to the chassis. In one position of the plug in the 120-volt socket, the chassis would be connected to the side of the 120-volt line which is not grounded to water pipes, conduits, etc., in the house or the repair shop. With this connection there would be 120 volts between the chassis and any grounded object. Some power supplies of this type are not isolated from the chassis, as many technicians have found out to their surprise.

This type of power supply often uses a 35W 4 rectifier tube instead of a selenium or silicon diode. The circuit action is the same in either case. The plate of the 35W 4 tube is connected to the tap on the filament of the tube. Thus, part of the filament resistance is in series with the plate of the rectifier and with the flow of B+ current. In this way, the resistance between filament pins 4 and 6 serves as the surge resistor. This means that a short across the output of the power supply will cause the filament of the 35W4 to burn out and remove all B+. This is a common failure in radios using this type of power supply. A pilot lamp is frequently connected across pins 4 and 6 to shunt some current around the filament.

Fig. 2-3. Transformer-powered full-wave power supply.

The tube filaments must be connected in series across the 120-volt line. This means that the total of all the filament volt ages must be approximately 120 volts, and that the tubes must all draw the same filament current. The tubes shown all draw 150 ma of filament current, and the voltage for each tube is given by the first numbers of the tube designation.

The Full-Wave Transformer-Operated Power Supply

The circuit is shown in Fig. 2-3. On one half of the input cycle, the lower plate of the rectifier is positive, and the solid line currents flow. On the other half cycle, the upper plate is positive, and the dotted-line currents flow. Both currents flow out of the transformer secondary through the center-tap, charging the filters, supplying the load, and returning to the cathode. Thus, there will be charging current for the filters and current supplied to the load on both halves of the input cycle.

This is one advantage of the full-wave system.

Since the filters are charged twice during each input cycle, the valleys in the voltage waveform are narrower and occur at a rate of 120 cycles per second instead of 60 cycles as with the half-wave systems. The filters are not required to support the load during an entire off-duty input cycle, and the DC volt age output can be made quite free from ripple even under heavy loads.

The use of the transformer in this power supply gives an other advantage-the line voltage can be stepped up to any desired value. Usually, the voltage at the ends of the secondary winding (which are connected to the plate of the rectifier) is about 600 volts rms. But since only half of the transformer is used at a time, about 300 volts DC is produced across the filters.

If the secondary voltage were 1200 volts rms, then with no load, the filters would be charged to about 846 volts on the peaks.

½ ( 1200) ( 1.41) = 846 volts

Since any amount of voltage output is possible with the proper turns ratio in the transformer and sufficiently high breakdown ratings of the capacitors, this type of power supply is used in high-power equipment. When the transformer is used in low-power equipment, it is because of the better filtering which is possible with heavy-current loads.

It is customary for the transformer to have a separate winding to provide voltage for the filaments of the tubes. The tubes are connected in parallel across this filament winding. This requires that all tubes have the same filament voltage, although the currents drawn by each may differ.

The Half-Wave Voltage Doubler

The circuit of a half-wave voltage doubler is shown in Fig. 2-4A. The 120-volt AC input is represented by the generator. In Fig. 2-4B, the charging half-cycle is shown with only one diode present because, with this polarity, D2 does not con duct, rendering the right half of the circuit inactive. Electrons flow, charging C1 to the generator voltage. In Fig. 2-4C the other half of the input cycle is shown with D2 and C2 in the circuit, but with D1 omitted because the polarity is such that it does not conduct. Electrons flow from the generator, charging C2 and passing through the load, returning to C1 through D2.

It is because C1 was already charged to approximately 120 volts that voltage doubling now takes place. The voltage previously on C1 during the first half cycle is now in series with the generator voltage. The voltage applied to C2 and to the load is therefore the sum of the generator voltage and the voltage stored on C1, or approximately 240 volts.

Theoretically, the output voltage would rise to twice the peak generator voltage if there were no load, but, in practice, there are losses in the diodes and capacitors. Thus, the output volt age supplied from C2 and C3 tends to drop during the off-duty input cycle because the load must be supported entirely from the charges on C2 and C3. With moderate loads, the output is seldom more than 285 volts DC; its value will be greater if the capacitances are made larger and the load current smaller.

(A) Basic circuit. (B) Charging half-cycle. Filters supplying load current. (C) Charge on C1 is in series with source.

Fig. 2-4. Transformerless half, wave doubler power supply.

Since the system is essentially a half-wave rectifier, the out put capacitors are charged only during one half of the cycle, so the ripple frequency is 60 hz. The voltage rating of C1 is usually 150V, and the rating of C2 and C3 is 350V to 450V. C3 does not play a part in the doubling, but is merely a ripple filter.

The advantage of the doubler is its ability to produce larger output voltages without a transformer. Thus, in order to avoid the filament winding necessary in the transformer-type models, the filaments are usually connected in series as they are in a regular half-wave transformerless system.

Full-Wave Doubler

The operation of a full-wave doubler, shown in Fig. 2-5, is similar to that of the half-wave doubler except that C1 and C2 are each charged separately on alternate halves of the input cycle. The ripple frequency is 120 hz and is thus easier to filter. In addition, the system is capable of somewhat more out put current before a serious decrease in the amount of output voltage occurs.

The full-wave doubler is not used often, however, because it is impractical to connect the tube filaments in series across the 120-volt line as they are in the half-wave system. This necessitates the use of a transformer to supply the filaments, and the main advantage of the doubler is lost.

Fig. 2-5. Full-wave doubler power supply.

The reason why it is not practical to connect the filaments in series across the line can be seen by a study of Fig. 2-5.

The cathodes of the tubes in the load are returned to B-, and their heaters must be returned to the 120-volt lines if no transformer is used. B- is connected to the bottom end of C2, and the heaters are actually connected to the top end of C2.

Thus, the voltage across C2 (about 120 volts) appears be tween the cathode and heater of the first tube in the series string. The second tube in the string would have a heater-to cathode voltage lower than the first by the amount of heater voltage dropped in the first tube, etc. Not many tubes are manufactured to withstand such high heater-cathode voltages, so the full-wave doubler is not widely used in the transformer less circuit. The full-wave circuit has been used, however, to double the voltage from a transformer secondary when very high DC voltages are required, such as those in radio transmitters, test equipment, and X-ray equipment.

Fig. 2-6. Full-wave bridge rectifier.

The Full-Wave Bridge

Besides the extra expense and bulk of the transformer used in the conventional full-wave circuit described previously, the circuit also has the disadvantage of producing a DC voltage from only one half of the full secondary voltage because the winding must be center-tapped. The bridge circuit in Fig. 2-6 rectifies the full secondary voltage and is capable of twice the DC output that the conventional circuit using the same transformer has. The arrows show that the system is a full-wave rectifier using two of the diodes on each half of the input cycle. This method of rectification is capable of greater output current at high voltages than the others, but it is expensive, requiring four rectifiers and a transformer.

The important points about power supplies are summarized here for convenience.



DC Output Ripple Voltage Type Voltage Frequency Regulation Under Load Under Load Half-Wave Approx.

Transformer less equal to 60 hz fair input rms

Full-Wave Approx.

Transformer equal to ½ 120 hz good therms secondary

Half-Wave Doubler Approx. 2 60 hz poor x input rms Full-Wave Doubler Approx. 2 120 hz fair x input rms Approx.

Bridge equal to 120 hz good secondary rms


In addition, the following facts should be remembered:

1. Rms is the effective value of AC as read on an ordinary meter.

2. With no load, filters will be charged to 1.41 times the input rms. This voltage can exceed the rating of the filters if the power supply is operated with no load.

3. In voltage doublers, the filters must have breakdown voltage ratings greater than 2.82 x the rms input.

4. The output voltage under load is dependent on the capacity of the filters. With heavier loads, larger capacitors are needed.

5. Hum is the result of the deep valleys in the output wave form. The frequency of the peaks is called the ripple frequency.

6. The filter capacitor connected closest to the rectifier is the input capacitor and serves mainly to support the output voltage. Capacitors connected later in the circuit are ripple filters.

7. Transformerless circuits can have a dangerous voltage between B- and grounded objects, such as water pipes or conduits. Therefore, the B- should be isolated from the chassis.

8. A surge resistor is always placed between the input filter and the rectifier to reduce the current when the supply is first turned on. This resistor is part of the filament in a 35W 4-type tube and is frequently shunted by a pilot lamp.


The preceding discussion of power supplies provides the background for understanding the troubleshooting method for a common and very simple symptom. A dead receiver is one in which there is no sound of any kind present when the volume control is turned to maximum. This trouble nearly always leads to repairs in the power supply. One of the exceptions is the case where the speaker itself has failed. Discussion of speaker failure will be reserved for later sections in which a more extensive analysis of the complicated symptoms will be presented.

In receivers having transformerless power supplies where all the tube filaments are connected in series, a burned-out filament will break the series string, and none of the tubes will light. This results in a completely dead receiver when the rectifier is one of the tubes in the series-filament string, because there will be no B+. The 35W4 rectifier tube is commonly used in these circuits, and part of its filament also carries B+ current, as was explained earlier. Any short in the load will cause excessive current in the 35W 4 filament and burn it out. This is the most frequent cause of a dead receiver and should always be checked first.

Often the filament of a new rectifier will also burn out immediately when inserted in the socket, due to the same cause that burned out the first one. So, it is advisable to measure the resistance between the rectifier cathode at the socket and B- before inserting a new tube. With the filter capacitors connected, the ohmmeter will swing to a very low resistance and gradually climb to 10K or more if there is no short. With a resistance greater than 10K it is safe to install the new rectifier. If the receiver operates normally for a few hours, the technician should conclude that the rectifier filament burned out through continued normal operation.

If the rectifier filament is good, or the receiver uses a selenium or silicon rectifier, the other tubes should be checked next. This can usually be done without removing the chassis by simply measuring across the filament pins (pins 3 and 4 on most seven-pin miniature types) with an ohmmeter. A deflection will indicate that the filaments are not burned out.

If all of the filaments are found to be good, the chassis should be removed to check the switch and line cord. This can also be done by using the ohmmeter. A good way is to short the prongs of the AC plug and, with the switch on, place the ohmmeter across the two ends of the cord where they are soldered in the receiver.

Instead of using the ohmmeter, some technicians short across the switch terminals with the receiver plugged into the outlet and watch for the filaments to light. If the filaments do not light, the line cord is to be suspected. If they do light, the switch is to be suspected.

In a few cases of dead-receiver symptoms, all the tube filaments as well as the line cord will be found good, and still the filaments fail to light. This can only be due to a break in the filament wiring between the tube sockets. An easy way to find such a defect is to use the AC voltmeter in the manner shown in Fig. 2-7. Starting at either end of the filament string, the meter probes are placed across successive points in the line where continuity is expected. If the points are actually connected, there will be no reading on the meter; but when the probes are placed across an open place in the line the meter will read the full line voltage if the line cord and the switch are both good.

When a receiver using a transformer-powered rectifier with parallel filaments is found completely dead, the analysis is much simpler. This is because it is very unlikely that all the filaments are burned out at the same time. The technician looks for one of three conditions:

1. All filaments, including that of the rectifier, are out, indicating that the trouble lies in the primary side of the transformer.

2. Only the rectifier filament is out, indicating that the transformer primary circuit is energized and that the trouble lies either in the rectifier tube itself or in its filament winding on the transformer, which is separate from the rest of the filaments.

3. The rectifier is lit, and one or more of the other filaments in the amplifier tubes are out, indicating again that the primary circuit of the transformer is energized. This means the rectifier circuit is working and that the trouble lies in the filament winding for the amplifier tubes, or in the filament wiring of one or more of the tubes, or per haps only in the tubes themselves.

Further testing will be governed by which one of these conditions exists. In the first case, the primary of the transformer, the line cord, the switch and the fuse (if one is present) should be examined for continuity with the ohmmeter.

In the second case, after the rectifier tube is replaced, the rectifier filament-winding of the transformer should be checked for continuity.

Fig. 2-7. Checking for continuity with voltmeter.

In the third case, if none of the amplifier tubes are lit, the voltmeter can be used to check the AC output of the filament winding. If only a few tubes are out, they should be replaced.

The ohmmeter will aid in finding breaks in the wiring between the tube sockets, but the technician must remember that the tube filaments are connected in parallel.

Fig. 2-8. A beam-power pentode output stage.


A typical output stage using a 50C5 beam-power pentode is shown in Fig. 2-8. The cathode is biased by allowing the cathode current to be drawn through R3 to produce the cathode voltage of +5. With the grid returned to ground (zero volts), there will be a potential difference of 5 volts between grid and cathode, with the grid being negative. The fact that no voltage is dropped across R2 permits the control grid to be at ground potential (zero volts). This may be confusing when first en countered, but it can be understood when one remembers that no current flows in R2.

The value of R3 is calculated as follows:

IRa =Ip+ Isc ha =35ma

ERa = bias desired = 5V

R3=Ea=~ ha 35ma = 143 ohms.

C2 prevents what is called cathode degeneration by keeping the cathode voltage constant as the cathode current changes with the signal. From the previous calculation it can be seen that when the incoming signal swings positive (increasing the current through the tube), the cathode voltage will also go more positive. This makes the grid more negative with respect to the cathode and tends to decrease the plate current. This action of decreasing the plate current while the incoming signal is trying to increase it is called degeneration and would take place if it were not for C2. Degeneration reduces the output.

C2 is a cathode bypass capacitor which permits AC current at audio frequencies to pass around the cathode resistor and thus cause no change in cathode voltage. The DC portion of the cathode current must pass through the resistor, and it is this current that produces the desired constant cathode volt age. Failure of the cathode bypass capacitor is a common cause of distortion or low volume, and it will be discussed later in detail.

Capacitor C3 serves the purpose of bypassing high-frequency audio currents around the primary of the output trans former. This action broadens and flattens the frequency response of the stage by keeping very-high-frequency audio out of the transformer. The inductive reactance of the transformer increases with the frequency of the current through it, and so it presents a higher impedance to the high-frequency audio.

This higher impedance gives rise to higher voltages across the primary and greater output from the speaker at higher audio frequencies. If it were not for C3, the radio would have a rather "tinny" sound due to the emphasis that would be given to the higher audio tones.

Three methods of connecting the capacitor are shown in Fig. 2-9. The results are about the same for either of the connections, but the symptoms caused by the capacitor if it shorts are different in each case.

In part A, a shorted capacitor merely removes all audio without seriously affecting any voltages. In part B, a shorted capacitor will remove the plate voltage and put the full B+ across the transformer winding. In part C, a shorted capacitor makes the plate and cathode voltages equal and effectively places the cathode resistor and the transformer in series across the power supply. The cathode bypass may sometimes be dam aged in this last instance because it may be subjected to a greater DC potential than it was designed to withstand.

Capacitor C4 in Fig. 2-8 blocks the DC voltage that is present on the plate of the preceding stage and passes on to the grid only the AC variations of the signal. This capacitor is a very common cause of distortion because a slight leakage of DC current through it changes the biasing of the stage. As shown in Fig. 2-10, electrons are drawn upward through R2, producing an unwanted voltage drop between the grid and ground. The tube may be damaged because the polarity of the voltage makes the grid positive, which increases the plate current.

Fig. 2-9. Three ways of connecting the plate bypass capacitor.

Push-Pull Output Stages

Fig. 2-10. The effect of a leaky coupling capacitor.

Power output of a stage can be doubled by using the circuit shown in Fig. 2-11. The name push-pull is derived from the fact that while one tube is increasing the current in its half of the output transformer, the other tube is decreasing the current in its half. This is indicated by the polarities of the waveforms shown.

The tubes are driven by equal signals which are 180° out of phase. That is, when the signal is going positive on one grid, it is going negative on the other grid. A positive-going signal on the grid of an amplifier increases plate current and de-creases plate voltage. The waveforms represent voltage changes at the respective points.

The increase in output power occurs in an interesting manner. When current is increasing through the upper tube, a magnetic field is built up around the upper windings of the transformer. The poles of this magnetic field are dependent on the direction of the winding on the core. Let us assume that the north pole is at the top and the south pole at the middle.

At the same time the current through the upper tube is increasing, the current through the lower tube is decreasing.

Fig. 2-11. Push-pull stage doubles power output.

This current is flowing in the opposite direction through the bottom half of the transformer. The two halves of the primary are wound in the same direction, and this would seem to pro duce magnetic poles in the lower half of a polarity opposite to those in the upper half. But this is not the case, because the current is decreasing in the lower half and a decreasing cur rent through a coil produces the opposite magnetic field from that caused by an increasing current. The result is that the field produced in the lower winding actually aids the field produced in the upper winding. With the field aiding, the transfer of power to the secondary is twice what it would be if the transformer were driven by only a single tube.

The push-pull arrangement is not the only way to double the power output by using two tubes, but it is the one most commonly used because of additional advantages over the other methods. Since B+ is fed to both tubes directly, any variations in the power-supply voltage, such as unfiltered AC ripple, appear at both plates simultaneously. This causes in phase changes across the transformer with resulting magnetic fields which oppose each other. For a similar reason, all second harmonics of the signal which are generated in the stage are cancelled in the output. Also, cathode degeneration, which was described in the single-tube output stage, does not occur, and no cathode bypass capacitor is necessary when the two cathodes are fed from a common resistor.

Low-power stages for home-type radios and phonographs are sometimes connected in push-pull to utilize these advantages. In these devices, one of the tubes is capable of furnishing more than enough power, and the tubes are operating well below their full capacity. Many technicians are surprised to find practically no change in the volume level when one tube is removed. However, there may be a slight increase in hum and distortion.

Fig. 2-12. A resistance-coupled push-pull stage.

What was previously said about low-power stages does not apply to high-power output stages of the type used in modulators for transmitters, or in large public-address systems.

When high power is involved, considerable damage can be done to the output transformer and the remaining tube when one tube is inactive.

A push-pull stage using resistance coupling in the input is shown in Fig. 2-12. The extra triode is necessary to give the phase reversal needed to drive the second tube in the push-pull combination. Note that the input to the phase-inverter triode is reduced so that one output tube will not be overdriven.

The First Audio Stage or Voltage Amplifier

The power-output stage is driven by a circuit which produces a large swing in its output voltage with only a small voltage swing at its input. Because only output voltage (not current or power) is coupled to the next stage, the first audio stage is called a voltage amplifier. The example shown in Fig. 2-13 includes some interesting extra features.

The grid of the stage is coupled through C1 to the volume control. Any signal voltage appearing across the volume control will be transferred to the grid and an amplified version will appear in the plate circuit. The sliding contact arm on the volume control allows any fraction of input signal to be selected. The very large resistor in series with the plate results in large changes in the plate voltage with only very small cur rent changes. An example is shown in Fig. 2-14.

Fig. 2-13. A voltage amplifier with volume and tone controls.

Fig. 2-14. A small plate-current change causes a large plate-voltage change

In general, the larger the plate-load resistance can be made, the greater will be the peak-to-peak swing of the plate voltage and, therefore, the greater will be the output signal. But the size of the plate-load resistor is limited by the amount of B+ voltage available. This is because the slight amount of no signal plate current causes a voltage drop across the resistor which is in series with the plate voltage. For example, if a 100K plate resistor were used with a B+ supply of 100 volts, a current of 1 ma through the tube would cause the entire B+ to be dropped across the plate resistor and leave no voltage applied between plate and cathode.

Tone Controls

Except for a few high-fidelity amplifiers, tone controls are merely filters which remove either the high frequencies or low frequencies. Even when the controls are labeled "boost controls," the circuit itself does not actually add highs or lows that were not present in the previous stages. When treble sounds are filtered out, the effect to the ear is like an increase in bass, and vice versa. In some circuits, an extra amplifier is added, which raises the signal level after filtering to give a more pronounced increase in those frequencies that were not filtered.

The principle of a change in capacitive reactance with a change in frequency is used in the tone-control circuit. Any capacitor which shunts the output or input of a stage will cause the stage to have very low gain at frequencies to which the capacitor presents low impedance.

In Fig. 2-13, assume that the 1-meg potentiometer is turned so the arrow is at the top, and that the signal through the tube contains some notes at 5000 hz.

159 X 10^-3 159 X 10^-3

Xe = fC = 5 X 10^3 X 1 X 10^-1 = 318 ohms

So the 5000-cycle note finds the output of the amplifier shorted to ground through 318 ohms. Only the AC signal is shorted, not the DC plate current, because no DC current can pass through a capacitor.

A signal at 50 hz sees 31,800 ohms from plate to ground, and thus is not shorted by this large resistance. The output to the next stage will then be about 100 times larger at 50 hz than at 5000. If the entire resistance of the 1-meg variable resistor were put in the circuit by turning the control so that the arrow is at the bottom, the shunting effect of the 0.1-mfd capacitor would be nullified by the 1 meg which would be in series. So the circuit is an adjustable treble filter.

A simple bass filter can be made by placing a capacitor between the grid resistor and ground. In this way, highs are not affected by the low impedance of the capacitor, since the grid resistor is normally grounded when no tone control is used. Bass notes, however, find the grid resistor increased by the amount of Xe which is in series. The control places an adjustable resistance in parallel with Xe, thereby reducing it to the desired value.

Fig. 2-15. A tone-control circuit for use in high fidelity equipment. In Fig. 2-15 is shown a very effective tone-control circuit which has become popular in high-fidelity equipment. C3, R3, and C4 form a voltage divider for highs only, having no effect on the bass. R4 selects the proportion of the high-frequency signal voltage to be applied to V2. R1, R2, and R3 form an other voltage divider which is adjustable for low-frequency signals. The highs are shunted around R2 by C1 and C2, so that changing this control does not affect the total impedance to high-frequency signals. C1 and C2 also give some compensation to the mid-range frequencies as R2 is moved from bass cut to bass boost.


Nearly every AM radio uses the combined osc/mixer stage shown in part A of Fig. 2-16. The signal from the station (720 khz, for example) is picked up by the loop antenna, which is tuned to the desired frequency by C1. This signal is then fed to the signal grid through pin 7 of the 12BE6. An amplified version of this signal appears in the plate circuit.

At the same time, the cathode current is carrying an oscillator signal generated by coil L1. The cathode tap on the coil causes a small amount of cathode voltage to be fed back to the control grid, causing oscillation to occur at the frequency for which LI and C2 are resonant. In this example, the oscillator frequency is 1175 khz.

Fig. 2-16. A combined mixer/oscillator circuit.

Both signals, 720 khz from the station and 1175 khz from the oscillator, appear in the plate circuit, and it is here that the mixing takes place. Whenever two signals are mixed in a non linear device, such as a vacuum tube, two additional signals equal to the sum and the difference of the originals will be produced. When the radio is tuned to receive 720 khz, there are four signals present in the plate circuit:

720 khz, from the station

1175 khz, from the oscillator

1895 khz, the sum

455 khz, the difference

Of particular interest is the fact that the same difference signal, 455 khz, is produced in the plate circuit for any incoming frequency. This is accomplished by tuning the oscillator coil and the loop antenna simultaneously with capacitors C1 and C2, both of which are varied by the same shaft. The circuits are arranged so that the combination of L1 and C2, the oscillator tank, is always tuned to a frequency 455 khz above the in coming signal.

If the radio were tuned to a new station, say 900 khz, C1 and the loop would resonate at 900 khz, and the oscillator tank would resonate 455 khz higher, at 1355 khz. The four frequencies in the plate circuit would then be: 900 khz, from the station 1355 khz, from the oscillator 2255 khz, the sum 455 khz, the difference For any incoming station, the first three signals vary, but the difference signal will always be 455 khz. Furthermore, this difference signal will have all the characteristics of the signal from the station; that is, it will carry the audio modulation from the studio.

This is the superheterodyne principle. Its advantage is that the output of the mixer stage can be fixed-tuned and the following stages designed for maximum efficiency at one frequency only. It does not matter where the receiver is tuned in the broadcast band: the station frequency will always be converted to 455 khz. This frequency is called the intermediate frequency and is fairly well standardized at 455 khz for broadcast band receivers, although other frequencies are used in other types of receivers.

C1 and C2 in Fig. 2-16 are the main tuning capacitors, and the dotted line shows that the rotors are connected on one shaft. Capacitors C3 and C4 are small screwdriver-adjusted units connected in parallel with the large capacitors to enable the two tank circuits to be initially adjusted 455 khz apart and to remain so as C1 and C2 are tuned with the main dial. Keeping the two circuits aligned over the entire dial presents a problem, and adjustments dealing with this are a part of the alignment procedure called tracking. Alignment is discussed in Section 6.

The circuit in part A of Fig. 2-16 shows a popular type of grid connection to the oscillator coil LL The little coil above the tapped tank coil is called a capacitor link. It is actually not used as a coil, but serves the function of a capacitor be tween the grid of the tube and the main tank coil. The little coil has only one end exposed as a terminal with the other end being hidden inside. One can see the problems confronting the unsuspecting technician who uses his ohmmeter to test for continuity from the grid terminal of the coil to ground.

To complicate matters, the alternate connections shown in parts B and C are used frequently, and in this case there should be continuity from the grid terminal to ground. With this type, a grid blocking capacitor C5 will be soldered in the circuit. Both types of coils have four external solder terminals, and a careful examination with the ohmmeter and reference to the schematic drawing is necessary to determine a failure in the oscillator coil.

Universal replacement-type coils (part D) are a combination of both types and have six solder terminals so that they can be used in either circuit. This type of replacement is also equipped with an adjustable core to aid in the tracking alignment.

This circuit will be referred to again in Section 4, where measurement of grid-leak bias voltage is used as a key test point to indicate the presence or absence of the oscillator signal in those cases where a receiver will not tune in any stations, but does pick up noise.

Fig. 2-17. A typical IF-amplifier circuit.


The output from the mixer/ osc stage is developed as a volt age in the first IF transformer at 455 khz. The secondary of this transformer is the grid circuit of the IF amplifier, as shown in Fig. 2-17, and is practically standard in every radio.

From the standpoint of servicing, it is the simplest and most trouble-free of the five stages in the receiver.

As mentioned in the section on the mixer/ osc stage, the advantage of the superheterodyne is that the major amplification can take place in a stage designed for maximum efficiency and operating at only one frequency.

Because its input and output are tuned to the same frequency and the stage has fairly high gain, there is occasion ally a tendency for oscillation to occur through mutual coupling between the plate and grid. A well-shielded tube is used, and the input and output wiring are kept separated under the chassis. Some receivers use a small metal shield under the chassis to isolate the input circuit from the output circuit.

Technicians often slightly detune one of the transformers in order to reduce the tendency of the stage to oscillate. This does not seem to impair the overall gain of the receiver seriously, and often will improve the fidelity by broadening the total bandwidth of the stage.

The screen-grid bypass capacitor, C2, is quite important in keeping the gain high and preventing oscillation. It may seem to be missing from some circuits but, in these cases, it will be found that the screen grid is returned directly to the power supply and that the output filter capacitor there serves the purpose.

Fig. 2-18. Sidebands produced by modulating a 720 -khz carrier with a 2000- hz note.

Fig. 2-19. The sidebands are reversed in the conversion process.


A few facts about the transmission of AM radio signals are needed in order to understand the operation of the detector.

When an RF amplifier in a transmitter is amplitude modulated with an audio frequency, two additional signals are produced.

They are called the upper and lower sidebands. The upper side band will have a frequency which is the sum of the main transmitter carrier and the audio frequency, and the frequency of the lower sideband will be the difference between the two.

As an example, suppose WGN in Chicago, which operates on 720 khz, is modulated with a 2000-cycle audio note. Fig. 2-18 shows that the total bandwidth of the signal will be 4000 cycles (or 4 khz). Also, from the diagram it can be seen that the original audio tone, 2000 cycles, is represented by the distance between the carrier and either sideband.

All these signals will pass through the tuner and each will be converted to an IF frequency, as explained earlier in this section. If the main carrier is converted to 455 khz, the side bands will be converted to 457 khz and 453 khz, respectively. The upper and lower sidebands are now reversed by the conversion process, as shown in Fig. 2-19. Note that the original audio (2 khz) is still represented by the distance between the carrier and either sideband.

Fig. 2-20. A diode detector circuit.

The three signals enter the diode detector, Fig. 2-20, which is simply a half-wave rectifier and filter. The reactance of capacitor C1 across load resistor R1 is very low for frequencies in the range of 400 khz to 500 khz. Thus, the load resistor is shorted out for currents at these frequencies. There is practically no output voltage produced for input currents above 400 khz.

In the detector circuit, the sidebands will combine with the main carrier to produce two beat frequencies which are the sum and difference. Using the upper sideband and the carrier, this gives:

Sum: 455 + 457 = 912 khz

Difference : 457 - 455 = 2 khz

The sum (912 khz) is shorted around the load resistor through C1, but the reactance of C1 at 2000 hz is quite high, and the beat frequency develops an output voltage across R1. In this manner, the difference frequency between the upper sideband and the carrier, which is the original audio signal, is produced at the output of the detector. The same action occurs between the lower sideband and the carrier and produces the same beat frequency equal to 2000 hz.

AVC Voltage

The current flowing through R1, the detector load resistor, produces a voltage at the top which is negative with respect to ground. Furthermore, this voltage will vary as the strength of the incoming signal changes, becoming more negative for stronger signals. This negative voltage is fed back to the grids of the IF and mixer stages and gives an automatic increase in negative grid bias when the incoming signal is strong. The increase in bias reduces the output signal from these stages, preventing a sudden burst of sound from the speaker when a strong station is tuned in. This automatic volume control (AVC) has the additional advantage of removing all bias from the controlled stages when no signal is present, thus making the receiver most sensitive when it is tuned between stations.

In this way, weaker stations are not overlooked when tuning across the dial.

Resistor R2 and capacitor C2 comprise a filter to remove the audio variations from the A VC voltage. The time required for C2 to charge and discharge through R2 is very long com pared to the time between cycles of the audio frequency. Thus, the voltage across C2, which is the AVC voltage, remains reasonably constant until the strength of the incoming signal changes.

Fig. 2-21. A typical 5-tube AC/DC receiver.

Fig. 2-22

The tuning capacitor across the loop antenna may be re turned to ground as in Fig. 2-22 or to the AVC line as in Fig. 2-21. In Fig. 2-21, the capacitor is insulated from the chassis, usually by rubber grommets, and a breakdown of this insulation results in a shorted AVC line. It is interesting to note the very great difference in the normal ohmmeter readings to be expected across the capacitor plates.

The AVC circuit used in most home-type receivers is not too effective. More elaborate circuits are used in short-wave, FM, and auto radios. In the automatic-tuning auto radios, the A VC voltage is used to control the tuning mechanism.

The Combination Detector, AVC, and First Audio Stage

The circuit of V3 in Fig. 2-21 combines several functions into one tube, and is the circuit used in nearly every broadcast band receiver. The 12AV6 tube contains a triode and a pair of diodes. The triode is used for the voltage amplifier, and one diode plate is used with the common cathode for the detector and A VC. The other diode is grounded. In Fig. 2-22, the diode plates are tied together. In other circuits, one plate is used for the detector, and one is used for AVC.


1. List five types of power supplies, and give one advantage and one disadvantage of each.

2. In the power supply shown below, what is the approximate voltage at terminals 1 and 2 under the load conditions shown, neglecting losses in capacitors and diodes? What is the resistance of the load?

3. Draw the schematic of a full-wave bridge-type power supply using 6V3 rectifiers, and show the filament connections.

4. What dangerous condition could develop if a technician were working on two radios next to each other on the bench and both used a transformerless power supply?

5. In the circuit shown below, what is the value of the cathode resistor?

6. What would be the effect on the sound of the radio if C3 in Fig. 2-5A were open?

7. When the cathodes in a push-pull stage are returned through a common resistor, the bypass capacitor can be omitted. Why?

8. Describe the effect of the control shown in the circuit below.

9. Draw a schematic of a universal-type replacement oscillator coil. If you were going to install such a coil, and the solder terminals had no identification, describe how you would determine which part of the coil is connected to each terminal.

10. List all the frequencies present in the detector circuit of an ordinary receiver when it is receiving a 1500-cycle audio note from the station.

11. Draw the detector circuit for Question 10, using a 1-meg load resistor and a .0001-mfd capacitor. Compute the reactance of the capacitor at each of the frequencies listed.

12. Describe two methods for preventing oscillation in an IF stage.

13. A radio using 455 -khz IF stages is tuned to a station at 920 khz. What is the oscillator frequency?

14. In Fig. 2-17, what is the purpose of C2? What do you think would be the result if it were shorted?

15. What is meant by "tracking"?

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