Q. I am studying transistor lab lessons. In them is a class B power amplifier that uses a PNP and an NPN transistor in series. I am having a hard time understanding how this circuit works. How can it work push-pull when the transistors are connected in series? With the diode connected between the bases, it seems only half of the signal waveform would get through.
A. The key to this circuit is the big electrolytic capacitor connected between the point where the emitters come together and the load. When the circuit is first turned on, this capacitor charges to 1/2 Vcc. It holds nearly all its charge through the positive and negative half cycles of the signal waveform.
When the input signal is zero, the driver transistor's collector is at about 1/2 Vcc. The current from the collector forward-biases the diode, so that the drop across it is about 0.6 or 0.7 volts. So the difference between the voltages on the bases of the output transistors is about one diode drop. This biases them just at the edge of conduction, which is what we want for Class B operation.
When the signal comes in on the base of the driver transistor, it makes the collector current rise above and fall below its no-signal level. So the current through the diode rises and falls in strength along with the signal, but it never reverses direction. The diode is forward-biased and conducts at all times.
The voltage with respect to ground at the ends of the diode, and thus on the bases, rises and falls along with the changes in the signal. When it rises, the base of the top transistor rises above its emitter voltage, turning it on. The base of the bottom transistor also rises above 1/2 Vcc. But since it is a PNP type, this reverse-biases its base-emitter junction and it turns off.
Note that the voltage across the output capacitor stays at about 1/2 Vcc. As the top transistor conducts, its emitter voltages rises. Since the voltage across the output capacitor stays about the same, most of the voltage increase on the top transistor's emitter appears across the load. Thus, the voltage at the top of the load swings positive as it follows the signal.
When the other half-cycle of the signal occurs, the driver transistor conducts more heavily, and its collector current increases. The current through the diode increases, but the voltage across it stays about the same. The voltage with respect to ground at the ends of the diode now falls below 1/2 Vcc. Thus turns off the top transistor, since its base-emitter junction is reverse biased. But the bottom transistor conducts, and its emitter follows the signal. Again, the voltage across the output capacitor stays about the same, and the voltage at the top of the load swings in a negative direction.
In this way, the output transistors take turns conducting, the top one conducting on positive half-cycles, and the bottom one on negative half-cycles. This would not be possible if the output capacitor did not hold its charge. In some amplifiers, the output capacitor is eliminated because a positive and a negative power supply are used, with the load connected to ground at their common point. You will also not that the bias diode conducts at all times.
The main advantage of this design is that no output transformer is needed, saving costs and making the amplifier much lighter in weight. The disadvantage is that the output transistors are connected in series across the power supply. There is really nothing to prevent them from shorting across the power supply if one of them fails. Often, this happens before the fuse has a chance to open and protect them. Very often, when an amplifier fails, it fails in this way, and both output transistors have to be replaced, along with their emitter resistors.