13.4. POWER AMPLIFIER WITH COMPLIMENTARY TRANSISTORS.

The following drawings show how a complimentary push-pull amplifier is working.

fig. 163

Fig. 163 shows two circuits. The upper one with a NPN transistor will have current through R1 while there is a positive signal at the input. The lower circuit will have current through R2 always when there is a negative input signal.

fig. 164

Fig. 164 shows the two circuits of fig. 163 combined. Now the two resistors are replaced by a single loudspeaker. Both currents - explained in fig. 163 - flow through the loudspeaker so causing an ac-current in the loudspeaker. This fits to our desire, to have a current flowing only if there is a signal voltage different from quiescence.

The big disadvantage of the circuit in fig. 164 is, that there are two batteries necessary for it.

fig. 165

Fig. 165 shows a circuit which gets rid of that disadvantage.

If there is a positive input signal, there will flow a current via the NPN transistor, the capacitor will have a capacity big enough so that it can be charged completely only during the longest possible half-waves (at lowest frequencies).

fig. 166a

fig. 166b

fig. 166c

So we will find after the positive halfwave at the capacitor a full positive charge as shown in fig. 166b. With a negative input signal the PNP-transistor gets conducting and there will flow a current - originated from the capacitor as a voltage-source - through the PNP-transistor and the loudspeaker as shown in fig. 166c.

The current flowing now through the loudspeaker is flowing backward - which means: there is flowing an ac-current through the loudspeaker.

Improvements of the simple circuit of a complimentary push-pull amplifier.

The circuit derived in fig. 166 has still two main problems which must be solved before it can be used for a receiver.

PROBLEM 1:

It is easy to see, that the voltage at the base of Tr1 can never reach a value higher than that of the supply voltage. But supposed Tr1 is made conducting (by a relatively high base-current) then the voltage at the emitter of Tr1 has a potential which is just 0.2 Volts lower than the supply-voltage.

fig. 167

As we know: to inject a base-current to Tr1 we need at least 0.7 V between Base and Emitter of Tr1.

This shows: With circuit shown in fig. 167 it will never be possible to make Tr1 fully conducting.

Consequence:

We have to find a method to supply point A with a potential about 0.6 V higher than the supply voltage. One possibility to achieve this is shown in fig. 168. But this is a very complicated and inconvenient way, because we need an additional cell for it.

fig. 168

Practical solution:

The most common way of solving that problem is the so called BOOTSTRAP CAPACITOR C2 shown in fig. 169.

fig. 169

Its function is like that: At “NO” input signal the voltage at point A will be about half of the supply voltage because there is flowing a medium current. The voltage at point C through the two collector resistors of Tr.3 will be about 75 % of the supply voltage. If now a negative signal occurs at the input the voltag at point A will be raised. Of course the voltage at point B is increased as well (base-emitter-voltage at Tr1 maximum 0.6 V). So C2 has a rather big capacity the voltage at point C reaches values higher than the full supply voltage.

PROBLEM 2:

If the bases of Tr.1 and Tr.2 are connected like shown in fig. 170 we would face a so called CROSSOVER-DISTORTION as shown in fig. 170 because it takes always at least 0.7V of voltage change until the transistors are starting to get conducting. A first solution would be to insert a resistor R2 as shown in fig. 171. But the dimensioning of R2 is extremly sensitive because:

- if it is too small there will be still a crossover distortion, and
- if it is too big there will be a lot of losses or even a short circuiting through Tr.1 and Tr.2

fig. 170

fig. 171

A better solution is to fit in two diodes like shown in fig. 172. In order to have an automatic adjustment in case of heating up of the transistors, very often is used an additional thermistor connected in parallel to the diodes as shown in fig. 163. This thermistor is mostly fixed to the heat sink of the transistor.

So transistors heat up, the thermistor which will cause a drop of its resistance and therefore a voltage-drop across the two diodes. This again causes a decreasing base-current for both transistors Tr.1 and Tr.2.

fig. 172

fig. 173

In order to avoid lude a crossover-distortion as far as possible, in practical circuits we find at last an adjustable resistor connected in series to our diodes. This adjustable resistor is used to preset the voltage across the two bases exactly to a condition where at quiescence a small collector current just starts to flow.

fig. 175

A rather advanced power amplifier of the COMPLIMENTARY PUSH PULL TYPE is shown in fig. 176.

fig. 176

CHECK YOURSELF:

1. Explain how a Push-Pull amplifier with transformers is working.
2. Explain how a complimentary Push-Pull amplifier is working.
3. Explain what each component in fig. 176 is useful for.