# Push pull amplifier

What's the aim of using a push pull amplifier instead of nothing? I mean, what is the difference between connecting the input signal to the final load instead of putting a push pull amplifier between the input and the final load? Are there any advantages? Because usually a push pull amplifier introduces problems such as the cross over distortion.

• Amplification. Impedance matching. Amplification. Buffering. Amplification. Jul 16, 2016 at 14:48
• Crossover distortion can be managed. Jul 16, 2016 at 14:49
• Try plugging a microphone or guitar straight into a loudspeaker. You won't hear anything. This must be obvious. What's your real question? Jul 16, 2016 at 14:50
• You may put a witch between your microphone and the loudspeaker if you don't like the push pull amplifier. Jul 16, 2016 at 16:05
• Don't be deceived by distortion statements. You get what you pay for. Class A-B Cerwin Vega Metron amps can put out 1,500 watts with .002% total distortion and 70V/uS slew rate. Some audio IC's have only .0002% distortion. Your still thinking of the way things "used to be".
– user105652
Jul 16, 2016 at 17:03

Suppose you had an input signal of 100 mV (capable of supplying 1 mA), and you needed to connect to a load of 8 ohms and deliver 3 watts to load.

Your input signal could only supply milli-watts of power.

Thus you need an amplifier in between the signal and the load (push pull or other)

It is used to provide power gain to a load. A simple wire will provide all load current through it to the load, and this may create a voltage drop due to nonzero output resistance of the driving circuit. A push-pull allows most of the current to flow from the power rails, and only a small amount (for BJT push pulls) flows from the driving circuit.

I've drawn two CE (common emitter) stages. The left is a single NPN stage, the right is a NPN + PNP push-pull stage.

simulate this circuit – Schematic created using CircuitLab

I want make a sinewave with an amplitude 10 V peak across the load.

Let's look at two situations, when the voltage across the load is +10 V and when the voltage across the load is -10 V. Since Rload is 10 ohms for +10 V I will need 1 A flowing into the load, for -10 V I need 1 A flowing out of the load.

At +10 V and the left circuit I will need to switch Q1 off completely.

At +10V and the right circuit I switch off Q2 completely and let Q3 supply 1 A.

So far so good but now -10V:

At -10 V and the left circuit I will need to make Q1 pull the 1 A from the current source Plus the current from the load, so 2 A in total.

At -10V and the right circuit I switch off Q3 completely and let Q2 pull 1 A.

Note how the left circuit is less efficient because half of the time Q1 needs to get rid of the current from I1 in addition to the output current it has to take care of.

You could replace I1 by a resistor, in this case you'd need about 10 ohm. With a resistor things get worse as the resistor will supply even more current as Q1 is doing it's best to pull the voltage (at its collector) down.

The right circuit is much more efficient because almost no current is wasted.

I've drawn CE stages but for CC (common collector or emitter followers) the same applies. Also the same applies if you would use MOSFETs instead of BJTs.

what is the difference between connecting the input signal to the final load instead of putting a push pull amplifier between the input and the final load?

A typical electret microphone will produce round about 5 mV RMS for 94 dB SPL at 1 kHz. 94 dB SPL is quite loud: -

The microphone output impedance will be about 1kohm at best so therefore it could dump 2.5 mV into 1k and produce a power of 6.25 nano watts.

This will barely tickle a loudspeaker. You just wouldn't hear it.

That is why we use a power amplifier (push pull or otherwise).