Electronic – How does this Push-Pull amplifier work

Let's start our with a single op amp and work out way toward a class AB design. Consider the voltage follower.

^{simulate this circuit – Schematic created using CircuitLab}

In this case \$V_{in}=V_{out}\$. Why? Negative feedback. Negative feedback forces the inverting pin's voltage to match the non-inverting pin. In other words, the op amp will do whatever it takes with its output to make \$V_{NI}=V_{INV}\$.

Let's take that a step further. We need more power to drive a low impedance speaker. Well, the average op amp will only have a few tens of milliamps of drive capability. That is where we add the power stage.

^{simulate this circuit}

By tying the inverting pin to the output of the power stage, we have created a voltage follower. \$V_{in}=V_{out}\$, but now the circuit has the ability to deliver much more current than the op amp output ever could. Because the op amp has negative feedback, \$V_{NI}=V_{INV}\$. The cross over distortion is eliminated by the op amp doing whatever it takes to satisfy that relationship. As an exercise, build this up on a bread board, put a sine wave into \$V_{in}\$, and observe the output of the power stage and output of the op amp. The two will not look anything alike!

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)