# Designing feedback push-pull amplifier

I'm trying to design this circuit:

With this information:
Rc = Rd = Ra = Rpot = 10kOhm
Vx/Vy = 11
Vx = 2.5V
Vin = 40mVp, 1kHz, sin

So, as Vx/Vy = 11, I assumed Vout = Vx and calculated Rf = 110kOhm
With, Vy = Vpot - Vin and Vx = 2.5V I calculated Vpot = 0.18V

This was the result of the circuit:

But this was the result of the simulation:

I was expecting a sin function as well, anyone could say what is wrong with this design?

• Re-check your math for Rf, using the equation for a non-inverting amplifier. Oct 26, 2021 at 12:11
• Bias and power supply of U1A look fishy. Mar 19, 2023 at 10:15

I think $$\R_C\$$ is in the wrong place. You have accidentally created a first stage with gain in the tens of thousands, or whatever the open loop gain of the opamp is!

Also 1:

If you consider the push-pull stage to be a voltage amplifier with unity gain, then U2B with feeback via the potential divider Rf & Ra produce a total gain of:

\begin{aligned} \frac{V_X}{V_Y} &= 1+\frac{R_F}{R_A} \\ \\ &= 1+\frac{110k\Omega}{10k\Omega} \\ \\ &= 12 \end{aligned}

You can expect any voltage $$\V_Y\$$ approaching $$\\frac{V_{CC}}{12}\$$ to saturate the opamp. That doesn't leave you with much room for manoeuvre; you need to provide a very precise DC operating point of $$\V_Y = \frac{V_{CC}}{24}\$$ to set your output at the midpoint between 0V and $$\V_{CC}\$$.

This means that you need to choose $$\R_8\$$ and $$\R_9\$$ very carefully, because they, in effect, control that quiescent level.

There are better ways to achieve this, which I'll describe if you want.

Also 2:

The transistor biasing resistors $$\R_1\$$ and $$\R_2\$$ are on the low side. If you think of them in terms of their Thevenin equivalent circuit (ignoring the diodes), the opamp effectively "sees" a 270Ω resistor to a voltage source at $$\\frac{V_{CC}}{2}\$$.

If, for example $$\V_{CC} = 12V\$$, then when the opamp output is at 0V, it is trying to sink a current of:

$$I = \frac{\frac{12V}{2}}{270\Omega} = \frac{6V}{270\Omega} = 22mA$$

That's a lot to ask from an LM324. If you set R1 = R2 = 1.5kΩ, this will drop to 8mA or so, a much more reasonable demand. That will help the opamp achieve a wider output voltage swing too. It will, however reduce the maximum output current that you can expect from the transistors.

An old LM324 is never used for audio because: It is noisy, it produces crossover distortion and its output high is 1.4V less than its positive supply voltage. Your amplifier produces a lot of distortion and a very low output power.

• This may not be an issue in the simulation, depending on how accurate the model used is; but these are important practical notes. At 5V, there are many newer and better op-amps than LM324 available, say TLV2372 for starters. Mar 19, 2023 at 19:32

The VCC voltage of 5V that biases Q1 and Q2 might be too low compared to the signal that drives the bases of Q1 and Q2 themselves.

Try VCC = 15 V.

Not an expert here, but diodes D1 and D2 do not look like they are in the right place. You want something like this:

Picture is from https://www.electronics-tutorials.ws/amplifier/amp_6.html.

Take them out and run your simulation. If everything else is OK, you should see a sine wave on your output with a bit of discontinuity at the crossover regions.

Rc should be in series with C1 as in the first circuit diagram. As you have it in the simulation the first stage is acting as a differentiator.

Insert a 2.2uF dc blocking capacitor in between Ra and ground.

Add small (1 Ohm) emitter resistors in series with each of the emitters of Q1 and Q2 to take up the voltage difference between the double diode drop and the Vbes of Q1 and Q2.

Set Vpot to mid-supply (Vcc/2) or just replace the pot with two equal value resistors. Then there should be a fairly large capacitor (10uF) across the lower resistor of the two to decouple supply noise/ripple.

These changes will then bias the entire circuit up to mid-supply with the input and output signals oscillating about 0V.

If you're expecting to output frequencies down to the bottom end of the audio range (20Hz) then you need to significantly increase the value of C2 to say 1000uF.