Increase amplification on single voltage class-AB amplifier

Using help from a previous post I've managed to build a functional class-AB amplifier. The problem that this one had was the requirement of two independent voltage sources for proper operation.

Since I don't have two 12 V power supplies but instead one 24 V supply, I decided it would be a good idea to hook up my speakers between two amps, one of them amplifying an inverted signal. The question I have is, how would I get more amplification out of this circuit?

I've already tried to just use a carbon-copy of the left half for the right half (except swapped inputs for U1A), as well as feeding the input signal to the inverting input coupled with resistors between the input signal, inverting input and transistor output.

• Your circuit diagram doesn't appear to match your description in the text. The right hand side of the schematic will produce a constant 12 V DC at its output. Jun 20, 2023 at 21:04
• What's the desired power output deliverable into the speaker and why? Whatever you want, justify it. That could be for learning experience --- that's fine --- but if so then start small, get that working and well-understood first before adding the additional complexities of more complicated power supply rails and heat dissipation issues and temperature compensation. Start one step at a time! Start with 300 mW, for example? Jun 20, 2023 at 22:35

As @Finbarr says, your right-hand side module is not a copy of the left, and will produce a constant +12V output, since that's its input.

I think your power stage (consisting of the MOSFETs Q1 and Q2, LEDs D1 and D2, and resistors R1, R2 and R5) looks OK, and from here I will assume that it works. I will show it as a symbol, for clarity, which makes it easier to draw and understand the system as a whole.

On the left, you have a system with gain $$\1+\frac{R_9}{R_8}\approx 2\$$ which is multiplying the input signal. On the right you have a voltage follower, with gain +1, and an input of +12V:

simulate this circuit – Schematic created using CircuitLab

Note: Gain 2 is approximate, and I only use it here to simply everything. Actual gain is 2.09, and you'll have to account for this. It's not clear why you need that gain; it certainly complicates the entire circuit, and things would be much simpler if it were 1.

The voltage across the speaker is

$$V_A - V_B = (2V_{IN}+12V) - (+12V) = 2V_{IN}$$

The right hand side contributes nothing, and this behaviour could have been achieved by the left side alone.

If, at node B, you made the signal $$\V_B = -2V_{IN} + 12V\$$, then you have:

$$V_A - V_B = (2V_{IN}+12V) - (-2V_{IN}+12V) = 4V_{IN}$$

This doubles the voltage across the speaker, and by $$\P=\frac{V^2}{R}\$$ you can see that this would quadruple the power delivered to the speaker.

Gains on the left and right sides must have opposite signs for this to work (note the minus sign before the 2), so that the signals at A and B are inverted with respect to each other, and of course both must be centered about +12V.

There are many ways this can be achieved, the most obvious being:

simulate this circuit

Note that the inverting stage here has an input impedance of 10kΩ (which you could of course increase), that places quite a heavy load on the source of its input, which you will also need to take into consideration. That's another issue though, beyond the scope of your question.

I'm sure you can think of other ways to achieve this same goal of having A and B inverted with respect to each other, and with the same amplitude. As I mentioned in the note above, you do not have to have gain above 1, and you could simplify this even further:

simulate this circuit

Here I have turned the left section into a plain old voltage follower (100% feedback), and reduced the gain of the right-hand section to –1. Now $$\V_A - V_B = 2V_{IN}\$$.

You will have noticed the reliance of all these designs on $$\V_{REF}\$$. Unfortunately, the way you have derived this reference is very poor, and as it stands, your design will have wild potential fluctuations on that node, as $$\V_{IN}\$$ varies, and since it is sourcing and sinking current to and from many different things connected to it.

It's best to either reduce the values of R10 and R11, to reduce the source impedance of that +12V reference, and simultaneously increase the impedances connected to it, or to produce a near-zero impedance source of +12V using another op-amp:

simulate this circuit

Lastly, since there could be many millivolts of difference between A and B, due to op-amp input offset voltage or other things, you will see a DC component of current through the loudspeaker, which is bad for a few reasons. To prevent that, place a capacitor in series with the speaker:

simulate this circuit

You aim to get a bridged output stage. That's a popular idea to get more power with lower operating voltage.

Without making ANY comments of the actual validity of your circuit design details I can tell a fact: The theoretical maximum output AC voltage to a speaker from a bridged output stage is 24V peak if the operating voltage is 24VDC and there's no transformer. That means max. 72 watts RMS power to a 4 ohm resistive load if the signal is a sinewave. One totem pole output stage with single 24VDC operating voltage could theoretically output only 12V peak i.e. 18 watts RMS to 4 ohm.

These are theoretical maximums. As said above, NO comment of the actual validity of your circuit details. Too much things that are totally undefined or left out of the drawing. For ex. I cannot see anything done to get enough voltage drive to the gates of the mosfets to utilize the full 24VDC operating voltage.

Bridging doubles the voltage gain, but the gain depends on the used feedback, too.