I'm writing this because some people might think that this is homework.

I have built a high power class-AB amplifier to connect it to a subwoofer of 2x2ohm 800W RMS. My main problem is that the class-AB amp only amplifies the current and not the voltage. I need something to also amplify the signal voltage.

I have been trying to amplify an audio signal with bandwidth of 10 Hz - 20 kHz. I am amplifying the input signal with +- 1 volt to +-12 volt with an op-amp (TL084.) Then, I use the signal from the op-amp to feed into a class-A amp to amplify the voltage to +- 32 volts but no matter what values I try at any frequency, I can't get a pure sign wave with the required amplitude.

I don't have access to high-voltage op-amps, the supply of the class-A amp is 64 volts and I need 20 mA output current.

Here is an example of what I have tried:

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Red is voltage before the output capacitor, green is input signal, blue is output after the filter capacitor.

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Update 1:

I tried the circuit with much higher voltage and different resistor values and I did get what I wanted, but it would be impractical to do so with 160 volts. What can I do to improve the efficiency or what other type of circuit I can use to achieve what I want?

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Update 2:

Here is a simplified schematic (I have some other capacitors and resistors in for protection and balancing to but they are not in the schematic) of my class-AB amplifier which I have tested at 800 W RMS and it worked fine. I'm only missing a driver for it.

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  • 1
    \$\begingroup\$ Be clear on what input voltage feeds the circuit and what output voltage you expect into your load resistor. Don't mix RMS and peak values up - be consistent. You should be aiming for closer to 32 volts DC quiescent on the collector so set your bias resistors accordingly. \$\endgroup\$
    – Andy aka
    Jul 12, 2021 at 9:48
  • 1
    \$\begingroup\$ If you want 64 Vpp at the output, you're going to need to power the circuit with significantly more than 64 VDC. Hint: plot the voltage at the emitter, too. \$\endgroup\$
    – Dave Tweed
    Jul 12, 2021 at 10:11
  • \$\begingroup\$ @Andyaka I have changed the voltage values so all of them are the same. I believe I have mentioned what is my input voltage level and what I require at the output. \$\endgroup\$
    – Oli
    Jul 12, 2021 at 10:29
  • \$\begingroup\$ In fact, with the values of R1, R2 and R5 shown, you'd need a supply voltage of at least 260 V. With a quiescent current of 40 mA, the circuit will be dissipating more than 10 W even with zero output, which demonstrates how inefficient this circuit configuration typically is -- the maximum output power is 320 mW. \$\endgroup\$
    – Dave Tweed
    Jul 12, 2021 at 10:37
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    \$\begingroup\$ @Colin, Thank you for helping and recommending the book. I'll try to make one out of discrete components. and update the post with results. \$\endgroup\$
    – Oli
    Jul 12, 2021 at 14:26

2 Answers 2


I'm afraid you're not paying attention to the appropriate issues.

Consider. With no signal in, your bias point at the base of the transistor is at about 7.8 volts. If you attempt to drive it with +/- 12 volts, the transistor base will be driven below ground for about 4 volts of the swing, and the signal will limit. And that is exactly what you're seeing.

Try driving the input with 5 volts and see what you get. It won't be high purity, since you don't have any negative feedback, but it will recognizably be a sine wave.

You need to change your R3/R4 values to give at least 12 volts as an operating point, and anything less than about 14 is playing with fire.

When you do this, you will also see your output start clipping due to your values for R1 and R2.

Consider the situation where you are at peak input voltage. At this point the emitter must be at about 24 volts. If, at this point, the transistor is completely turned on, the collector will also be at 24 volts. Since the greatest output voltage you can get is 64 volts, you cannot possibly get a swing greater than 40 volts.

You can derive an expression for your required supply voltage in terms of your output gain and input signal.

Vin(PP) is the peak-to-peak input voltage. Assuming an ideal transistor (zero base-emitter drop and the ability to drive Vc down to Ve), the required supply voltage is

Vs = Vin(PP) x (1 + R1/R2), with

(R3/(R3 + R4)) x Vs = Vin(PP) / 2

Since you don't have access to ideal transistors, you'll need a slightly higher Vs and a slightly higher (at least a volt) bias point set by R3 and R4.

You will also need to take into account the loading of C1 and R5, which I have ignored.


After a lot of dissociation, I realized, me being a beginner at audio amplifier design and analog electronics, I need a lot more studying to make a proper high-power audio amplifier.

Using a class-A amp as driver for my class-AB output stage is out of my league and very inefficient with lots of distortion. So I used a high-voltage op-amp as non-inverting amplifier (OPA454AIDDA) as my driver stage for class-AB amplifier.

I have made a "monstrosity" that just "works" with no audio distortion at max power, so any one at my level who wants to build one can do so. I will also update this answer with scope screen shots of the circuit working and probably schematics of the final design.

Class-AB stage max RMS power is around 500 Watts @ +-32 volt output voltage with 2x2 ohm subwoofer connected in parallel.


  • Keep in mind that this design is highly unstable and could fail for many reasons but it works with no audio distortion even at high output power.
  • If for any reason any one the power BJTs fails short It will destroy the power supply and could result in a fire. So be sure to put some fuses in between and use the design at your own risk.
  • \$\begingroup\$ Your many emitter-follower transistors have a voltage loss then the 64V supply will only produce an output of maybe 52VP-P or less. Then The RMS output voltage is 52 /2.828= 18.4V RMS. The power is 18.4V squared (338V/1 ohm= (338/1 ohm)= 338W. \$\endgroup\$
    – Audioguru
    Jul 13, 2021 at 16:47
  • \$\begingroup\$ @Audioguru, good point but peak voltage of the power supply is actually around +- 37.2 volts which drops to +-36.8 around full load. I used 32 volts because it’s nice round number. Thanks for pointing it out, I’ll include that in the final answer too. \$\endgroup\$
    – Oli
    Jul 13, 2021 at 19:08
  • \$\begingroup\$ @Audioguru, I have also adjusted the gain of the driver op-amp so I get around +- 32 volts at the speaker output. \$\endgroup\$
    – Oli
    Jul 13, 2021 at 19:12
  • \$\begingroup\$ 28.3V RMS into 1 ohm is 800.9W. 23.3V RMS is 65.9V peak-to-peak. The supply to the class-AB bunches of emitter-followers must be about +37V and -37V. \$\endgroup\$
    – Audioguru
    Jul 14, 2021 at 20:32
  • \$\begingroup\$ Another issue which I just noticed. You need to insert small resistors at the emitters of your 4 final power transistors in your output stage. If you don't, whichever of each pair has a slightly higher gain (for a given base emitter voltage) will draw more current. That means that it will dissipate more power. This will cause the emitter-base voltage to drop, giving even more base drive, more collector current, and more power dissipation. When multiple BJTs are connected in parallel without taking this into account, the classic term is "firecracker mode". I think you can figure it out. \$\endgroup\$ Jul 18, 2021 at 23:06

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