I'm concerned about input/output capacitors in power amplifiers and their destructive impact on the sound-wave.

In my understanding, for simple transistor DIY amplifiers you'll need a really big capacitor (normally available only as electrolytic capacitor) to prevent square wave at low frequencies from being bent towards saw-like shape.

I am also interested in building my own transistor amplifier (10W should be more than enough) with simple schematic. I'd like to use small amount of parts.

I came up with a plan and would like to ask if I made any serious mistake here. (I understand that I'll need to add some small capacitors between op-amp V+ and V- to ground to reduce noise.)

Drawing of planned power amplifier:

drawing of planned power amplifier


Adding new drawing according to comments (removed second OP-amp, added diodes between bases hope that helps for temperature compensation).

new drawing of planned power amplifier

  • 3
    \$\begingroup\$ Building a class B amplifier without having looked into existing designs and how they deal with crossover distortion will have far more destructive impact on the sound wave than an electrolytic capacitor theoretically could, which it doesn't really. Have you tried simulating it! It helps everybody to draw the schematic with +ve voltages to the top, and bias currents flowing downwards. \$\endgroup\$
    – Neil_UK
    Commented Oct 2, 2021 at 19:14
  • \$\begingroup\$ I expect this to work as a class AB amplifier, not class B. \$\endgroup\$
    – Testik
    Commented Oct 2, 2021 at 19:16
  • 4
    \$\begingroup\$ Ah! I see. The extra voltage from R3/R4 R6/R7 puts both output devices into conduction. There's no control of the overlap against tempertaute and rail voltage. There's no overall feedback to control against VBE non-linearity. Have you simulated the distortion? I ask again, have you looked at any amplifier designs and the theory of solving crossover problems. You are simply reinventing a square wheel that many designers have already spent decades hammering into a round shape, and publishing designs and commentary on them. \$\endgroup\$
    – Neil_UK
    Commented Oct 2, 2021 at 19:20
  • 2
    \$\begingroup\$ Without temperature compensation, it will go in thermal runaway and burn the output devices. \$\endgroup\$
    – bobflux
    Commented Oct 2, 2021 at 19:34
  • 2
    \$\begingroup\$ "it behaves so perfectly that I'm in doubts". Well, I have doubts: without any feedback, the crossover distortion will be tremendous. \$\endgroup\$ Commented Oct 2, 2021 at 19:47

1 Answer 1


No. That circuit is not valid might work perfectly only in a simulation. Real world components have tolerances such as op amps have offset voltage - both op amps might want to drive their own input pins to different voltages but they can't do that now.

And since you have a dual voltage to begin with, you don't need output capacitors as the output can be DC coupled. Just use a single op-amp and include the emitter followers at the output to boost current and take the feedback from there. Usually good results are achieved by proper biasing of the transistor bases and finding a suitable quiescent emitter bias current via emitter resistors to minimize crossover distortion.

  • \$\begingroup\$ thanks, I updated the post with a new picture please tell if that's better (I understood that I'll need to change 3k resistors later to fix crossover distortions) \$\endgroup\$
    – Testik
    Commented Oct 2, 2021 at 21:07
  • \$\begingroup\$ That buffer stage is still driven in open loop, outside the feedback. Also you can't expect the Vbe drops to match the diode drops. Is there a reason you don't drive the middle point of diodes with the op amp directly? \$\endgroup\$
    – Justme
    Commented Oct 2, 2021 at 23:16
  • \$\begingroup\$ thanks, I found a scheme of emitter follower where middle point driven with op amp directly: nutsvolts.com/uploads/wygwam/NV_0801_Marston_FIGURE11.jpg I think it's not complete (no diodes or resistors), so this is my version: pasteimg.com/images/2021/10/03/3rd.jpg I hope now I've got the general feedback loop now? I can explain why I'm placing 47k/1k resistors: it's because I expect input signal to be of 250mV amplitude so I need some gain, I hope this doesn't "kill" the feedback. And what should I place instead of diodes to match the "Vbe drop"? \$\endgroup\$
    – Testik
    Commented Oct 3, 2021 at 0:05
  • \$\begingroup\$ @Testik That last diagram at the bottom of your question also has problems. Tell me what happens when you try and drive towards either rail, for example. Not hand-waving. Work out the details of how your output stage gets its recombination currents. I want to see numbers and sources. Also, you won't get \$10\:\text{W}\$ from \$\pm 12\:\text{V}\$ rails into \$8\:\Omega\$. (That's just a basic sniff-test result.) \$\endgroup\$
    – jonk
    Commented Oct 3, 2021 at 3:48

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