I am trying to make a simple audio amplifier using an LM386 audio amplifier and a class AB amplifier push-pull output stage. Since I couldn't find something that explicitly combines the two (LM386 for voltage amplification and transistor for supplying the required current), I thought I might give combining the two a shot. The LM386 application was taken from the chip datasheet, and the class AB design from here. The schematic is here. I couldn't put this on the schematic but pin 7 of the IC is connected to a 100 nF capacitor to ground, and between pins 1 and 8 there is a 10 uF capacitor in series with a 10 k potentiometer in order to set the gain from 20 to 200. The circuit will run from a single 9V battery.

My questions are

  1. In the LM386 datasheet the electrolytic capacitor on the output has the plus towards the amp, but on the AB amplifier the electrolytic capacitor on the input has the plus towards the transistor, ie in the opposite polarity. Could someone help me understand how these capacitors should go and what should I do here?

  2. I thought since the current might get high, I used resistors rated for 1 W. The question is regarding the 10k pot on the input, is 4W an ok rating? Or if the input signal is too high might it get burnt out? Also, between pins 1 and 8 of the op-amp I used a small vertical potentiometer in series for the capacitor, which is rated for really low power. Is this ok?


Could someone also please verify the overall circuit and tell me if it is Ok and whether it will work or not?

Later edit: Thank you for the answers. I understand that both the LM386 (with an internal push-pull output) and the amplifier together have too much gain so I will have to remove one of them in the final project. I have made the modifications suggested. I only have one question left, whether I should leave the 2 caps back to back from the LM386, or should I just use one with the + towards the LM386 output, and why? Also, are the modifications correct?

modified circuit

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    \$\begingroup\$ Why don't you simulate it? \$\endgroup\$
    – Andy aka
    Commented Aug 1, 2017 at 15:51
  • \$\begingroup\$ Make a screenshot or export an image of it and include that here. I'm not about to agree something on digikey to view your schematic. \$\endgroup\$ Commented Aug 1, 2017 at 15:52
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    \$\begingroup\$ Transistor Q3 has collector and emitter reversed. \$\endgroup\$ Commented Aug 1, 2017 at 16:57
  • \$\begingroup\$ R3 isn't going to vary input signal much with an open-end. Perhaps you intended it to be grounded? 4W is overkill here. \$\endgroup\$
    – glen_geek
    Commented Aug 1, 2017 at 17:27

2 Answers 2


You don't need the Class AB external BJT amplifier at all. Notice in the datasheet for the LM386 that the output stage is already a Class AB amplifier. Given the voltage you want to work at you do not need any external add on amplifier for the LM386.

enter image description here

There are several mistakes in your circuit:

  1. Q3 is inverted. You will draw excessive current connected this way (and it won't work).

  2. You only need one capacitor coupling the LM386 to your output stage. Change the 250 uF (C2) to 47 uF and eliminate C3. The plus sign goes to the LM386 output because this will sit at approximately VCC/2.

  3. R3 need to have the lower end of the pot grounded, it won't act as a volume control connected as shown. It will also upset the bias in the LM386.

  4. You need a capacitor to isolate the input signal source. I'd recommend a 10 uF in series with R3.

  5. There is no feedback around both amplifiers so you have two independent amplifiers. The gain for your LM386 is 20 so overall gain is going to be in the hundreds, which is way too high for your application.

  • \$\begingroup\$ Why is that the correct polarity for C3? In other words, I know the LM386's output will swing around Vcc/2, but the way I'm thinking isn't there some voltage on the other side via C4 and then R5? If not, where am I wrong here? Also in these kinds of situations how do you determine the proper polarity for the cap? \$\endgroup\$ Commented Aug 1, 2017 at 18:50
  • \$\begingroup\$ Also shouldn't I eliminate C3, instead of C4? \$\endgroup\$ Commented Aug 1, 2017 at 19:46
  • \$\begingroup\$ Absolutely right....eliminate C3. The cap to the output of the LM386 needs to be positive to the output because at steady state (no signal) the output of the LM386 will be at VCC/2 and the base of Q1 at about 0.7 V. This is all mute of course since you don't need the BJT stages at all. \$\endgroup\$ Commented Aug 1, 2017 at 23:10
  • \$\begingroup\$ Thank you for the reply! It makes sense now, the difference between Vcc/2 and the base-emitter voltage in the transistor. \$\endgroup\$ Commented Aug 1, 2017 at 23:19
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    \$\begingroup\$ @andreas.vitikan. ....because the signal input is at 0 V and the base of the input transistor is at 0.7 V. This assumes that the input signal is AC coupled and centered around 0 V. \$\endgroup\$ Commented Aug 3, 2017 at 3:44

The circuit should be ok if you fix the polarity on Q3. Electrolytic caps conduct if they are back biased, so running C2 and C3 back to back is ok to drive Q1 base, and C4 is biased properly to drive the speaker. 1 W is overkill for the resistors, as is 4 W on the pot. You can use regular 1/4 W or 1/8 W resistors everywhere. Add some bypass capacitance on the LM386. You might want to try it without the LM386 - you have a lot of gain in both stages. The voltage swing required on your class AB input is only millivolts.

  • \$\begingroup\$ Thank you very much for your answer. You're right, Q3 is indeed a PNP, my bad on the schematic. What value should I use for the bypass cap to LM386? Also how do I know they're ok back to back, and when to use such configurations? What is the theory or the general rule? \$\endgroup\$ Commented Aug 1, 2017 at 18:33
  • \$\begingroup\$ Also it slightly confuses me. Does the PNP-NPN push-pull output also step up voltage? \$\endgroup\$ Commented Aug 1, 2017 at 18:35
  • \$\begingroup\$ The push pull stage (Q2 and Q3) of the class AB amplifier does not provide gain, but Q1 does. \$\endgroup\$ Commented Aug 1, 2017 at 18:40

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