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This is the view of my circuit related to the question. My intention was the opAmp will adjust its output so that the voltage on the speaker will be equal to the input waveform at the noninverting terminal of the opAmp(V(out_adder)).

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Here you can see that(see red arrows), the output of U5 opAmp and the power on the transistors change while time passes and stabilizes at some level. However, this is not desired. I like circuit as when it is in its initial situation.

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Here you can see the very first results in a closer look.

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Here you can see the final results in a closer look. The distortions on the output of U5 opAmp result in(I assume) distortions(please focus on the peaks, there is a little bit clipping at some points which alters speaker voltage from V(out_adder)) at my speaker(load) voltage.

Why does the output of U5 opAmp changes in that way? I assume the changes in the power consumption of the transistors are tied to this change.

Edited to up.

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    \$\begingroup\$ A DC path should be made from op-amp output, through to the transistor emitters, and back to op-amp inverting input. Capacitors C6, C7 and C8 block this DC path now. You now have extremely high DC gain, which will drive output toward one of the power supply rails after a time. \$\endgroup\$ – glen_geek Dec 6 '19 at 0:45
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    \$\begingroup\$ Where did this schematic come from? It has several design errors involving DC feedback and output bias current. Almost every stage is incorrect. \$\endgroup\$ – Sparky256 Dec 6 '19 at 1:24
  • \$\begingroup\$ @Sparky256 It is something I myself trying to do, so yes it probably has errors. Any advices? \$\endgroup\$ – muyustan Dec 6 '19 at 4:52

It's a poor design.

  • LM741 current limiting is causing clipping also has 2V saturation from Vcc,Vee . Get Darlingtons
  • If these 9V were batteries the 1 uohm would be >=10 ohms
  • The caps are not needed , except ADD Zoebel RC Snubber.
  • C8 is not required.

  • 2SB1258 2SD2642 may be good options for Darlingtons or DIY equiv

  • below I showed quasi PNP using power NPN, signal PNP
  • power NPN 10A for example

For more output, use Rail to Rail CMOS OP Amp with Darlingtons

strong text 6W pk 3W RMS 8 Ohms REV C enter image description here 200R represents current limit on Op Amp

Rev D Gain =5 (fixed err.)

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If still using poor LM741 OA , you can reduce Darlington saturation voltage using Quasi-power pairs with Power PNP,NPN to make the opposite NPN-PNP Simulation HERE enter image description here

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    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$ – Dave Tweed Dec 6 '19 at 21:59
  • \$\begingroup\$ Is that 2-2.5 W of power dissipation on the power transistors normal for this kind of applications? Or is it too much? \$\endgroup\$ – muyustan Dec 6 '19 at 22:00
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    \$\begingroup\$ Normal linear efficiency 5.8/7.7W= \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 6 '19 at 22:02

Your basic approach is OK...a linear buffer stage accepting an input signal, followed by discrete transistor NPN/PNP current amplifier of type Class-B.
If you're new to this kind of design work, best to approach it in small steps. Eventually, you can push this basic approach to a full Class AB high-quality amplifier with gain, and overall feedback.

1st step: simple gain-of one Class B

Here, the op-amp is used as a simple buffer. Gain of +1. Local feedback only. The NPN/PNP transistor pair allows significant current gain, but has slightly less-than-1 voltage gain. This design yields significant cross-over distortion, but is otherwise reasonably linear. If you use 9V batteries, a simple Thevenin battery model is used: Vt=9V, Rt=2 ohms. Be aware that those 2-ohm resistors are not to be added to your circuit.

A generic LM741 opamp is shown. There are better choices. Be careful to choose an opamp that allows high DC supply voltages (many have a limited supply range). A better choice would be one allowing rail-to-rail output swing...you'd like output voltage to be able to approach +9V on its positive swing, and also approach -9V on its negative swing when you increase signal amplitude. The Zobel network is not shown for simplicity:


simulate this circuit – Schematic created using CircuitLab

2nd step: overall gain-of-one feedback

Looks similar, but the overall circuit feedback has a subtle result: crossover distortion from the Class-B output transistors is much-reduced by the op-amp. Voltage gain should now be very close to +1.0. This can actually be an acceptable circuit operating at low input frequencies. At higher frequencies, many opamps slew too slowly to properly adapt to the cross-over glitch:


simulate this circuit
There are other ways to deal with cross-over distortion. Every one involves separating the bases of Q1 and Q1 with a DC bias voltage. This is risky, because you can add too much DC bias, which tends to overheat those two transistor with shoot-through current that is wasted (doesn't flow through the load).
Adding bias moves the output stage from straight Class-B to Class AB, and is an advanced step (not going to address it here). It is risky because most bias circuits need to be temperature-compensated. Those transistors often run hot.

  • \$\begingroup\$ Thanks for the answer, my strategy was just like you explained. From the outer view, opAmp just adjusts its output in such way that the voltage on the speaker(load) will be equal to input at non-inverting terminal. Right? So what should be my considerations regarding the output limits of the used opAmp? It's output current is limited as far as I know, where does that make a problem? \$\endgroup\$ – muyustan Dec 6 '19 at 19:02
  • \$\begingroup\$ One more question : Is 2 ohms of ESR valid for those commonly used commercial 9V batteries? \$\endgroup\$ – muyustan Dec 6 '19 at 19:23
  • \$\begingroup\$ With a high current-gain NPN & PNP, current required from the opamp should be within limits, but not by much. So shift focus on voltage drops...you lose about a volt from the 2-ohm battery internal resistance. You lose 0.6V from transistor base-emitter junction. You lose about 2V because op-amp can't swing all the way to the DC supply rail. All these subtract from 9V - so you're left with about 5.4V peak AC voltage delivered to 8 ohms. Is that enough? you decide. \$\endgroup\$ – glen_geek Dec 6 '19 at 19:43
  • \$\begingroup\$ I did a crude experiment with a fresh 9V Alkaline battery. \$ R_T=2 ohms\$Forget what was the test current. A near-dead 9V has much higher internal resistance. Increasing those two 10uf capacitors to 1000uf may help a little. If you possibly can, replace with a stiff (stiff meaning Thevenin resistance is near zero) DC supply of 9V. \$\endgroup\$ – glen_geek Dec 6 '19 at 19:52
  • \$\begingroup\$ thanks, but why do I lose base-emitter voltage drop? To me, it looks like it should be collector emitter voltage which should be omited. \$\endgroup\$ – muyustan Dec 6 '19 at 20:26

With the help of my imagination and grandson's fiber pens:) I visualized the invisible electrical quantities voltage and current to make clearer the circuit operation. In this geometrical interpretation, voltages are represented by vertical bars (like water columns) in red and currents - by loops (like water flow) in green.

I have shown only the positive half of this complementary circuit; the rest part is below the sheet. I have illustrated the 2-transistor Darlington version as more sophisticated... and since it is needed here. It would be interesting to show also the 1-transistor version... and to compare the op-amp behavior in both cases. Then we can see how the "clever" op-amp increases less or more its output voltage to compensate one or two base-emitter voltages VBE.

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Below is a screenshot of an animated Flash movie dedicated to this circuit phenomenon. I hope administrators have nothing against to put here a link to this useful resource - Strange things can be put into feedback loop (select the transistor from the library on the left and "hide the dog" on the right:)

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    \$\begingroup\$ Thanks for such a nice answer. I will have a look it, when I am available. \$\endgroup\$ – muyustan Dec 8 '19 at 9:34

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