First of all this is not related to a HW or anything; I just wanted to build an audio amplifier and test my skills. I really don't see why my design is not working as intended, without the speaker load it amplifies the input signal just right but when I connect the speaker load (4 Ohm) the power of the speaker is limited to microwatts. Can anyone help? Sorry I didn't sketch the circuit in the editor since I already had drawn it in LTspiceamp circuit P.S. I actually have built the circuit but I can barely hear something coming from the speaker when I blow on the mic.

  • 3
    \$\begingroup\$ Your values for R4 and R5, especially R5, limit the current available to the low resistance of the load, so the load power is very low. Ignoring C2 for a moment, the maximum voltage across the speaker is 24 x (4/2204), or about 44 mV, for a maximum peak power of 47 uW. \$\endgroup\$ Jun 16 '16 at 15:24
  • \$\begingroup\$ Please space out your schematic, it's very hard to read. Also, you might consider linking us the LTSpice file so we can verify your simulation. Finally, the 1 mF capacitors look suspicious, are those the intended value? \$\endgroup\$
    – uint128_t
    Jun 16 '16 at 15:25
  • 5
    \$\begingroup\$ This circuit cannot be used as a POWER amplifier. Its a voltage amplifier but it has virtually no current driving capability. Connecting a 4 ohm speaker at the final stage is like adding a short circuit to the output. \$\endgroup\$ Jun 16 '16 at 16:15
  • 2
    \$\begingroup\$ Did you even look at any audio power amplifier design before you started? \$\endgroup\$ Jun 16 '16 at 16:50
  • 1
    \$\begingroup\$ Suggested reading: Nelson Pass Zen amplifier, or Douglas Self's various writings on audio amplifiers. Your output stage is not practical for a power amplifier. All the current that goes to the speaker has to pass through R5. \$\endgroup\$
    – mkeith
    Jun 16 '16 at 16:54

The gain of the last stage is approximately equal to: $$Gain_3 = {{R8}\over{R4}} = {{4}\over{2200}}=0.001818$$ You must use a push-pull emitter follower. Transistors desirable to take a large gain. With such a supply voltage using two stage voltage gain enough (\$Gain=1000 \div 5000\$) output and push-pull emitter follower (gain approximately equal 1). The last stage can be done on a single-ended power transistor with the radiator cooling. But modal current in this case is very high.


When you take an output signal from the collector you are buying in to a circuit that has a signal gain that is governed by collector resistance divided by emitter resistance. This is because the signal on the base is also the same amplitude (within reason) as the signal on the emitter.

Because collector current largely equals emitter current it then becomes a triviality to say that gain = Rc/Re.

So, your final stage would have a gain of 1 because Rc = Re but you then put a 4 ohm speaker in parallel with Rc (via the output capacitor) and gain rockets down to a fraction of what it was.

Virtually all common-place amplifiers buffer the output (the sensitive colelctor node) with an emitter follower (push pull on more powerful amps).

Another way of looking at it is by saying the signal at the collector is sourced from an impedance equal to the collector resistor and of course if you have 1Vp-p being fed via 2k2 to a 4 ohm speaker your get 1.81 mVp-p.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.