I am testing a simple circuit in which an electret mic is hooked up to two transistor preamps, then to an 8ohm speaker. I am using a 9v battery, and the circuit looks like this. This circuit is meant to be simple enough to teach to kids at school. enter image description here

I have followed it exactly, and it works... somewhat. As soon as I plug in power, a low noise comes from the speaker. When I try to speak into the mic, the sound changes, but you can't actually hear works. Just change in volumes/sound change. I really want a "megaphone" like circuit, without the excess noise and clear amplification.

please note I prefer this transistor amp over a lm386 chip for this particular project as it is for use in a school

What issues could be causing the excess noise?


Made purpose clearer.

  • \$\begingroup\$ The easiest fix for this circuit is to delete it and start over. I could explain various problems, but to get what you seem to want (microphone input, low power 8 Ohm speaker output), it would be easier to start with a different topology. \$\endgroup\$ Dec 16, 2016 at 17:01
  • \$\begingroup\$ Could you explain a little more by "deleting"? Start over with what? \$\endgroup\$
    – agillbraun
    Dec 16, 2016 at 17:02
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    \$\begingroup\$ I can't imagine what part of "start over" is confusing. This circuit is crap, silly, a joke. "Deleting" it could mean tossing out the paper it's scribbled on, burning it, or ripping it into small shreds and throwing them up in the air while screaming like a chicken. \$\endgroup\$ Dec 16, 2016 at 17:15
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    \$\begingroup\$ Here's a link to a similar, but properly done, circuit here on Stackexchange.. The link goes to an answer that includes the diagram and an explanation of the circuit. \$\endgroup\$
    – JRE
    Dec 16, 2016 at 17:37
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    \$\begingroup\$ For "teaching" purposes, it might be good to start with a statement of the problem: "I have a mic with an xxx ohm impedance and xxx mV of signal to drive a speaker with an xx ohm impedance and an xxx mV of signal desired." Then explore solutions after the problem has been stated first. Let's not teach folks to jump to the first solution they come upon without understanding the problem. \$\endgroup\$
    – scorpdaddy
    Dec 16, 2016 at 18:39

2 Answers 2


As others have mentioned, that is not a very good circuit. Alas, we can't expect that just because we found it on the internet it will be a decent design. I wold use a circuit more like this one.

transistor intercom circuit

Source: http://www.circuitstoday.com/transistor-intercom-circuit

This is billed as an "intercom circuit". But it is a good example of a circuit with sufficient stages of gain to take an electret microphone signal and boost it up to speaker level. The switch (S2) will create a tone to signal the other end of the intercom. You can omit S2 and C3 if you aren't using it for an intercom.

  • \$\begingroup\$ Thank you very much, i'll give it a try. So you're saying the other circuit just didn't provide ample amplification? Anyway thanks for the answer. \$\endgroup\$
    – agillbraun
    Dec 16, 2016 at 17:37
  • \$\begingroup\$ The other circuit has several problems as already observed by others. It takes A LOT of gain to get from an electret mic capsule to drive a speaker. The second transistor in your circuit provides ZERO gain (because it is an emitter follower). So that puts the ENTIRE burden on the first transistor which is very poorly biased. That circuit is practically guaranteed to fail. Just because you find a circuit on the internet doesn't mean it even works at all. Consider the source and the community feedback. \$\endgroup\$ Dec 16, 2016 at 17:49
  • \$\begingroup\$ He's too polite to say that the other circuit straight up sucks canal water - but it does, and Olin is right about how bad it is. This is a much better circuit, but not because it provides more amplification. It was designed by someone who knew how things work. The other looks like something a teenager tossed together after reading a short description of amplifiers and transistors. \$\endgroup\$
    – JRE
    Dec 16, 2016 at 17:51
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    \$\begingroup\$ This circuit is better in that it at least has a chance of enough gain. However, it still suffers from very poor biasing of the transistors. The first two are biased completely open loop, and the third almost completely. Whether this circuit works at all is highly dependent on the gain of the particular transistors used. This circuit is still in the "sucks" category. \$\endgroup\$ Dec 16, 2016 at 18:19
  • \$\begingroup\$ It is harder to find good, simple discrete transistor circuits anymore because it is ancient technology here in the age of integrated circuits. It is not clear why a discrete transistor circuit is desired here? Are you teaching archaeology? Or modern technology? Olin's legitimate beef with the transistor biasing could be resolved with a couple of resistors and a re-design of the circuit. But the question "why discrete?" remains. \$\endgroup\$ Dec 16, 2016 at 18:32

This mess isn't worth trying to salvage. I'll briefly mention a few problems:

  1. There are no component designators in the schematic. This makes it hard to talk about the circuit. Remember that people are volunteers here, so not including them is rather rude when you expect a discussion.

  2. There are many ways to draw a neat schematic, including the built-in schematic editor. There is no excuse for hand-scribbled slop. You can draw a schematic neatly and scan it, but sloppily as you did is a insult to anyone you ask to look at it.

  3. The first transistor is poorly biased. Where it ends up is highly dependent on its gain.

  4. The gain of the first stage is indeterminate. It will probably be "large", but will also vary largely from part to part.

  5. The second transistor is used as emitter follower, so the B-C resistor makes no sense.

  6. DC coupling the speaker can work for small amplifiers like this, but you have to be prepared for the significant quiescent current. That doesn't work well with one of those clip-on 9 V batteries.

  7. The current thru the speaker puts a significant load on the battery, which will cause the battery voltage to vary. This is actually the source of the problem you are asking about. That's how the output is feeding back into the input. 10 µF for trying to stabilize the 9 V rail against this is a joke.

  • \$\begingroup\$ Hey thanks for the great answer. This isnt actually my circuit. I found it off the web. \$\endgroup\$
    – agillbraun
    Dec 16, 2016 at 17:15
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    \$\begingroup\$ There isnt a need to be sarcastic. I was asking a simple question, and being a beginner, there is no need to be rude and indirect. \$\endgroup\$
    – agillbraun
    Dec 16, 2016 at 17:21
  • \$\begingroup\$ @Aaron: Actually, for someone that expects a circuit they "found on the internet" somewhere to work, there is. \$\endgroup\$ Dec 16, 2016 at 17:23
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    \$\begingroup\$ I apologize for being new electronics. Please don't become a teacher, as you turn people off by being arrogant. \$\endgroup\$
    – agillbraun
    Dec 16, 2016 at 17:25
  • \$\begingroup\$ @Aaron: It's OK to be new to the subject, but then you don't start out by asking about some imagined solution instead of the real problem. \$\endgroup\$ Dec 16, 2016 at 17:29

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