0
\$\begingroup\$

I found this circuit in many places:

Circuit with 12 V and different resistance

The same resistance (68 kΩ) works with 3.7 V, 9 V, and 12 V. The LED flashes with the mic input (say with music).

I tried with an old Samsung earphone mic, didn't work, I tried connecting the mic output of my working earphones, didn't work. I am not sure where am I wrong.

I don't have a 68 kΩ resistor. What I have is a 5 V power supply and a 56 kΩ resistor. I tried joining two in series.

Can anyone explain how this circuit works so that I can calculate the resistor values I need to make this work with a different mic, resistance and voltage? Is this circuit working as a switch or as an amplifier?


Added - background: I have a electret mic from old earphones.
I want to glow the led with music beats, that in turn would fall on an LDR that is attached to a BT136 triac's MT2 and Gate, in series with a resistance. Eventually it will glow or turn off an ac light with music beats.

My BT136 circuit is working fine, but I am not able to accomplish the DC part of this project. I am searching for circuits with least components because I have access to limited components in my area. Can my goal be achieved anyhow by using the BT136 and the mic, with minimal components?

\$\endgroup\$
7
  • 2
    \$\begingroup\$ That's a terrible circuit, I wouldn't expect it to work except maybe with a specific LED and a specific microphone, and maybe a resistor selected for the specific gain of the transistor. \$\endgroup\$
    – John D
    Commented Sep 2, 2022 at 20:44
  • \$\begingroup\$ Much depends on the details of the microphone. If it has a very low DC resistance, it essentially shorts base to emitter, and would need to put out more than about 600 mV to start forward biasing the transistor. It may work better if you put a capacitor in series with the mic, and maybe a 5k resistor from base to emitter. \$\endgroup\$
    – PStechPaul
    Commented Sep 2, 2022 at 22:00
  • \$\begingroup\$ What is your goal? Learn about transistors? Blinking light to music? For the later, start with an LM3915. \$\endgroup\$
    – Mattman944
    Commented Sep 2, 2022 at 23:47
  • \$\begingroup\$ @Mattman944 the goal is to know if this circuit works or not. If it works, how do I choose the right resistance to make it work with 5v dc. \$\endgroup\$
    – Xigstan
    Commented Sep 3, 2022 at 3:52
  • 1
    \$\begingroup\$ @Mattman944 I lightly edited your entirely OK comment because some people have a penchant for misunderstanding terminology and I'm trying to head off flags at the pass, as it were. I'm getting a bit twitchy about such things. Not your fault :-) \$\endgroup\$
    – Russell McMahon
    Commented Sep 9, 2022 at 12:19

1 Answer 1

5
\$\begingroup\$

I found this circuit in many places.

I hate to be the one sapping all the fun out of internet-based circuit building, but a lot of those circuits are just filler material on sites full of filler material.

I assume that any circuit diagram that has a site watermark going right through it will not work as shown. There are so many non-working or barely working circuits out there that it's not even funny anymore. It's sad all around.

That circuit will "work" only with a dynamic microphone or a speaker used as a microphone, and then the resistor has to be adjusted so that the LED is just glowing very faintly. For best effect, use a high-efficiency red LED. When you then scream into the microphone, or tap it strongly with your finger, you might see the LED blink momentarily. This may be easier to see in a dark area than in full light.

When the "microphone" is a large enough speaker, this circuit is a wonderful tapping detector: it detects when you tap the center of the speaker cone.

If the idea is to have a LED that lights up when some reasonable threshold of sound pressure is exceeded, then you'll need a couple more transistors to form a comparator, and something to extend the pulse length so you can actually see it well.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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