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Speaker output to electret mic input

Thread starter #1
Newbie question. I want to power a small motor from the output of a speaker. The motor will turn on when the speaker output is loud. Like syncing lights to music. It is mainly intended to be driven by voice output. I have ordered one of these little LM393 sound detection modules from eBay: https://www.ebay.com.au/itm/273571705679

I am assuming that the output is just on/off as described here: http://henrysbench.capnfatz.com/hen...nd-detection-sensor-tutorial-and-user-manual/ . The plan is to hook up a mosfet to the output and drive the motor with it.

I don't want to act on environment sounds so my intention is to remove the electret mic and run two wires directly from the speaker terminals to the mic inputs, with a 10k series resister and 100 ohm parallel resistor like this: http://www.epanorama.net/circuits/line_to_mic.html

Is that going to work?
Will it have any significant impact on the sound output from the speaker?

I guess the other alternative would be to change the mic to a directional (cardioid) one if it is an omnidirectional and put some foam around it to block out other noise.
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Well-Known Member
The only thing you need to be careful of is the fact that the electret mic has a small power supply via a resistor (I suppose technically it is a load resistor), which powers it's internal transistor, so when you remove the mic you will also need to disconnect this resistor from the coupling capacitor. To be honest I don't think it will make a noticeable difference to the signal but it might present an unacceptable load on the microphone's supply.
You may also need to modify the resistor values circuit you linked to - experiment and see.


Well-Known Member
Hi, yes, R1.
It might be easier to cut the trace that connects the resistor to VCC, which also makes the modification more easily reversible. Though now I look at it you don't need to bother. If you connect the L-pad via a coupling capacitor, then it makes no difference what's going on in the mic circuit with regards to DC. Try maybe 1uF to 10uF.

All that said, R1 is quite high and the module doesn't have an on-board regulator, so you are most likely perfectly ok doing your original plan anyway. Just account for R4 and R1 on the module being in parallel with R2 in the L-pad.

There might be a better way to do this however, if you are prepared for something a little more complicated, in the shape of an opto-isolator. The reason being the opto-isolator has an output transistor which you can connect directly in place of the mic and will probably provide adequate signal with no further modification. The other side of the opto is an LED which you can connect across the speaker, and depending on how powerful the signal is will need either a reverse diode across it and series resistor, or a small amplifier (1 transistor, coupling cap and a few resistors which would also need to connect the module's Vcc supply) Don't connect it direct it could blow the LED and distort the sound.

It all depends on how powerful your signal source is. Basically, the louder it is, the less effort you need to make!


Well-Known Member
Most Helpful Member
You show an attenuator that converts a line level to a mic level but you do not have a line level, instead you have a speaker level that you did not say how loud it will be.
Of course, ebay knows nothing about electronics so they do not say the sensitivity of the module.

Try it with your 10k to 100 ohms attenuator. If it is too sensitive to noises then increase the value of the 100 ohms or replace the 100 ohms resistor with a 1k trimpot.

dr pepper

Well-Known Member
Most Helpful Member
You'll also need a decoupling capacitor, like C1 is further down on that page, speaker o/p's may have a small dc component.
You wont get great performance, but then you wont need it driving a motor.
Thread starter #7
So I think I am hearing that I should just try it out and see what happens. I have control over the speaker volume so I can increase it gradually until I get what I want. I can likewise play with the trimpot on the module. (It is only a little 40mm speaker)

Do I really need a second decoupling capacitor or will the existing C1 suffice? Bear in mind I am not looking for a "clean" audio signal. The only requirement would be that any DC component does not cause the output to stay on permanently.

Rather than replacing 100 ohm with 1k trimpot would it not be safer to add 1k trimpot in series with 100 ohm?
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Well-Known Member
Question would have been easier if we'd known it was only a 40mm speaker up front! Otherwise we just have to imagine what it might be!

The extra coupling cap (I think that's what DP actually meant to say!) is "just in case". There is a DC level present at the module's input after you remove the mic, albeit via a high resistance, and as DP says there is a small chance of there being DC on the speaker wires, even though there shouldn't be. Best to isolate them with a cap.

No just put in the 1k trimpot in place of the 100R, you can adjust the level all the way down to near zero that way.. The other resistor is big enough to stop anything bad happening.


Well-Known Member
Most Helpful Member
I imagined a sound level of at least 20V peak (25W RMS into 8 ohms).

A 40mm speaker (squeaker) probably has 0.5W into 8 ohms which is 2V RMS with a peak voltage of 2.8V. The 10k to 100 ohms attenuator reduces the peak voltage to 28mV which is low enough to prevent damage but might be too low for the detection to occur then the 100 ohms must be increased.

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