I'm working on a small audio project and I need a circuit that does the following: take a signal from an electret mic, rectify the signal, amplify it, and remove the DC offset. I have tried many circuit designs but they are not working fine with me (too much noise and with DC offset). Should I mention that the electret mic's resistant is approximately 2.8 kΩ. Thanks in advance. any help will be appreciated 🙂
Electret Mic signal amplification without DC offset
amplifierelectret
Related Solutions
I think you may be overcomplicating this a little bit - why don't you just connect the electret microphone (observing polarity) to the microphone wires via a 3.3Kohm resistor and see what happens.
Your circuit won't work as it stands because the signal you are trying to put back onto the line gets fed back to the microphone via the 10k and it'll all end up very low gain in my humble opinion.
I don't think there is a substitute for trying it out and if it's not loud enough, putting a 10uF capacitor across the 3k3 resistor I mentioned and if that doesn't sound like the right volume maybe drop the 3k3 to 1kohm.
As for matching the impedance - this won't make a great deal of difference to the kit that recieves the signal. It's likely got an unbalanced input and is probably not differential so I don't think you need to jump thru hoops on this.
If you still believe the mic on its own won't work, try this: -
Digital Signal Processors can do the FFT in real time. They have a particular architecture which means that certain instructions (MAC - multiply accumulate) can be done very quickly.
But then this turns in to an embedded software project too as well as the electronics design, if you're ok with that.
You can get development boards from companies such as Texas Instruments which contain a processor and break-out box arrangement for experimentation.
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Best Answer
This is what wiki says: -
This will work fine providing you AC couple your microphone to the input and AC couple the output. Simple series capacitance will do the decoupling of the DC signal.
PS it's only half wave rectification shown above.
This is a full wave circuit