# Electret mic amplifier circuit

#### abdulwahab.hajar

Joined Jun 14, 2016
93
So I'm trying to take an audio input from the electret microphone, and connecting the output to an oscilloscope to check if my input is being taken correctly.
I have a voltage of 5V with a 1K resistor to power the mic, and then I have a 1uf capacitor to stop the DC offset voltage from going into the amplification stage, after that I have an amplifier (741) with a gain of 10 ( as I have a 10K resistor through feedback, whereas the input is through a 1K resistor)...
However, when connecting my voltage out to the oscilloscope the signals are very weak as if they are not being amplified??
Where did I go wrong with this circuit??
Thank you

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#### Bordodynov

Joined May 20, 2015
3,012
See

#### Dodgydave

Joined Jun 22, 2012
10,483
Your input is wrong, you need to use the Non-inverting pin for the mic.....

#### AlbertHall

Joined Jun 4, 2014
12,156
Where are the supply pins of the '741 connected?
Neither the inputs nor the output of the '741 will work close to the supply rails. Note the circuit in post #2 uses positive and negative supplies for the '741.

#### Robin Mitchell

Joined Oct 25, 2009
819
Edit: Im an idiot, missed the dual supply rail

Yeah, your best bet with this circuit is to do the following:

Determine your power rail voltages (I am going to assume a single supply, no negative voltages).
Determine your gain (for example, 10)
Determine your max vpp of your electret (in this example, +- 80mV ~ 160mV)
Create a divider that adds enough voltage to make the coupled signal all positive ( in this case, a divider that will output 80mV with no signal input)
Take your coupled signal after C1 and feed it into that potential dividers output to add the voltages.

Sorted! At -80mV the signal from the couple stage will be 0V and at 80mV will be 160mV. A multiplication of 10 makes this a signal with 1.6V peak to peak with a DC offset that can easily be removed (or kept) at the output.

#### AlbertHall

Joined Jun 4, 2014
12,156
Sorted! At -80mV the signal from the couple stage will be 0V and at 80mV will be 160mV. A multiplication of 10 makes this a signal with 1.6V peak to peak with a DC offset that can easily be removed (or kept) at the output.
If either input goes to the negative supply rail the '741 won't work. With a 5V supply you'd be pushing it to get a 1.6V p-p output. It is a long way short of a good op-amp.

#### atferrari

Joined Jan 6, 2004
4,666
Edit: Im an idiot, missed the dual supply rail

Yeah, your best bet with this circuit is to do the following:

Determine your power rail voltages (I am going to assume a single supply, no negative voltages).
Determine your gain (for example, 10)
Determine your max vpp of your electret (in this example, +- 80mV ~ 160mV)
Create a divider that adds enough voltage to make the coupled signal all positive ( in this case, a divider that will output 80mV with no signal input)
Take your coupled signal after C1 and feed it into that potential dividers output to add the voltages.

Sorted! At -80mV the signal from the couple stage will be 0V and at 80mV will be 160mV. A multiplication of 10 makes this a signal with 1.6V peak to peak with a DC offset that can easily be removed (or kept) at the output.
Hola Robin.

The above, is it the follow up to what post? Confused to say the least.

#### Robin Mitchell

Joined Oct 25, 2009
819
@atferrari Have you ever had those times when you go mad and bonkers in the head? I think I was like that when I made that post. Not only did I NOT answer the question but I waffled about how to amplify a signal using components (like the LM358) that I failed to mention. I think I have had a stressful day

While typing this there is a nurse saying in the background "can someone please send mr.mitchell back to his room"

#### AnalogKid

Joined Aug 1, 2013
10,122
Where did I go wrong with this circuit??
As above, without knowing the power connections, it is difficult to say. However, the DC bias resistor and opamp input resistor combine for a load on the mic output of 500 ohms. This is relatively low, and will reduce the signal going into the opamp. A better total load impedance is 2.2 K to 10 K, depending on the electret element used.

ak

#### ian field

Joined Oct 27, 2012
6,536
Your input is wrong, you need to use the Non-inverting pin for the mic.....
The whole point of an op-amp is that it can be inverting or non-inverting. Either input is OK as long as you do the bias right.

Some single rail power amps work with a grounded input - but the op-amp requires split rails for a grounded input to work, or a "virtual earth" provided for that input by a decoupled voltage divider.

#### Bordodynov

Joined May 20, 2015
3,012
Output mic PP=1.35mV OUTput PP=1.35mV*100=135mV at loudness of the sound 64dB. 90dB OUTput PP=2.7V.
See