If I want to make a simple Mic and amplifier circuit, and if I used this circuit for the mic:

Would this work and then connect the output to a suitable op amp circuit?

Also how do I determine what values to use for R1 and R2?

 

R1 is a little small, I think it should be 10k or so. You might not even need R2.
Do you have specs on the microphone?

 

R1 is a little small, I think it should be 10k or so. You might not even need R2.
Do you have specs on the microphone?

I bought them from goldmine electronics and this is all they have on them:

Primo electret microphone made for non directional applications features 100-10,000Hz frequency response, 2V to 10VDC standard operating voltage. Size: 10mm (.39") Dia. x 5mm (.20") High. Metal case. Sensistivity -44dB@1kHz, S/N Ratio 60dB, current 650uA max (Vcc=6V), Impedance 1Kohm (1KHz). Brand new prime. Primo# EM-100 (H). Package of 4.

 

It says 650uA max at 6 volts so aim for 600uA. With a 12V supply you will need 10k for R1. (12V-6V)/600uA

 

Increase R2 to 47k. Add R3, also 47k, from Vcc (your + supply) to the junction of R2 and output. That will keep the average voltage on the output at Vcc/2.

If you are planning on using a TL07x, TL08x or LF353 opamp, you will need to use a 9v to 12v supply, and either make R3 a little smaller or R2 a little larger than 47k, as those opamps can go to Vcc-1.5v, but cannot input or output within 3v of their negative supply.

 

Increase R2 to 47k. Add R3, also 47k, from Vcc (your + supply) to the junction of R2 and output. That will keep the average voltage on the output at Vcc/2.

If you are planning on using a TL07x, TL08x or LF353 opamp, you will need to use a 9v to 12v supply, and either make R3 a little smaller or R2 a little larger than 47k, as those opamps can go to Vcc-1.5v, but cannot input or output within 3v of their negative supply.

Is this what you mean?

 

vindicate,
Yes, you understood me correctly.

Note that you should increase the value of R2/R3 to roughly 47k-100k. Otherwise, it will load the electret mic signal too much.

Metal film resistors are suggested. They will generate much less noise than carbon or carbon film resistors.

 

Would anyone care to explain the purpose of R2 and R3 to me? I'm building this more as a learning purpose than functional so it would be nice to know

Also, the the tl081 and not only getting within 3V of the Negative rail. Why does makeing R3 bigger help that?

 

Would anyone care to explain the purpose of R2 and R3 to me? I'm building this more as a learning purpose than functional so it would be nice to know

Ahhh, I already did. It keeps the output biased at Vcc/2.

If you are going to use this with a single supply opamp, you will need to keep the signal in the middle of the range that the opamp can see.

Also, the the tl081 and not only getting within 3V of the Negative rail. Why does makeing R3 bigger help that?

No, you increase R2 or decrease R3 to keep "output" (the input to the opamp) biased at the middle of the voltage range of the opamp.

If you were using a TL071 or TL081 or LF351, the range is Vcc-1.5v to Vee/GND + 3v.

So, in the case of a 9v supply, that's 3v to 7.5v. The average of those two is 5.25v.

So, if R2 is 47k, what do you think R3 needs to be, in order to keep the average at 5.25v?

If you don't know how to calculate it, look in our E-book.

 

65625 ohms?

Ok so R2/3 sets the output voltage. I assume that the mic(the part of the circuit to the left of the cap) outputs voltage as well. So is that voltage "added" to the 5.25 volts?

I'm not really sure how mics work, so this is really a guess.

 

65625 ohms?

Nope.
Guess again. How did you come up with that number, anyway?

Start with 47k. You know already from my previous calculations that you want 5.25v at the node labeled "output".
How much current will flow through a 47k resistor when 5.25v is applied across it?

Once you figure that out, what resistance will you need from the junction of output & R2 to get 5.25v when the supply is 9v?

Ok so R2/3 sets the output voltage. I assume that the mic(the part of the circuit to the left of the cap) outputs voltage as well. So is that voltage "added" to the 5.25 volts?

The mic doesn't "output" voltage. It acts more or less like a variable resistor. You have a fixed 10k resistor to +9v.

I'm not really sure how mics work, so this is really a guess.

That was a guess all right.

 

Oh, by the way - instead of using a 10k resistor from the battery to the high side of the electret mic, use a 1k from the battery, then a 9.1k in series with it to the mic. Then connect a 1uF cap from the junction of those two resistors to ground. This will help to eliminate feedback problems.

 

I should have checked my work...oops.

Is the correct answer 34090 ohms?

The mic doesn't "output" voltage. It acts more or less like a variable resistor. You have a fixed 10k resistor to +9v.

OK, so it doesn't output, but it does create fluctuations in the output voltage by acting as a voltage divider with the fixed resistor?

So if this is correct, does that voltage then get "added" to the 5.25v? Or how does that work?

Oh, by the way - instead of using a 10k resistor from the battery to the high side of the electret mic, use a 1k from the battery, then a 9.1k in series with it to the mic. Then connect a 1uF cap from the junction of those two resistors to ground. This will help to eliminate feedback problems

Sounds good.

 

I should have checked my work...oops.

Is the correct answer 34090 ohms?

5.25v/47,000 Ohms = 111.7uA current through R2.
9v-5.25v = 3.75v to drop across R3.
Since R=E/I, and E=3.75 and I=111.7uA, R=33,572 Ohms.

34,090 is close enough. You can use a 33k resistor for R3.

OK, so it doesn't output, but it does create fluctuations in the output voltage by acting as a voltage divider with the fixed resistor?

Yes, that's the idea.

So if this is correct, does that voltage then get "added" to the 5.25v? Or how does that work?

No, actually the voltage across the mic and the fluctuations will be quite small, and near 0v. That's why you need the 10uF cap; it blocks the DC level, but couples the effects of the AC signal through. On the right side of the cap, the DC level is established by (R3+R2)/R2.

 

So if I do something like this:

Should I be seeing fluctutions on the output of this circuit? The source is a saw tooth generator that goes up to 600nA and down to 0V. This should be pretty close to the the mic circuit we have been discussing.

 

You don't have a ground for the 40Hz signal source.
It'll take a while for the cap to charge after power-up.

The lower resistor should be 47k.
The upper resistor should be 33k.

There should be a 9.1k resistor - oh, forget it.

 

You don't have a ground for the 40Hz signal source.
It'll take a while for the cap to charge after power-up.

The lower resistor should be 47k.
The upper resistor should be 33k.

There should be a 9.1k resistor - oh, forget it.

Sorry I should have explained. I wasn't trying to duplicate the circuit we have been talking about. I was just trying to simulate a microphone circuit, just so I could understand how the output works.

Thanks for the help though, I didn't mean to be frustrating

 

Would a circuit like this:

http://electronics-diy.com/electronic_schematic.php?id=753

Work as an amp for the mic? I have TL081 and LM358's to use with it, so I would just swap one of those in for the LM386.

 

Use this circuit instead:

AudioGuru posted it.

Change R2 to either 68k or 75k, one or the other will work fine - better than the 100k, as it will keep the signal better centered in the TL081's input/output range.

 

If you want to connect a speaker to the above preamp, then use an LM386 connected as shown in the datasheet.

National Semiconductor has a datasheet for the LM386 on their site.