I would like to sample the output of a microphone. I am using the following set-up in order to tune the DC offset at the microphone output.


simulate this circuit – Schematic created using CircuitLab

The aim is to get exactly 1.65V (half the follower supply voltage). This is why I used a 100k potentiometer (measured value is 97k) in parallel with the microphone.

While using this set-up, I got a kind of surprise. Without the microphone, I get \$ V_{out} = 8.8 \cdot \frac{P x}{R1 + P}\$, where \$ x \$ is the potentiometer wiper value (0 = wiper to ground), which leads to \$ x = 0.205\$ to get 1.65V at the output.

When the microphone is connected, the output voltage drops down to 1V, which is correct if I suppose that the microphone acts like a resistor \$ Rm\$.

Given the equation \$ V_{out} = 8.8 \cdot \frac{P x}{R1 + P\left( 1 + \frac{R1}{Rm} \right)}\$, which I suppose to be correct, I get \$ V_{out} = \$1\$\,\$V when \$ Rm = 17.5\,\$k\$\Omega\$.

Then, I changed the potentiometer wiper position in order to get 1.65V at the output when the microphone is connected. Following the previous equation, this gives me \$ x = 0.315 \$, which is correct when I check the potentiometer value using an ohmmeter.

So, finally the microphone parallel resistance seems to be 17.5\$\,\$k\$\Omega\$. However, I (of course) do not get the same value when I use the ohmmeter directly connected to the microphone (without polarizing its FET). I get about 1\$\,\$k\$\Omega\$, which seems to be normal.

My questions are: Why do I get this value of 17.5\$\,\$k\$\Omega\$? What does this value means? Is it relevant to assume the microphone acts like a parallel resistor? (I suppose it can be, because this is like considering the DC component of the microphone FET drain current).


Your electret microphone probably uses a JFET transistor to provide hi-impedance to the condenser element whilst acting as a constant current circuit to the output. The constant current might be 0.3mA for instance and if you measure the current at two different voltages you'll get approximately the same current but, if you calculated the supposed resistance at these two different voltages you will get different values.

When powered by your circuit via the 10k resistor what you might find is that the (guessed-at) 0.5mA drawn by the microphone drops 3V across the 10k ohm and leaves about 5V across the microphone - this would appear to be like a 16.7kohm resistor where the microphone is. Obviously you got 17.5kohm and I've tailored the numbers to try and match this. Try measuring at the top of the pot and the 10k to see what voltage it actually is.

Now, when you measure with a meter you won't be supplying 5V - it'll be more like 1V (another guess) and the JFET may be less than perfectly biased and it will start to conduct more than 0.2mA - remember, when a JFET loses its bias it conducts heavily compared to the biased condition. So this might mean 1V and 1mA (not 0.2mA) which equals 1kohm.

  • \$\begingroup\$ Ok, I see the point: the mic has to be considered as a constant current source. But using a 0.5mA current drawn by the mic, I found 3.4V across the mic. So, the real drawn should be about 0.3mA (datasheet says 0.5mA max.). Also, why isn't the appearing mic resistance given by the ratio between the voltage across the mic and the 0.5mA current drawn (5V/0.5mA = 10k according to your values) ? \$\endgroup\$ – Spin Mar 20 '14 at 21:11
  • \$\begingroup\$ I think the problem is your formula. In your first formula you have P then in your 2nd formula you have tried to put this in // with Rm but the formula doesn't stack up because you have an R1 term in there and it shouldn't be there. I think P becomes \$\dfrac{P\cdot R_M}{P + R_M}\$ \$\endgroup\$ – Andy aka Mar 20 '14 at 21:24
  • \$\begingroup\$ I don't think I made a mistake in my calculations... You can check here, but I think this is correct. \$\endgroup\$ – Spin Mar 20 '14 at 22:01

5 years late on this.

Your calculation of \$R_m\$ appears to be incorrect.

When you first set your wiper without the microphone (i.e. no microphone), you got 1.65 V at \$V_{out}\$ (after wiper is set). For this condition, the resistance between wiper and ground would be 20062.5 Ω. After this, you mentioned you connected the microphone, and you wrote that \$V_{out}\$ then became 1 V. Now, this 1 V will be at the wiper terminal. This means that the bottom terminal of resistor \$R_1\$ (and the top of the microphone and the top of the potentiometer resistance) would be at 4.8349 V, simply because you already know in advance the resistance between wiper and ground, and you also know the 1 V voltage across wiper to ground (which means you know the current through the full potentiometer resistance, \$R_P\$).

If full potentiometer resistance is \$R_P\$ (which you labelled as \$P\$), then resistance \$R_m\$ would be calculated as:

$$R_m = \frac{R_1.R_P}{0.82R_P - R_1} = 13948.8\space\Omega$$


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