i am referring to this datasheet



As you can see, there is extra circuity needed apart from the microphone. One thing i don't understand is the output impedance...the specs lists output impedance of the mic to be 2.2 Kohm and in the circuit a 2.2 Kohm resistor is connected to terminal 1. I know the definition of impedance but my question is how is output impedance useful?

Thanks for the help!


I don't see anything to indicate that RL is \$2.2k\Omega\$. That resistor is there as part of the power supply to the microphone. If the resistor weren't there, then the output of the microphone would be shorted to the power supply, and then you'd have no output signal; the output signal would just be \$V_s\$.

RL together with the capacitor form a capacitive coupling. The signal we care about from the microphone is always changing, and mostly we care about the frequencies from 20 Hz to 20 kHz, because that's the range of human hearing. Higher frequencies can pass through capacitors, while lower frequencies can not. DC, being the lowest frequency (0 Hz) can't pass through a capacitor at all.

So, the capacitor blocks \$V_S\$ from the output. The signal from the microphone, which is AC, is less blocked by the capacitor than it is by the resistor, so most of the current goes to the output.

  • \$\begingroup\$ Thank you Phil for adding that image. I was unable to add that image. So, if i understand it correctly the microphone needs a power supply to power the transistor inside the microphone package..is that correct? \$\endgroup\$ – user734861 Jun 20 '13 at 23:49
  • \$\begingroup\$ @user734861 that's exactly it. The transistor is necessary in fact, to convert the very high output impedance of the electret element, which is basically a capacitor, to a low output impedance, so this signal can be transferred over a long wire effectively. \$\endgroup\$ – Phil Frost Jun 21 '13 at 1:57
  • \$\begingroup\$ what happens if output impedance is high and we are transferring signal over a long wire? \$\endgroup\$ – user734861 Jun 21 '13 at 16:12
  • \$\begingroup\$ @user734861 The answer isn't short. It would make a good new question, though. \$\endgroup\$ – Phil Frost Jun 21 '13 at 18:43

how is output impedance useful

It's not always useful, in fact it's a hindrance sometimes but you have to live with what the circuit is and does.

The circuit inside the microphone is just a JFET amplifier and that converts the tiny microphone voltage to current and this current flows through RL and up to the supply voltage Vs. The article linked to for the microphone is for a 2 pin microphone that would require an external "2k2" going to a battery or supply voltage. Here is wiki showing a typical circuit. The "2k2" is in series with the drain of the JFET and this is what the output voltage develops across.

So the tiny input voltage is turned to current and the resistor RL turns it back to voltage BUT with a decent chunk of amplification. The simplicity of this circuit comes with the drawback that there is an output impedance that should not be shunted too much by the input impedance of the amplifier it connects to. If the input impedance of the amp were 2k2, the output voltage from the microphone would be halved compared to an amp with +100kohm input impedance.

There are three terminals on the microphone so why didn't they use something like an op-amp internally with the third terminal being the output. The op-amp would have very little output impedance and you could probably run it into an amp with input impedance as low as 100 ohms without too much trouble.

But, on the other hand who cares; nobody intentionally designs audio amps with low input impedances but if they need to be that way then that's just as easy as making them +100k. Using a JFET in the mic is dead easy and it's suprisingly good on hifi being class A with no chance of cross-over distortion or feedback (that supposedly causes intermodulation distortion we can apparently hear LOL).

  • \$\begingroup\$ Maybe they did use something "like" an op-amp, namely an open-collector comparator. :) :) \$\endgroup\$ – Kaz Jun 20 '13 at 23:43
  • \$\begingroup\$ Thank you Andy! I would like to ask the same question. So, if i understand it correctly the microphone needs a power supply to power the transistor inside the microphone package..is that correct? \$\endgroup\$ – user734861 Jun 20 '13 at 23:50
  • \$\begingroup\$ @user734861 That is quite correct. \$\endgroup\$ – Andy aka Jun 21 '13 at 10:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.