I am coupling an electret microphone to the input of an audio codec which does all the amplification, so all I need is biasing and filtering. I picked up the circuit from a colleague and I'm having some trouble understanding an aspect of the circuit (attached).

The issue I have is the resistors on the right hand side of the drawing - why are these needed. This circuit was adapted from an analogue solution where the output was to the inverting input of an opamp, where clearly these 22k resistors would be required, but is that normal in a passive filter?

Also, the 1k resistor in parallel is confusing, I'm guessing it may be to set input impedance, but this is defined in the datasheet of the codec, so could this be fine to remove?


1 Answer 1


Usually, the output level from an electret microphone is quite high compared to regular moving coil microphones so some form of attenuation makes a lot of sense. If you find it's too much then lower or short the resistors. It's better have provision in the circuit if a PCB is being made rather than get the scalpel and glue out.

  • \$\begingroup\$ The attenuation is provided by the 22k series resistors? The codec is fitted with a digital attenuator, but I can see that this would make sense too. What might the 1k in parallel be for? \$\endgroup\$
    – droseman
    Jul 24, 2014 at 12:43
  • \$\begingroup\$ The 1k is needed for the attenuator to work - if the amplifiers input impedance were a reliable 1k then I'd say use that but, also consider that you may want to increase attenuation by a few dB and changing the 1k to something lower is twice as easy as changing 2 x 22k \$\endgroup\$
    – Andy aka
    Jul 24, 2014 at 16:57
  • \$\begingroup\$ Reducing impedance increases attenuation? I thought it would be the other way round. Whats the theory behind that ? \$\endgroup\$
    – droseman
    Jul 24, 2014 at 17:24
  • \$\begingroup\$ It divides the incoming voltage to the 22k resistors - it's called a potential divider. \$\endgroup\$
    – Andy aka
    Jul 24, 2014 at 17:34
  • \$\begingroup\$ Of course... I should have seen that. Thanks for your help Andy \$\endgroup\$
    – droseman
    Jul 25, 2014 at 6:04

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