I'm new to audio and I want to ensure if it's proper circuit for integrate MAX9860 codec and EMG-610P microphone?

Microphone schematic EMG-610P details: https://www.monacor.com/media/FLE/EMG610P.pdf

MAX9860: https://datasheets.maximintegrated.com/en/ds/MAX9860.pdf

Thanks for any answer.

  • 2
    \$\begingroup\$ Did you notice that your microphone has three connections, but your circuit only has two? \$\endgroup\$ – JRE Jun 5 '19 at 11:16
  • 1
    \$\begingroup\$ Why aren't you using the MICBIAS output of the MAX9860? \$\endgroup\$ – JRE Jun 5 '19 at 11:18
  • 1
    \$\begingroup\$ 10k. Why that value? \$\endgroup\$ – JRE Jun 5 '19 at 11:18
  • 1
    \$\begingroup\$ Did you look at the microphone datasheet how to connect it? Also the codec datasheet gives information how to connect single-ended microphones. \$\endgroup\$ – Justme Jun 5 '19 at 11:19
  • \$\begingroup\$ @JRE Yes I have noticed that, but as I read other notes,I suppose schematic from EMG610P.pdf is for preamplifier circuit not for codec. According to other MAX design guide I should use 2k2 resistor instead 10k. \$\endgroup\$ – Fasset Jun 5 '19 at 11:28

Data sheet circuit for microphone: -

enter image description here

From this you can somewhat conclude that the microphone might need a minimum DC current of about 3 volts / 680 ohms = 4.4 mA. This means that using 2 x 10 kohm to feed DC to the microphone is likely to be insufficient.

I would configure the MAX9860 for single ended inputs to match the circuit shown in the microphone data sheet.

  • \$\begingroup\$ Thanks for answer. If MAX will be in single ended configuration (MICLN input in series with 1uF to ground according to datasheet), then MICLP should be connected signal from microphone by 1uF and 10uF in series or only 1uF? \$\endgroup\$ – Fasset Jun 5 '19 at 12:05
  • \$\begingroup\$ 1 uF in series with 10 uF is about 0.9 uF so I would not bother with the 10 uF and just use the 1 uF. 1 uF will give a cut-off frequency that is substantially sub-audio. \$\endgroup\$ – Andy aka Jun 5 '19 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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