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I'm self-taught in electronics and while I can do the digital stuff fairly readily, analog throws me for a loop. I'm looking at an audio CODEC and the reference schematic has this on the input to the ADC:

enter image description here

I understand that C11 is there to capacitively couple the input to the ADC and to prevent any DC offset from the input to feed into the op-amp front end of the ADC.

What purpose(s) do C12, R1, and R2 serve?

(In case anyone is wondering, this is the reference schematic from the Wolfson WM8731 CODEC. http://www.wolfsonmicro.com/products/audio_hubs/WM8731/)

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1 Answer 1

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The resistors form an attenuator (0.5 or - 6dB) and the 220pF capacitor forms a low pass filter with the resistors, to roll off the unnecessary high frequencies.

The low pass filter cutoff frequency (the -3dB point, so effectively -9dB from input) can be determined by the thevenin resistance of the resistors and the capacitor value:

(1 / (1/5.6k + 1/5.6k)) = 2.8k

Roll off:

1 / (2 * pi * 2.8k * 220pF) = 258.3kHz

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  • \$\begingroup\$ Why did you use a parallel of the two resistors to calculate the resistance? They look series to me, and only R2 should matter for the C12 charge? Thus, R should be 5k6? \$\endgroup\$
    – Jon Watte
    Aug 28, 2013 at 4:30
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    \$\begingroup\$ It's hard to explain well in a comment - have a read of Thevenin's Theorem, link 2, link 3. SPICE the circuit to see for yourself it's correct. \$\endgroup\$
    – Oli Glaser
    Aug 28, 2013 at 4:44

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