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I'm trying to drive a speaker from an ESP8266 for a project, and found an article suggesting (ab)using I2S as a 1-bit Pulse Density Modulation DAC. Apparently that's pretty noisy, so the article gives the following circuit diagram for the low pass filter:

How is this a low pass filter?

If I'm reading Wikipedia right, the 1KΩ resistor and 10nF capacitor to ground are a first order RC filter cutting off 16KHz. What does the 10µF polarised capacitor do?

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    \$\begingroup\$ removes any DC offset \$\endgroup\$
    – PlasmaHH
    Commented Jul 2, 2018 at 10:27

2 Answers 2

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The first RC does indeed act as a low pass filter.

The second capacitor (the one on the right), combined with the load impedance presented by whatever is connected to the audio out, acts as a high pass filter.

Together, these two filters act as a band pass filter.

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  • \$\begingroup\$ Your answer is not wrong, exactly, but the band is so broad that "band pass" is probably not the word one would use for it. At 10 uF, that series capacitor looks pretty big. I wonder what would happen if you just shorted it. \$\endgroup\$
    – thb
    Commented Jul 2, 2018 at 23:08
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    \$\begingroup\$ Well, if you're using headphones with a resistance of 25 ohms (which is pretty standard) then your LF cutoff is about 600HZ. Now to be fair, this clearly isn't designed to directly drive anything (due to the output impedance of the first RC) and its more likely that the load impedance will be on the order of 10k-100k, giving a LF cutoff of 1.5HZ-0.15HZ, but that's definitely still a band-pass filter, although it's also a DC blocking capacitor. \$\endgroup\$
    – BeB00
    Commented Jul 3, 2018 at 9:49
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It's simply an AF (audio frequency) coupling capacitor, removing any DC that is on the signal.

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    \$\begingroup\$ Cheers! Looking those up, there's a relationship between capacitance and the frequency you want, and 10µF suggests a ~100Hz frequency… does this end up as a crude band pass filter? \$\endgroup\$
    – Garth Kidd
    Commented Jul 2, 2018 at 11:35
  • \$\begingroup\$ @GarthKidd exactly. \$\endgroup\$
    – user20088
    Commented Jul 2, 2018 at 12:16
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    \$\begingroup\$ @GarthKidd The cut off frequency of the second C will depend on your load \$\endgroup\$
    – BeB00
    Commented Jul 2, 2018 at 14:04

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