# Cirrus Logic CS4334 DAC Low Pass Filter?

I have a Cirrus Logic CS4334 audio DAC chip working (datasheet here). One thing that confuses me is the "typical connection diagram" circuitry from the datasheet shown below.

In my experience with audio DACs, the R and C values on the output lines are for a typical resistor/capacitor (RC) low pass filter where part of the calculation is the cutoff frequency (F sub S). However, I'm used to the circuit and calculation used for the RC filter being different than what is used in the circuit diagram below.

I'm used to an RC low pass filter circuit being like this where the signal being filtered first goes through a resistor and then there is a capacitor to ground following the resistor. However, in the circuit below the signal first has the capacitor to ground then the resistor to ground (???). Maybe this is a different kind of low pass filter I'm not aware of?

Also, note the equation for the C value in the diagram below. Say I was a low pass filter with frequency cutoff of 20KHz and I'm going to use a 22 ohm resistor. I take it I should use a capacitor around 188 nanofarads as shown in my math below, agree?

C = (22 + 560) / (4π * 20000 * 22 * 560) = 1.879e-7 = 188nF


Any insights anyone might have to help me understand this better would be much appreciated. Thank you.

• That's an oversampling DAC so usually you would put the filter much higher than 20 KHz (but lower than half the actual sampling rate it oversamples to). Commented Apr 25, 2021 at 18:56
• @user1850479 Thanks for the info. I don't understand though, I thought human hearing is typically at a maximum of 20 KHz, what would be the point of allowing higher frequencies than that? Commented Apr 25, 2021 at 20:12
• @TerenceD That DAC oversamples the audio and already has built-in lowpass filters, both digital and analog lowpass filters. Therefore the external analog filter does not need to be set to cutoff rate of 20 kHz, because if you do, you will already be cutting away 3dB at 20 kHz, and you can get better bandwidth by setting the cutoff frequency higher. It basically is not used any more for limiting the audio bandwidth, but filtering away the residuals from the internal filters, oversampling blocks and sigma-delta modulators. Commented Apr 25, 2021 at 20:34
• @Justme - Awesome!!! I appreciate the info, and it's more clear to me now that I have a bit more to learn about DAC output circuitry :) Much appreciated! Commented Apr 25, 2021 at 20:38
• @TerenceD I simply read the datasheet. Please note that using 22 ohms and 188 nF as RC filter may not fall into the region where the output is stable, and the selection of 20 kHz cutoff frequency may not be optimal. Simply use the datasheet values or evaluation kit examples, and if you need something else, then buffer the output with an op-amp. Commented Apr 25, 2021 at 20:48

Since this is a single supply DAC, the output has a DC bias.

First there is 267k of DC load resistance.

Then there is a 3.3uF DC bias blocking capacitor to remove DC bias.

Then there is a 10k resistor to ground to keep DC bias at 0V, regardless of if there is a load connected or not.

Then you have the traditional RC lowpass filter.

Finally the load at the input of next device is modeled as resistor to ground.

• Thanks @Justme for the explanation. So, if I understand correctly, the 560 ohm resistor is the resistor of the traditional RC lowpass filter, correct? Commented Apr 25, 2021 at 20:10
• Yes, correct, 560 ohm resistor is the R of the traditional RC filter. Commented Apr 25, 2021 at 20:25

The output is likely CMOS OpAmp negative feedback and 267k provides some minimal DC load current

The 1st CR filter is High Pass DC blocking at 5Hz $$\\omega =33=1/(3.3uF*10k) ~~~ \$$ assuming no load, but will shift upwards with the same load towards 10 Hz.

The 560 Ohm is to prevent oscillations with a large capacitive load reducing the Op Amp phase margin.

The RC filter may be selected for your -3B BW and is dominated by 560 in parallel with load * C out. That's all the formula does is assume 0 Ohm source impedance with RC LPF shunted by load. I would choose a load >=10k only

## Suggestion

C= 10 nF , Rload=10k with passband attenuation 0.5 dB from 560 ohms and -3.5dB cuttoff = ~ 25 kHz You can't hear that but it causes a 45 deg phase shift at break point and about a 30 deg phase shift at 15kHz which you can hear. Old TV's had a noisy flyback sweep oscillator for NTSC that you could hear on some sets in the 70's and early 80's just above this freq. Unless you care about linear phase of stereo imaging , it won't matter that much.

• Awesome info!! Thank you!! Commented Apr 26, 2021 at 1:54