How to calculate expected RMS noise voltage out of digital audio system?

Question

I have a PCB which includes an SOC with a built-in DAC that connects to an external active high-pass filter which then connects to a Class AB amp IC to drive an audio load.

Using a spectrum analyzer, I measured the idle noise (20 Hz - 20 kHz bandwidth) out of my Class AB amp IC of my total system at ~400 μV_RMS. I am trying to calculate this on paper to better understand it.

Referencing equation 8.21 from here, I believe the equation for calculating the expected rms output noise would be:

$$V_{noise_{RMS}} = \sqrt{BW} \cdot \sqrt{ e_{n_{DAC}}^2 \cdot A_{HPF}^2 \cdot A_{AMP}^2 + e_{n_{HPF}}^2 \cdot A_{AMP}^2 + e_{n_{AMP}}^2}$$

Where BW is the system bandwidth, e_n is the voltage noise density of the component, and A is the gain of the component.

BW has an upper limit of 20 kHz since this is an audio application and a lower limit of 2 kHz (the cutoff of the highpass filter). So,

$$ BW = \sqrt{20\ \mathrm{kHz} - 2\ \mathrm{kHz}} = \sqrt{18\ \mathrm{kHz}} $$

The HPF's noise voltage density is supplied in the datasheet. To find values for the noise voltage density of the DAC and the AMP IC, would it be sufficient to measure their idle RMS output voltages and divide by the square root of the spectrum analyzer measurement bandwidth? For example, I measured 10 μV_RMS out of the DAC across 20 kHz bandwidth. Would the DAC noise voltage density then be:

$$ e_{n_{DAC}} = 10\ \mathrm{\mu V_{RMS}} / \sqrt{ 20\ \mathrm{kHz} } = 71\ \mathrm{nV_{RMS}}/\sqrt{\mathrm{Hz}}?$$

My current numbers are off by orders of magnitude so I'd appreciate any help!

Edit: mistakenly forgot to put the 18kHz bandwidth number under a square root

Answer 1

First off, draw a block diagram of your system.

The total noise output is as you wrote, but there are clarifications to the equation to make is easier to read and show that the noise for the individual components is the output noise spectral density.

$$ e_{no_{SYSTEM}} = \sqrt{BW} \sqrt{(e_{no_{DAC}} \cdot A_{HPF} \cdot A_{AMP})^2 + (e_{no_{HPF}} \cdot A_{AMP})^2 + (e_{no_{AMP}})^2 } $$

This is valid if the noise density is constant over the bandwidth.

\$ \sqrt{BW} = \sqrt{20kHz - 2 kHz} = 134.2 \$ if the HPF is a brick-wall filter. Otherwise, you need to know the noise bandwidth of the HPF.
Perhaps a source of your error is in the bandwidth calculation which is messed up.

Question: How did you measure the output noise and get 400uV RMS with a spectrum analyzer? When measuring total noise you use a voltmeter that has true RMS functionality, then divide the total noise by the square root of the noise bandwidth to get spectral noise density. Perhaps you have a fancy analyzer that can read out total noise over a selected bandwidth, or, you used an oscilloscope's Vrms math function.

The reliable method of quantifying noise with a spectrum analyzer is to set the analyzer to read out in noise density (i.e., \$ V/\sqrt{Hz} \$).

Regarding your question of calculating the noise density of the DAC, your calculation is valid if the noise density is constant over frequency. If the noise and/or gain changes over frequency and/or the noise floor changes when a signal is present, the job gets harder to come up with total noise over the bandwidth number.