Can someone please explain exactly what happens to the input signal in the QA402 when different input ranges, 0-18 are set as opposed to the attenuation in the 2nd line of that dialogue? Thx for any help.
Hi @Moto, as you change the input levels, different gains and/or attenuations are being applied to the incoming signal. There are two blocks to think about: The first block is the “big attenuator” and it happens before any active circuitry is encountered. Given the QA402 supply rails are ~+/- 14V, a signal can never be larger than +/- 11Vp (8Vrms) before it enters the active circuitry. So, the “big attenuator” (aka front end attenuator) handles that.
The ADC wants to see an input voltage of about 1Vrms. So, we need another round of adjustments to get from 8Vrms (max) to the 1Vrms that is suitable for the ADC.
So, if you input a 0 dBV signal, it can go straight into the ADC. If you input a 6 dBV signal, it needs the gain cut by 6 dB. If you input a 12 dBV signal, it needs the gain cut by 12 dB.
If you enter a 24 dBV signal, it needs the gain cut by 24 dBV. The front-end attenuator is 24 dB.
Thx Matt. I did see elsewhere in one of your docs that input range should be 20dbv above the signal level. is that correct?
Yes, @Moto, that is correct–generally for best THD on the QA402 you’ll want the input signal to be about 15-20 dB below the max input range. That is a function of the converter and was present in the AKM converters too on the QA401. The first converter I’ve seen that doesn’t have this limitation is the ES9822. It’s really pretty happy right up to the -1 dBFS level.
Ok. Thx. On a related,maybe, question, I’m testing a dac at 1khz from REW ->usb->dac->QA402 with the QA402 reading 1.02vrms.
At an input range of 12db the 2nd harmonic is -110 and the 3rd is -104.
At an input range of 18db the 2nd is -118 and the 3rd is -133.
Why just changing the input range on the QA402 6 db do the harmonics completely flip order of size?
Hi @Moto, the answer is best conveyed looking at Fig 6 from the PCM4220 ADC data sheet. This is for THD+N, but it’s a good proxy for THD too since THD+N is limited by harmonics near 0 dBFS.
This graph is a common representation for converters, and if you aren’t familiar with it, it’s usually referred to as a THD+N Level. What you normally see is a THD+N ratio (which looks more like a bathtub)
This plot will tell you the level of the THD+N for a given input. With -80 dBFS input, the THD+N level will be about -120. So the THD+N ratio will be -40 dB. With -40 dBFS input, the THD+N will still be about -120. So the THD+N ratio will be -80 dBFS input. At -2 dBFS input, the THD+N will be -110, and that’s how you get the best-case -108 dB THD+N the PCM4220 claims.
But the point here is that converter harmonics are very low up until -18 dBFS or so, and then they start rising very, very quickly.
Now, back to your question: When you input 1Vrms = 0 dBV and the input range is 12 dBV, you are inputting a signal that is effectively -12 dBFS, and you can see the bump there has a peak. With -12 dBFS in, you would expect -108 for harmonics, or roughly -96 dB relative to the peak.
When you change the input range to 18, suddenly your 0 dBV input signal is hitting the ADC at -18 dBFS, and you are back to the region where the harmonics aren’t going crazy due to overload.
This is how the guidance to target an input level that is 15-20 dB above the actual input signal was determined. it should also explain your observation. Please let me know if not clear. Thanks!
Matt thx for your very thorough and patient response!