0
\$\begingroup\$

I keep looking for Audio specific ADCs and the majority of these devices work at low voltages on the analog circuitry (5V for example).

Lets say we want to process an higher voltage audio input signal (let´s say 10Vrms), then a signal conditioning circuit must be placed between the input and the ADC to reduce the input voltage.

Then, after DAC, the signal must be restored to nominal level (in our example 10Vrms).

  • Won´t this gain reduction at the input and signal restoration at the output increase the overall SNR of the circuit?

  • Why do manufacturers use this low voltage at the analog input of the ADC?

\$\endgroup\$
  • \$\begingroup\$ I think you mean "decrease the overall signal to noise ratio." If doing you describe could increase the SNR, then you could clean the noise from your signal just by running it through the ADC/DAC conversion a few times. \$\endgroup\$ – JRE Nov 25 '17 at 9:15
  • \$\begingroup\$ In terms of audio signal levels 5V is already high. It looks like you are trying to take the output of a amplifier and convert that. Can you perhaps describe or put up a simple diagram of what you are trying to achieve with your system. Generally less circuitry is better in terms of noise. \$\endgroup\$ – RoyC Nov 28 '17 at 9:49
  • \$\begingroup\$ Many pro audio equipment is capable of giving up to 26dBu (check for example the classic ams neve 1073). It is a microphone preamp and suppose you want to convert it to digital. Imagine you have 20Vpp input signal, you should first add a signal conditioning circuit before de ADC reducing SNR heavily and degrading the overall system performance. \$\endgroup\$ – user6127833 Nov 28 '17 at 21:18
0
\$\begingroup\$

Why use a 5 volt ADC range? 1) the fewer circuits, the cleaner the digitization results

2) the fundamental noise floor is the sqrt(KT/C); a large large sample-hold capacitor thus sets the floor; yes, the comparator (in the ADC) also contributes; any opamps used in a oversampling ADC also contribute.

Often any precision amplification is achieved using Cin/Cfb ratios in switched-cap opamp circuits. Given the large areas needed for low-noise amplification, the ADC designers (and marketing and Apps engineers) may decide to use ALL that capacitance in the ADC sample-hold and in any binary-weighted feedback DACs, etc.

\$\endgroup\$
0
\$\begingroup\$

The ADCs work with typical voltage levels for the circuitry they must interface with.

You often have 5V or 3.3V as a power rail in digital circuitry. The analog sections of the ADC can also only work within the available power rails. If you've only got +5VDC as a rail, then the opamps and other analog parts of the ADC cannot work with anything above 5V or below ground.

To work with a 10V signal, you need a power rail of 10V for your analog circuitry. Generating that is a nuisance, and can introduce more noise into other parts of the system - getting 10V out of 5V involves a switching power supply, and they are notoriously noisy. Yes, you can clean up the noise. Better if you don't have it there to begin with.

As for "increasing the signal to noise ratio" - that doesn't happen. I think you meant to say "decrease the signal to noise ratio." There is a bit of confusion in the wording. If it were expressed as a fraction, then I think you'd understand better.

A signal to noise ratio of -20dB means that the noise is 0.1 times the signal level. So, for a signal level of 1Vpp, you would have a noise level of 0.1Vpp.

If you want to increase the signal to noise ratio, you must either lower the noise or increase the signal. Lets take the example above, and improve the signal to noise ratio to -40dB. That means that the noise is 0.01 times the signal level. So, for a signal of 1Vpp, the noise would be 0.01Vpp.


Now that you understand what the signal to noise ratio is, lets go back to your question.

Every time you amplify a signal, you add noise to it. Every amplifier makes a little noise, so when you pass a signal through it the signal becomes noisier. The signal to noise ratio gets worse because the fraction that is noise increases.

Take our -20dB SNR example. Say our signal has a SNR of -20dB and we pass it through a crappy amplifier with a gain of 1 that adds 0.1 Vpp of noise to the signal. We already have 0.1Vpp of noise because the signal to noise ratio is -20dB. We add another 0.1Vpp of noise from our crappy amplifier, and the signal to noise ratio changes to -13dB. We have a higher number, but it means that the ratio is worse.

So, every time you pass your signal from 10Vpp down to 5Vpp, then amplify it back to 10Vpp, you will add noise to it and make the signal to noise ratio worse.

This isn't usually a problem. Attenuators (to go from high level to low level) and amplifiers (to go from low level to high level) can be designed so as to not add a noticeable amount of noise to the signal.

Of course, the more times you attenuate or amplify, the worse the signal to noise ratio gets. You avoid that by only doing those things when you must, and by using good (low noise) amplifiers in those cases.

\$\endgroup\$
  • \$\begingroup\$ Thanks for that correction. As for the answer you have only written what I have already wrote in my question. Adding attenuation/amplification stages to the signal chain will only deteriorate the SNR. So, why aren't ADCs designed capable of processing higher level voltage input signals? \$\endgroup\$ – user6127833 Nov 26 '17 at 15:36
  • \$\begingroup\$ Read the first section again. It tells you you would need a power source with a voltage range as large as the signal range. Digital circuits don't generally have access to voltages that large. \$\endgroup\$ – JRE Nov 26 '17 at 15:59
  • \$\begingroup\$ Of course digital circuits work with lower voltages. But the question is why aren't most ADCs designed to be power with higher voltages in their Analogue part... \$\endgroup\$ – user6127833 Nov 28 '17 at 20:57
  • \$\begingroup\$ Using a digital power source to obtain a higher voltage for an analogue input is not good...in fact, independent power supplies should be used to power the digital and analogue parts of the ADC if I'm not wrong. \$\endgroup\$ – user6127833 Nov 28 '17 at 21:10
  • \$\begingroup\$ I think you got signal and noise the wrong way around. -20dB SNR means 1Vpp noise for 0.1Vpp signal. \$\endgroup\$ – immibis Feb 19 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.