# Complete ADC SNR formula when signal is not full scale

I am very in pain to find a (any) paper giving more than the traditional SNR for full-scale wave sampled by ADC

$$SNR(Full Scale Waveform) = 6.02N + 1.76 (dB)$$

As far as I can remember if the signal if not full-scale there is an additional negative term in the form of 20log(V/Vfs)

$$SNR(V) =? 20log(V/Vfs) + 6.02N +1.76 (dB)$$

• Is this last equation true ?

• Why is it not often shown/written (circuits without Automatic Gain Control are not so rare, leading to small waveform sampling)

• Is there an open source document exposing this issue ?

• For a smaller waveform (let's say 50%) it's equivalent to using an N-1 bit ADC on full scale. Dunno what this has to do with AGC but hey ho. Commented Oct 7, 2020 at 20:50
• Wiki answers this in SQNR and SNR. AGC improves SNR and the designer must be aware of worst case SNR Commented Oct 7, 2020 at 20:57
• @Andy aka: that's sound logic then, 20log(1/2) = -6.02 and it is like 6.02(N-1) in the simple formula. AGC to boost the signal closer to full-scale (if not adding more noise...)
– NGI
Commented Oct 7, 2020 at 21:09
• @Tony Stewart Sunnyskyguy EE75 . This is not so obvious for me but reviewing the external links of the wiki articles you pointed I found the Analog Devices MT-003. Eq2 which express the same but with SINAD and ENOB ( which for a first rough approach are for me equal to SNR and N). Will read the complete paper
– NGI
Commented Oct 7, 2020 at 21:18
• a similar question was asked today in SQNR the voltage range is $$\delta$$ Commented Oct 7, 2020 at 21:19

You're overthinking this making things (in your mind) more complex than they are.

Suppose we have a 10-bit ADC, then the will be roughly 60 dB (yes, theoretically 61.96 dB but I like round numbers, also: feel free to show me an ADC that works exactly according to theory, but I digress). So SNR = 60 dB

Let's say that this ADC has a full scale of 0 dBV, that's 0 dB Volts and in my world that's an amplitude of 1 Vpeak.

So then the noise level will be 60 dB below that 0 dBV so noise = -60 dBV (that's 1 mVp).

Now if we feed the ADC a smaller signal of for example 100 mVp = -20 dBV, the noise will still be at -60 dBV. So that means the SNR (for that smaller signal) will be -20 dBV - -60 dBV = 40 dB. The SNR became 20 dB smaller, which makes sense as the signal was made 20 dB smaller.

Does that match with your second formula?

SNR(V) =? 20log(V/Vfs) + 6.02N +1.76 (dB)

Yep, I think it does.

This isn't written in articles because it is (in my opinion) quite obvious.

An AGC would complicate things but not much, the AGC would simply add gain and increase the amplitude of the signal (not the noise). But an AGC will affect the dynamic range of your signal (it will be lowered), often that can harm your signal.