The circuit has been simplified as shown in the picture. Signal source is connected with an analog amplifier, which amplifies the ac signal and also raise it by 1.5V to match with the range of the 0-3V ADC. Some characteristics of the circuit:

  • Signal of interest: 10kHz sine wave with maximum amplitude 8mV
  • Signal noise density: \$5 \times 10^{-4} V/ \sqrt{Hz}\$
  • Signal bandwidth: center at 10kHz, width of 200Hz
  • ADC effective resolution: 9 bit
  • ADC sampling rate: 96kHz
  • Calculated ADC quantization noise: \$ 6.8 \times 10^{-6} V/\sqrt{Hz}\$ Circuit structure

I want to measure the amplitude of the signal. The ADC cannot be changed. My question is, how to determine the gain of the amplifier? Under above conditions, the low-resolution of the ADC does not seem to be a problem any more, since it operates at a high frequency. And the analog noises dominate the ADC quantization noise.

The ADC quantization noises will not worsen the SNR, since the analog noises dominate. And the amplifier will not improve SNR either. Could I make the amplifier gain =1, assuming the noises added by the amplifier is negligible?

However, since the signals are very small, I am not sure if the voltages can be detected by the ADC actually. Under this case, how should the gain of the amplifier be designed? Are there any theories behind this? I dont want to amplify the voltage to the full range, since that will lead to addition of several OpAmps and it does not improve SNR at all.

Thank you very much!

  • \$\begingroup\$ Some ADCs have built-in amps that might help you. Also, don't forget to put a well-designed RC filter between the amp and ADC to prevent kick-back noise and limit the noise generated by the amplifier. \$\endgroup\$
    – Damien
    Commented Mar 22, 2015 at 0:27

4 Answers 4


Your signal is buried in the noise. Assume you band limit your input in a 200Hz bandwidth.

$$ V_{n(rms)}=5 \times 10^{-4}V/\sqrt{Hz} \times \sqrt{200Hz} \approx 7mV $$

Your input SNR

$$ V_{in} = 8mV\\ SNR = 20log(\frac{V_{in}/\sqrt{2}}{V_{n}}) = -1.9 $$

When you amplify your signal, the noise will be amplified too. You may need a lock-in-amplifier.


Thanks @Brian Drummond, i think i should complete the math for you :).

Assume the noise are white noise, the amplitude should be Gaussian distribution. It's common practice to take the peak-to-peak value of Gaussian noise to be 6.6 times the rms value, since the instantaneous value is within this range 99.9% of the time.

$$ V_{n(p)} = 3.3 \times V_{n(rms)} \approx 23mV $$

The gain allowed without make the ADC input saturated:

$$ G_{max} = 1.5V / ( V_{n(p)} + V_{in} ) \approx 48 $$

Because the 3.3 is statistical value, you may choose a gain less than this.

  • \$\begingroup\$ +1. I confess I'm not inclined to do all the work for a freebie, but at least pointed in the right direction! \$\endgroup\$
    – user16324
    Commented Nov 29, 2014 at 11:01
  • \$\begingroup\$ Thanks for your reply! You pointed out that another limitation is to avoid clipping of the noises. And we are using some post processing techniques to deal with the noises. My main concern is, what is the minimum amplification ratio before weakening the performance of the system? Because I want to use less OpAmps.. \$\endgroup\$ Commented Nov 29, 2014 at 11:03
  • \$\begingroup\$ @Brian Drummond: :), agree with you, sometimes it's just enough to point out the right direction and leave some space for the OP. When i wrote half of my equations, i find i've to leave for other things. So, there is a not "completed" answer. I found it's not good to leave such an answer, so i deleted it, but i think i should point it out, barely amplifier it may not enough. So, your answer actually completed it. \$\endgroup\$
    – diverger
    Commented Nov 29, 2014 at 11:15
  • \$\begingroup\$ @richieqianle: Um, you mean the 'minimum' ratio or 'maximum' ratio before weakening the performance of the system? I think you mean 'maximum', right? \$\endgroup\$
    – diverger
    Commented Nov 29, 2014 at 11:20
  • \$\begingroup\$ If you really need less amplification, that means your amplified signal may much less than the input range of the ADC, then you waste your ADC's SNR. Your ENOB will be reduced. \$\endgroup\$
    – diverger
    Commented Nov 29, 2014 at 11:34

We really need to know the noise bandwidth to answer. If this is the same as the signal BW then you simply need gain, but as "diverger" pointed out in his (now deleted) answer you have 7mv (rms) noise or S/N about -2dB.

Your signal is buried in the noise.

$$ V_{n}=5 \times 10^{-4}V/\sqrt{Hz} \times \sqrt{200Hz} \approx 7mV $$

Your input SNR

$$ V_{in} = 8mV\\ SNR = 20log(\frac{V_{in}/\sqrt{2}}{V_{n}}) = -1.9 $$

above formulae quoted from "diverger"'s answer : quote and Mathjax aren't playing nice together for me!

If the noise BW is a full 20kHz you need a tight bandpass filter to reduce the noise BW to 200Hz as Andy alludes to, in addition to the gain, thus limiting the noise voltage to 7mVrms instead of Andy's 60+mv.

Then, you want gain (to make best use of the limited resolution of the ADC) but not too much (to avoid clipping the noise.) With 7mv rms noise, assumed white, the peak voltage will be above n times the mean for m% of the time. Either (the right way) search for peak-mean ratio statistics of white noise or (hand-waving) adopt a fairly generous peak-mean ratio of 5:1 so you want to allow 35mv peak or 70mv pk-pk, giving you a gain of 3000/70 or about ... 42. (In practice, 40 or 50 to make the post-process scaling easier).

And remember that poor S/N ratio ... you will need some post-processing - filtering or a digital implementation of a lock-in amplifier ... to recover the wanted signal from all that noise.

EDIT : Update in response to a comment to Diverger's post...

Also for the narrowband bpf, we found that it is not easy to tune the center frequency. Therefore we put the filter inside the processor.

NO! If you mean the 200Hz BW filter, that won't work. Specifically, it can put you back in the regime described in Andy's answer, with a signal/noise ratio of about -20dB, and a maximum permissible gain of about 6.

Reduce the noise bandwidth as far as you can in the linear domain before the non-linear process of sampling.

  • \$\begingroup\$ Thanks for your reply! This is actually the kind of answer I am looking for. Could I ask what the minimum amplification ratio is without hurting the performance? I do not want to amplify to much to save the OpAmps.. \$\endgroup\$ Commented Nov 29, 2014 at 11:04
  • \$\begingroup\$ I see no reason to go below the sort of gains calculated by Diverger and I - a gain of 40 at 10kHz is a GBW product of 400kHz; a single opamp with a GBW of 5x this value or more should be adequate, and I don't think that will be hard to find. \$\endgroup\$
    – user16324
    Commented Nov 29, 2014 at 11:11
  • \$\begingroup\$ The question was a simplification. I understand that such an Opamp can be found. And I would like to understand the minimum gain possible. Since the analog noises dominate ADC noise, it seems to me that we do not even need to amplify the signal... \$\endgroup\$ Commented Nov 30, 2014 at 10:02
  • \$\begingroup\$ Could I have your opinion on it? \$\endgroup\$ Commented Nov 30, 2014 at 10:28

If you have a signal noise density of \$5×10^{−4}V/\sqrt{Hz}\$, over a (say) 20kHz bandwidth, this is an RMS noise of 71mV and much bigger than your 8mV signal so, I'd recommend filtering the signal first in order to remove as much noise outside the 200Hz bandwidth that the signal occupies.

If you didn't filter the signal you have a noise that has a 99.9% probability of it having a p-p amplitude within 6.6\$\sigma\$ of the RMS value i.e. it has a typical p-p amplitude of 6.6 x 71mV = 469mVp-p.

Compare this with the p-p value of your wanted signal (22.6mVp-p)

This limits the amount of gain you can apply to 3V/(0.469 + 0.023) = 6.1.

This calculation assumes that you can "live" with clipping 0.1% of the time on the basis that you probably won't be seriously "damaging" the measurement of the signal you want.

Do yourself a favour and pre-filter the signal with a couple of op-amp stages OR live with a gain that gives you grainy resolution in your measurements. On the plus side (of using with a gain of 6.1), because your sampling rate is 96kHz you get some process gain in that you can average samples thus reducing noise above 10kHz.

  • \$\begingroup\$ Thanks Andy! For you help as always. The noise bandwidth is 200Hz actually. What I want to ask is what is the minimum ratio of amplification before hurting the system performance, since I want to use less OpAmps.. \$\endgroup\$ Commented Nov 29, 2014 at 11:05
  • \$\begingroup\$ If the signal noise bandwidth (3dB points) is 200 Hz, the actual noise is 7.1mV. This equates to a p-p value of 46.7mVp-p (6 sigma or 99.9% confidence). Your signal is 22.6mVp-p so, playing a little conservatively I'd add this to the noise to get ~70mVp-p. Your ADC noise over a 50k bandwidth is 1.52mV so this is trivial and forgettable. MaxVp-p is 3 so 3/0.07 implies a gain of 43 max before your signal and noise starts over-ranging the ADC. That's where I'd pitch the gain stage and I might go for a bit less say 30. \$\endgroup\$
    – Andy aka
    Commented Nov 29, 2014 at 14:42

Amplification may not help SNR (in any real-world system it will usually make it worse) but it will allow you to sample the signal at a useful resolution.

Given that 8mV is barely 1 count of a 9-bit ADC at 3.3V (and less than 1 count at 5V), at the moment your sampling resolution isn't very useful.

I'd say you have very little choice but to make the trade - drop your SNR a little in exchange for actually being able to capture the signal. A voltage gain of between 100 and 200 should give a reasonable balance.

  • \$\begingroup\$ Could I ask the reason for 100 - 200? Can the ability of capturing signal be modeled by quantization noises? And by the way, voltage gain will not decrease SNR in my case, since the gain stage contributes little noises. What I want to do is to reduce the OpAmps actually. :P Thanks! \$\endgroup\$ Commented Nov 28, 2014 at 7:34
  • \$\begingroup\$ @richieqianle A gain of 100-200 would give you 800-1600mV to work with, which would give you enough sampling granularity to be useful. That said, my answer doesn't take any notice of your noise problem. As the others have said, you must filter the signal, and you must amplify it, or it's useless to you. \$\endgroup\$
    – markt
    Commented Nov 28, 2014 at 21:32
  • \$\begingroup\$ @richieqianle Forget quantization noise, SNR, etc. The main problem is that the resolution of the ADC is 5.8mV. This means with an 8mV max signal, if you didn't have any gain the only thing the adc would ever read is 000000000 or 000000001 assuming no noise. You cant do anything with that. \$\endgroup\$ Commented Nov 24, 2015 at 2:20
  • \$\begingroup\$ @THEMuffinMan7 if im correct, oversampling can increase resolution. \$\endgroup\$ Commented Nov 24, 2015 at 2:27

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