How to select gain for an amplifier before ADC?

Question

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}\$

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!

Answer 1

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.

---

Update:

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.