Imagine you have an input signal s(t). And the signal of course has some noise n(t). So lets say the SNR value is x.

Now if we attenuate this signal by a voltage divider by 100 lets say, will the SNR of the attenuated signal remain same or?

Consider the following two scenarios:

1-) I log voltage data of a signal varying between -100mV to +100mV.

2-) And in the second case I log the voltage with same resolution "after amplifying this signal by factor 10" which means I log as -1V to +1V.

Neglecting the amplifier's noise contribution, which way is better in practice? Why?

  • \$\begingroup\$ or???????????????? what? \$\endgroup\$
    – Andy aka
    May 19, 2017 at 9:50
  • \$\begingroup\$ it would be "remain same or not?" \$\endgroup\$
    – user1245
    May 19, 2017 at 10:01
  • \$\begingroup\$ How do you 'log the signal' when it is contaminated with noise? \$\endgroup\$
    – Chu
    May 19, 2017 at 10:20
  • \$\begingroup\$ @Chu By a data logger device. My question is what is the benefit of amplification 10 times if the ADC resolutions are the same in both cases. Is that because SNR increases? \$\endgroup\$
    – user1245
    May 19, 2017 at 10:23

4 Answers 4


Here is the thing, noise is actually a misnomer, signal distortion is what it should really be called.

Every time you pass a signal through any component or even a trace of wire, the signal that comes out the other end WILL have some distortion compared to the original signal. Some of that is "filtering" and "reflections" some of it in cross-coupling from other signals outside of the intended signal path (noise).

Voltage Divider

If you can pass your signal with a signal to noise ratio of X through an ideal divide by 10 resistor divider, your signal would still have the same "noise ratio" of X.

Unfortunately, in reality there is no such thing as an ideal anything.

After division the signal will be distorted a little by inductances and capacitances and the physics of the resistor itself. Further, you just built a signal mixer to add in whatever noise is on the ground side. How much distortion is introduced depends on the quality of the circuit and the nature of the signal. The "noise" distortion could be lower but usually it will be higher.

Do It Late

Reducing a signal level is generally something to be avoided. As I already mentioned, you are actually mixing in the ground noise, but you are also producing a signal that is now more SUSSEPTIBLE to the ambient noise in the system. As such, if it is absolutely necessary to reduce a signal to feed into some device, like an ADC, it is prudent to do that reduction as late in the signal processing chain as possible, and physically as close as possible to the ADC.


The same goes for amplification. The entering signal along with the noise will again be distorted on the way through the amplifier. Different frequencies will again be distorted by different amounts. We actually design circuits to take advantage of that and call them filters.

As for feeding into an ADC.

ADCs compare an input signal with a reference signal. Obviously, if the reference is noisy you will get LSB comparison errors. If there is an ambient noise level, then that can couple into and produce a larger "noise component" on a low level reference compared to a larger reference. As such, ADCs work better in general at the larger end of their acceptable signal range.

HOWEVER: That does not necessarily mean amplifying the signal so you can maximize the reference of the ADC is the right thing to do. If that same noise is coupled into the signal you are trying to measure before you amplify it, you are back where you started, but now the signal carries the added distortion of the amplification.


There is a balance in there somewhere. Ultimately, the best method is to limit the number of times you have to mess with the signal on it's way into the ADC and keep the ADC's reference at a level that you can tolerate the reference noise. Less "messing" also limits the effects of component tolerances. And of course, keep the signal and reference as "quiet" as you can. Some circuit tuning is often required to optimize the ADC chain.


Cost can often be also be a limiting factor. More accuracy generally means more cost. Part of the design process also involves deciding how much error you can tolerate and how much extra cost you can afford to get there.

  • 1
    \$\begingroup\$ Interesting point. I think you mean if ADC range is so small and so reference signal is too low the SNR will have more affect on measurements. Makes really sense. \$\endgroup\$
    – user1245
    May 19, 2017 at 13:32
  • 1
    \$\begingroup\$ interesting but invalid. Noise is defined by current or some load impedance so V(SNR) is irrelevant until I or R is defined. (-1) \$\endgroup\$ May 19, 2017 at 13:40
  • \$\begingroup\$ @TonyStewart.EEsince'75 you are correct in terms of ultimate SNR measurement, but in terms of being able to identify the signal from the noisy distorted mess of a measurement at any point in the circuit it is entirely relevant. \$\endgroup\$
    – Trevor_G
    May 19, 2017 at 13:49
  • \$\begingroup\$ @TonyStewart.EEsince'75 however, I removed the specific SNR references to avoid the conflict in terms. \$\endgroup\$
    – Trevor_G
    May 19, 2017 at 13:58
  • \$\begingroup\$ SNR and resolution are both very relevant. If SNR is such that resolution of ADC contributes some quantization noise to SNR then amplifying to full scale is always better, otherwise no difference. capiche? my reference to current noise depends on R and Noise figure but neglecting this for high impedance inputs, it comes down to Noise amplitude and quantization level to determine optimum signal input level. sorry about my confusing you. \$\endgroup\$ May 19, 2017 at 14:44

With a signal that is ten times larger, the quantisation noise from the ADC will be about ten times lower. On the other hand, if the x10 amplifier has an error of 2% then you will have to factor this into the bigger picture because it may represent a more significant error than just ADC quantisation noise. Amplifiers may not be perfect spectrally either so again, it's another error to consider. Amplifiers can also produce offset voltages that add an error.

In short, the most appropriate measures are specific to the application and generalisations are just that.

  • \$\begingroup\$ By quantisation error do you mean matching? If ADC is set 12bit for both for -10mV 10mV and -1V +1V range scenarios, is there still quantisation error? \$\endgroup\$
    – user1245
    May 19, 2017 at 11:10
  • 1
    \$\begingroup\$ Quantisation error has nothing to do with matching. I think you need to google it. \$\endgroup\$
    – Andy aka
    May 19, 2017 at 11:14
  • \$\begingroup\$ Oh if so, thats a great reason to amplify. Now I have to figure out why amplification reduces this error. I hope can find an answer to this \$\endgroup\$
    – user1245
    May 19, 2017 at 11:17
  • \$\begingroup\$ If you look up what quantisation error is, I think you'll find it rather obvious why amplification reduces it. \$\endgroup\$
    – Hearth
    May 19, 2017 at 11:33
  • \$\begingroup\$ Imagine 1V signal is fed to a matched ADC range set to 0 ... 1V with 12bit resolution. And also second case, imagine 0.01V signal is fed to a matched ADC range set to 0 ... 0.01V with 12bit resolution. In the first case the step error is: 1V/(2^12); in the second case: 0.01V/(2^12). And the relative error remains the same. This is opposite to your arguements. Could you explain this please? I dont get it. \$\endgroup\$
    – user1245
    May 19, 2017 at 11:38

Generally if quantization error is much less than 10% of Vpp noise (best case) then gain does not help significantly as SNR only degrades a bit.

( a wee bit not a binary bit;)

But if Quantization error= Signal noise error (pp) equal then SNR degrades 6dB. Quantization errors include gain offset and monotonicity or missing codes

eg. if SNR is 10:1 or 20db and quantization error is 10% of noise peak then SNR data out becomes 10:1.1 or 9.1:1 or 19.2dB so not much different from 20dB SNR

e.g. if SNR is 100:1 or 40dB and Q error is 10% of noise peak then SNR input is 100:1 and output is 100:1.1 or 39.17db

So a 10% quantization error of noise peak contributes a loss of 0.8dB in SNR in both cases.

(which is my definition of a wee bit)

The more technical solution is defined as follows;

\$ ENOB = \frac{SINAD − 1.76}{ 6.02 }\$

where all values are given in dB and

  • SINAD is the ratio indicating the quality of the signal,

    • \$SINAD=\frac{P_{signal} ~~~ + ~ P_{noise} ~~~ + ~ P_{distortion}}{P_{noise} ~~~+~P_{distortion}}\$
  • the 6.02 term in the divisor converts decibels ( log10 ) to bits ( log2 )

  • the 1.76 term comes from quantization error in an ideal ADC

Effective number of bits (ENOB) is a measure of the dynamic range of an analog-to-digital converter (ADC).


Here is 200milliVolt peakpeak into 20bit 5vpp ADC. SNR is 78dB, even tho 0.2/5 is 1/25th of fullscale.

Do you need to amplify?

enter image description here

If we amplify by 10:1, SNR becomes 90dB.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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