0
\$\begingroup\$

Normally rms value and the mean value of a perfect DC voltage is same.

So lets say I want to measure the gain of an amplifier.

I first apply an offset Vin_off to the inputs, then I measure Vout_off.

Then I apply Vin to the inputs, then I measure Vout.

So I calculate the gain as:

\$G_{dB} = 20 \ \log_{10}([Vo_{mean} - Vo_{off-mean}]\ / [Vi_{mean}-Vi_{off-mean}])\$

Basically I measure the gain by calculating the ratio of the change in output voltage to the change on input voltage.

But as you see I use mean values..

What if the input and output signals also have some noise on them?

Should the mean values or instead rms values be used as:?

\$G_{dB} = 20 \ \log_{10}([Vo_{rms}-Vo_{off-rms}]/[Vi_{rms}-Vi_{off-rms}])\$

\$\endgroup\$
5
  • \$\begingroup\$ This is a very difficult and non-standard approach in my opinion. If the amplifier can work with DC then DC gain and AC gain will be the same. I do not see the point of subtracting DC (Vout_off) from the AC (Vout), also not seeing why you would use the mean. Go read a book explaining how this is done: apply AC at input so that output does not clip/distort. Then ratio gain is Vpp_out/Vpp_in Since the signal is then large noise can be ignored. Measuring noise is a different game, I suggest you first do the gain properly, meaning, how everyone else measures it. \$\endgroup\$ Commented Nov 24, 2017 at 15:02
  • \$\begingroup\$ There is no Vpp this is about the gain when the input is constant DC with some noise. \$\endgroup\$
    – floppy380
    Commented Nov 24, 2017 at 15:05
  • \$\begingroup\$ The gain at 1Hz is different than the DC gain. So I need to find the gain at exactly 0Hz i.e at DC by experiment \$\endgroup\$
    – floppy380
    Commented Nov 24, 2017 at 15:07
  • \$\begingroup\$ Use a function generator and oscilloscope averaging to measure input and output voltage with greatly reduced noise. \$\endgroup\$
    – user57037
    Commented Nov 24, 2017 at 16:48
  • \$\begingroup\$ If you are trying to measure DC levels, you DEFINITELY don't want to use RMS. Mean is a better choice. If you are using an oscilloscope, use averaging to reduce noise. I am referring to the ability of many digital scopes to sample multiple waveforms and average them together. This is part of the "acquisition" function of the oscilloscope. \$\endgroup\$
    – user57037
    Commented Nov 24, 2017 at 17:04

2 Answers 2

1
\$\begingroup\$

The term "mean value of a perfect DC voltage' is redundant. A perfect DC voltage is always the same, so the mean, mode, and median averages are all identical to the instantaneous voltage at any time. However in practice a 'perfect' DC voltage is impossible, because at some time the voltage had to change to become what it is. Therefore any measurement will always be an average over some time period.

You can average out high frequency noise to get a more accurate 'DC' reading. However if you take the mean average the result will be lower than the rms value. If you have a 1V DC signal with 1V peak sine wave 'noise' voltage on it (so the peaks are 0V and 2V) the mean average will still be 1V but the rms voltage will be 1.2152V.

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

Gain would be easily measured with a BPF to reject the noise such as a spectrum Analyzer.

\$\endgroup\$

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.