0
\$\begingroup\$

I have been trying to find an answer to this question and I am sure I probably just don't know what the circuit is called so I am unable to find it. Basically I want to take two input signals that for example are exactly the same (ex. 1Vp-p @ 1hz) and put it through an op amp to only get the differences. Image below of what I am kind of thinking:

enter image description here

So for the above circuit the two buffer opamp's create 2 signals that are the exact same. Then the second opamp stage add's a gain of 2 to each individual signal. The thought on the third stage is to take the the two signals and cancel each other out to have a 0Vp-p output signal. To do this I thought just putting each signal to the input of an opamp with no gain would give this result since the voltage on each input terminal would be the exact same. But when I simulate it with the circuit above it only acts like a buffer. The through is that I can then add noise to one (or both) of the signal paths and only have the 3'rd opamp output the noise and no signal, which then I could sum the two similar signals together and remove the non-common noise.

Sorry for the question I am not an electronic engineer (just a hobbyist) and I am just trying to learn things by trying and then looking up/asking questions.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ What you are looking for is a "differential amplifier". \$\endgroup\$
    – user16324
    Commented Aug 3, 2020 at 17:36

1 Answer 1

Reset to default
1
\$\begingroup\$

But when I simulate it with the circuit above it only acts like a buffer

Try adding the resistor in the green box below: -

enter image description here

You need to make the right hand op-amp a proper differential amplifier as used inside such devices as Instrumentation Amplifiers: -

enter image description here

Note that R3 in the above circuit is the addition I made to your schematic.

\$\endgroup\$
5
  • \$\begingroup\$ That worked perfectly. Like I said I am an electronics noob but If you could tell me if I am on the right track? That worked because the negative feedback create's a "virtual ground" and you need to create the same thing on the positive end? (Also it say's I cannot except your answer for 4 more minutes but I will do so when it lets me!) Also you wouldn't happen to know if what I am doing is pointless or not? I get a differential signal is meant to remove common mode noise from the signal.... \$\endgroup\$
    – Boeggs
    Commented Aug 3, 2020 at 17:36
  • \$\begingroup\$ But is there any way to remove differential noise (if that's even a thing) between two signals that should be exactly the same? So for example you could split a audio signal into a differential pair, amplify them then use a differential amplifier like you included to add common mode rejection, but also split each differential pair into two similar pairs and re-combine them after amplification to remove noise created in the circuit via resistances, etc? \$\endgroup\$
    – Boeggs
    Commented Aug 3, 2020 at 17:36
  • \$\begingroup\$ You are on the right track to evolving an instrumentation amplifier!!! Unfortunately they are fully evolved already so no kudos for invention. Yes, this method is often used to eradicate common-mode noise. You also don't need the two op-amp buffers on the left because the op-amps with a gain of two will do the same job. \$\endgroup\$
    – Andy aka
    Commented Aug 3, 2020 at 17:39
  • \$\begingroup\$ Thanks, I am trying to learn my way through analog design (I find it much more interesting and challenging than digital circuit design). I also just ordered a bunch of books from douglas self as I have been told he is someone that has actually evolved analog design. I get the idea behind a differential pair to reject common mode noise but is there any way to remove "differential" noise (I am just calling it that for lack of knowing the proper term) or noise that is on one line and not the other? \$\endgroup\$
    – Boeggs
    Commented Aug 3, 2020 at 17:46
  • 1
    \$\begingroup\$ Differential noise is the bane of the EE. If it's spectrum coincides with the wanted signal we get problems that might only be reduced by increasing our signal level. \$\endgroup\$
    – Andy aka
    Commented Aug 3, 2020 at 17:55

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.