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I have this circuit and I was asked to determine V2=f(V1) and V3=f(V2) which I determined below. The next step is to figure out what the purpose of this circuit is but I'm not sure how to do it from the equations I got.

circuit

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The first op-amp is a differential amplifier set up to measure its common mode rejection ratio, and possibly its common mode input range.

The second op-amp is an inverting, dampened low pass filter. Real-world application is not known but possibly used to dampen a peak in the frequency response of the first op-amp.

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  • \$\begingroup\$ How did you arrive at that conclusion \$\endgroup\$ – Pedro Oct 24 '18 at 12:22
  • \$\begingroup\$ Is it not obvious? I have see hundreds of these over the last 40 years. \$\endgroup\$ – user105652 Oct 24 '18 at 13:59
  • \$\begingroup\$ It is not obvious for me \$\endgroup\$ – Pedro Oct 24 '18 at 14:25
  • \$\begingroup\$ Then you need to study more op-amps wired with differential inputs that are shorted together to perform common mode testing. \$\endgroup\$ – user105652 Oct 24 '18 at 14:30
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    \$\begingroup\$ For this purpose R1=R2 and R3=R4. \$\endgroup\$ – user105652 Oct 24 '18 at 14:34
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It's hard to see what the first half of the circuit does. It could be a differential amplifier with the inputs tied together, depending on the res values. The second half is an inverting amplifier, as your equations show, with a pole at some frequency determined by C and R7 and a zero at a higher frequency determined by R6, R7 and C. So some sort of equaliser?

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