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Alright, I'm currently learning more about circuit analysis and op-amps using an electrical engineering and measurement book; the problem is that most questions don't have answers, and I don't really understand what they want me to do during this problem.

I've done some OP-amps examples before, but I've always been given the Uin value for that, which I don't have here, so don't really know how to obtain my desired values.

"Uout = A*Uin + B (where Uin is the sensor output voltage), figure out A and B" is the question in the book, and I don't really understand how to obtain the "A" and "B" value. Usually you can just take the two resistor values such as R1/RF, and multiply by the voltage source which in this case is 0; however taking (1.5k/4k)*0 would just result in zero. And even if I did manage to find Uout, I still don't understand how I would get the A and B constants from it.

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    \$\begingroup\$ and multiply by the voltage source which in this case is 0 No, the input is Vin, not zero. That first opamp is non-inverting. Calculate its gain, and you can get V1. It looks like the second opamp adds 1V to V1 (but check the gain). You probably recognize the last one as a plain inverter. Put it all together into a big equation, then simplify down to a*Vin + b. \$\endgroup\$
    – aMike
    Commented May 26, 2020 at 12:45
  • \$\begingroup\$ The first op-amp is nothing more than an ordinary non-inverting amplifier. Do you remember the gain equation for the non-inverting amplifier? Next, we have inverting amplifiers do you remember the gain equation for an inverting amplifier? \$\endgroup\$
    – G36
    Commented May 26, 2020 at 14:42
  • \$\begingroup\$ Oh, right! Forgot about the non-inverting amp equation which is 1 + r1/r2 for instance. And the gain eq for inverting op amp is simply - r1/r2 for instance. \$\endgroup\$
    – user224075
    Commented May 26, 2020 at 14:45
  • \$\begingroup\$ Also do you know superposition theorem? \$\endgroup\$
    – G36
    Commented May 26, 2020 at 14:46
  • \$\begingroup\$ If I recall it's when you disable all other "sources" and calculate each individual source for the circuit, and then do the same for all sources and sum it all in the end? \$\endgroup\$
    – user224075
    Commented May 26, 2020 at 14:49

2 Answers 2

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You are trying to find the transfer function, not the actual voltages.

To figure that out, find the gain of the path from Uin to Uout.

U1 has a gain of \$ \frac {Rf} {R1} +1\$ which yields a non-inverting gain of 3.67

U2 has a gain (from U1) of -1 and so does the final amplifier, so the total gain (from Uin) is 3.67.

Then we take Uin2 which goes to a summing junction and the total gain to Uout from Uin2 is +1 (from -1 * -1).

So the transfer function is Uout = (Uin * 3.67 ) + Uin2 which is (Uin * 3.67) + 1V

Therefore A = 3.67 and B = 1

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To start things off Uout, its a gain of -1, you have 2 resistors with the same value, and it is trying to keep the - node at 0V, so any positive voltage in = an equivalent negative voltage out.

Next up U1, but this time its trying to keep the - node at Uin, It is a resistor divider to ground, so to have that node at the same voltage it needs to be a higher voltage of the same polarity, U1's output will be "Uin * (4000 + 1500) / 1500"

This leaves U2 which is summed, and also has a negative gain, as it is trying to keep its - node at ground but will leave solving that gap to you

At the end you derive this down to the relationship Uout = Uin1 +-/ something +-/ Uin2 some addition and multiplication of the 2 inputs gives you your output.

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