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I'm doing a lab work on a circuit that takes a square signal in the input and first amplify it, then shape it and the amplify it again. So the circuit is actually composed by: an amplifier (inverting), a shaper, a second amplifier (non-inverting).

The problem is in the last part, where our professors asked us to bias DC the second amplifier because our signal is between [0;0.256]V, then it is amplified to [0,1.8]V and then we want the bias to offset our signal so that it stays in [-1;+1] with a bias voltage of -0.9V. Our professor said that we should take our 5V generator (that's the only one we have left to use), use a split resistor to make it to 0.9V and then use a buffer to separate that part of the circuit from the last amplifier (so that the gain of the last amplifier depends only on the two resistors R1 and R2). When I do the simulation with LTspice without the bias part (putting to ground R1), the signal is correctly amplified to [0;1.8]V. But, when I try to insert the bias DC, the signal is shifted to [-5;-3]V (more or less) which is too much. I just want it to be shifted with a factor of 0.9 to the bottom (so that it stays between [-1;1]V).

Can you help me to understand what I am doing wrong? I'll just link the simulation, the file and photo I get

Circuit

signal after shaper

signal after bias and amplifier

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  • \$\begingroup\$ Why use a "bunch" of power supplies? Just 2 are needed for powering all opamp's. \$\endgroup\$
    – Antonio51
    Nov 22, 2023 at 20:52
  • \$\begingroup\$ yeah I know, actually this is a circuit that i took from a friend of mine and they used all those power supplies and didn't get the chance to change them into one \$\endgroup\$
    – Gabriele
    Nov 22, 2023 at 21:06

1 Answer 1

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You did nothing wrong.
Just choose the value of your resistors R3 and R4 (My R9 and R10).

Like this ...

enter image description here

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  • \$\begingroup\$ thanks for the quick answer! I'm just a bit confused, shouldn't I choose R3 and R4 so that the output from the buffer is 0.9V? Shouldn't that be the Vbias I need to insert in the circuit? I'm just asking because unfortunately I don't know that much abut bias voltages. Also because we were asked to put a bias voltage of -0.9V \$\endgroup\$
    – Gabriele
    Nov 22, 2023 at 21:51
  • \$\begingroup\$ Is it because the output voltage from the buffer goes to the opamp (that for that signal works as an inverting opamp amplifier), and so the actual voltage to sum to the input signal of 1.8V (the one going in the non inverting input) would be Vbias=Vbuffer*(-R7/R8), where Vbuffer is the output voltage from the buffer? \$\endgroup\$
    – Gabriele
    Nov 22, 2023 at 22:51
  • \$\begingroup\$ You could use R3 and R4 to output the 0.9 V. You must then change another resistor (R1 or R2) to adapt the output to the result you want. But the gain of the last amplifier would then change. \$\endgroup\$
    – Antonio51
    Nov 23, 2023 at 8:46

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