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CIRC

I am having trouble with my circuit as shown above. The capacitor has zero initial values. Why is the output voltage shifted by 3V?


The original purpose of the circuit is to implement a PI controller. I have graphed the inverse Laplace transform of the output voltage and the difference between the theoretical computation and simulation is a shift of 3V.

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Your amplifier has a capacitor in the negative feedback loop so it will behave as an integrator. Any input bias current will be integrated causing the output to ramp up or down, depending on the current, until it saturates.

The OP07's output characteristics are that it can reach ±13 V on a ±15 V supply so it can probably reach ±3 V on a ±5 V supply. It appears that yours has maxed out at +3 V.

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  • \$\begingroup\$ Ideally, the current should be zero from time 0. How do I fix that stray current bias ? That bias must be a spike of dirac multiplied by a very large value \$\endgroup\$
    – Han
    Sep 22 at 6:05
  • \$\begingroup\$ I fixed it by placing a resistor to discharge the capacitor. I may need to change the theoretical computation to account for the additional impedance \$\endgroup\$
    – Han
    Sep 22 at 6:25
  • \$\begingroup\$ That's the right approach. You are giving a path for the DC so you should find two things: 1. There is a small output offset of \$ I_{BIAS} \times R_f \$ and you've affected the DC and low-frequency response of your integrator as now you have created a high-pass filter. \$\endgroup\$
    – Transistor
    Sep 22 at 7:05

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