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Consider this circuit. enter image description here

Since a constant current source is present, the output voltage will always be $$V_o=i_sR_f$$

However, you can also drive $$i_s=i_c+i_R=C\frac{dv_c}{dt}+v_c/R$$

Then you can say that the output voltage is a function of capacitor voltage and is an exponential function.

The two approaches seem contradictory. Which one is correct? Is the output voltage constant or not? Why?

Edit: Assume the current source is initially turned off, then turned on at t=0. So I am interested in the transient situation just after turning on.

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  • \$\begingroup\$ Given enough time, C will charge to vc=is*R and d/dt vc=0. Where is the conflict? \$\endgroup\$ Commented Feb 28 at 6:38

2 Answers 2

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In steady state (DC) the capacitor is an open circuit so \$i_C = 0\$. Put another way, in steady state the voltage across the capacitor does not change so $$\frac{dv_c}{dt} = 0$$. That means that your second equation is $$i_S = i_R = \frac{v_C}{R} = \frac{v_R}{R}$$ (since \$v_C = v_R\$) and therefore \$v_O = -i_SR_f\$ (watching your signs). If \$i_S\$ is a constant then \$v_O\$ will also be constant.

The situation is more complex during a transient as some current from the current source will flow through the capacitor (and will be diverted away from the resistor). However, Kirchhoff's Current Law dictates that the current flowing into the left side of the RC network is \$i_S = i_C + i_R\$ (with \$i_C\$ and \$i_R\$ defined as flowing to the right) and the current flowing out of the right side of the network (into the node at the op amp's inverting input) is \$i_C + i_R\$ -- but that's just \$i_S\$ as well. \$i_S\$ flows through \$R_f\$ to produce \$v_O\$ by Ohm's Law. It doesn't matter that \$i_S\$ as seen by the op amp is composed of two time-varying currents in this case, as those two currents sum to the current source. All that matters is the current seen by the op amp at its input. You could replace \$i_S\$ with a varying current -- a sine wave, a ramp, etc., and \$v_O\$ will simply be \$i_S\$ multiplied by \$-R_f\$ (assuming the op amp can behave linearly).

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The two approaches seem contradictory. Which one is correct?

Any impedance in series with a current source is of no consequence in the derivation of a transfer function and, that series impedance can be removed and replaced with a short circuit. This rule is basic 101 circuit analysis. Just as any impedance in parallel with a voltage source is irrelevant.

Is the output voltage constant or not?

It is constant for the reasons given above.

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