# Transient analysis of op-amp circuit

What is the expression of V if V(0) = 4V?

simulate this circuit – Schematic created using CircuitLab

I am unable to proceed further after forming the following equation: -v/R1=(v-V)/R2+C1dv/dt

• Instead of slapping some formulas on the circuit, do you understand what will happen? First assume C1 isn't there, what are the voltages and currents? Then I connect C1 and it is charged to 4 V, what happens now? Commented Feb 17, 2018 at 14:39
• Like @Bimpelrekkie says then apply a step on the output, so you get dv/dt feedback inverted so the result is an integrator. in+ - Vin- =0 Commented Feb 17, 2018 at 15:01
• Overload recovery of an opamp is a piece-wise function and best left to a spice simulator. At $$t = 0+\eps$the opamp's output will slew and then hit the negative rail... Commented Feb 17, 2018 at 16:15 ## 2 Answers Let the op amp output voltage be$v_o$At the inverting input,$v^-=0$since it's a virtual earth. Therefore, KCL at the$ v^-$node gives: C\frac{dv_o}{dt}+\frac{v_o}{R_2}=0 or \frac{dv_o}{dt}+\frac{v_o}{R_2C}=0 Solving this ODE: v_o=Ae^{-t/R_2C} Initial condition: t=0,\: v_o=4 Hence$A=4\, and:$$ v_o=4e^{-t/R_2C}=4e^{-2t}

If the op-amp output is 4 volts (and stable at that value) then, because an op-amp has a massive open-loop gain, the differential voltage between the inverting and non-inverting inputs must be close to zero volts i.e. in the realm of 1 mV. Given that the non-inverting input is connected to 0 volts, the inverting input is probably within a milli volt or so of 0 volts.

This means that one side of the capacitor is at 4 volts (because it connects to the output) and the other side is around +/- 1 mV away from 0 volts.

In an ideal op-amp the gain is infinite and there are no input errors so Vc = Vo.

It's got nothing to do with R1 or R2 theoretically.

• Ah a down vote.... any comment to explain why? Commented Feb 17, 2018 at 19:43