I think I know how to solve this, but I'm not sure about passive sign convention. Let me run though my reasoning.
- We are told that the capacitor has no initial energy, so I can assume that V0 in the capacitor voltage equation is 0.
- Since the terminal voltage=0, V(capacitor)=Vd, where V(capacitor) = (1/C) [integral][i(t)] + 0
- By inspection, i(t) = 1/125k = 8E-8 amps
- From all of this, I calculate that the initial voltage at t=>infinity is 8v.
- I can reuse the capacitor voltage equation, integrating from 0 to t and subbing in the dc biases (+20v, -10v) for V(capacitor) to find a t where the op amp is about to saturate.
I think that's how I solve the circuit(correct me if I'm wrong).
My question is, between the 2 time intervals (before and after the switch moves), I think passive sign convention necessitates a negative sign in the capacitor voltage equation in the second half, and no negative in the first half, because of the new voltage source. Is that right? Or am I forgetting something?
- t=20ms-: V(cap)= + (1/C) [integral][i(t)] +V0
- t=20ms+: V(cap)= - (1/C) [integral][i(t)] +V0
If my guess is right and there is a negative sign in the second half, I find that t= -.03, .0045s. So that would mean that the Op amp never saturates. Would that be right?