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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?

Like this:

  • t=20ms-: V(cap)= + (1/C) [integral][i(t)] +V0
  • Versus
  • 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?

  • \$\begingroup\$ When you say "the initial voltage at t => infinity", that makes no sense at all. \$\endgroup\$
    – The Photon
    Commented Nov 27, 2016 at 5:41
  • \$\begingroup\$ FWIW, passive sign convention says that whatever terminal of a device you label as '+', then current going in to that terminal should be considered positive, and current going out should be considered negative. Since you haven't labeled any of your devices with + and - to indicate which direction you consider to be positive voltage, passive sign convention can't tell you anything about which direction of current to consider as positive. \$\endgroup\$
    – The Photon
    Commented Nov 27, 2016 at 5:45
  • \$\begingroup\$ I meant that the value for the initial voltage in the capacitor voltage equation is the one I found in the previous time interval \$\endgroup\$
    – JohnDoe
    Commented Nov 27, 2016 at 20:47

2 Answers 2


We're going to have to start at the beginning.

First, in the US the long bar in a battery is conventionally assigned to be positive, and the short ones negative. This means that when switch position A is connected the input is +1 volt through 125 k. When switch position B is closed the input is -8 volts through 100 k. You have calculated the magnitude of the A current appropriately.

The output of the op amp is properly described by your equations, except that you have the signs wrong. Keep in mind that the op amp is set up in an inverting configuration, so a + input provides a - output, and vice-versa. However, since the input currents in both cases are constant, the integral of the current is not constant. I leave it to you to determine what it really is. This should suggest to you that the op amp will, in fact, reach saturation.

If you want to be a smart-ass, you might mention on your homework that in general it is not a valid assumption that the saturation voltage of an op amp equals the power supplies, but it's clear that you are expected to assume this.

  • \$\begingroup\$ So in considering the effects of Passive Sign Convention on Op Amps, should I always consider the direction going out of the output terminal to be positive? Or is it arbitrary? \$\endgroup\$
    – JohnDoe
    Commented Nov 27, 2016 at 20:50
  • \$\begingroup\$ @JohnDoe - No, it's not arbitrary. Consider current to flow from positive to negative. Out of the positive terminal and into the negative. At some point you will learn that the physical realization of current is electron flow, and electrons move from negative to positive. For the purposes of circuit analysis you must ignore this. Current flows from positive to negative. \$\endgroup\$ Commented Nov 28, 2016 at 1:01
  • \$\begingroup\$ So in the problem I described above, I'd pick the direction of the current from either Vsource and stick with it for all calculations. Is that right? \$\endgroup\$
    – JohnDoe
    Commented Nov 28, 2016 at 2:27
  • \$\begingroup\$ @JohnDoe - No. You have to pick the direction which is consistent with the different supply voltages, since the saturation voltages are different. +1 and -8 for A and B. \$\endgroup\$ Commented Nov 28, 2016 at 4:17

\$i(t) = 8\times10^{-6}\text{A for } 0 \le t \le .02\$

\$i(t) = -80\times10^{-6}\text{A for } t \gt 0.02 \$

You can make this assumption as long as the op-amp does not saturate.

The output will first go negative (from 0V) at a rate of 8\$\mu\$A/C, then (from that voltage) ramp positive at a rate of 80\$\mu\$A/C. When it reaches either supply voltage it will be considered saturated (since it is 'ideal'). This follows directly from the behavior of a capacitor: dv/dt = i(t)/C.

Since it ramps positive without bound it will most certainly saturate at some time.

You should explicitly check it doesn't run into the negative supply first though (it doesn't).

  • \$\begingroup\$ To check that it doesn't run into the negative supply first, I would determine the voltage just before the switch, right? \$\endgroup\$
    – JohnDoe
    Commented Nov 27, 2016 at 20:52
  • \$\begingroup\$ That's right- you need it anyway to determine the time, just compare to the relevant supply voltage. \$\endgroup\$ Commented Nov 27, 2016 at 21:10
  • \$\begingroup\$ so I see here that if i take the signs of the sources to be passive sign convention, i.e. in part 1, current is 1v/125k, why does the voltage of the capacitor turn out negative? Photon tells me that it's b/c it's in an inverting setup. Do I have to recognize that an inverting op amp yields a negative output from a positive input, or is there something I'm missing? \$\endgroup\$
    – JohnDoe
    Commented Nov 27, 2016 at 22:18
  • \$\begingroup\$ Take the positive current situation. Current is flowing into the side of the capacitor connected to the inverting input and out of the other side. The voltage measured from output to input will increase in a positive direction, so the output must be doing negative to balance the op-amp inputs. Clear (as mud)? \$\endgroup\$ Commented Nov 27, 2016 at 22:28

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