enter image description here

Vo = Vin, so the rise time for both should be the same, but the capacitor takes some time charging.

How can I find the charging time without any other information here?

  • \$\begingroup\$ The charging time is proportional to R*C. What is the R of an ideal opamp output? \$\endgroup\$
    – Ste Kulov
    Jul 18 at 5:31
  • \$\begingroup\$ All the information is in the word "ideal". What is the dv/dt of an ideal op-amp? \$\endgroup\$
    – Justme
    Jul 18 at 5:32
  • \$\begingroup\$ Since the input rise time is not zero, you don't have a paradox. You can calculate the required output current from your ideal op-amp. \$\endgroup\$ Jul 18 at 5:47
  • \$\begingroup\$ For ideal opamp rise time is 0, so as soon as the input changes output will change. That means answer should be 0. Is that correct? \$\endgroup\$
    – sheetal
    Jul 18 at 6:09
  • \$\begingroup\$ @sheetal No. The rise time is not zero for the ideal op amp, but the max. slew rate is infinite. This means that the ideal opamp follower always will satisfy Vo = Vin as you correctly write. \$\endgroup\$
    – tobalt
    Jul 18 at 6:16

1 Answer 1


As @tobalt pointed out, if "slew-rate" is infinite for an ideal op-amp, the output will always "follow" the "input" (Also, output impedance = 0).

Note that, in the "real world", the output follows input ... until the "capacitor" become too big. And then, the output impedance of the op-amp also "slew" down the rate of output ... as can be seen in this simulation.

enter image description here

Note also that op-amp can become "unstable" ... As in this case ...

enter image description here


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