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There is a given op amp circuit with an open loop gain of 100. Rest of the parameters of the op amp is ideal. The voltage rail of the op amp is +/- 5V. The input wave form Vi(t) = cos(100t).

So now the question asks to find the rise time (0-100%) of Vo(t). Can anyone please help me with this question?

enter image description here

So my attempt was

At t = 0 sec, my output should be 100xcos (0) = 100V But since my output voltage cannot go beyond +5V (Vsat), Vo = 5 V.

So I can calculate the final value, but I can’t figure out how to find the rise time. I was thinking for an ideal op amp it should be 0s. But that’s a wild guess and I’m not really sure.

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  • \$\begingroup\$ You need to try and explain what you have tried to do in order to solve the problem. This site isn't a free homework answering service. \$\endgroup\$
    – Andy aka
    Commented Aug 1, 2018 at 16:25
  • \$\begingroup\$ @Andyaka Sorry was my first questions in SE. I edited it to add my attempt. Thanks for pointing it out. \$\endgroup\$
    – Saptarshi Ghosh
    Commented Aug 1, 2018 at 16:40
  • \$\begingroup\$ It doesn't matter if it's a sine or cosine waveform so, assume a sine and ask yourself how many degrees of phase must pass (from 0 degrees and 0 volts) for a sin wave with peak amplitude 100 volts to reach 5 volts. Convert that phase angle to time. \$\endgroup\$
    – Andy aka
    Commented Aug 1, 2018 at 16:45
  • \$\begingroup\$ Is the frequency response of the opamp given? This might be a Laplace Transform based question \$\endgroup\$
    – ijuneja
    Commented Aug 1, 2018 at 16:52
  • \$\begingroup\$ @ijuneja no unfortunately, this is all that’s given \$\endgroup\$
    – Saptarshi Ghosh
    Commented Aug 1, 2018 at 16:54

1 Answer 1

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I'm not giving a full answer because the OP needs to learn.

Saying that the input is a cosine wave is confusing you. The output will have the same rise and fall times if it were a sine wave except now you can start at t = 0, whence the output waveform begins at 0 volts and 0 degrees of phase angle.

If the amplified waveform was unbounded by the +/- 5 volt rails it would reach peaks of +/- 100 volts. If you superimpose the 5 volt limits you get something very similar to this: -

enter image description here

It's a simple matter of using your brain/equipment to calculate arcsin(0.05) to give you the degrees that have passed to hit a level of 5 volts.

Can you take it from here?

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  • \$\begingroup\$ *begin nitpicking* It's not just arcsin(0.05) though, the input to the sine wave was 100 t, not 1 t. *end nitpicking* \$\endgroup\$
    – Harry Svensson
    Commented Aug 1, 2018 at 17:43
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    \$\begingroup\$ @HarrySvensson arcsin(0.05) gives you degrees based on attaining 5% of the peak i.e. 5 volts in 100 volts. You've then got to convert to time by accounting for the frequency. \$\endgroup\$
    – Andy aka
    Commented Aug 1, 2018 at 18:12

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