# How to calculate rise time of an Op Amp

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 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.

• 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. – Andy aka Aug 1 '18 at 16:25
• @Andyaka Sorry was my first questions in SE. I edited it to add my attempt. Thanks for pointing it out. – Saptarshi Ghosh Aug 1 '18 at 16:40
• 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. – Andy aka Aug 1 '18 at 16:45
• Is the frequency response of the opamp given? This might be a Laplace Transform based question – ijuneja Aug 1 '18 at 16:52
• @ijuneja no unfortunately, this is all that’s given – Saptarshi Ghosh Aug 1 '18 at 16:54

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: - 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?

• *begin nitpicking* It's not just arcsin(0.05) though, the input to the sine wave was 100 t, not 1 t. *end nitpicking* – Harry Svensson Aug 1 '18 at 17:43
• @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. – Andy aka Aug 1 '18 at 18:12