op amp question stuked... help needed

Thread Starter


Joined Jun 14, 2013
An amplifier using an opamp with slew rate SR=1v/μsec has a gain of 40db.if this amplifier
has to faithfully amplify sinusoidal signals from dc to 20 KHz without introducing any slew-rate
induced distortion, then wht is the max input signal value??????? :confused:


Joined Oct 18, 2012
You need to start by telling us what you have done to solve the problem, or, if you feel you can't, where you think you should start.

We do not do your homework for you.


Joined Mar 31, 2012
Q1) What signal (input or output) is going to hit the slew-rate limit first?

Q2) What part of that signal requires the highest slew rate?

Q3) How does that required slew rate relate to the amplitude and frequency of the input signal?


Joined Apr 24, 2007
Define your input signal to be Vin = (A)sin (2π f t)

Then Vout = (gain)(A)( sin (2π f t))

The output slew rate at any time is the rate of change with respect to time. i.e. the derivative!

You also know that dVout/dt can have a maximum of 1V/usec.

Using the Chain Rule to differentiate, Slew Rate is

dVout/dt = -2π f(gain)Acos(2π f t)

Generally, you would find the values of t where max/min occur by differentiating again and setting the derivative to zero and solving for t, but in this simple case we can see that when cos = -1, it will give the maximum dVout/dt, which gives

Max dVout/dt = 2π f(gain)A

We need the worst-case frequency. Then we could solve for A, with the max dVout/dt value given.

By inspection, larger f will give larger Max dVout/dt.

So worst case would be

1 V/usec = 2π (20000)(gain)A

Solve for A.