An OPAMP has a slew rate of 5 V/μS. The largest sine wave output voltage possible at a frequency of 1 MHZ is
[Hint: Slew rate is defined as the max. rate of change of output voltage. Its unit is V/μS.]
A. $$10\pi {\text{V}}$$
B. $$5{\text{V}}$$
C. $$\frac{5}{\prod }{\text{V}}$$
D. $$\frac{5}{{2\prod }}{\text{V}}$$
Answer: Option D
This Question Belongs to Electrical Engineering >> OP Amp
Join The Discussion
Comments ( 4 )
Related Questions on OP amp
Current cannot flow to ground through .
A. a mechanical ground
B. an a.c. ground
C. a virtual ground
D. an ordinary ground
When a step-input is given to an OP-amp integrator, the output will be
A. A ramp
B. A sinusoidal wave
C. A rectangular wave
D. A triangular wave with dc bias
I'm getting 5v.
Isn't slew rate the maximum voltage change possible per second(or here us).
So the maximum change in voltage the sin wave can have is 5v/us.
if i differentiate the sin wave( A*sin(10^6*t) ):
I get A*10^6cos(10^6*t)
The maximum change in voltage is when cos()=1.
Thus,
A*10^6
5/2pi
formula is
SR = 2 * pi * f * Vo
Idiot answer is 5/2pi
how the maximum output voltage is calculated