need a help with op amp problems

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
135
hi
i have searching on for more problems to help me with the op amp especially with slew rate
i found these problems and i couldn't solve them can you please help me
10.png
i couldn't understand the solution or how was it found
13.png
also this problem i didn't know how to solve it
 

crutschow

Joined Mar 14, 2008
25,379
The maximum slew rate of a sinewave is the derivative of the function ia as stated in the problem:

1597686200711.png

Thus you plug the known values into the formula (dV/dt = Vm*2πf) to find the unknown, where dV/dt is the slew-rate, Vm is the peak value of the sinewave, and f is it's frequency.
You do know how to solve for an unknown in such an equation?
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
135
The maximum slew rate of a sinewave is the derivative of the function ia as stated in the problem:

View attachment 214982

Thus you plug the known values into the formula (dV/dt = Vm*2πf) to find the unknown, where dV/dt is the slew-rate, Vm is the peak value of the sinewave, and f is it's frequency.
You do know how to solve for an unknown in such an equation?
of course i know how to solve such equation but actually i don't understand how it was derived
it is easy to find the maximum frequency for a given slew rate and Vm but i don't know i find it complicated to solve for Vm i feel like i'm missing something so i don't understand how this came
 

crutschow

Joined Mar 14, 2008
25,379
i know how to solve such equation but actually i don't understand how it was derived
The rate-of-change of a function is found by taking it's derivative.
Thus you take the derivative of the sine function to get its rate-of-change (slew-rate).
The derivative maximum occurs when the COS equals 1.
What's difficult about that?
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
135
The rate-of-change of a function is found by taking it's derivative.
Thus you take the derivative of the sine function to get its rate-of-change (slew-rate).
The derivative maximum occurs when the COS equals 1.
What's difficult about that?
ok i'll solve it here and if u please show me what is wrong with my answer
dv/dt = Vp 2pi f cos(2pi f t )
T= 1/f = 1us and t at Vp = 0.5us since it is symmetrical wave then cos(2pi f t)=1
=> dv/dt = vp 2pi f
5v/1us = vp 2pi 1M
and when i solve for vp it is equal to 0.79v NOT 5/2 so where did i make a mistake
 
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