"2.97 What is the highest frequency of a triangle wave of 20- V
peak-to-peak amplitude that can be reproduced by an op amp
whose slew rate is 10 V/us (us-microseconds)? For a sine wave of the same frequency, what is the maximum amplitude of output signal that
remains undistorted?"


My answer:

I assume that Sedra is considering a wave without DC component (he does not give any info about this...), so if the wave is alternating between the x axis with 10 V at maximum peak and -10 V at minimum peak it will give a slope of

slope = 20/ T

Well, slope = SR hence 20/T = 10^7 ( since SR = 10 V/us) hence T=2microseconds...

f=1/T == f = 1/2 us ==> f = 500 kHz (Sedra gave the 250 kHz as the final answer?? Maybe he considered 20 V as a peak voltage and not peak-to-peak??)

Then

for the sine wave, considering the frequency 500 kHz it gave

v=vo sin (2*pi* 5*10^5 * t) [V]
dv/dt = v0 * 2*pi*10^5 cos (......) -- but cos (..) is at maximum with 1.. so

dv/dt| (max) = v0 * 2* pi*10^5

SR= dv/dt == 20 *10^6 = v0 * 2*pi*5*10^5 ==

v0 = 3,18 V , (Sedra gave 6,37 V - DOUBLE the mine value)

Please see if I'm right there. Thanks!

 

Hello,

The triangle wave is 20 Volts, the slewrate is 10 V / μS:
The triangle wave takes 2 μS to go up AND 2 μS to go down.
So the total time is 4 μS wich leads you to the 250 kHz.

Bertus

 

Hello,

The triangle wave is 20 Volts, the slewrate is 10 V / μS:
The triangle wave takes 2 μS to go up AND 2 μS to go down.
So the total time is 4 μS wich leads you to the 250 kHz.

Bertus

hmm.. 20 volts peak-to-peak... not 20 volts-peak... that's why I told the slope must be 20/T.. (I considered 10 V-peak without any dc component.. sedra seems to consider 0 V to be the minimum peak, in this way, I can understand why he has the given values.)... Where is my mistake?

Where you got the 2 microseconds?

Thanks again.

 

Hello,

A voltage change of 20 volts takes 2 μS at a slew rate of 10 V/μS.
As the voltage has to go up and down you need to double the time so 4 μS for a full cycle.

Bertus

 

Check out this graph:

Count the Slew Rate. Do you see that it is actually
\(S.R.=\frac{V_{\small{p-p}}}{T/2}=\frac{2 \cdot V_{\small{p-p}}}{T}\) ?
Compare it to your own result and see why Sedra is correct.

 

Thanks a bunch!! Now I know where I mistaked! Now it is perfectly clear.
This is a great site!