slew rate problems

Discussion in 'Homework Help' started by jorgemotalmeida, Jul 7, 2010.

  1. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
    "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!
     
    Last edited: Jul 7, 2010
  2. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,346
    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
     
  3. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
    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.
     
    Last edited: Jul 7, 2010
  4. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,346
    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
     
    jorgemotalmeida likes this.
  5. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Check out this graph:
    Slew Rate.png
    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.
     
    jorgemotalmeida likes this.
  6. jorgemotalmeida

    Thread Starter Member

    Oct 23, 2007
    23
    2
    Thanks a bunch!! Now I know where I mistaked! Now it is perfectly clear.
    This is a great site!
     
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