Op-amp analysis

Discussion in 'General Electronics Chat' started by mentaaal, Sep 28, 2008.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Hey guys, ok i have been pondering this for ages, and cant understand it:

    In my notes, the phase relation of output to input is given by: theta = atan(f/fc) where fc is the frequency corresponding to Aol = 1/b

    So finding the phase angle at any frequency f is done by the above formula.
    But why is this so? Why must the frequency be calculated with respect to fc? What does having the circuit in a closed loop have to do with the phase of the output?

    I have another nagging annoyance as well:
    The performance of an opamp can be represented by the following equation:

    Actual performance equation = ideal performance * 1/(1+(1/BAol(jf)))

    What is the performance then when Baol is = 1? The reason why i dont understand this is because to eliminate the j in the equation, my lecturer found the modulus of the complex number only (which i also dont understand why this can be done)

    For example: gain error was found for some circuit to be: 1/(1+0.1J)
    to eliminate the j, lecturer got the modulus of the denominator like: sqrt(1^2 - 0.1^2)

    How and why can this be done? And following this line of thought, at BAol = 1, wont the gain error be: 1/(1-1) which is infinity?

    If j is in the equation, then its a complex number and a real number divided by a complex number is a complex number! How can just finding the modulus of the comlex number be used here?

    Thanks!
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Do you have an example circuit that you can post?

    hgmjr
     
  3. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Ok, this example was for an opamp with a unity gain frequency of 1 mhz and the closed loop gain was 100 (40db)

    on the bode plot, 100 db corresponds to 10khz. So if you want to find the phase angle at 10khz you'd find atan(10k/10k) = 45 derees

    for a phase angle at 800 khz = atan(800/10) = atan(80) = 89 degrees

    for a phase angle of 100 hz = atan(100/10k) = atan(0.01) = 0.57 degrees

    I dont understand why the closed loop gain should matter in finding the phase angle?
     
  4. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Have i given enough information or am i missing something? Because what i have included here is all that i have in the notes pertaining to this problem...
     
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