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Class A push-pull

Discussion in 'Electronic Design' started by [email protected], Dec 11, 2008.

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  1. Guest

    I have simulated a class A push-pull amplifier, but i don't understand
    what i've obtained...
    In a class A push pull the maximum current in load is Imax=2*Iquies;
    so if load is R//C :
    1)Voutmax=R*Imax
    2)slewrate=Imax/C
    Is it right?
    If i use as load only R, the first condition is met; but if i consider
    only C (without parallel resistor), slew rate is very high and i don't
    understand why.....can you help me?
    Thanks in advance
     
  2. MooseFET

    MooseFET Guest

    What simulation program are you using? You can display the current in
    the capacitor to see what is charging it up and down.
     
  3. Bob Eld

    Bob Eld Guest

    Based on my meager understanding of what you have said it looks like the
    following may be what's happening:

    The "R" is in parallel with the "C". Depending on frequency, the "R" shunts
    current away from the cap reducing the slew rate. When the "R" is removed
    all of the current is available to slew the cap increasing the rate.

    Do a paper calculation of the of the impedance of the situation, the
    capacitive reactance and its affect with and without the resistor to get a
    feel of how the current splits at various frequencies.
     
  4. Eeyore

    Eeyore Guest

    Absolutely !

    Graham
     
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