Class A push-pull

Dec 11, 2008 2 Replies

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

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

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.

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