# Class A push-pull

L

#### [email protected]

Jan 1, 1970
0
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?

M

#### MooseFET

Jan 1, 1970
0
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?

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

B

#### Bob Eld

Jan 1, 1970
0
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?

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.

E

#### Eeyore

Jan 1, 1970
0
Bob said:
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.

Absolutely !

Graham

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