Class ab amplifier dc biasing

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paulmdrdo

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Hi can you tell me what KVL loop in the circuit should I use to prove that VCE1 = VCE2 = VCC voltage. Thanks.
 

Obviously, the statement is not true, except for Vout = 0. Then the answer is trivial.
 

FvM said:
Obviously, the statement is not true, except for Vout = 0. Then the answer is trivial.
Click to expand...
I forgot to mention that Vs is ignored. Can you tell me that trivial answer? Do we just say that it's equal VCC becuase it's labled VCC?
 

Vs = 0 doesn't strictly cause Vout = 0. R1, R2, Q1, Q2 can have different parameters.

You should know if labels "+VCC" and "-VCC" implicate voltage levels of +/- VCC, I just guessed so. But if you ask me if it's true...

To make it clear in circuit analysis terms, you would sketch voltage sources. Seriously, I don't understand what's exactly the objective of your question if it involves simplifications and inaccuracies.
 

for that to be true the xtors must be very close in gain ...
 

The output KVL applies to -Vcc, Vce(Q2), Vce(Q1), + Vcc. We have:

-Vcc+Vce(Q2)+VC(Q1) = Vcc that is:
Vce(Q1)+Vce(Q2) = 2Vcc

if Vout = 0 we have (set Vout=0 and apply KVL to both lower and upper transistors):

-Vcc+Vce(Q2) = 0 ==> Vce(Q2) = Vcc
and Vce(Q1) = Vcc

This condition can be reached when Vs=0 if the npn and pnp behave exactly the same (as already said by FvM). Under this condition, if Vs <> 0 then
Vce(Q1) <> Vce(Q2) and of course Vce(Q1)+Vce(Q2) = 2Vcc applies.

This means the output voltage can span from -Vcc+Vce(Q2sat) to +Vcc-Vce(Q1sat)
 

Hi. albg, thank you! This is the answer i was looking for!
 

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