Hero999
Banned
Now I think your circuit will have a very wide operating point for its transistors since its emitter voltages are so low.
One transistor has a Vbe of 0.7V and it will be almost cutoff. Another transistor has a Vbe 0f 0.6V and it will be almost saturated.
Temperature will also affect the operating points. The transistors will be thermometers.
Sorry I didn't bother to label my components, sat the lef transistor is Tr1 and the right is Tr2.
What you're saying doesn't make any sense. I know simulations aren't always accurte but still.
Hfe = 50
Vc(Tr1) = 12V
Vc(Tr2) = 9.75V
Hfe = 500
Vc(Tr1) = 6.7V
Vc(Tr2) = 6.8V
If the output voltage is 3Vp then so what?