- Joined Nov 19, 2019
As you can see the current in Rout and Vout are zero.
The following is without the Cout capacitor:
1) Why is the I(Rout) current constant? Shouldn't it be ac? I only found one way to explain this:
a) The ac starts only in the collector's path (after vout, Rc node).
b) The Iac(R1) will not split because Vcc is shorted in AC analysis so Iac(R1) will fully go into the ground.
c) This means that I(Rc) will be fully DC.
d) Which means that when Idc(Rc) splits, one current will be the I(Rout) = 26mA and the other will lift Ic up.
The emitter's current is also dc is this normal?
1 is also explaining why I(Rout) and Vout are zero in the first image. It's because there is only dc there so the capacitor
So what am I missing? Am I taking the wrong node as output?
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