# Output resistance of CE amplifier small signal model

I was studying input/output impedance of CE amplifier small signal model and I wanted to know why the output impedance (looking from the collector) is just Rc in parallel with dynamic output resistance of the transistor (ro).

When calculating output impedance why are RB and rbe neglected? and also is the current through Rc just Ic if so why ?

• An ideal current source has infinite resistance. So for impedance "looking in" you can just throw it away. At that point it's pretty obvious that $r_o$ is in parallel with $R_\text{C}$ and nothing else is there anymore to worry about. So what else would the impedance look like?
– jonk
Mar 17, 2020 at 22:24

Well, we are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

Using Mathematica, I wrote the following code:

In[1]:=FullSimplify[
Solve[{Ix == I1 + I2, I1 == Ix + I0, I2 == Iy + I6,
0 == Iy + I0 + I5, I6 == I3 + I4, I5 == I3 + I4, I1 == (Vx)/R1,
I2 == (Vx - V1)/R2, I3 == (V1)/R3, I4 == (V1)/R4}, {Ix, I0, I1, I2,
I3, I4, I5, I6, V1}]]

Out[1]={{Ix -> (Iy R1 R3 R4 + (R1 + R2) R3 Vx + (R1 + R2 + R3) R4 Vx)/(
R1 R2 R3 + R1 (R2 + R3) R4),
I0 -> -((Iy R3 R4 + (R3 + R4) Vx)/(R3 R4 + R2 (R3 + R4))),
I1 -> Vx/R1, I2 -> (Iy R3 R4 + (R3 + R4) Vx)/(R3 R4 + R2 (R3 + R4)),
I3 -> (R4 (-Iy R2 + Vx))/(R3 R4 + R2 (R3 + R4)),
I4 -> (R3 (-Iy R2 + Vx))/(R3 R4 + R2 (R3 + R4)),
I5 -> -(((R3 + R4) (Iy R2 - Vx))/(R3 R4 + R2 (R3 + R4))),
I6 -> -(((R3 + R4) (Iy R2 - Vx))/(R3 R4 + R2 (R3 + R4))),
V1 -> (R3 R4 (-Iy R2 + Vx))/(R3 R4 + R2 (R3 + R4))}}


When calculating output impedance why are RB and rbe neglected?

Because the lower end of the current source and rE are grounded. End of story.

The current through Rc is modified / dependent on external load currents.