# How to find the output impedance for the unbypassed Common-Emitter Emitter bias configuration when ro presents?

I have found an equation in the book of Robert Boylestad & Louis Nashelsky, as (27) on page 277 of the 2014 edition:

Which pertains to the circuit depicted in Fig.29-30 on page 275 of the book.

But that equation is not matching with the equation which one gave solve for this type of question previously.

$$Z_o=R_c \parallel \left[r_o+ \frac{\beta (r_o+r_e)}{1+\beta r_e/R_E}\right]$$

How has this equation been derived?

• It is the equation of Zo when ro presents. Commented Jan 28, 2022 at 13:15
• No equation is correct by 100% . Each equation contains simplifications/neglections. Therefore, slight deviations between formulas are normal - and you should tell us which equations you are referring to.
– LvW
Commented Jan 28, 2022 at 13:17
• @LvW Zo=Rc|| [ro+ {Beta*(ro+re)}/{1+(Beta*re)/RE}] How this equation has been derived? [Edited by a moderator to add a notification to the site member being responded to.] Commented Jan 28, 2022 at 13:21
• Your equation assumes a linearized small signal model for a transistor (not sure, which model) but it at least skips all slowness phenomenas and probably also feedback which happens purely inside the transistor. You should show the used model. How an amp looks electrically as seen from the output can be decided after the schematic of the amp, its signal source and the transistor small signal model are all known.
– user136077
Commented Jan 28, 2022 at 16:02

The equation you give matches perfectly equation (27) from the book that you presumably are referring to. However, there seems to be a problem with either that equation or the circuit drawn in which the 'output current' $$\I_o\$$ seems to be the current through the collector's resistor $$\R_C\$$.
Normally the output impedance refers to an output current (which would then be called $$\I_o\$$) which results from applying a voltage ($$\V_o\$$) to the output of the circuit while the input (the base) is shorted (connected to signal ground).
In the picture where $$\Z_o\$$ and $$\V_o\$$ are drawn where one would expect them, $$\I_o\$$ is given as the collector current. Later (p277) they (Boylestad and Nashelsky) argue that as $$\R_E >> r_e\$$, $$\Z_o=R_C\$$, which seems a bit obvious given their definition of $$\I_o=I_C\$$.