# Zₒ for small signal analysis with BJT for unbypassed emitter and rₒ in place I try to solve Zout for this BJT small-signal model but I find

RC || (ro + RE)

because Zout is calculated with Ib = 0 => βIb = 0.

From this link too Vin was calculated.

But the expression for Zo given in the book by Boylestad and Nashelsky is:

Zo = RC || (ro + β (ro + re) / (1 + β re / RE)])

Can you explain how to derive this?

• Put the schematic in your question. Jun 14, 2020 at 19:12
• Just a question for you. What happens to $Z_o$ when $R_\text{E}=0\:\Omega$ from your last equation? Is it sensible?
– jonk
Jun 14, 2020 at 21:05
• If RE=0 Ω (short circuit) from the last equation Zo = RC || [ ro + β (ro + re) / (1 + infinity)] = Zo = RC || [ ro ] = RC . It is sensible but considering ib = 0 Jun 14, 2020 at 21:35
• Vi = 0 not => ib = 0 Jun 14, 2020 at 22:28
• But what do You want to show me by thinking about "RE= 0 Ω" jonk ? Jun 14, 2020 at 23:39

Solution : Vi = 0 not => ib = 0 If you put the equations correctly:

point C :   io vo ve ib
point E :      vo ve ib
ib def  :         ve ib


3 unknowns : (vo OR io) AND ve AND ib

to get vo = function(io) OR io = function(vo) 