0
\$\begingroup\$

enter image description here

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?

\$\endgroup\$
5
  • \$\begingroup\$ Put the schematic in your question. \$\endgroup\$
    – Andy aka
    Commented Jun 14, 2020 at 19:12
  • \$\begingroup\$ Just a question for you. What happens to \$Z_o\$ when \$R_\text{E}=0\:\Omega\$ from your last equation? Is it sensible? \$\endgroup\$
    – jonk
    Commented Jun 14, 2020 at 21:05
  • \$\begingroup\$ 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 \$\endgroup\$ Commented Jun 14, 2020 at 21:35
  • \$\begingroup\$ Vi = 0 not => ib = 0 \$\endgroup\$ Commented Jun 14, 2020 at 22:28
  • \$\begingroup\$ But what do You want to show me by thinking about "RE= 0 Ω" jonk ? \$\endgroup\$ Commented Jun 14, 2020 at 23:39

2 Answers 2

1
\$\begingroup\$

Solution : Vi = 0 not => ib = 0 enter image description here

\$\endgroup\$
1
0
\$\begingroup\$

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)

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.