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I have been studying the small signal analysis of BJT with re model with the emitter terminal not bypassed. I am able to derive those Zi, Zo,Av, Ai long as ro is assumed to be infinite. But I am unable to find Zi with a finite ro. Following is the circuit diagram:

enter image description here

Below is the equivalent re model: enter image description here

I tried with google, but have not been able to find any. Everywhere the solution is done omitting ro.Kindly help me with the analysis or provide some helpful link. Thanks in advance.

EDIT:

let, beta=B

From my analysis, Vb=Ib[Bre+(B+1)RE]

So, Zb=Vb/Ib= [Bre+(B+1)RE]

Zi=RB||Zb

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

Zb=Bre+[{(B+1)+RC/ro}/{1+(RC+RE)/ro}]RE

which I am not able to understand. Please explain it.

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    \$\begingroup\$ The issue is that your first equation ignores r_o. It assumes the current I_E is (beta+1)I_B, which is only true if you assume r_o is infinite. If you consider r_o, you need to consider the current that goes through it (meaning some of the beta*I_b from the dependent source goes through r_o). \$\endgroup\$ – Shamtam Nov 19 '12 at 8:15
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enter image description here
Let the current through r0 be Ir. Applying KCL,

1.  Ie = (B*Ib) + Ib + Ir.
2. Io = Ir + (B*Ib).

Applying KVL,

(Io*Rc) + (Ir*r0) + (Ie*Re) = 0.

From eq 2:

Ir = -[(B*Ib*Rc) + Ie*Re]/r0+Rc.

Ie = (1+B)*Ib - [(B*Ib*Rc) + (Ie*Re)]/(r0+Rc).

Ie(1 + (Re/r0+Rc)) = (1+B)Ib - (B*Ib*Rc)/r0+Rc.

Vb = Ib(B*Re) + Ie*Re.  

Vb= Ib(B*Re) + Ib*Re[(B+1) + (Rc/r0)]/[1 + (Rc+Re)/r0]

Zb = B*Re + Re[(b+1) + (Rc/r0)]/[1+(Rc+Re)/r0]

Please pardon me I don't know which software to use for formatting. I tried to upload an image, but it failed. I have skipped some of the steps which could be easily arranged.

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This problem can be solved by converting the current controlled current source to current controlled voltage source by using source transformation technique. So applying source-transformation technique, this circuit will look like this.

schematic

simulate this circuit – Schematic created using CircuitLab

We will obtain two equations, when we apply KVL rule both in input and output side.

Note:

Current through resistor \${\beta}r_e\$ is \$ i_b\$;

Current through \$r_O\$ and \$R_c\$ is \$ i_c\$;

Current through resistor \$R_E\$ is \$ i_e\$ = \$i_c\$ + \$i_b\$

Analysis

For input side,

\$V_b = i_b.{\beta}r_e + {(i_b+i_c)}R_E\$

=> \$V_b = i_b.({\beta}r_e + R_e) + i_c.R_E\$ --------------->equation-1

For output side,

\${i_b}{\beta}{r_o} = {(i_b+i_c)}R_E + {i_c}{R_E} + {i_c}{r_o}\$ --------------->equation-2

From equation-2, we can derive an equation for \$i_c\$ in terms of \$i_b\$.

\$i_c = i_b.\cfrac{({B}r_o-R_E)}{(R_E+R_c+r_o)}\$

Substituting it in equation-1 yields

\$V_b = i_b.({\beta}r_e + R_E) + {({i_b}{.\cfrac{({\beta}r_o-R_E)}{(R_E+R_c+r_o)}})}.{Re}\$

So Ib is common in right hand side. Taking it outside as common,

\$V_b = {i_b}.({{\beta}{r_e} + R_E + {\cfrac{(B.ro-Re)}{(Re+Rc+ro)}}.R_E})\$

Taking \$R_E\$ common from the second and third terms in the RHS, then

\$V_b = {i_b}.({\beta}r_e + ({1 + {\cfrac{({\beta}.r_o-R_E)}{(R_E+R_c+r_o)}}).R_E})\$ ---------->equation-3

\$V_b = {i_b}.({\beta}r_e + (\cfrac{(R_c+(1 + {\beta})r_o)}{(R_E+R_c+r_o)}.R_E)\$

Dividing numerator and denominator of second term in RHS by \$r_o\$ results

\$V_b = {i_b}.({\beta}r_e + (\cfrac{( (1+{\beta})+{\cfrac{Rc}{ro}})}{( 1+{\cfrac{(R_E+Rc)}{ro}})})\$

\$=>Z_B=\cfrac{V_b}{i_b} = {\beta}r_e + (\cfrac{( (1+{\beta})+{\cfrac{Rc}{ro}})}{( 1+{\cfrac{(R_E+Rc)}{ro}})}\$

so circuit input impedance will be \$Z_i=R_b || Z_b\$

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  • \$\begingroup\$ This result cannot be OK. The second term is dimensionless and must not be added to a resistance. \$\endgroup\$ – LvW Jul 17 '15 at 8:21
  • \$\begingroup\$ which is the second term ?? I didnt get \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 9:10
  • \$\begingroup\$ The second term of the last formula Zb=... \$\endgroup\$ – LvW Jul 17 '15 at 9:58
  • \$\begingroup\$ you are wrong.. in that equation first term and second term are resistance itself.. second one is not constant. its a pure resisrance term \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 10:13
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    \$\begingroup\$ @LvW I referred a book yesterday and u r right. pardon me. i am an electronics graduate.. but i work in totally different field. when i realize i forgot this very basic.. i became shocked. i applied the same rule of addition and substraction for division too.. it's my big mistake. sorry for all troubles. in order to review my answer, i need to study signal analysis again ( since electronics is fully out of touch now). but will spend some time and try to correct it. if u can spot the pblm in my analysis, then pls point it out. i will correct it. again sorry for all troubles.. thanks for ur tym \$\endgroup\$ – Rajeev K Tomy Jul 18 '15 at 4:12

protected by Community Nov 26 '15 at 14:30

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