# BJT high frequency open circuit time Constant PROBLEM

Discussion in 'General Electronics Chat' started by acelectr, Apr 26, 2011.

1. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
Hi, I am working on high frequency models of the bjts and I am having problems with some calculations. I've attached 2 pics. As it is seen Th, high freq. time constant, equals to the sum of each capacticor multiplied with their seen resistance from their terminals. My problem is with the Cu capacitance. Firstly when checking for equivalent resistance shouldn't we short the voltage source. If so than the cotrolled current gmVpi should be 0 and thus become open. Then how Cu's terminals sees this resistance value R'sig(1+gmR'L)+R'L. What is it that I am missing here???? Is there some other way for estimating this equivalent resistance issue? I also put an Itest(test current source) replacing with Cu and got quite strange result same as the given Cu's equivalent resistance with only R'sig and R'L interchanged positions.

Appriciate any help

File size:
24.7 KB
Views:
36
File size:
21.8 KB
Views:
39
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
It is not allowed to short voltage-controlled current source.
As for Cu look in Google Miller effect

3. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
I did not mean the conctorlled source but the other small signal voltage source. I know to solve this particular circuit with the miller effect but I am trying to look for another way around. Miller effect may not always save you.

4. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
The question is actually, when shourting the voltage source and leaving open all Cpi and CL what is the equivalent resistance looking from Cu??????????????????????

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
To find equivalent resistance looking from Cu we need to solve this very simple circuit.
And our equivalent resistance is equal:
Req = Voltage across current source (Vcs = Va - Vμ) divided by current source current .

Reg = (Va - Vμ) / Iμ

So we can write one nodal equation

Iμ = ( (-Iμ*Rsig)*gm) + Va/RL (1)

Solving for Va

Va = RL*(Iμ + gm*Iμ*Rsig)

But to find Req we need to know voltage across the current source Vcs

Vsc = (Va - Vμ) = ( RL*(Iμ + gm*Iμ*Rsig) ) - (-Iu*Rsig)

Vsc = Iμ*(RL + Rsig + gm*RL*Rsig)

And Req = Vsc/Iμ = RL + Rsig + gm*RL*Rsig = Rsig (1 + gm*RL) + RL

So we find the equivalent resistance seen by Cμ

• ###### Cu.PNG
File size:
7 KB
Views:
50
Last edited: Apr 27, 2011
acelectr likes this.
6. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
you rule man thnx. I've so got in to these complex transistor stuffs that I think I've forgat some important basic circuit theory :S