Open / Short circuit time constants for pole approximation

Here is "trivial" source on the method for you.
Try it yourself so you can learn,the explanation is clear and the method not that difficult .
Post here if you have problems...I promise we shell help

 

I think they just used the relation re=1/gm.
Makes any sens?

 

I substituted gm with 1/re
This the circuit you need to solve for Rpi

Here is the correct Circuit:

 

Since you made a valiant try at determining the resistance that C_pi "sees", you deserve to see how it is done. Unlike the book, you can ask me to explain any facet of the solution you don't understand. You are going to have fun studying the solution.

Ratch

 

At the start, I am calculating the open circuit voltage at B' and Vo. That is where the capacitor Cpi is connected. I should have said I removed rpi instead of Vpi. I need an open circuit to calculate the Thevenin impedance. Thevenin impedance is open circuit voltage divided by short circuit current. After the Thevenin impedance is calculated, then rpi is added in parallel to get the total impedance Rpi . I don't know why you would short Vsig. That is one of the voltages that activates the transistor.

Later edit: I think I see now why you would short out Vsig. I think you are trying to find the Thevenin impedance by shorting the voltage sources, opening the current sources, and calculating the resistance between the two opened points of the circuit. That method is simple, but only works if no dependent sources are present. Since the voltage controlled current source gm*Vpi is a dependent current source, that method will bite you in the butt if you use it. By calculating the Thevenin resistance by (open circuit voltage)/(short circuit current), all sources including dependent sources are taken into account, and a correct resistance is always obtained.

Ratch

 

george2525,
It is clear that the solution Ratch gave you is not according to the method discussed (quote: "I don't know why you would short Vsig"...),it is the conventional way.


1.from my previous drawing I replaced vbe back to be the original Vpi.
2.Since rpi is a parallel resistor it will be added in parallel at the end.
3.The current source Vpi/re and R'L are replaced with their Thevenin's equivalent
4.The resistance R'pi is the voltage Vpi divided by the current i(blue loop).

solving for i (KVL):

i= (Vpi*R'L/re+Vpi) /(R'sig+R'L)=Vpi*(R'L/re+1)/(R'sig+R'L)

R'pi=Vpi/i = (R'sig+R'L) / (R'L/re+1)

now we add rpi in parallel
Rpi=R'pi || rpi ==> the solution.

See no Vsig,.as advertised...

 

Nice to hear it finally worked for you
BTW,
which book is it from?

 

Ratch,
This Open-circuit time constant method is supposed to be an intuitive approximate way to analyze frequency response(quick and dirty in a sense).

And, you are absolutely correct :
only independent voltage/current sources can be "eliminated".
This isn't mentioned in the "trivial source" below,
nevertheless it is solvable and somewhat simpler than the conventional way,
as I have shown above

Read about the method here("trivial source")
Open-circuit time constant method

 

,

I make no representations for the Open-circuit time constant method. I only calculated the Thevenin resistance. as requested by the OP and the book, of the B'--Vo terminals with rpi removed. I did this by using the combination Thevenin-Norton theorem method, which looks at the voltage and current outward from the circuit instead of inward toward the circuit like the single Thevenin or Norton theorem does. The advantage is that the combination Thevenin-Norton method can be used for dependent sources, whereas the Thevenin or Norton theorem cannot. Merely divide the open circuit voltage across the terminals by the short circuit current through the terminals to get the Thevenin resistance.

Ratch

 

Your result doesn't appear to be correct.

To make it easier to compare your solution with the book solution, put both over a common denominator:



Your solution isn't the same as the book solution; it has an extra term, re*R'L, in the denominator.

 

The Electrician,
You are correct my result in this form is an approximation:
Lets divide the last eq. by re*rpi we get:
[R'L+R'sig]/ [1+R'sig/rpi +R'L(1/re+1/rpi)] equation - ***-

Note that rpi=re(beta+1) thus 1/re>>1/rpi
so the 1/rpi term can be omitted and we get the book's result.

[R'L+R'sig] / [1+ R'sig/rpi +R'L/re].

Note also that the approximation started in my reply #7 with
"I substituted gm with 1/re" .

To get the exact result:
replace back 1/re with gm in the equation -***-

(1/re+1/rpi)= gm+1/rpi
note that gm=beta/[(beta+1)*re] ; 1/rpi=1/[(beta+1)*re]
inserting gm and 1/rpi we get:
beta/[(beta+1)*re] + 1/[(beta+1)*re]=1/re

the exact term.

 

Are you still interested in a reasonably full explanation?