Hey there
the Exercise questions are:
1)Express all the currents in the circuit using Ib1(given that :hfe=50,Vbe=0.7)
2)find Rb,Given that Vc=10v(and Of course hfe=50,Vbe=0.7,Vc is labeled in the circuit) and Rc=1 kilo-ohm

Well, I have tried to solve the following attached questions but I encountered a few problems :
1)Is my equivalent small signal circuit correct ,more specifically are my currents directed as well?( second picture)? my analysis is based on short circuiting the capacitors and Vcc.
2) If so , what is the meaning of question 2 -- in which circuit should I label Vc=10v,should I still apply small signal model ? is this on the place I marked with hollow dot ?
(if it should be there , something is not correct with my currents since 2551Ib1 goes through Rc and it should be
the exact opposite!!! ) ?

Thanks in advance!

 

Is there really a need to post 1.6 MB and 1.8 MB pictures when ~20 KB images will do just fine:



Your small signal equivalent picture is so messy that it is impossible to tell much from it. But the transistors need to be replaced by their small-signal equivalent circuits.

As for Question #2, that has nothing to do with small signal. It is asking about the DC quiescent operating point, which is large signal.

 

First of all thank you,
Second -how should I apply large signal model on the original circuit ? how does a capacitor behave in that situation?
I couldn't find any information about large signal analysis involving capacitors (Is there a book dealing with it ?)
How actually should my circuit look like for question 2?

 

Just use the original circuit with a DC input (it doesn't matter what the value of the input, so you can just tie Vin to 0V if you like).

The question is merely asking what value of Rb will make Vc equal to 10V. Don't make it harder than it is.

 

Ok
,I still can't see how should the cacpacitor act (open circuit?!)
May I ask you to draw the final circuit for me?

 

Yes, at DC a capacitor acts like an open circuit (infinite impedance).

So redraw the circuit with the capacitors replaced by open circuits.

And, no, I won't draw the final circuit for you. Make your best effort and we can discuss it and make adjustments if needed.

 

To solve for "RB" refer to the original circuit.
1) solve for "IC" by using the "VC" value given.
2) then divide that value by the combined betas of the darlington pair. Note: (combined values is not the sum) to solve for "IB"
3) solve for the voltage drop across "RB" by using the values given for "vbe".
4) now solve for "RB" by using ohms law with "IB" and voltage across "RB".

This will get you in the ballpark, for a value for "RB".