# Casecode Configuration

Discussion in 'Homework Help' started by rainman, Oct 20, 2007.

1. ### rainman Thread Starter New Member

Oct 20, 2007
4
0

i dont know how to find the following unknown for the circuit below:

1. VB1
2. VB2
3. IC1 AND IC2
4. VOLTAGE GAIN

plz help me ... the circut is saved as a picture file ...

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2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Some of the circuit is cutoff at the bottom. Can you fix that and repost?

hgmjr

Oct 20, 2007
4
0

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Ok, but when I blow up the picture to get a closer look at the component values they become too fuzzy to clearly read.

hgmjr

5. ### rainman Thread Starter New Member

Oct 20, 2007
4
0

okay i will post anther one ... thanks for ur concern

6. ### rainman Thread Starter New Member

Oct 20, 2007
4
0
okay i have reupload the picture ... know its clear so plz help me

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
To start with, where do you think you need to start to obtain the values the problem asked you to find?

hgmjr

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,255
311
You haven't said just how exact your solution must be, but an approximate solution is fairly easy. First, assume the base current of each transistor has a negligible effect on the bias network (beta is 200, so this is reasonable).

Treat the three resistor network of Rb1, Rb2 and Rb3 as an unloaded voltage divider and calculate the voltage at the base of each transistor. If the transistors are silicon, then the emitter voltages are about .7 volts lower than the corresponding base voltage.

Knowing the voltage at the emitter of Q1, you can calculate the emitter current of Q1. Since the betas are high, the collector currents and emitter currents are almost equal (more exactly, the collector is the emitter current minus the base current, but remember that beta is 200).

In reality, the collector currents are a little smaller than the emitter currents, and the base currents do slightly load the bias network, so an iteration or two will improve the accuracy of the solution.

Once you have the currents, you should be able to calculate the gain. If you have trouble with that, come back here, and show what you are able to calculate.