# Transistor Circuit Analysis

Discussion in 'Homework Help' started by blake, Mar 11, 2011.

1. ### blake Thread Starter New Member

Mar 11, 2011
1
0
This is for a 300 level electronics lab class and I'm wondering if I'm close:

The problem:

If the following silicon transistor (V(BE) = 0.6 V) has a β value of 100, determine I(C) and V(CE).

(See attached PDF)

My attempt at a solution:

First off, I'm wondering what sort of circuit this is exactly - at first I thought common-base, but with only the -15 V source, I'm thinking common-emitter.
Then, does the capacitor have any effect as far as circuit analysis and solving this problem? I figured "no" and attempted to solve as follows:

I found an expression for α (0.99) and noted that
V(BC)=V(CE)-0.6

Then I used Kirchoff's laws for the BE loop and the "loop" from the -15 V down to ground:

BE Loop:
-15 + 2000I(C) + V(CE) + 1000I(E) = 0
-15 + 2990I(C) + V(CE) = 0

15 V to Ground:
-15 + 40000I(B) +V(BC) + 2000I(C) = 0
-15.6 + 2400I(C) + V(CE) =0

I then solved these equations to get:
I(C) = 1 mA
V(CE) = 18.04 V

Any tips would be greatly appreciated.

Also, if anyone can recommend a good electronics text it would be a huge help. Mine (Principles of Electronic Instrumentation by Diefenderfer) is filled with errors and is short on example problems.

Thanks,
BL

• ###### circuit1.pdf
File size:
42.6 KB
Views:
48

Dec 26, 2010
2,147
302
These results are incorrect. You have found VCE to be greater than the supply voltage, which is not possible. Also, Ic = 1mA is clearly wrong.

Note that the base supply chain includes a 10kΩ resistance to ground. It does not seem to be included in your calculations, but it needs to be.

3. ### saqib altaf New Member

Apr 7, 2011
2
0
its a type of voltage devider biased common emitter in DC analysis of this circuit RE I-e
1k ohm resistance is included but in AC analysis this resistance is bypassed means it is removed from the circuit make ur calculations according to this configuration.
for this
first check the condition βRE>10R2 if it is

than use the equation:

VB=R2 Vcc/R1+R2

VB-VBE-VE=0

VE=VB-VBE

IE=VE/RE

for collector emitter loop

-IERE+VCE-IcRc+Vcc=0

VCE=-Vcc+Ic(Rc+RE) IE is nearly equal to IC as we know