# Simple DC Circuit Analysis with Transistor

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1. Apr 17, 2017

### Dan97

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
All loops are clockwise;

KVL Loop around 1V, 120kOhm and VBE (I1 is assigned to the loop):

-1V + 120kOhm * I1 + VBE = 0
I1 = IB = 0.3V / 120kOhm = 2.5 * 10-6 A;

Loop around 10kOhm, vCE (I2 is assigned to the loop);

KVL Loop around 10kOhm, vo, and 20V (I3 is assigned to the loop):

10kOhm * I3 + 20V - vo = 0
vo = 10kOhm * (I2 - I3)
10kOhm * I3 +20V + (I3 - I2) * 10kOhm = 0 (1);

At the node connecting the collector part of the transistor, Io and 10kOhm:

I3 + (I2 - I3) + IC = 0
since Io = I2 - I3

IC = B * IB = 80 * 2.5 * 10-6 = 2 * 10-4 A

I2 = -2 * 10-4 A (2);

Subbing (2) into (1):

10kOhm * I3 +20V + (I3 + 2 * 10-4 A) * 10kOhm = 0

20kOhm * I3 = - 20V - 2 * 10-4 A * 10kOhm

I3 = -22 / 20k A = -1.1 * 10-3

I2 - I3 = 9 * 10 -4 A ! = 6 * 10 -4 WRONG

And I've been stuck redoing the same equations for a day now... Would be nice if someone could help find the problem in my "solution" so I can finally move on haha.

Last edited: Apr 17, 2017
2. Apr 17, 2017

### Staff: Mentor

Hi Dan97.

clockwise Σ of voltages: vo + 10k × I3 + –20 = 0

EDITED

Last edited: Apr 17, 2017
3. Apr 17, 2017

### Dan97

Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - vo +I3 * 10kOhm?

I'm not sure what you're trying to say...

4. Apr 17, 2017

### cnh1995

The given values of Vo and Io do not follow Ohm's law.
I am also getting Io=9*10-4 A.

5. Apr 17, 2017

### Dan97

Hi @cnh1995 ,

Could you show your working (if it's not too much of a hassle)? :)

6. Apr 17, 2017

### Staff: Mentor

Sorry, typo. Corrected.

Following your path, it will be +20V - vo – I3 * 10k = 0

Current through a resistor (in fact, through any element that's not a source) flows from its more-positive end to its less-positive end. If you draw an arrow showing this PD across the resistor, the arrow's tail is at the less-positive end and the arrow head is at the more-positive end.

7. Apr 17, 2017

### Dan97

@NascentOxygen ,

Why would it be - I3 * 10kOhm? Isn't that breaking Ohm's Law?

8. Apr 17, 2017

### cnh1995

The given answers are 12V and 600uA, which are incorrect. They don't follow Ohm's law.
Collector current Ic=2x10-4A.

Using KVL you can write,
104*Io=20-104*(Io+2*10-4).

Solving this for Io, we get Io=9×10-4 A.

9. Apr 17, 2017

### Dan97

@cnh1995 ,

Oh actually I've made a big fuss for nothing! I've just look at the author's errata and the answer is indeed 9 * 10 -4!

10. Apr 21, 2017

### e0ne199

what book is this anyway?

11. Apr 23, 2017

### Dan97

Alexander's Fundamentals of Electric Circuits