# Homework Help: Kirchoff voltage law to transistor circuits?

1. Aug 14, 2011

### logearav

1. The problem statement, all variables and given/known data
I want to know how to arrive at the equation for the base emitter side and collector emitter side of a transistor using kirchoff voltage law.

2. Relevant equations
Kirchoff voltage law is $\sum$V = $\sum$ IR.
Please refer my attachment in which i have to arrive an equation for base emitter side. So starting from the point O i proceed in the direction given in red arrows which gives
IERE+I1R2 = -VBE
Am i correct? Similarly help is needed to arrive the same to the collector emitter side.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### IMG1_0001.jpg
File size:
5.5 KB
Views:
418
2. Aug 14, 2011

### Staff: Mentor

Since the base is drawing some (small) current, the voltage at the base will not be precisely the value given by a simple voltage divider action of the the base bias resistors R1 and R2. If R1 and R2 happen to be relatively small in value and they carry a current between Vcc and ground that is much greater than the base current, then the approximation may be adequate. Usually this is not the case (in a practical circuit it would be a wast of power).

You might want to look for a simple equivalent circuit model with which to replace the transistor in the circuit so that you can write the loop equations. Maybe something like this:

[Image source: http://people.seas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_models/bjt_models.html ]

#### Attached Files:

• ###### Fig1.gif
File size:
12.5 KB
Views:
1,646
3. Aug 14, 2011

### fleem

From the diagram it appears I1 is defined as flowing downward, yet your (red arrows) loop defines it flowing upward. So you need to change its sign.

Secondly, you did not make clear what values are given and what are unknown. From your text it is clear that Vbe is unknown, but what about the other symbols on the diagram? Are we to assume all other symbols on the diagram are given?

4. Aug 14, 2011

### logearav

Mr. fleem, the arrow mark in the red is the direction i proceed in a closed loop for arriving the kirchoff voltage law.
This is a voltage divider bias used in npn transistor.

5. Aug 14, 2011

### fleem

Yes, but I'm just saying be careful to keep track of the sign of the current. Since the red arrows point opposite the direction of the current defined in the diagram, so (assuming) you've been given the value for that current in the diagram, you must change its sign to match the definition used by the red arrows.