Transistor Base Current problem.

[Cameron1993]

Hello, this is one of my assignment questions and I don't really know where to begin.

QUESTION

Any help would be appreciated as to how to begin and eventually find out the values of IB1 and IB2.

Please be as detailed as possible as I have never analysed a transistor circuit before.


Thanks

 

[audioguru]

1) What is the emitter voltage of transistor #1?
2) What is the emitter voltage of transistor #2?
3) Use Ohms' Law to calculate the current in RL.
4) What is IB2?
5) What is IB1?
6) What is the total current gain?
It is a simple assignment.

 

[Cameron1993]

I got to step 4 before, but thought I had gone wrong.
How do I work out IB2 and IB1?
Do I just divide the current at RL by 50 and then 100?

 

[Cameron1993]

So transistors control current from collector to emitter. What is the significance in VBE?

 

[ronsimpson]

So transistors control current from collector to emitter. What is the significance in VBE?

This is the voltage drop from Base to Emitter.
Base1 is at what voltage?__________
Emitter1 is less than Base1 by VBE. What is the voltage of Emitter1?________
Emitter1 is connected to Base2. The Base2 is the same as Emitter1.
Emitter2 is less than Base2 by VBE. What is the voltage on Emitter2?________

Now you know the voltage of Emitter2, and the voltage across Rl.
What is the current in Rl.

 

[Cameron1993]

And for B2 and B1 I just divide this number accordingly?

 

[MrAl]

Hi,

Do you know how to solve this if the first transistor was not in the circuit?
That is, the input signal (+5v) would connect directly to the base of the second transistor and the first transistor is removed from the circuit.

Also, specified voltages do not change so they can be treated as constants. The base emitter voltages are specified as 0.7 volts each so they do not change. The input voltage is also specified as +5v so that does not change either. Since the two base emitter diodes are in series, the emitter of the second transistor will be the input voltage minus the two voltage drops of the base emitter diodes. Thus the voltage across the emitter resistor is known with only a little calculation which only involves subtraction.

The single transistor circuit may be easier for you to start with. The dual transistor circuit is almost the same except the gains combine and the voltage drops add up.

The collector current is the base current times the Beta. The emitter current is the base current times Beta plus 1. So the emitter current is:
Ie=Ib*(Beta+1)

so the base current is not Ie/Beta, it is Ie/(Beta+1).

The reason for this is because the emitter current is the sum of the collector current (Ib*Beta) plus the base currrent Ib, which equals:
Ib*Beta+Ib

which factored algebraically is the same as:
Ib*(Beta+1)

Once you calculate everything required you should run through it and make sure everything works out to what you have actually calculated.

 

[audioguru]

I got to step 4 before, but thought I had gone wrong.
How do I work out IB2 and IB1?
Do I just divide the current at RL by 50 and then 100?

Yes.

 

[MrAl]

Cameron1993 said:
I got to step 4 before, but thought I had gone wrong.
How do I work out IB2 and IB1?
Do I just divide the current at RL by 50 and then 100?

Yes.

Hi audioguru,

That's not exactly correct. We have to divide by 51 to get the second transistor base current. Once we have that then we have to divide by 101 to get the first transistor base current. The emitter current is Ib*(Beta+1) not Ib*Beta.

 

[audioguru]

The beta of a transistor is a RANGE of numbers usually higher than 200 and it is different at different currents so I do not bother adding "1" to the beta to calculate the emitter current.

 

[MrAl]

The beta of a transistor is a RANGE of numbers usually higher than 200 and it is different at different currents so I do not bother adding "1" to the beta to calculate the emitter current.

Hi,

I understand completely, but we usually have to go by the exact theory when a test comes up or something like that where they want the most accurate answer. On a test like this when the Beta is specified it is not subject to change unless that is also specified.

For a Beta of 100 it's only 1 percent more, but for a Beta of 10 it is 10 percent more. So it is better to know the exact formula and later estimate than to estimate first and later find out it didnt work quite right. For instruction is is better to instruct with the most accurate answer and then instruct with a method of estimation.

So just in case he needed the most theoretically correct answer he now has it, as well as an estimation