# Current flow direction in BJT

Discussion in 'Homework Help' started by mau80, Dec 15, 2010.

1. ### mau80 Thread Starter New Member

Oct 24, 2010
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Hi, I've found some problem finding out the voltage of o node in a very simple circuit. u node
I cannot find out how the current in emitter in a NPN BJT could have negative sign.

I'd like to ask your help in this analysis.

I've attached the circuit simulated by Pspice, and my computation.

Could you give to me some input?

Thank you very much for your help!

Maurizio

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well the current is "positive", only simulation assume that current is "negative" when he comes out form the element terminal.

Vu = I2*R2 = ( Ie + (Vcc - Vu)/R3) * R2

Ie = (β+1) *Ib

Ib = (Vcc - Vu+Vbe)/R1

Or

I2 = I3 + Ie

I2 = Vu/R2

Ie = [(Vcc - (Vu+Vbe))/R1 ] * (β+1)

I3 = (Vcc - Vu)/R3

And if we solve this

Vu = (R2* (R1*Vcc - R3*(Vbe - Vcc) (1 + β)) / (R1*(R2 + R3) + R2*R3 (1 + β)) =

=( R1*Vcc - R3*(1 + β)*(Vbe - Vcc)) / ( R1 + R3 + (R1 R3)/R2 + R3*β ) = 2.5551V

Or we could use superposition for this circuit:

And then by inspection we can write

Vu = Vcc * ( R1/(β+1) || R2) / ( R3 + ( R1/(β+1) || R2) + (Vcc - Vbe) * R2||R3 / ( R1/(β+1) + R2||R3 )

And since R3>> R2 and R1 then

Vu ≈ (Vcc - Vbe) * R2||R3 / ( R1/(β+1) + R2||R3 ) ≈ (Vcc - Vbe) * R2 / ( R1/(β+1) + R2 ) = 2.54816514V

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3. ### mau80 Thread Starter New Member

Oct 24, 2010
21
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Dear Jony,

thank you very much for your reply.
I'd like to ask if you can tell me how you've write following formula:

Vu = (R2* (R1*Vcc - R3*(Vbe - Vcc) (1 + β)) / (R1*(R2 + R3) + R2*R3 (1 + β)) =

=( R1*Vcc - R3*(1 + β)*(Vbe - Vcc)) / ( R1 + R3 + (R1 R3)/R2 + R3*β ) = 2.5551V

I can follow you solution until:

Vu = I2*R2 = ( Ie + (Vcc - Vu)/R3) * R2

Ie = (β+1) *Ib

Ib = (Vcc - Vu+Vbe)/R1

Or

I2 = I3 + Ie

I2 = Vu/R2

Ie = [(Vcc - (Vu+Vbe))/R1 ] * (β+1)

I3 = (Vcc - Vu)/R3

Thank you very much for your great help!

Bye
Maurizio

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,590
1,285
Have you try solve it by yourself ?

I2 = I3 + Ie

Vu/R2 = [(Vcc - (Vu+Vbe))/R1 ] * (β+1) + (Vcc - Vu)/R3

In this step it is end of the electronic, but the beginning of mathematics.

5. ### mau80 Thread Starter New Member

Oct 24, 2010
21
0
I've reached the correct result following your suggestion.

I'd like to ask last thing. Where is the error in the attached computation.
Why I don't find correct result?
I think taht Vu can be computed a Vcc - I3R3, right?
I've spent a lot of time trying to find out this but I've not found.

Thank you very very much!

Maurizio

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6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Equation looks OK, it looks like you made a mistake in calculation.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,590
1,285
I only provide help via forum or Private Message

Last edited: Dec 16, 2010
8. ### mau80 Thread Starter New Member

Oct 24, 2010
21
0
Thank yuo for your help. Finnally I was able to find out the error.
This error is in considering:

Vcc - Vu + G_gamma

The correct form is:

Vcc - ( Vu + V_gamma )

Vcc - Vu - V_gamma

This give correct result.

Thank you very much for your suggestion and help!

Bye
Maurizio