# Bipolar common emitter voltage

Discussion in 'Homework Help' started by ham3388, Oct 15, 2016.

1. ### ham3388 Thread Starter Member

Jul 3, 2012
97
3
Hi again dear friends. , hope all are doing well.
Please could some one evaluate my work and advise me in case any correction isrequired.
Have a good day

2. ### WBahn Moderator

Mar 31, 2012
18,069
4,905
What steps have you taken to check your results?

Which do you expect to have the higher Vbe, the answer to (i) or the answer to (ii)?

3. ### ham3388 Thread Starter Member

Jul 3, 2012
97
3
I think the answer to (ii) sould have higher Vbe since Ic is more

4. ### Bordodynov Active Member

May 20, 2015
669
194
See

Last edited: Oct 18, 2016
5. ### ham3388 Thread Starter Member

Jul 3, 2012
97
3
Thanks Bordodynov...
The Vbe voltage is not matching with the values that obtained by me...

So what has gone wrong, I'm wondering.

Can you suggest where is my mistake?

6. ### Bordodynov Active Member

May 20, 2015
669
194
i. 25e-3*ln((5e-3/0,98e-13)+1) = 0,6163872887428
exp(Vbe/25mV)-1=5e-3/0,98e-13
exp(Vbe/25mV)=5e-3/0,98e-13+1
Vbe/25mV=ln((5e-3/0,98e-13)+1)= 24,65549154971
24,65549154971*25e-3 = 0,6163872887428
I shall understand with program LTspice, why program gave the value-added result.
ii. if Vce=const=5V ==>
Vbe(9.95mA)=25e-3*ln((9,95e-3/0,98e-13)+1) =0,633590654711
but Vce != 5V Vce=(10V-1000*9.95)=50mV Vbe-Vbc=50mV !
That must be taken into account for the calculation of both terms of the equation
Make the equation. Unfortunately for accurate calculation of the gain "ar" needed.

7. ### ham3388 Thread Starter Member

Jul 3, 2012
97
3
I appreciate you
I don't understand the hi lighted once.
Which formula you have used.
Can you explain this for me because I'm confused little bit since they are not according to my formula

May 20, 2015
669
194
See

9. ### Bordodynov Active Member

May 20, 2015
669
194
I figured out with LTspice.
With regard to my calculations. In the beginning I wrote the final result, and then brought the detailed steps.
Do not pay attention to the second calculation result.
In the case of 9.95 mA compose equation Ebers-Mole. Take as I ar = 0.5 (af = 0.98). Drop unit in parentheses (because of their smallness in comparison with the exhibitors). Add the equation Vbe-Vbc = 50mV. Use the properties of the exponential exp (A-B) = exp (a) * exp (-b)

ham3388 likes this.
10. ### ham3388 Thread Starter Member

Jul 3, 2012
97
3
Thanks very much Bordodynov
You are great
Finally I got the correct values
Again I say thank you very much since you paid attention to my thread.
Godby for timbieng.