# Bipolar junction transistor analysis at emitter.

Discussion in 'Homework Help' started by lam58, Jan 12, 2014.

1. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Hi for the question in the attached image below, would

Ve/Vs = [hfe*Re]/[Rb + hie + (hf'e*Re)], with hf'e being the value I found for Vc/Vs?

File size:
72.2 KB
Views:
41
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I would rather have:

$\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}$

where hfe is the value per the original problem statement.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
What's the difference between hfe and hf'e? I never did see an explanation for that in the other thread.

4. ### Efron Member

Oct 10, 2010
81
15
t_n_k is right.

$\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}$

5. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Ok but would Vc/Vb stay as (-hfe*Rc)/{Rb+hie + (Re*hfe)} or would this also be Re*(hfe + 1)?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Do you know why we use (hfe+1) ??

7. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Not entirely, I think it's because it's the resistance Re + the resistance of the emitter Re affected by the hfe of the junction?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
We use hfe+1 because the emitter current is equal to
Ie = Ib + Ic = Ib + Ib*hfe = (hfe + 1)*Ib
And this is why Re resistor is seen at the base as (Hfe +1)*Re

lam58 likes this.
9. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
So to clarify my initial question is yes it would be (Re*hfe+1)/{Rb+hie+(Re*hfe+1)}?

Feb 17, 2009
3,957
1,097

11. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Sorry just spotted it (Rc*hfe)

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
But you in post 9 show equation for the voltage gain

The correct one is Ve/Vb = (Re*(hfe+1))/{Rb+hie+(Re*(hfe+1))

But now Vc/Vb = ?

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
Sometimes you are asking about Ve/Vs and Vc/Vs, but other times it seems to morph into Ve/Vb and Vc/Vb.

What do you understand to be meant by Ve/Vb as opposed to Ve/Vs? Wouldn't Vb mean the signal voltage at the base directly, whereas Vs is the signal at the left end of Rb? So, Ve/Vb is a different gain ratio than Ve/Vs?

14. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
It's supposed to be Vc/Vs, it's just a typo.

Vc/Vs= (hfe*Rc)/{Rb + hie +(Re*1+hfe)}

Although technically wouldn't Vc/Vs = Vc/Vb since R1, and R2 are ignored?

Last edited: Jan 13, 2014
15. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
If there were a capacitor across Re that grounded/shorted Re, would Ve/Vs still = $\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}$ ?

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097

OK, now every thinks looks good except one small detail.
The voltage gain Vc/Vs is negative because transistor in common emitter configuration have 180 degree phase shift.
When the base current increases the collector current also increases.

As you can see when input voltage is increasing ( base current also) the Vce voltage decrease. So we have 180 degrees out of phase (negative gain).

lam58 likes this.
17. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Ahh ok, I see. Thanks

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
No ,try to use small signal analysis and you will know.

19. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Would there be no gain since Re is grounded meaning Ve/Vs would = Vs?

Last edited: Jan 13, 2014
20. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Yes, how can we talk about the voltage gain in circuit in which we short our output terminal to ground.