Emitter-Follower Question..

Discussion in 'Homework Help' started by Permia, May 6, 2010.

1. Permia Thread Starter New Member

May 6, 2010
8
0

Assuming the transistor is conducting :. VVE = 0.7V,

And that the transistor is operating in its active region :. IB is porportional to IC..

IC = β x IB

IE = IB + IC

and so

IE = IB(1+β)
= 101 x Ib

80x10^3xIB + 1.2x10^3xIE + 0.7 = 20

and after working that out I get:

201200IB = 19.3

Then using the above equations:

IB = 95.92μA
IE = 9.69mA
IC = 9.59mA

VE = 11.628

but

VC = 26V ?

I've been over it and over it but cant figure out what I've done wrong. Can anyone explain why I get this?

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
This equation is not correct. You are assuming that the only current flowing in the 80K is the base current - which won't be the case. Keep in mind that the 80K and 8K in combination with the 20V supply are forming a voltage divider from which the base current is drawn.

3. Ethan New Member

Feb 7, 2009
6
1
You need to use: Ie=(1+B)Ib. You will have: -Vth +IbRth +Vbe +ReIe then you can solve for the collector current and Vc.

4. Permia Thread Starter New Member

May 6, 2010
8
0
I follow what you guys are saying.

I've calculated Vth as 1.81

and Rth as 7.27

and so I get

-1.81 + 7.27xIB + 0.7 + 1.2x10^3x101xIB = 20

am I right so far?

This gives IB as 174.16μA

5. Ethan New Member

Feb 7, 2009
6
1
You are applying KVL in the Base-Emitter-Source loop where did you get the =20 from?

-Vth +IbRth +Vbe +Re(1+B)Ib=0

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
No ...

-1.81 + 7.27x10^3xIB + 0.7 + 1.2x10^3x101xIB = 0

7. Permia Thread Starter New Member

May 6, 2010
8
0
yeah figured that after I typed it out. Once I wrote it down on paper I realised I was being an idiot.

Time for bed methinks..

IB = 8.64μA?

..can get the rest from there.

Thanks guys.