# BJT Beta

Discussion in 'Homework Help' started by Agonche, Aug 31, 2012.

1. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
If $\beta =30$

* Find $V_B, V_E, V_C$.

* If $R_B$ changes to $270k\Omega$, find $V_B, V_E, V_C$.

* For what value of $\beta$ will voltages V(B),V(E),V(C) get back to previous values.

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Here's what I got.
$I_B=0.074977mA$
$I_C=2.24932mA$
$I_E=2.3243mA$

$V_B=2.0244V$
$V_C=-2.9268$
$V_E=6.65757$

If R(B) changes to 270k:

$I_B=0.02347mA$
$I_C=0.70398mA$
$I_E=0.72745mA$

$V_B=6.3358V$
$V_C=-7.09924$
$V_E=7.03588$

Here comes the 'problem', finding Beta...
I calculated $I_B$ using the previous (first) V(B) value.
$I_B=0.0074977mA$
So If V(C) gets back to the previous value, I(C) should be the same.
And $\beta = \frac{I_B}{I_C}=300$.
But I realized I can find Beta with another method, using K-laws.
$9-2.7kI_E-0.7-270kI_B=0$
using:
$I_B=\frac{I_E}{\beta +1}$
and the value of I(C) found first, I get:
$\beta =309$

Can anyone tell me if any of the solutions is right or wrong and why.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,480
1,264
Do you use a simulation program to find Ib,Ic and Ie ??

3. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
No.
$I_B=\frac{9-0.7}{27k+2.7k\cdot 31}$
Ic=Beta*Ib
Ie=(Beta+1)*Ib

=)

4. ### mlog Member

Feb 11, 2012
276
36
Check your original VE. It's inconsistent with your emitter current. The emitter voltage should be ~0.7 V above the base voltage.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,480
1,264
For

Hfe = 30 and Rb = 27K

Ib = (9V - 0.7V)/ ( 27K + 31*2.7K) = 74.977μA

Ic = 30 *Ib = 2.249mA

Ie = 31 *Ib = 2.324mA

Ve = 9V - Ie*Re = 2.725V

Vc = Ic*Rc - 9V = -2.927V

Now for Rb = 270K and β = 30

Ib = (9V - 0.7V)/ ( 270K + 31*2.7K) = 23.466μA

Ic = 30 * Ib = 703.986μA

Ie = 31 * Ib = 727.452μA

Ve = 9 - Ie*Re = 7.035V

Vc = Ic*Rc - 9V = 7.099V

And I don't see any problem ?

6. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
@mlog - Yeah you're right. Ve should be 2.724V.

@Jony130 - Thanks for the confirmations, but I'm having problems with the third part of the problem.

so with Rb=270k, I have to find the value of Beta so the terminal voltages get their previous values.

I think Beta should be 300, because Rb changed from 27k to 270k.
So if Beta was 30 first, it should be 300 when Rb=270k and the terminal voltages get back to their previous values.

I tried another method (you can see in the first post) and I get Beta=309.
What am I missing ?
Is Beta 300 or 309 or none?

Feb 17, 2009
4,480
1,264