BJT Beta

Thread Starter

Agonche

Joined Aug 26, 2011
30
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.



==================================================

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.
Thanks in advance.
 

mlog

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

Jony130

Joined Feb 17, 2009
4,925
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 ?
 

Thread Starter

Agonche

Joined Aug 26, 2011
30
@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.

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

Jony130

Joined Feb 17, 2009
4,925
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?
I think that both of your answer are correct.
Simply it is impassible to get exact voltage for R = 270K.
The the reason for this is that emitter current is equal
Ie = Ic + Ib but for R=270K the Ib is ten times smeller than previous values. And this cause the "error".
 
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