# Help rearranging an equation

Discussion in 'General Electronics Chat' started by Ronscott1, Nov 10, 2009.

1. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
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Here is the original equation:
Ie = Vcc - Vbe / ((Rc+Re+Rb) / Bdc)
I need to solve for Rb. Could you guys help me?
Best,
Ron

Jul 7, 2009
1,585
141
Rb = [(Vbe*Bdc)/(Vcc - Ie)] - Rc - Re

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I think your starting equation is incorrect - probably just misplaced parenthesis.

Assuming you have made the approximating assumption Beta = (1+Beta)

I would have written

Ie = (Vcc-Vbe)/(Rc+Re+(Rb/Beta))

or more correctly (i.e. not using the approximation)

Ie=(Vcc-Vbe)/(Rc+Re+(Rb/(1+Beta))

To re-arrange the approximate version

Ie(Rc+Re+Rb/Beta)=Vcc-Vbe

Rc+Re+Rb/Beta=(Vcc-Vbe)/(Ie)

Rb/Beta=(Vcc-Vbe)/Ie-(Rc+Re)

Rb= Beta(((Vcc-Vbe)/Ie)-Rc-Re)

Nov 5, 2009
15
0

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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While someonesdad has correctly re-arranged your equation it should be pretty obvious that the term

[(Vbe*Bdc)/(Vcc - Ie)]

is not a resistance - it's a nonsense.

6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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I'll write it nicely...

$I_e = \frac{V_{cc} - V_{be}} {{R_c+R_e+} \frac{R_b}{1+\beta}}$
...
Hand Waving here.... (See t_n_k's post)
...
$Rb=\beta (\frac{V_{cc}-V_{be}}{I_e})-Rc-Re$

Readable is good. Click on the Ʃ in reply box to get the Tex Editor/helper.

7. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
15
0
T_N_K,
When I set my work up using your equation I am getting very different results from what I calculated before and what multisim is telling me. However when I drop that pair of parenthensis I am getting similar answers. I do not have a scanner so I cannot scan my schematic or calculations

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784

$Rb=\beta(\frac{(Vcc-Vbe)}{Ie}-Rc-Re)$

9. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
15
0
I will post my work tomorrow when I get to school and have access to a scanner but thank you guys so much for you help

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Is this what we are imagining?

File size:
22.7 KB
Views:
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11. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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My bad for getting the parenthesis wrong:

$\large{Rb=\beta (\frac{V_{cc}-V_{be}}{I_e}-Rc-Re)}$

It looks correct to me.

12. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
15
0
t_n_k,
That is exactly what I am working on that same exact circuit. I was given the problem to design the resistor values Rc, Rb, and Re to meet the following specifications:
Vcc=9V
Vce at midpoint
Ic=1 mA
βdc=300

Here is my work until the point we are at now:
Ve=0.1*Vcc
Ve=(0.1)(9V)
Ve=0.9V
Re=Ve/Ie Since Ie≈Ic
Re=0.9V/1mA
Re=900Ω
Rc=4Re
Rc=4(900Ω)
Rc=3.6kΩ
And here is where I got stuck. I am following VDB guideline from Albert Malvinos Electronics Principles Book.
Is there something that I am doing wrong? If so how should I go about calculating these resistor values?

Best,
Ron

13. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
15
0
In other words, Are there certain steps that someone should take when designing a collector-emitter feedback amplifier? Please help