Designing a bias circuit

  • Engineering
  • Thread starter lisp
  • Start date
  • Tags
    biasing bjt
  • #1
1
0

Homework Statement


It is required to design the bias circuit as shown in figure for a BJT whose nominal [tex]{\beta} [/tex] is 100.
upload_2015-2-8_13-37-21.png
upload_2015-2-8_13-37-43.png



a.) Find the largest ratio [tex]{(R_B/R_E)}[/tex] that will guarantee [tex]{I_E}[/tex] remains within 10% of its nominal value for B as low as 50 and as high as 150.

b.) If the resistance ratio found in a.) is used, find an expression for the voltage [tex]{V_(bb) = V_(cc) *(R_2/(R_1+R_2)) }[/tex] that will result in a voltage drop of [tex]{V(cc)/3}[/tex] across [tex]{R_E}[/tex].


Homework Equations



[tex]V(bb) = V(cc) * (R2/ (R1+R2))[/tex]
[tex]Rb = R1*R2 / (R1+R2)[/tex]

The Attempt at a Solution



a.) Using the stability factor equation found somewhere else (not sure if it is a relevant equation).

[tex] S = (1+\beta) (1+ (R_B/R_E))/(1+\beta+(R_B/R_E)) [/tex]

For largest ratio, [tex]{\beta}[/tex] is low and S is high.

[tex] 1.1 = 51 * (1+(R_B/R_E))/(51+(R_B/R_E)) [/tex]
[tex] R_B/R_E = 0.1022 [/tex]

b.)
[tex] V_(bb) - R_B*I_B-0.7-I_E*R_E = 0 [/tex]
[tex] V_(bb) = R_B* (I_E)/(B+1) + 0.7 + V_(cc)/3 [/tex][/B]
 

Attachments

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
b.)
V(bb)−RBIB−0.7−IERE=0​
Hi lisp. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

You might be able to use this approach:

Vbb = RB*IE/(β+1) + IE*RE + 0.7

= RE*IE (1 + RB/(RE(β+1))) + 0.7

A decision has to be made: what value to use for β?

Can you post a link to the derivation of the Stability Factor? That might assist someone who may be able to help with that question.

P.S. with the name "lisp", you should be comfortable with lots of nested parentheses!
 
Last edited by a moderator:
  • #3
LvW
836
222
Hi lisp,

I don`t know where your stability formula comes from. Nevertheless, I have used another formula in ealier times, involving simply the max and min values for Ic~Ie:

K=Ic,max/Ic,min=(Numerator N)/(Denominator D) with
N=[(1+B1)/B1 + RB/RE*B1] and D=[(1+B2)/B2 + RB/RE*B2]
with B1=Bmin and B2=Bmax.

Now you can solve for (RB/RE)max. if you know the value of K.
Based on the given 10% Ic variation we have K=(1+0.1)/(1-0.1)=1.222.
 
Last edited:

Related Threads on Designing a bias circuit

  • Last Post
Replies
7
Views
26K
  • Last Post
Replies
7
Views
837
  • Last Post
Replies
1
Views
573
Replies
7
Views
4K
Replies
4
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
860
  • Last Post
Replies
2
Views
866
Top