BJT emitter degeneration problem

  • Thread starter Thread starter likephysics
  • Start date Start date
  • Tags Tags
    Bjt
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
likephysics
Messages
638
Reaction score
4

Homework Statement


a)Find R1
b)Find the change in IC if Re varies by 5%.
Given:
Ic = 0.25mA
β=100
(see attached ckt)

Homework Equations





The Attempt at a Solution



Ic=0.25mA, Ib=0.0025mA
Vbe = 0.6956V
Ie=Ic+Ib = 0.2525mA
Re = 200*0.2525 = 50.5mV

Found R1 by applying KCL at base.
(Vcc-Vbe-Vre)/R1 = (Vbe+Vre)/10k + Ib
R1=22.75K

For part b) Re be +5% = 210 ohms
Vre = 210*Ie= 53mV
Is this even correct, I took the old Ie. What next?
I can't use any equations having Vbe or Ib since they change with increase in Vre.
 

Attachments

  • bjt re.png
    bjt re.png
    3.3 KB · Views: 679
Physics news on Phys.org
likephysics said:
Ic=0.25mA, Ib=0.0025mA
Vbe = 0.6956V
Ie=Ic+Ib = 0.2525mA
Re = 200*0.2525 = 50.5mV

Found R1 by applying KCL at base.
(Vcc-Vbe-Vre)/R1 = (Vbe+Vre)/10k + Ib
R1=22.75K
How did you determine Vbe = 0.6956V? For such low currents, I would have used 0.6v. It's simply a case of going with 0.7 or 0.6, usually. But the difference is not very important.

For part b) Re be +5% = 210 ohms
Vre = 210*Ie= 53mV
Is this even correct, I took the old Ie.
Well, you can't use the old IE. You know that is going to change.
You know everything about the circuit biasing except IB. So form a couple of equations in a similar way to how you did before and solve for IB. This time you know R1, and the only unknown is IB.
 
NascentOxygen said:
How did you determine Vbe = 0.6956V? For such low currents, I would have used 0.6v. It's simply a case of going with 0.7 or 0.6, usually. But the difference is not very important.
Ic is given, so I used Vbe = Vt ln (Ic/Is).

Well, you can't use the old IE. You know that is going to change.
You know everything about the circuit biasing except IB. So form a couple of equations in a similar way to how you did before and solve for IB. This time you know R1, and the only unknown is IB.

Well, to find Ib, I have to assume Vbe. can I assume the old Vbe (0.695)?
 
likephysics said:
Well, to find Ib, I have to assume Vbe. can I assume the old Vbe (0.695)?
I think it is perfectly reasonable to make that assumption, but you should be guided by worked examples that you have done in class. The usual justification is that the circuit is going to be constructed using preferred value resistors, anyway, and on top of this, they typically have a 5% tolerance. (Yes, I did look askance at the 3kΩ collector resistor, and wonder about preferred values ...)

If you do make the assumption that Vbe barely changes (to avoid having to use numerical techniques to arrive at the 'solution'), you can always go back to the Vbe log equation and demonstrate that the revised value of IC makes little difference to Vbe.

At such low currents, I still think 0.7v is a bit high, maybe 0.65 is more realistic.