# Homework Help: BJT emitter degeneration problem

1. Feb 5, 2012

### likephysics

1. The problem statement, all variables and given/known data
a)Find R1
b)Find the change in IC if Re varies by 5%.
Given:
Ic = 0.25mA
β=100
(see attached ckt)
2. Relevant equations

3. 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.

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2. Feb 7, 2012

Anyone?

3. Feb 7, 2012

### Staff: Mentor

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.

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.

4. Feb 8, 2012

### likephysics

Ic is given, so I used Vbe = Vt ln (Ic/Is).

Well, to find Ib, I have to assume Vbe. can I assume the old Vbe (0.695)?

5. Feb 9, 2012

### Staff: Mentor

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.