# (NPN) Transistor

1. Sep 15, 2009

### uzair_ha91

1. The problem statement, all variables and given/known data
In circuit (figure), there is negligible potential drop between B and E, where β (current gain) is 100, calculate
(i) base current (ii) collector current (iii) potential drop across Rc (iv) Vce
(Ans: 11.25 µA, 1.125 mA, 1.125 V, 7.875 V)
http://img147.imageshack.us/img147/6084/183h.png [Broken]

2. Relevant equations

In this case, $$\beta$$=Ic/IB and Kirchhoff's Voltage Law.

3. The attempt at a solution

Applying KVL to this circuit,
Vcc= IBRB + VCE + ICRC
9= (100,000)IB + VCE + (1000)Ic
Because IB= Ic/$$\beta$$
Therefore: 9= 1000.Ic + 1000.Ic + VCE
9-2000.Ic=VCE
---------------------------------------This is where I get stuck..Am I doing this right?

Last edited by a moderator: May 4, 2017
2. Sep 15, 2009

### Staff: Mentor

That doesn't look right in the first equation. You should have two equations -- one for the base circuit, and one for the collector circuit.

Last edited by a moderator: May 4, 2017
3. Sep 15, 2009

### skeptic2

Before going any further, can you explain how you got this?

Vcc= IBRB + VCE + ICRC

4. Sep 15, 2009

### uzair_ha91

By applying KVL... (did I do it wrong?)

You mean I should take two loops of currents? Can you elaborate please?

5. Sep 15, 2009

### waht

I don't know why this problem assumes that Vbe is negligible, it's never negligible in a real transistor.

What is IBRB is equal to? (assuming Vbe is ground)

6. Sep 15, 2009

### uzair_ha91

Hmm, so what you are trying to say is that if we neglect Vbe, IBRB=0......, that doesn't make sense.

7. Sep 15, 2009

### uzair_ha91

8. Sep 15, 2009

### skeptic2

I don't think KVL is the best thing to use in this situation. Since Vbe is essentially 0, you have a path from Vcc to Gnd through Rb. Can you calculate the current through Rb? That current is your Ib. Can you take it from there?

9. Sep 15, 2009

### waht

What's the voltage difference between Vcc and Vbe? (assume Vbe is zero, as asked by the problem).

Knowing that, can you find Ib using ohm's law?

10. Sep 15, 2009

### uzair_ha91

By applying ohm's law,
V=IR
or Ib= Vcc/Rb = 9*10^-5 A
But the answer given at the back of my book is = 11.25 µA

11. Sep 15, 2009

### waht

The book is correct......

What's the value of Vcc, and Rb?

12. Sep 15, 2009

### skeptic2

Try using a value for Rb of 800k.

13. Sep 15, 2009

### uzair_ha91

oh yeah sorry... Ib=V/R = 9/800,000 = 11.25 µA........
Ic= Ib*beta= 1.125 mA
Vc=Ic*Rc= 1.125V
These were easy...thanks.

14. Sep 15, 2009

### skeptic2

If Vcc = 9V
and IcRc = 1.125V,

what does Vce have to be?

15. Sep 15, 2009

### uzair_ha91

Vcc= -Vce - IcRc ( So we do get to apply KVL after all )
Vce= Vcc-IcRc
= 9-1.125
=7.875 V
Again, thanks for all the help.