Solve NPN Transistor HW: IB, Ic, V Rc, Vce

[uzair_ha91]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Applying KVL to this circuit,
Vcc= I_BR_B + V_CE + I_CR_C
9= (100,000)I_B + V_CE + (1000)Ic
Because I_B= Ic/\beta
Therefore: 9= 1000.Ic + 1000.Ic + V_CE
9-2000.Ic=V_CE
---------------------------------------This is where I get stuck..Am I doing this right?

 

[berkeman]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Applying KVL to this circuit,
Vcc= I_BR_B + V_CE + I_CR_C
9= (100,000)I_B + V_CE + (1000)Ic
Because I_B= Ic/\beta
Therefore: 9= 1000.Ic + 1000.Ic + V_CE
9-2000.Ic=V_CE
---------------------------------------This is where I get stuck..Am I doing this right?

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.

 

[skeptic2]

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

Vcc= IBRB + VCE + ICRC

 

[uzair_ha91]

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

Vcc= IBRB + VCE + ICRC

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

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.

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

 

[waht]

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

Vcc= I_BR_B + V_CE + I_CR_C

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

 

[uzair_ha91]

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

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

 

[uzair_ha91]

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

What do you think about this?

 

[skeptic2]

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



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

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?

 

[waht]

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

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?

 

[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

 

[waht]

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

The book is correct...

What's the value of Vcc, and Rb?

 

[skeptic2]

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

Try using a value for Rb of 800k.

 

[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.
Vce=? (What about this one?)

 

[skeptic2]

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.
Vce=? (What about this one?)

If Vcc = 9V
and IcRc = 1.125V,

what does Vce have to be?

 

[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.