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

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Discussion Overview

The discussion revolves around solving a homework problem related to an NPN transistor circuit. Participants are tasked with calculating the base current (IB), collector current (IC), potential drop across the collector resistor (Rc), and the collector-emitter voltage (Vce). The conversation includes theoretical considerations, application of Kirchhoff's Voltage Law, and Ohm's Law.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant presents an initial attempt at applying Kirchhoff's Voltage Law (KVL) to the circuit but expresses uncertainty about their approach.
  • Another participant suggests that two equations are needed: one for the base circuit and one for the collector circuit.
  • Concerns are raised about the assumption that the base-emitter voltage (Vbe) is negligible, with one participant arguing that it is never negligible in a real transistor.
  • There is a discussion about calculating the base current (IB) using Ohm's Law, with a participant initially calculating it incorrectly.
  • After some back-and-forth, a participant correctly calculates IB as 11.25 µA and subsequently derives IC and the voltage drop across Rc.
  • Participants discuss the implications of the values derived and how they relate to Vce, with one participant confirming the final calculation for Vce as 7.875 V.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of Vbe and the application of KVL. While some participants agree on the final calculations for IB, IC, and Vce, the discussion reflects ongoing uncertainty regarding the assumptions made in the problem.

Contextual Notes

There is an unresolved discussion about the validity of neglecting Vbe and its implications for the calculations. The accuracy of the initial equations and the assumptions made about the circuit components are also points of contention.

uzair_ha91
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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/IB and Kirchhoff's Voltage Law.

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:
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uzair_ha91 said:

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/IB and Kirchhoff's Voltage Law.

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?

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

Vcc= IBRB + VCE + ICRC
 
skeptic2 said:
Before going any further, can you explain how you got this?

Vcc= IBRB + VCE + ICRC
By applying KVL... (did I do it wrong?)

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

uzair_ha91 said:
Vcc= IBRB + VCE + ICRC

What is IBRB is equal to? (assuming Vbe is ground)
 
what said:
What is IBRB is equal to? (assuming Vbe is ground)

Hmm, so what you are trying to say is that if we neglect Vbe, IBRB=0..., that doesn't make sense.
 
uzair_ha91 said:
You mean I should take two loops of currents? Can you elaborate please?

What do you think about this?
 
uzair_ha91 said:
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?
 
uzair_ha91 said:
Hmm, so what you are trying to say is that if we neglect Vbe, IBRB=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?
 
  • #10
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
uzair_ha91 said:
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?
 
  • #12
uzair_ha91 said:
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.
 
  • #13
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?)
 
  • #14
uzair_ha91 said:
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?
 
  • #15
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.
 

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