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(NPN) Transistor

by uzair_ha91
Tags: transistor
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uzair_ha91
#1
Sep15-09, 01:38 PM
P: 92
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 ľA, 1.125 mA, 1.125 V, 7.875 V)


2. Relevant equations

In this case, [tex]\beta[/tex]=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/[tex]\beta[/tex]
Therefore: 9= 1000.Ic + 1000.Ic + VCE
9-2000.Ic=VCE
---------------------------------------This is where I get stuck..Am I doing this right?
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berkeman
#2
Sep15-09, 01:47 PM
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P: 41,281
Quote Quote by uzair_ha91 View Post
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 ľA, 1.125 mA, 1.125 V, 7.875 V)


2. Relevant equations

In this case, [tex]\beta[/tex]=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/[tex]\beta[/tex]
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.
skeptic2
#3
Sep15-09, 01:47 PM
P: 1,822
Before going any further, can you explain how you got this?

Vcc= IBRB + VCE + ICRC

uzair_ha91
#4
Sep15-09, 01:54 PM
P: 92
(NPN) Transistor

Quote Quote by skeptic2 View Post
Before going any further, can you explain how you got this?

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

Quote Quote by berkeman View Post
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
#5
Sep15-09, 01:55 PM
P: 1,636
I don't know why this problem assumes that Vbe is negligible, it's never negligible in a real transistor.

Quote Quote by uzair_ha91 View Post
Vcc= IBRB + VCE + ICRC
What is IBRB is equal to? (assuming Vbe is ground)
uzair_ha91
#6
Sep15-09, 02:01 PM
P: 92
Quote Quote by waht View Post
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
#7
Sep15-09, 02:03 PM
P: 92
Quote Quote by uzair_ha91 View Post
You mean I should take two loops of currents? Can you elaborate please?
What do you think about this?
skeptic2
#8
Sep15-09, 02:14 PM
P: 1,822
Quote Quote by uzair_ha91 View Post
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
#9
Sep15-09, 02:14 PM
P: 1,636
Quote Quote by uzair_ha91 View Post
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?
uzair_ha91
#10
Sep15-09, 02:23 PM
P: 92
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 ľA
waht
#11
Sep15-09, 02:45 PM
P: 1,636
Quote Quote by uzair_ha91 View Post
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 ľA
The book is correct......

What's the value of Vcc, and Rb?
skeptic2
#12
Sep15-09, 02:46 PM
P: 1,822
Quote Quote by uzair_ha91 View Post
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 ľA
Try using a value for Rb of 800k.
uzair_ha91
#13
Sep15-09, 02:59 PM
P: 92
oh yeah sorry... Ib=V/R = 9/800,000 = 11.25 ľA........
Ic= Ib*beta= 1.125 mA
Vc=Ic*Rc= 1.125V
These were easy...thanks.
Vce=? (What about this one?)
skeptic2
#14
Sep15-09, 03:17 PM
P: 1,822
Quote Quote by uzair_ha91 View Post
oh yeah sorry... Ib=V/R = 9/800,000 = 11.25 ľA........
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
#15
Sep15-09, 03:23 PM
P: 92
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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