# (NPN) Transistor

by uzair_ha91
Tags: transistor
 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 µA, 1.125 mA, 1.125 V, 7.875 V) 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?
Mentor
P: 41,096
 Quote by 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) 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?
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.
 P: 1,814 Before going any further, can you explain how you got this? Vcc= IBRB + VCE + ICRC
P: 92
(NPN) Transistor

 Quote by skeptic2 Before going any further, can you explain how you got this? Vcc= IBRB + VCE + ICRC
By applying KVL... (did I do it wrong?)

 Quote by berkeman 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?
P: 1,636
I don't know why this problem assumes that Vbe is negligible, it's never negligible in a real transistor.

 Quote by uzair_ha91 Vcc= IBRB + VCE + ICRC
What is IBRB is equal to? (assuming Vbe is ground)
P: 92
 Quote by waht 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.
P: 92
 Quote by uzair_ha91 You mean I should take two loops of currents? Can you elaborate please?
P: 1,814
 Quote by uzair_ha91 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?
P: 1,636
 Quote by uzair_ha91 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?
 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 µA
P: 1,636
 Quote by 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
The book is correct......

What's the value of Vcc, and Rb?
P: 1,814
 Quote by 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
Try using a value for Rb of 800k.
 P: 92 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?)
P: 1,814
 Quote by 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?)
If Vcc = 9V
and IcRc = 1.125V,

what does Vce have to be?
 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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