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(NPN) Transistorby uzair_ha91
Tags: transistor 
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#1
Sep1509, 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/I_{B} and Kirchhoff's Voltage Law. 3. The attempt at a solution Applying KVL to this circuit, Vcc= I_{B}R_{B} + V_{CE} + I_{C}R_{C} 9= (100,000)I_{B} + V_{CE} + (1000)Ic Because I_{B}= Ic/[tex]\beta[/tex] Therefore: 9= 1000.Ic + 1000.Ic + V_{CE} 92000.Ic=V_{CE} This is where I get stuck..Am I doing this right? 


#2
Sep1509, 01:47 PM

Mentor
P: 41,096




#3
Sep1509, 01:47 PM

P: 1,814

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


#4
Sep1509, 01:54 PM

P: 92

(NPN) Transistor



#5
Sep1509, 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.



#6
Sep1509, 02:01 PM

P: 92




#7
Sep1509, 02:03 PM

P: 92




#8
Sep1509, 02:14 PM

P: 1,814




#9
Sep1509, 02:14 PM

P: 1,636

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


#10
Sep1509, 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 


#11
Sep1509, 02:45 PM

P: 1,636

What's the value of Vcc, and Rb? 


#12
Sep1509, 02:46 PM

P: 1,814




#13
Sep1509, 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?) 


#14
Sep1509, 03:17 PM

P: 1,814

and IcRc = 1.125V, what does Vce have to be? 


#15
Sep1509, 03:23 PM

P: 92

Vcc= Vce  IcRc ( So we do get to apply KVL after all )
Vce= VccIcRc = 91.125 =7.875 V Again, thanks for all the help. 


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