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

In summary, the conversation discusses a circuit with negligible potential drop between B and E, where the current gain is 100. The task is to calculate the base current, collector current, potential drop across Rc, and Vce. The equations used are Kirchhoff's Voltage Law and beta=Ic/IB. By applying Ohm's Law, the values for base current, collector current, and potential drop across Rc are found to be 11.25 µA, 1.125 mA, and 1.125 V respectively. The value for Vce is then calculated using KVL to be 7.875 V.
  • #1
uzair_ha91
92
0

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 [Broken]

Homework Equations



In this case, [tex]\beta[/tex]=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/[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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  • #2
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 [Broken]

Homework Equations



In this case, [tex]\beta[/tex]=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/[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.
 
Last edited by a moderator:
  • #3
Before going any further, can you explain how you got this?

Vcc= IBRB + VCE + ICRC
 
  • #4
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?
 
  • #5
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)
 
  • #6
waht 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.
 
  • #7
uzair_ha91 said:
You mean I should take two loops of currents? Can you elaborate please?

What do you think about this?
 
  • #8
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?
 
  • #9
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.
 

1. What is the purpose of solving for IB, Ic, V Rc, and Vce in an NPN transistor circuit?

The values of IB, Ic, V Rc, and Vce are important parameters that help us understand the behavior of an NPN transistor in a circuit. These values give us information about the base current, collector current, collector resistance, and collector-emitter voltage, respectively, which are essential for analyzing and designing transistor circuits.

2. How do I calculate IB, Ic, V Rc, and Vce in an NPN transistor circuit?

To calculate these values, you will need to know the values of the base-emitter voltage (Vbe), base resistor (RB), and collector resistor (RC), as well as the transistor's beta (β) value. You can use Ohm's Law and the transistor's characteristic curves to determine IB, Ic, V Rc, and Vce.

3. What is the relationship between IB, Ic, and β in an NPN transistor circuit?

In an NPN transistor, the collector current (Ic) is equal to the base current (IB) multiplied by the transistor's beta (β) value. This relationship is represented by the equation Ic = β * IB. This means that the base current controls the collector current, and the beta value determines the amplification factor of the transistor.

4. How does Vce affect the operation of an NPN transistor?

Vce, or the collector-emitter voltage, is an important parameter that affects the operation of an NPN transistor. It is the voltage difference between the collector and emitter terminals, and it determines the output voltage of the transistor. If Vce is too low, the transistor will not be able to amplify the input signal effectively. On the other hand, if Vce is too high, the transistor may enter saturation and not function properly.

5. Can I use the values of IB, Ic, V Rc, and Vce to determine the operating point of an NPN transistor?

Yes, the values of IB, Ic, V Rc, and Vce can be used to determine the operating point, or Q-point, of an NPN transistor. The Q-point is the stable operating point of the transistor where it amplifies the input signal without distortion. By analyzing these values, we can determine if the transistor is operating in the active or saturation region, which is crucial for proper circuit design.

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