Determine hFE of PNP-Transistor

Discussion in 'Homework Help' started by S.Sphereson, Apr 20, 2014.

  1. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    The resistor over Ucb is giving me a hard time...

    I need to find hFE for given transistor circuit but get stuck at evaluating Ib and Ic.

    [​IMG]

    [​IMG]
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    The question gives you the Collector voltage and the Collector resistance, so it should be easy.:)

    What do you calculate the Collector current value.??
     
  3. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    10mA.... Not correct though, I know. What am I missing?
     
  4. ericgibbs

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    Ohms law,,, Ic = Vc/Rc , so why do you think 10mA is wrong.?

    Where Vc= 2.3v and Rc = 230R
     
  5. Jony130

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    Feb 17, 2009
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    As you know current that is flow through Rc resistor is equal to
    IRc = Vc/Rc = 10mA But IRc current is not equal to Ic current. Try to think about base current and 20K resistor.

    [​IMG]
     
  6. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    OK so I have Ic, it seems. I suspected I had it... now I know I had it and hence Ib is my problem.

    In a previous problem Ib, as was Ic, went to ground through a resistor Rb. Now base and collector are connected. How do I approach something like that?
     
  7. Jony130

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    Feb 17, 2009
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    Look at this diagram

    [​IMG]

    Where in red you have the IB current path. And in blue you have Ic current path. Also notice that you know the voltage across 20K resistor. So if you know voltage across resistor, you can easily find current that is flow through this resistor.
     
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  8. ericgibbs

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    The Collector to Base resistor provides negative feedback, you must assume the voltages given in the question are steady state.

    Your working calculations so far are on the right track.:)
    Consider what Jony is pointing out so that you can complete the calculation.
     
  9. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    Ubc = 2V

    So is the current running through the 20k resistor I20k = 2V/20kohm = .1mA ??

    If so, how would that help me??
     
  10. ericgibbs

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    Jan 29, 2010
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    So what do you consider is the actual Collector current Ic value.?

    Given Irc=10mA and Ib=0.1mA

    EDIT:
    hFE = Ic/Ib
     
    Last edited: Apr 20, 2014
  11. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    I'm thinking Collector current Ic value is the sum of Ib and Irc...

    When I do that I get hFE = Ic/Ib = 0.99
     
  12. S.Sphereson

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    Apr 17, 2014
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    Is that correct?

    The answer should be hFE=99.

    I get 0.99 => I must be on to something..?
     
  13. Jony130

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    Are you sure about this ? Take a look at this diagram

    [​IMG]

    And apply KCL for IRc current.
     
  14. ericgibbs

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    Look at this expanded diagram of your circuit, showing the individual transistor currents.
    E
     
  15. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    h_{FE}=\frac{0.01A-0.0001A}{0.0001A}=99

    So Ic = Ic - Ib because Ib provides negative feedback Collector -> Base?
     
  16. ericgibbs

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    Your equation should read Irc = Ic+ Ib.:)

    Both the Ic and Ib currents 'flow' thru the 230R

    Your hFE looks OK to me.

    You will cover negative feedback in your future studies, for the time being consider this method of collector to base of biasing sets the DC operating conditions of the circuit.
     
  17. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    OK. If hFE = Ic/Ib... is hFE = Irc/Irb ok? If I do that I get hFE = 101.

    I want hFE = 99

    Quite confused at moment.
     
  18. ericgibbs

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    I did not say that...:)

    hFE is not Irc/Irb

    hFE= Ic/Ib = 99, where Ic = Irc-Ib.

    Its Collector Current/ Base Current

    E

    You said earlier Ic = Ic - Ib .!
    It should read, Ic= Irc-Ib
     
  19. S.Sphereson

    Thread Starter New Member

    Apr 17, 2014
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    OK. Not getting it right at all. :)

    Can you give me the collector and base currents so that I can compare to my calculations?

    For some reason it is not working out.
     
  20. Jony130

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    Ib = (Vb - Vc)/Rb = (4.3V - 2.3V)/20kΩ = 100μA
    IRc = Vc/Rc = 2.3V/230Ω = 10mA

    And from KCL we have

    IRc = Ib + Ic and Hfe = Ic/Ib

    Ic = IRc - Ib = 10mA - 0.1mA = 9.9mA and Hfe = 9.9mA/0.1mA = 99
     
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