Base current in transistor circuit

Discussion in 'Homework Help' started by circuit2000, Mar 10, 2007.

  1. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
    33
    0
    1) For the circuit shown in figure, calculate emitter current(I<E>),Collector current(I<C>),base current(I<B>) and the collector emitter voltage.(Beta=current gain=[I<C>]/[I<B>]=50).Neglect the voltage drop across base emitter terminal.
    I solved it in the following way:
    Applying Kirchhoff voltage law to the loop SBEPS,
    25+5-{(I<B>)R1 = 0
    I<B> = 30/R1
    = 30/200 m Amp
    = 0.15 m Amp
    But the answer given in my book is 0.1 m Amp.
    I took the voltage across the loop as 25 + 5=30 volt because, the emitter connected to -5 volt is similar to the emitter being connected to the negative terminal of a 5 volt battery. Similarly, the base connected to +25 volt, is similar to the base being connected to the positive terminal of the 25 volt battery. So, as we traverse along the loop SBEPS, we would find that we move from negative terminal to positive terminal of the 2 sources and hence both positive.
    If I take the voltage across the loop as 25 – 5=20 volt, I would get the book answer. Which is right?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The voltages add, as you would expect. From teh base source to emitter is 30 volts, from the collector source to emitter is 25 volts. I believe your result for base current is correct.
     
  3. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
    33
    0
    Thanx for the guidance.
     
  4. antseezee

    Active Member

    Sep 16, 2006
    45
    0
    Ib = (25 - (-5)) / 200k = 0.15mA
    Ic = 50 * Ib = 7.5mA
    therefore, 7.5mA = (20 - Vce - (-5)) / 2k
    Vce = 10V
    Ie = Ic + Ib
    Ie = 7.65mA

    Hope that's right. Haven't done transistors in a year.
     
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