Base Current

Discussion in 'Homework Help' started by jlcstrat, Mar 6, 2010.

  1. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    How would you find base current for a class A amplifier with no base resistor?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    For me the question is a little cryptic - are you able to elaborate further?
     
  3. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    I don't have the exact numbers with me, but basically it was a voltage divider biased transistor. I have the two biasing resistor values and VCC, but that's it. I was asked for base current, but couldn't figure out how to get their answer.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    A schematic would be helpful.

    hgmjr
     
  5. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    [​IMG]

    This sin't the exact one, but this is basically what I was given.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Were you given the Beta of the transistor as part of the problem statement? If so what is it?

    Were you also given the Vbe to assume in the problem statement? If so, what is it?

    hgmjr
     
  7. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    Vbe was .6 if I remember correctly, but that was all I they gave.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    You actually do have a base resistor embodied in the resistor divider R1 and R2. Calculate the Thevenin equivalents of the base voltage divider.

    hgmjr
     
  9. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    It's just R1 in parallel with R2 right? At least that's what I did, but it didn't work out.
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    That is because you probably overlooked that you have to use Thevenin's equivalent voltage as well. Not 12V.

    hgmjr
     
  11. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    I used the drop on R2 for the voltage. Is that correct?
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    Yes. The voltage across R2 is the Thevenin equivalent voltage for the network consisting of R1 and R2. Was your answer off by much?

    hgmjr
     
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  13. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    Yeah, over double! I just took the voltage across R2 divided by the parallel combination of the two biasing resistors. Oh well...I don't know what else I could have done. Thanks for helping. At least I know I'm not crazy now!
     
  14. hgmjr

    Moderator

    Jan 28, 2005
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    Did you consider the input impedance of the base? Can you post your calculation that yielded the incorrect answer? I should be able to spot where you went off the rail.

    hgmjr
     
  15. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    I think that R1=4.7K and R2=1k with 9Vdc. I found Vr2 to be 1.579V. I divided that by 824.56 to get 1.91 mA. The correct answer was supposed to be around .7 mA
     
  16. hgmjr

    Moderator

    Jan 28, 2005
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    You forgot to subtract the value of Vbe from Vth before you divided.

    hgmjr
     
  17. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    That still leaves me with 1.187 mA . I know the answer was supposed to be .7 because I've been going over and over it.
     
  18. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
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    If I divide Vbe by Rb I get close, but why would I do that?
     
  19. hgmjr

    Moderator

    Jan 28, 2005
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    is there an emitter resistor or is the emitter tied to ground?


    hgmjr
     
  20. jlcstrat

    Thread Starter Active Member

    Jun 19, 2009
    58
    3
    There is a resistor, but I wasn't given the value.
     
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