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Where does this voltage come from?

  1. Jan 24, 2014 #1
    Hello guys,

    I had a quick question about one of the exercises of my electronics book. Step #1 says that I should do:

    (5 - 0.7)/(10 + 101 * 1) = 0.039 mA

    However, I can't seem to find where the 0.7V comes from. I simulated the circuit in Multisim and i'm measuring the voltages everywhere and can't seem to find where those 0.7V come from. I even measured Vbe of Q1 and got 808.574mV instead of the 0.7 I am expecting.

    Could someone help?

    Thank you

    Attached Files:

  2. jcsd
  3. Jan 24, 2014 #2


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    What do you reckon is the base-emitter drop of Q1 ?
  4. Jan 24, 2014 #3

    Good point.
  5. Jan 24, 2014 #4
    I am assuming it is 0.7 V because that is the Vbe when the BJT is active, but could you give a brief the process of how to analyze this circuit?
  6. Jan 25, 2014 #5


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    I would just start with the assumption that Q1 is going to be on. Seems to me everything else follows from there (assuming that the base-emitter drop of Q1 is .7v, which is a reasonable assumption although really it should be part of the problem statement)

    EDIT: and of course if things DIDN'T follow from that assumption, then it would have to be considered a bad starting assumption and you'd have to try something else, but since it seems obvious to be likely, it is clearly the place to start, and of course in this case it does work out right.
  7. Jan 25, 2014 #6


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    This is how they developed that equation:

    You do KVL from +5 to ground under the 1k ohm resistor. +5-10000(Ib)-0.7-betaIb(1000)=0
    Solve this for Ib and you will get (5-0.7)/(10000+101*1000) (in amps)

    Multisim used a diode with a different volt drop, not 0.7
  8. Jan 26, 2014 #7
    BTW, don't expect diode (or base-emitter) voltage drops to be exactly 0.7V every time. It depends on many things... current through the junction, type of diode, etc. For instance, high voltage rectifiers and switching diodes are quite different.
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