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Circuit with BJ transistor.

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the base current of the following circuit


    2. Relevant equations

    3. The attempt at a solution

    The solution to the problem is 4 micro amperes. Why? I get base current = 0 A because the transistor is satured and colector current=emissor current=1 mA
  2. jcsd
  3. Jun 8, 2013 #2
    Base current zero in saturation mode. you mean beta is infinite!. Sure?
  4. Jun 8, 2013 #3
    base voltage = 7-1,5=5,5
    emissor voltage=base voltage-0,6=4,9
    emissor current=1 mA
    If it is saturated then
    colector voltage = 5,1 and current also 1 mA

  5. Jun 8, 2013 #4


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    There are a lot of assumptions being made, including the assumption that the transistor is in saturation.

    Suppose that you assume that it is not saturated, and that IC = β IB. What values do you calculate for IB and IC? How about VCE? Are the results self-consistent, or is saturation confirmed?
  6. Jun 9, 2013 #5
    I have done the math since the beggining, I just want to know if the solution is wrong or not. If is not satured then ib= 2 micro Amperes and ic=998 micro Amperes. Colector voltage becomes 5,11 and Vce=0,21. It is in the active zone. solution is wrong right?
  7. Jun 9, 2013 #6


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    Staff: Mentor

    It does seem to me that a base current of 4 μA is too high for this circuit, and that your 2 μA is a more reasonable value.
  8. Jun 10, 2013 #7


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    I agree with 2uA calculated as follows...

    The base voltage is fixed at 5.5V so the emitter voltage will be 4.9V (cannot be lower than 4.9V).


    Ie = 4.9/4900 = 1mA

    Ie = Ib + Ic

    Ic = Ib * hfe

    Ie = Ib + hfe*Ib

    Ib = Ie/(1+hfe)

    Substitute values

    Ib = 1mA/501 = 2uA

    Ic = Ie - Ib = 0.998mA

    That gives a collector voltage of 5.11V. Not quite saturated.
  9. Jun 10, 2013 #8


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    On reflection I think it unlikely that the LED would drop 1.5V at 2uA. So the base voltage would be pulled higher and transistor probably would be saturated.

    Will try and have another look at it.
  10. Jun 10, 2013 #9
    That is a good point but i don't that the question is asking for the non-linear operation of the diode. But go have a look, i am also interested in knowing the behaviour of the ciruit.
  11. Jun 10, 2013 #10


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    Ok try this. Lets not assume the transistor is saturated for the moment. Lets replace the LED with an unknown resistor value Rb and assume Vce is unknown. Then I think you can write the following three KVL equations...

    (Ib+Ic)*Re + Vce + (Ic*Rc) - 10V = 0

    (Ib+Ic)*Re + Vbe + (Ib*Rb) -7V = 0

    7V - (Ib*Rb) + Vbc + (Ic*Rc) = 0


    Ic = Ib*hfe

    Vbe + Vbc + Vce = 0

    I believe there are 5 unknowns, Ib, Ic, Rb, Vbc, Vce.
    Assume Vbe = 0.7V.

    Then solve for all unknowns.

    I've not tried it !
  12. Jun 10, 2013 #11
    I don't think one can take take the resistance of a diode unless it is given there in the question itself. We just replace it by a voltage source for forward biased mode and
    open circuit for reversed biased mode. Even if we have to take the resistance of diode we have to take the voltage source of 1.5V with it in series.

    So the equation 7V - (Ib*Rb) + Vbc + (Ic*Rc) = 0 would modify to 7V - (Ib*Rb) - 1.5 + Vbc + (Ic*Rc) = 0.

    But still i don't agree with the notion of taking Rb, without being given in the question.
  13. Jun 10, 2013 #12


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    I wasn't really suggesting you calculate a fixed value for the diode resistance. There is no need to. It should be possible to solve for Ib without actually calculating an equivalent value for Rd. I dissagree on Vd. You would not need to take the Vd=1.5V into account. My point was that the base current is so low that it's not valid to use a model that has a constant voltage drop.

    If you substitite Ib*Rb with 1.5 V that means you have 5 equations and 4 unknowns. Is it possible that would give you multiple answers?
  14. Jun 11, 2013 #13

    rude man

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    The transistor is most definitely saturated.

    So to get the base current, use ie = ib + ic. If saturated, what is Vc? And therefore ic? And ie? And then the grand finale : ib?

    Hint: ib is way, way bigger than 2 uA! And beta has nothing to do with the problem.
  15. Jun 11, 2013 #14
    If you could tell us how you got the saturation mode. Because i don't find any problem with the solution given by CWatters.

    Now Vcb = 5.11-5.5 = -0.39, which is greater than -0.4.
    If you could tell us where CWatters is wrong becouse i agree with CWatters.
  16. Jun 11, 2013 #15

    rude man

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    ie = (V - 2Vd)/Re = 5.5V/4.9K = 1.12 mA
    ic = (10-5.5)/Rc = 4.5V/4.9K = 0.92 mA
    ib = ie - ic = 1.12 - 0.92 = 0.20 mA

    EDIT: I did not notice that the drop across the LED is 1.5V. That makes C Watter's analysis essentially correct. (I would assume Vbe = 0.7V rather than 0.5V).

    There is about 0.4V headroom for Vce so the transistor is not in saturation and the use of beta = 500 is correct.

    Sorry folks.

    Q is saturated because the computation ic*Rc ~ ie*Rc = 5.5V which is > (10 - Ve) and so impossible.
    Last edited: Jun 11, 2013
  17. Jun 12, 2013 #16


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    I used Vbe = 0.6 as that's stated in the problem.

    Personally I'm still undecided how best to answer the question..

    If you take the question at face value and use 1.5V for the LED drop then the transistor is not quite in saturation and you get Ib=2uA

    However i think the LED drop is unlikely to be 1.5V at 2uA in which case Ib is likely to be higher. See my post #10 for a way to calculate Ib in that case.

    Will be interesting to see what the official answer is.
  18. Jun 12, 2013 #17

    rude man

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    I agree, 1.5V for an LED runing only 2 uA is unlikely. For another thing, no way is it putting out visible light.

    Per the OP's post the 'official' answer is 4 uA.

    I have shown how to compute ib if the transistor is saturated in my previous post, wherein I assumed a drop of 0.8V for the LED (1.5V total LED plus Vbe).
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