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How to solve this transistor problem?

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data
    The circuit has two transistors 'Q1' and 'Q2' as shown in the figure with 'Beta1'=100 and 'Beta2'=50. The circuit figure is shown below and Vbe1=Vbe2=0.7 volts.


    Need to calculate I1,Ib1,I2,Ib2,Ie1,Ic1,Ic2




    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 15, 2012 #2
    Imagine that Q2 didn't exist, and the base resistor was hooked up directly to Q1. What would your first step be in that case?
     
  4. Apr 16, 2012 #3
    The approach i used was that i considered I2=Ie1 since its series current.And used kvl equation around Q2 [ .7 + .7 -Ib2(100 kohm) -I2(10 kohm) -Ie1 (100 ohm)]

    Then i converted the currents in the form of Ib2 by considering that [(Beta2 + 1)Ib2 = Ib1] and further using equation [(Beta1 + 1)Ib1 = Ie1]. So i converted whole equation in form of Ib2 and calculated Ib2 from there.

    From Value of Ib2 I calculated Ib1 and Ic2 using Formula's [(Beta2)Ib2=Ic2] and [(Beta2 + 1)Ib2 = Ib1]

    Now i have the value of Ib1,Ib2 and Ic2 so i calculated the Value of Ie1 using equation [(Beta1 + 1)Ib1 = Ie1] and also the value of Ic1 from [(Beta1)Ib1 = Ic1]

    Now again using Kvl in the loop formed by 24 volt, Vcb2 ,10 kohm, 100 ohm and 1Kohm i calculated Vcb2.

    By using KVL in top left loop (24V,Vcb2,82Kohm and 100Kohm i calculated I1)

    I am confused that approach i used is it correct ? or what else should i do to solve this problem in minimum amount of time ?
     
  5. Apr 16, 2012 #4
    The thing is my book doesn't has end results so if anyone could help me with this ?
     
  6. Apr 16, 2012 #5

    NascentOxygen

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    Staff: Mentor

    I2 drains to earth and is lost, it doesn't go through Q1.
    KVL: the sign of each term is dependent on your assigned current directions.
    .7 + .7 + Ib2(100 kohm) -I2(10 kohm) + Ie1 (100 ohm) = 0
     
  7. Apr 16, 2012 #6
    So on that basis I2 is not equal to Ie1

    Thanks

    So how should i calculate all the current values ?
     
  8. Apr 16, 2012 #7
    You should be able to use KCL to write a relationship between I1, I2, and Ib2. After that, you should be able to use the fact that you know the voltage across the 82k + 10k combo to get a second equation, at which point you should have two equations in two unknowns.
     
  9. Apr 16, 2012 #8
    Ya i can use the kcl but how to use the kvl to my advantage ?
     
  10. Apr 16, 2012 #9
    Remember that the Vcc at the top is technically connected to ground via a voltage source, so that gives you another loop in your schematic. Using KVL on that gives you an equation in terms of I1 and I2. Your existing KVL gives you an equation in terms of I2 and Ib2. Using KCL, you can describe I1 in terms of I2 and Ib2, so you can eliminate that variable from the equations. What you're left with is two equations in two unknowns.
     
  11. Apr 17, 2012 #10
    Then two equations will be like this ?

    Equation (1)

    24 - I1(82 kohm) - I2(10 kohm) = 0

    Equation (2)

    0.7 + 0.7 + Ie1(100 ohm) + I2(10 kohm) - Ib2(100 kohm) = 0

    Then using Kcl :- I1 = Ib2 + I2

    and getting Ie2 = Ib2 [(Beta1 + 1)(Beta2 + 1)]

    ? Should i solve like this ?
     

    Attached Files:

  12. Apr 17, 2012 #11
    That looks pretty close. Check your signs in the second KVL equation, though--I think some of them may be backwards. Also, since Ie1 is going in the other direction from the base currents, you'll need another minus sign in the bottom equation as well.
     
  13. Apr 17, 2012 #12

    NascentOxygen

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    Staff: Mentor

    Did you not like my corrections to this?

    I calculate IE1 = 10.14mA. but have not checked my working. See whether you disagree.
     
  14. Apr 17, 2012 #13
    If you're going to do it like that, the 0.7's need to be negative, don't they?
     
  15. Apr 17, 2012 #14
    The method over here is correct then ?

    And i am always confused with the sign convention ! the voltage drop is from emitter to base right ?

    I am considering my current flow in the direction opposite to "I2" and in the direction of "Ib2" and taking Voltage rise = +ve and Voltage drops = -ve then won't 0.7 volt be positive, and voltages across 10 kohm and 100 ohm be +ve while voltage across 100 kohm be -ve ?
     
  16. Apr 17, 2012 #15
    The current in the emitter lead shows the actual current right ? But the question showed the current direction of Ie1 opposite to the that "Direction shown in transistor 1"
     
  17. Apr 17, 2012 #16
    Let's back up...the sign conventions here are turning what should be a relatively simple problem into a huge mess that I think is confusing all of us.

    First, flip [itex]Ie_1[/itex] around. Now it goes the same direction as [itex]Ib_2[/itex], so your equations involving the Betas don't have any negative signs in them.

    Second, do the KVL by going clockwise. That means that every stage along the way is a voltage drop, except for the 10k resistor (the voltage drop is from base to emitter in an NPN transistor). I always find it easier to sum voltage DROPS in KVL equations, because then you're following the arrow of the current, instead of going against it.

    By doing it this way, I get the following equations:

    [tex]
    82k\Omega\cdot I_1 + 10k\Omega\cdot I_2 - 24V = 0\\
    100k\Omega\cdot Ib_2 + 0.7 + 0.7 + 100\Omega\cdot Ie_1 - 10k\Omega \cdot I_2 = 0\\
    I_1 - I_2 - Ib_2 = 0\\
    Ie_1 = (\beta_1 + 1)(\beta_2 + 1)Ib_2
    [/tex]
    Mixing all of those together should give you two equations in the unknowns [itex]I_2[/itex] and [itex]Ib_2[/itex]. NascentOxygen, do you agree?

    (P.S. Once you've solved it, don't forget to flip [itex]Ie_1[/itex] back around to agree with the original definition of the problem. This is a really bizarre convention that I think is responsible for most of the confusion here--are you sure you copied that down right?)
     
    Last edited: Apr 17, 2012
  18. Apr 17, 2012 #17

    NascentOxygen

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    Staff: Mentor

    Over where? I can't see where you are pointing! :tongue:
    Then you should spend some time on this, to adopt a technique that overcomes this failing. No one is going to accept that as an excuse for getting circuit analysis wrong.
    For an NPN the base is more positive than the emitter. Picture that emitter arrow as the arrow in an ordinary diode.
    The voltages across the 100kΩ and 100Ω will have the same sign.

    EDIT: I just noticed that you have the arrow for IE1 pointing in an illogical direction. This of course changes what I've previously written. I sharn't change now, or I'll be confused, too!
     
  19. Apr 17, 2012 #18

    Ok thanks alot !

    I will be working on sign conventions then !
     
  20. Apr 17, 2012 #19
    I posted a response that details how to do this, but it got put on the first page right before it rolled over to page 2. Did you see it?
     
  21. Apr 18, 2012 #20
    Thank you
    :)
     
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