need help to prove this equation in transistor

Discussion in 'Homework Help' started by saif-aljanahi, Apr 16, 2016.

  1. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    Hello, i have an idea about this equation but i tired of try solving it and it's wrong

    sketch-1460738182664.png
     
  2. panic mode

    Senior Member

    Oct 10, 2011
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    show your steps.
     
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  3. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    Here are my steps DSC_0051.JPG
     
  4. panic mode

    Senior Member

    Oct 10, 2011
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    so far so good, but final equation only uses It and Ie1, startc combining what you have.
     
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  5. WBahn

    Moderator

    Mar 31, 2012
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    The supposed answer makes no sense.

    You have a three terminal device and thus know that Ie1 = It + Ib2 (where Ib2 is the base current into Tr2). Therefore we know that the emitter current must be equal to or greater than It.

    Now consider that alpha is something a little less than one and, in an ideal transistor, it is exactly equal to one. If the transistors are ideal, the there is no base current and Iee = It. But your target answer would claim that It = 2Ie.

    EDIT: I jumped the gun and when mentally looking at the target answer assumed the middle term went to zero in the extreme of alpha going to unity. Clearly that is not the case. The target answer is correct.
     
    Last edited: Apr 25, 2016
  6. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    i have solved this question by myself about two days ago and i love to share it here for help any one .
    i will call alpha symbol as " a "

    IT= IC1+IC2
    IC= aIE
    IT= a1 IE1 + a2 IE2
    IE2= IB1
    :. IE1= IC1+IB1
    IB1= IE1-IC1
    IB1= IE1- a1 IE1
    IT=a1 IE1+a2 IE2
    IT=a1 IE1 + a2 (IE1 - a1 IE1)
    IT=a1 IE1 + a2 IE1 - a2 . a1 IE1
    IT= (a1-a2.a1+a2) IE1
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Good job.

    A slightly more streamlined approach might be the following:

    It = Ic1 + Ic2
    It = α1·Ie1 + α2·Ie2 (since Ic ≡ α·Ie)
    It = α1·Ie1 + α2·Ib1 (since Ib1 = Ie2)
    It = α1·Ie1 + α2·(Ie1 - Ic1) (since Ie = Ic + Ib)
    It = α1·Ie1 + α2·Ie1 - α2·Ic1
    It = α1·Ie1 + α2·Ie1 - α2·(α1·Ie1)
    ∴ It = (α1 - α1·α2 + α2)·Ie1
     
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  8. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    can anyone help my , Derivation of RMS voltage and current , i found it on google but it Difficult
    and i have an exam tomorrow
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Simply start with the definition of what you are trying to find.

    You have a waveform v(t). If that v(t) is applied to a resistor, it will result in power being dissipated in the resistor. What is that power on average? What DC voltage, Veff, would need to be applied to that same resistor in order to dissipate that same average power?

    That's all there is to it. Find Veff in terms of v(t) and you have the derivation of Vrms.
     
  10. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    ok, give me an example, and i will save the equation in my mind, not apply it, im not professional in math
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Then you are just digging yourself a hole because you want to just memorize things instead of understanding concepts so that you can apply them. Engineering is about problem solving and to be any good at solving engineering problems you have to be able to apply engineering concepts to problems that you have never seen before (indeed, perhaps that no one has ever seen before). The reason why the definition you found on line is confusing to you is that you previously took the short cut and just memorized a bunch of things to pass a test without bothering to understand them well enough to apply them. As a result, you don't understand the prerequisite concepts involved in the derivation. So now you want to double down and just memorize something else. That process will not end well.

    So take it one step at a time.

    If you have a voltage v(t), what is the instantaneous power, p(t), that is dissipated in a resistance R?
     
  12. saif-aljanahi

    Thread Starter New Member

    Feb 26, 2016
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    Well I'm agree words , but how to understand thing I can not understand
    , in Iraq , Education is very primitive, so do not teach everything but overlook some steps

    ok how i will found v(t)
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    v(t) is any arbitrary voltage waveform. It can be anything. Today it might be

    v(t) = 325 V · sin(2pi · 50 Hz · t)

    and tomorrow it might be

    v(t) = 14 V + 20 V · sin(2pi · 188 MHz · t + 37.2°)

    For your purposes here it is just v(t). The goal is to derive a formula that you can use to calculate the RMS voltage associated with any voltage waveform you might ever be interested in.
     
  14. Papabravo

    Expert

    Feb 24, 2006
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    There are two thing which may hep you:
    1. v(t) = i(t)*R, or the instantaneous voltage v(t) is equal to the instantaneous current i(t) times the CONSTANT resistance R. This is also known as Ohm's Law.
    2. The instantaneous power P(t) in a circuit is v(t)*i(t)
    This is all you need to know. You may not be professional in math, very few people are, but you should know arithmetic and algebra.
     
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