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Number of electrons making it through an NPN transistor

  1. Mar 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose ##n## electrons attempt to move through an ##NPN## transistor, there's a probability that some of the electrons traversing the ##P## area will recombine and and not make it to the other side. The infinitesimal probability in a region dx is given by ##dP=\frac{dx}{\lambda}##. Let ##n(x)## represent the remaining electrons at the point ##x##. Show that if ##n_0## is the initial number of electrons and the ##P## area goes from 0 to L that ##n(L)=n_0e^{-\frac{L}{\lambda}}##

    2. Relevant equations

    3. The attempt at a solution

    I would think that ##n(x)## would just be the initial number of electrons minus the integral of the probability function from 0 to x times ##n_0##, i.e. ##n(x)=n_0-n_0\int_{0}^{x}dP=n_0(1-\int_{0}^{x}\frac{dx}{\lambda})## but this can't be correct since this is decaying linearly as opposed to the exponential decay in what I'm trying to show.
     
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  3. Mar 11, 2016 #2

    ehild

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    T
    The number of recombining electrons is not constant during the whole length L. If P is the probability that an electron recombines in the region dx, and there are n(x) electrons at x, what is the number of electrons that recombine in the interval dx?
     
  4. Mar 11, 2016 #3
    Oh so the number of electrons that recombine in an interval dx will be n(x)*dP?

    Hmm: then ##n(x)=n_0-\int_{0}^{x}\frac{n(x)dx}{\lambda}=n_0-\frac{n(x)^2-n(0)^2}{2\lambda}## however this results in a quadratic. Edit: forgot lower limit

    Okay I think I got it now, I'm supposed to take the derivative of that function and then using the fundamental theorem of calculus I get:
    $$
    \frac{dn(x)}{dx}=-\frac{n(x)}{\lambda} \hspace{3mm} \mbox{Is it correct that I removed the dx from the integral?}
    $$
    which after integrating gives $$
    \ln(n(x)/n_0)=-\frac{L}{\lambda}+C
    $$
    but now I'm confused again because according to this I should have ##C=+\frac{L}{\lambda}## since at ##n(0## we have ##0=-\frac{L}{\lambda}+C##
     
    Last edited: Mar 11, 2016
  5. Mar 11, 2016 #4

    ehild

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    Yes.

    Yes, it should be ##\frac{dn(x)}{dx}=-\frac{n(x)}{\lambda}##
     
  6. Mar 11, 2016 #5
    I'm still not understanding why I'm getting ##C=\frac{L}{\lambda}##, seems like ##C## should be equal to zero?
     
  7. Mar 11, 2016 #6

    ehild

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    The integral is from x =0, so you should write ##\ln(n(x)/n_0)=-\frac{x-0}{\lambda}##
     
  8. Mar 11, 2016 #7
    Ah yea I thought it would be a dumb mistake like that, forgot since it's a definite integral I don't need to add a constant lol.

    For the next part of the question I'm supposed to now solve it in terms of time and an initial speed v which I managed to do using ##dP=\frac{dx}{\lambda}=\frac{dx}{dt}\frac{dt}{\lambda}=v\frac{dt}{\lambda}##, this gave where ##t_f## is the time to traverse: ##n(t_f)=n_0e^{\frac{-vt_f}{\lambda}}## but now I'm supposed to assume that ##t_f<<\frac{\lambda}{v}## so it is clear I should taylor expand about 0 after substituting ##a=\frac{\lambda}{v}## to obtain ##n(t_f)=n_0e^{-\frac{t_f}{a}}## so ##n(t_f)\approxeq n_0(1-\frac{t_f}{a})## but they want me to use this to show that this gives ##\frac{i_b}{i_c}=\frac{t_f}{a}## with ##i_b## being the base current so I would assume that ##i_c## is the circuit current. This implies ##\beta=\frac{i_c}{i_b}## is a large number.
     
  9. Mar 11, 2016 #8

    ehild

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    Ic is the collector current, and yes, β=Ic/Ie is a large number usually (in the range of 100) . Can you copy the original text?
     
  10. Mar 11, 2016 #9
    If ##0<x<L## is small then ##t_f<<\frac{\lambda}{v}=a##, Prove that this results in ##\frac{i_b}{i_c}=\frac{t_f}{a}## if ##i_b## is defined as the base current. The (DC) current gain ##\beta=\frac{i_c}{i_b}## is large, this is one characteristic of an NPN transistor.
     
  11. Mar 11, 2016 #10

    ehild

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    If dn(0)/dt is rate electrons arriving at the base and dn(L)/dt is the rate they live it, dn(0)/dt =IE and dn(L)/dt = Ic. What is the base current then? :)
    The electrons recombining in the base make the base negative. That extra negative charge flows out the base as the base current.
     
  12. Mar 11, 2016 #11
    Ok so ##i_E=\frac{dn(0)}{dt}=-\frac{n_0}{a}## and ##i_c=\frac{dn(t_f)}{dt}=\frac{-n_0(1-t_f/a)}{a}## then ##i_b=i_E-i_C=-\frac{t_f}{a^2}## which doesn't give the correct result when I plug in ##\frac{i_B}{i_C}##

    Ok I managed to get it I was making an algebra mistake. Ill add it when I get home since latex isn't that fun on mobile.
     
    Last edited: Mar 11, 2016
  13. Mar 11, 2016 #12
    ##i_E=\frac{dn(0)}{dt}=-\frac{n_0}{a}## and ##i_c=\frac{dn(t_f}{dt}=-\frac{n_0(1-\frac{t_f}{a})}{a}##, since ##i_b=i_E-i_c## we have $$i_b=-\frac{n_0}{a}+\frac{n_0(1-\frac{t_f}{a})}{a}=-\frac{n_0t_f}{a^2}$$

    Now to determine ##i_b/i_c## we have
    $$
    \frac{i_b}{i_c}=\frac{-\frac{n_0t_f}{a^2}}{\frac{-n_0(1-\frac{t_f}{a})}{a}}=\frac{t_f}{a(1-\frac{t_f}{a})}=\frac{t_f}{a}(1-\frac{t_f}{a})^{-1}\approxeq \frac{t_f}{a}(1+\frac{t_f}{a})=\frac{t_f}{a}+\frac{t_f^2}{a^2}\approxeq\frac{t_f}{a}
    $$
    (since ##t_f<<a## we have that ##\frac{t_f^2}{a^2}\approxeq 0##) Had some fun solving this, reminded me of the algebra we did when we solved the free falling particle question which also involved quite a few approximations.
     
  14. Mar 12, 2016 #13

    ehild

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    Congratulation!
    It was not clear what was n. If it is the concentration of the free electrons at a position x in the Base, the current is this concentration multiplied by the speed of the electrons, with their charge and the cross-section area of the base. Anyway, the current is proportional to n, the base current is proportional to n(0) - n(L)=n(0)(1-e-L/λ) , and Ic is proportional to n(0)e-L/λ. ##\frac{I_B}{I_C}=\frac{1-e^{-L/λ}}{e^{-L/λ}}=e^{L/λ }-1##, which can be approximated as L/λ.
     
  15. Mar 12, 2016 #14
    Ohhhh, I see that your method is a lot more efficient! We can just use the fact that ##n(L)=n(t_f)\Longrightarrow n_0e^{-\frac{L}{\lambda}}=n_0e^{-\frac{t_f}{a}}\Longrightarrow \frac{L}{\lambda}=\frac{t_f}{a}## and then substituting into the approximation you made above we get the desired result.
     
  16. Mar 12, 2016 #15

    ehild

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  17. Mar 12, 2016 #16

    davenn

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    that should be In a β = Ic / Ib or more fully β = Ice / Ibe


    Dave
     
  18. Mar 12, 2016 #17

    ehild

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    Yes, it was a typo. β = Ic / Ib is a large number.
     
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