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Derivation of voltage for vacuum diode space charge region

  1. Feb 18, 2009 #1
    I have found the 1D differential equation relating voltage and position for a vacuum diode in the space charge region, which is

    [tex] \frac{d^2V}{dx^2} = constant * V^{-1/2} [/tex]

    and I know the solution to be

    [tex] V(x) = V_0 \left(\frac{x}{d}\right)^{4/3} [/tex]

    which is found by multiplying both sides by [itex] V' = \frac{dV}{dx} [/itex] and then integrating the following expression with homogeneous boundary conditions:

    [tex] \int V' dV' = constant*\int V^{-1/2} dV [/tex]

    What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

    [tex] V(x) = V_0 \left(\frac{x}{d}\right)^{4/5} [/tex]

    The two answers are different so obviously it's a mistake to separate variables, but for the life of me I can't tell where it is. Could anyone enlighten me? Thanks.
     
  2. jcsd
  3. Feb 18, 2009 #2

    gabbagabbahey

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    I'm not exactly sure what you mean by "separation of variables" in this case. To apply separation of variables, [itex]V[/itex] would have to be a function of more than one variable; but in this case, it is a function of only [itex]x[/itex]. SOV is typically used to turn PDEs into ODEs, not to solve an ODE.

    I can only guess that what you meant to say was that you tried treating it as a "separable differential equation"; but you have a second order DE....how exactly does one integrate [tex]V^{1/2} d^2 V[/tex] and [tex]dx^2[/tex]???
     
  4. Feb 19, 2009 #3

    lanedance

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    yeah wasn't totally sure what was meant, however a trick to reduce it to 2 first order DEs as follows:

    First write
    P = V' = [tex]\frac{dV}{dx}[/tex]
    then
    P' = V" = [tex]\frac{dp}{dx}[/tex] = [tex]\frac{dp}{dV}[/tex].[tex]\frac{dV}{dx}[/tex] = p.[tex]\frac{dp}{dV}[/tex]

    otherwise just assume solutions V~xy diff & equate coeffs
     
    Last edited: Feb 19, 2009
  5. Feb 19, 2009 #4
    Thanks for your replies. My mistake was that I though I could solve [itex] V^{1/2} d^2 V = dx^2 [/itex], which stemmed from an incomplete understanding of the method of separation of variables (I took the fact that you could solve something like [itex] d^2x/dt^2= F/m [/itex] using double integration, and incorrectly applied that to the differential equation above). I reviewed my notes, and now see why that isn't possible.
     
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