Simple Derivation Of Diode Equation

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I am looking for the simplest possible derivation of the diode equation. I need it to reference to it in my advanced higher project. Basically I have to dissect it. I unfortunately don't know how to derive it and I can't find one simple enough to use.

This is what I am trying to end up with:

http://www.freeimagehosting.net/uploads/0d2e6cbc92.jpg
 

Answers and Replies

  • #2
I don't think it's possible to get to that equation through a simple derivation. I remember that the continuity equation must be used (in one dimension, for simplicity) and then incorporate boundary conditions and these definitely are not simple. so you might need to reference the whole derivation, though you can skip obvious steps.

Anyway, I got two pages of simple derivation. I can send these later to you as I haven't got a scanner here?

Good luck
 
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  • #3
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Well ideally the simplest way possible. Or even just an explanation of the steps. I am so glad that I don't have to understand it at this point. :smile:
 
  • #4
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Please anyone? The report is due in at 3.30 today and I would really like to have the analysis of the derivation included in it. Its not essential but it would essentially round the whole thing off nicely.

Thanks
 
  • #5
Please
anyone? The report is due in at 3.30 today and I would really like to
have the analysis of the derivation included in it. Its not essential
but it would essentially round the whole thing off nicely.
Thanks
Here's the derivation; I hope it's not late
[tex]{L_p}[/tex] = diffusion length of holes ([tex]{cm}[/tex])
[tex]{D_p}[/tex] = diffusion constant ([tex]{cm^2}/s[/tex])
[tex]{L_p}[/tex] = average carrier life time ([tex]s[/tex])

Firstly, we find [tex]\Delta{p}[/tex]

where [tex]\Delta{p}[/tex] is the minority carrier concentration at the
edge of the depletion region (DP)

we know that the built-in voltage is given by

[tex]{V_{bi}} = \frac{kT}{q}\ln({\frac{{N_A}{N_D}}{n_{i^2}}})[/tex]

Applying the law of mass action [tex]{n_{i^2}} =

{n_{no}}\times{p_{no}}[/tex]

we get

[tex]{V_{bi}} = \frac{kT}{q}\ln({\frac{{n_{no}}{p_{no}}}{n_{i^2}}})[/tex]

Rearranging

[tex]p_{po} = p_{no}\exp(\frac{qV_{bi}}{kT})[/tex] => eqn.1

For non-equilibrium situation, i.e. when there's forward bias voltage

[tex]{V_f}[/tex]

[tex]p_{p(0)} = p_{n(0)}\exp(\frac{q(V_{bi}-{V_f})}{kT})[/tex] => eqn.2

or
[tex]p_{po} = p_{no}\exp(\frac{qV}{kT})[/tex]

where [tex] V = {V_{bi}}-V_f[/tex]

assuming low injection level, i.e. [tex]{p_p}\approx{p_{po}}[/tex]

eqn.1/eqn.2 by doing this we get [tex]\Delta{p}[/tex]

therefore

[tex]\Delta{p} = p_{no}\exp(\frac{qV}{kT}-1)[/tex]

Using the continuity equation, we get an expression for the current density

[tex]J_{p(x)} = q\frac{D_p}{L_p}\delta_{p(x)}[/tex]

since [tex]\Delta{p} = \delta_{p(x=0)} [/tex]

so [tex]\delta_{p(x=0)}= p_{no}\exp(\frac{qV}{kT}-1)[/tex]

[tex]J_{p(x)} = q\frac{{D_p}{p_{no}}}{L_p}\exp(\frac{qV}{kT}-1)[/tex]

Doing the same with electrons, we get a similar expression, hence

[tex]J_{total} = J_p + J_n = J_s\exp(\frac{qV}{kT}-1)[/tex]

This is the standard way of expressing the diode equation. However, if we multiply the above expression by the cross-section area, we get the current I.
 
  • #6
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August_2007, after the statement "Applying the law of mass action we get ...", the expression for Vbi computes to zero. Since ni^2 = nn0*pn0, then ni^2/(nn0*pn0) must equal 1. So, the ln (1) is zero. So Vbi becomes zero.

Could you clarify? Thanks.

Claude
 

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