- #1

- 22

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This is what I am trying to end up with:

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

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- Thread starter 03myersd
- Start date

- #1

- 22

- 0

This is what I am trying to end up with:

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

- #2

- 3

- 1

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

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

Last edited:

- #3

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- #4

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Thanks

- #5

- 3

- 1

Here's the derivation; I hope it's not late

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

[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

- 1,177

- 86

Could you clarify? Thanks.

Claude

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