Simple Derivation Of Diode Equation

In summary: E. Shannon, an Irish-American mathematician and physicist, is best known for his work on communication theory, information theory, and digital signal processing. In this article, Shannon discusses the mathematical foundations of digital communication.In summary, Shannon discusses the mathematical foundations of digital communication, and explains how the diode equation can be used to calculate the current through a diode. He also discusses how to calculate the cross-section area for a diode, and how this information can be used to calculate the voltage across a diode.
  • #1
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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
 
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  • #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
 
Last edited:
  • #3
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
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
03myersd said:
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.
 
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  • #6
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
 

1. What is the diode equation?

The diode equation is a mathematical expression that describes the current-voltage relationship of a diode, a semiconductor device that allows current to flow in only one direction. It is also known as the Shockley diode equation, named after physicist William Shockley who first derived it.

2. How is the diode equation derived?

The diode equation is derived using basic principles of semiconductor physics, specifically the diode's current-voltage characteristics and the Shockley diode equation. It involves solving for the current that flows through a diode at a given voltage, taking into account factors such as the diode's voltage drop and temperature.

3. What is the significance of the diode equation?

The diode equation is important in understanding the behavior and performance of diodes in electronic circuits. It helps in predicting the amount of current that will flow through a diode at a certain voltage and temperature, which is crucial in designing and analyzing electronic systems.

4. Can the diode equation be simplified?

Yes, the diode equation can be simplified in certain cases, such as when the diode is operating in a specific range of voltages or temperatures. For example, the Shockley diode equation can be simplified to the ideal diode equation in ideal conditions, where the diode behaves like a perfect conductor or insulator.

5. How accurate is the diode equation?

The accuracy of the diode equation depends on various factors, such as the parameters used in its derivation, the type of diode, and the conditions under which the diode is operating. In general, the diode equation provides a good approximation of the behavior of diodes, but it may not be completely accurate in all cases.

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