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I am trying to build a simple model of a pn junction in order to show high school students how it works. I am not looking for something precise, so I use really simple discretization and I discard units. I get satisfying results for equilibrium, but when bias is applied, I reach an impasse.

My diode model is made of 2 slabs of p doped semiconductor (blue) and 2 slabs of n doped semiconductor (pink). Density of acceptors in the p region (NA) is equal to density of donors (ND) in the n region. The figure below shows the conditions at t = 0, when the p-region and the n-region come in contact. The total charge Q of each slab is 0 initially.

enter image description here

I suppose complete symmetry for holes and electrons (same diffusion constant D and same mobility $\mu$). This implies that n1 = p4, n2 = p3, n3 = p2 and n4 = p1 at all subsequent times, which allows me to concentrate on holes only.

The equilibrium condition is that the total current of holes (diffusion + drift) must be equal to zero at each interface separating adjacent slabs.

The diffusion current of holes (IhD) at an interface is taken to be equal to the density of holes in the left adjacent slab minus the density of holes in right adjacent slab (taking D = 1). For example, the initial hole diffusion current in the interface between slabs #2 and #3 would be 10-0 = 10.

The drift current of holes (IhE) is the product of the density of holes, the mobility of holes and the electric field at the interface. Taking the slabs to be thin and very large, the electric field at an interface must then be equal to the total charge of slabs to the left of the interface minus the total charge of slabs to the right of the interface.

There is an equation for the total current for each of the 3 interfaces and an equation for conservation of charge (p1 + p2 + p3 + p4 = 20), since recombination is neglected . This allows me to find the values of the four unknowns p1, p2, p3 and p4. I obtain the following results at equilibrium (for D = 1 and $\mu$ = 0.1):

enter image description here

We can see that the total hole current at interfaces is zero. For example, for the interface between slabs #1 and #2:

E12 = -1.52 – (-4.56 + 4.56 +1.52) = -3.04

Diffusion hole current: IhD = 9.77-7.19 = 2.58

Drift hole current : IhE = paverage$\times$$\mu$$\times$E12 = 0.5$\times$(9.77+7.19)$\times$0.1$\times$-3.04 = -2.58

Therefore Ih = IhD + IhE = 2.58 – 2.58 = 0.

We can see that an electric field exists inside the diode at equilibrium. The built-in voltage Vbi can be calculated from the values of the electric field: Vbi = -(-3.04 -12.16 -3.04) = 18.24. We can also have a glimpse at a depletion region in slabs #2 and #3, which are farther from neutrality that slabs #1 and #4.

So far, so good. Now, let’s try to extend this simple model to see what happens when forward bias is applied to the junction by metal electrodes at both extremities.

For example, let’s suppose that forward bias causes a steady state (hole) current Ih= 1 to the right. I find that the following density configuration is compatible with such a current:

enter image description here

Of course, in order to have steady state there must be a hole current Ih = 1 at the interface between the left/right electrode and slab #1/#4. I suppose infinite conductivity for the electrodes (ohmic contact), so no electric field is needed to produce this total current (E01 = E45 = 0). Therefore the voltage across the diode must be V = -(-1.5 - 10.02 - 1.5) = 13.02.

This is where the following questions arise:

  1. Is there a relation between this steady state voltage (13.02), the equilibrium built-in voltage (18.24) and the applied voltage in forward bias?

  2. Is the hole-electron symmetry assumption (n1 = p4, n2 = p3, n3 = p2 and n4 = p1) also valid for steady state?

  3. Will hole-electron generation/recombination have to be considered to be able to see the current asymmetry in forward and reverse bias?

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  • $\begingroup$ This looks to be a bad model. For example, what you call a "depletion" region isnt depleted. This seems more complicated than the normal intro to diodes. $\endgroup$
    – Matt
    Commented Aug 23, 2021 at 11:02
  • $\begingroup$ @matt I know there are more direct ways to explain diodes. My main goal with the model is to give students a chance to apply basic notions that they learned previously (electric field by a charged plate, drift current, holes, steady state, etc.). I thought it could be rewarding to be able to understanding the pn junction in passing. $\endgroup$
    – courno
    Commented Aug 23, 2021 at 22:20

1 Answer 1

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I'm having some trouble following your model. I would suggest you stick to the normal introduction to diodes most textbooks have. I think they are more easily understood than this.

To address your questions:

Is there a relation between this steady state voltage (13.02), the equilibrium built-in voltage (18.24) and the applied voltage in forward bias?

No, you can apply whatever bias you want. Your power supply doesnt know or care what the built in voltage is.

Is the hole-electron symmetry assumption (n1 = p4, n2 = p3, n3 = p2 and n4 = p1) also valid for steady state?

With all your assumptions about symmetry between electrons and holes then this too should be valid.

Will hole-electron generation/recombination have to be considered to be able to see the current asymmetry in forward and reverse bias?

It shouldnt. I cant tell if your model requires it or not. But if your model requires it then you have a bad model.

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  • $\begingroup$ Of course I can apply whatever bias I want, but different values of applied voltage will lead to different values for current. In my example the current is +1, and I am looking for the applied voltage that is responsible for it (18.24-13.02=5.02?). $\endgroup$
    – courno
    Commented Aug 23, 2021 at 22:10

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