Fundamentals of the PN Junction

[TJonline]

TL;DR Summary

I'm posting a brief description of the operation of a PN junction (diode) and inviting anyone to give me their opinion of anything significant that I've missed or misunderstood. Thanks.

Unbiased circuit (no driving voltage source) **************

Fig 1

The N-type material has a tendency to donate electrons to the P-type (i.e. absorb holes from) across the PN junction via diffusion current exclusively within the conduction band of both materials. It does so after those electrons have made the small jump from the N-dopant's extrinsic donor energy level (black dots in Fig 1), through its extrinsic Fermi energy level (dotted line in Fig 1) and to the conduction band just above due to thermal energy/agitation.

That tendency (the Coulomb effect) is due to the relative abundance of electrons (majority carriers) in the N-type and is analogous to diffusion from high to low concentration in liquids. Before the join, the N-type is electrically neutral throughout with an equal number of electrons and protons for every atomic nuclei including its dopant nuclei.

Those nuclei contribute the donor energy level near the conduction band that accommodates its 'extra' electron that doesn't fit within its valence band that is already filled by it's own remaining four electrons and an electron from each its four silicon neighbors in the covalent bonds (shared electron pairs) of the silicon crystalline lattice to form a stable 'octet' (filled valence band).

Fig 2

The P-type material has a tendency to donate holes to the N-type (i.e. accept electrons from) across the PN junction via diffusion current exclusively within the valence band of both materials. It does so after some of its valence electrons have made the small jump up from its valence band, through its Fermi energy level (dotted line in Fig 2) and to its extrinsic acceptor energy level (white circles in Fig 2) just above due to thermal energy/agitation.

The jump up of the electron can be viewed as a jump down of a hole from the acceptor band to the valence band. That tendency is again due to the Coulomb effect in the relative abundance of holes in the P-type. Before the join, the P-type is also electrically neutral throughout including its dopant nuclei. Those nuclei contribute the acceptor energy level just above the valence band such that there is one too few electrons for each dopant atom to complete an 'octet' with the help of its silicon neighbors.

The donor band is close to the valence band in the sense that it is relatively easy for holes in the valence band to move between all nuclei in the crystalline lattice and thermal agitation is sufficient to provide that energy. I use the terms 'energy' and 'band' more or less interchangeably (represented as the vertical direction in the included Figures).

Fig 3

The two diffusion current components are additive in effect and continue until the number of excess electrons and negative ions in the P-type's conduction and valence bands (respectively) and the number of positive ions and holes in the N-type's valence and conduction bands (respectively) are great enough that any further increase in that generated charge imbalance (voltage) across the PN junction is countered by opposing drift current induced by that voltage.

At that point, the PN junction is at equilibrium with a constant, finite, intrinsic junction voltage and with (continuing) equal and opposite diffusion and drift currents. Essentially, that voltage is the result of the competition between the tendency of atoms to seek electrostatic stability (form a stable octet) acting over shorter distances and the electrical influence of the induced voltage acting over longer distances according to Ohm's law.

Fig 4 (note that light and dark grey denote N and P type materials as opposed to energy levels/bands as in the other figures included herein)

Electrons and holes in the P-type's valence and conduction bands and the N-type's valence and conduction bands can also enter and leave both materials through their respective attached conductors. The conduction and valence bands in conductors overlap such that electrons are free to move between both, their Fermi levels being somewhere within that overlap (Fig 5).

Fig 5

Forward bias driving voltage applied ********

Fig 6

The applied voltage's positive terminal causes the conductor attached to the P-type to attract electrons from (i.e. inject holes into) the P-type's valence and conduction bands, and its negative terminal causes the conductor attached to the N-type to inject more electrons into the N-type's conduction and valence bands.

If that forward bias voltage is greater than the junction voltage (i.e. is able to resupply all lost electrons from the N-type and resupply lost holes from the P-type), then the junction voltage is overcome (Fig 6) and current continuously flows through the diode from attached conductor to attached conductor.

Reverse bias driving voltage applied ********

Fig 7

The applied voltage's negative terminal causes the conductor attached to the P-type to inject electrons into (i.e. absorb holes from) its valence and conduction bands, and its positive terminal causes the conductor attached to the N-type to absorb holes from the N-type's conduction and valence bands, increasing the depletion voltage (Fig 7), and continuing until diffusion and drift currents are again equal and opposite at a new equilibrium such that the driving voltage is exactly opposed by the depletion voltage (like two batteries in series and their positive terminals in contact) and no net current flows through the circuit.

***********************
* Do the reverse bias injected charges migrate towards the depletion region and if so, by what mechanism? Or does the depletion region actually extend linearly across the entirety of the diode from attached conductor to attached conductor?

* Electrons occasionally fall from the conduction band to the valence band, resulting in the annihilation of an electron/hole pair and the emission of a photon (the basis of LEDs). Electrons occasionally leap from the valence band to the conduction band after absorption of a photon resulting in an electron/hole pair (the basis of solar cells). In a garden variety diode the rate at which those occur is insignificant. Those are relatively high energy interactions (as opposed to low energy interactions like thermal effects and voltage and current flow in wires and circuits) such that only certain photons have sufficient energy to produce them according to the quantum model of atoms in which low energy infrared photons generally don't have sufficient energy to induce anything interesting at all (except heat).

* Think of the Fermi level as the surface of a sea in a water analogy. The body of the sea and its surface are all the occupied atomic energy states at the lowest energy levels that their associated electrons can find, just as water always seeks a local depression (pool) below which surface topology dictates that water can flow no further. The Pauli Exclusion Principle dictates that no two electrons can occupy the same energy state, just as water molecules can't occupy the same space. Ripples upon that surface are analogous to low energy thermal and other effects. Quantum leaps (photonic effects) are 'unusual' events akin to a rock thrown into the surface that can sometimes induce a splash, or a water drop falling from a significant height causing another drop to pop up in response and inevitable ripples as well, since most all dynamic/energetic phenomena create or absorb some heat energy as a byproduct and are thus less than perfectly efficient.

* The Figures were borrowed from the HyperPhysics website, which I highly recommend and am sure many of you are already familiar with.

* Here is a Windows RTF version that anyone may feel free to mark up as they please. I only ask that you do so with strikethroughs and with red color to suggest what to omit or modify, leaving the original text. Thanks.

http://afafa.org/Unlinked/Diode_model_notes.rtf

 

[berkeman]

Hi TJ,

I haven't read your post in detail, but I had a couple of questions...

How would you explain the similar graphs for solar cells / photodiodes? What is different and the same about them? Why are they different?

 

[TJonline]

Hi berkeman,

I'm actually getting acquainted with the nitty gritty of silicon devices myself and my post was really a request for anyone to show me if I got anything really wrong in my understanding of the basic diode. But if you point me to a graph or two, I may be able to provide a little insight. According to Wikipedia, "The common, traditional solar cell used to generate electric solar power is a large area photodiode." The two are, of course, designed and optimized for entirely different applications - power generation and detection respectively, so I'd expect them to have vastly different characteristics. That's about all I could tell you about them without doing further research. I just haven't worked with such devices.

Ted

Hi TJ,

I haven't read your post in detail, but I had a couple of questions...

How would you explain the similar graphs for solar cells / photodiodes? What is different and the same about them? Why are they different?

 

[eq1]

Summary:: I'm posting a brief description of the operation of a PN junction (diode) and inviting anyone to give me their opinion of anything significant that I've missed or misunderstood. Thanks.

It's kind of a big wall of text so I'll admit to just briefly scanning it but it seemed you only mentioned the concentration gradients once, and that was basically in passing. And your description of reverse bias kinda makes me suspect you're missing the forest for the trees.

The 10,000ft view is: Because there is a concentration gradient we get diffusion. This is true of any system, like the liquid system you mentioned. If there was nothing to oppose the diffusion, which comes about from the carrier concentration gradient, eventually the piece of silicon would stop having a P and N sides and it would have carriers uniformly distributed. But that's not what we observe so something is opposing it, forming a barrier. That something is the depletion region e-field which comes into being because some charged diffused before the e-field came into being. The depletion region e-field is superimposed with the external e-field. So negative external e-field, barrier goes up, no charge can sneak across in the forward direction. Positive e-field, the barrier goes down and the flood gates open, unchecked charge diffusion.

The details in your post are how we quantify and add calculation to the high level concepts. They allow us to engineer the system.

I highly recommend Chenming Hu's, the inventor of the FinFET, text on this subject.

https://www.chu.berkeley.edu/modern...-integrated-circuits-chenming-calvin-hu-2010/

I think you would benefit a lot from the material in chapters 2 and 4.

 

[TJonline]

Thanks, eq1. this is the sort of feedback I'm looking for. I'm no authority and am coming up to speed myself. I'll study the source material you provided with great interest.

Specifically, I'd like to ask you or anyone if I was correct in stating that both valence and conduction bands in both N and P types pass charge to and from the conductors attached to both.

 

[TJonline]

berkeman, as a 'Mentor', are you able to allow me the privilege of continuing to edit my original post, as I'm finding that improvements are needed (in addition to your own - thanks) as I apply others' suggestions? Or is that ability not available (or desired if it tends to lead to confusion).

 

[eq1]

Specifically, I'd like to ask you or anyone if I was correct in stating that both valence and conduction bands in both N and P types pass charge to and from the conductors attached to both.

I think the answer is probably yes (because once current is flowing charge has to come and go somewhere and if not the attached conductors then where) but I’m not totally sure I understand the question.

Maybe you’re wondering if the metal / semiconductor interface also creates a barrier?

 

[TJonline]

Of course charge is passed to and from the conductors. The essence of the question is whether both valence and conduction bands contribute to that current in both N and T types. Thanks.

 

[eq1]

I’m confused because the valence and conduction bands are energy levels, check the y-axis on the band diagram, so they don’t contribute to a current. So I suspect you’re mixing terminology or I’m misunderstanding you. But if you’re asking if particles with that amount of energy or more, in the presence of an electric field, contribute to current in both types of semiconductor then the answer is yes.

 

[TJonline]

eq1, do you have a good knowledge of the physics of semiconductors and the PN junction or are you speculating? I am hoping to hear from someone who knows the physics intimately. I'm talking about electrons moving to and from those bands into the attached conductors. I'm familiar with the band diagrams of the P and N types and their junction. Just not those of the P and N types relative to the attached conductors or the precise nature of current flow between them. If you don't know, it's ok to say so. Thanks.

 

[eq1]

I'm familiar with the band diagrams of the P and N types and their junction. Just not those of the P and N types relative to the attached conductors or the precise nature of current flow between them.

The energy levels of conductors touch and are lower in magnitude on average. That’s what makes them conductors, as opposed to semiconductors or insulators. Because it takes less energy to move in a conductor the carriers are free (you'll notice what I did there) to move in the conductor once they reach it.

Things are not so simple though. You can't just touch any old conductor to a semiconductor. For example notice the barrier in this figure:
https://en.wikipedia.org/wiki/Metal...on#/media/File:Schottky_barrier_zero_bias.svg

In an actual IC, unless we actually want to make a rectifying contact, we never connect metal directly to the substrate, or well for that matter. The foundry checks will make you use polysilicon for that and the metal will not be allowed to connect to polysilicon without silicide in place.

 

[TJonline]

According to one source, if the metal's work function is greater than the semiconductor's (its Fermi level is lesser), a (Schottky) barrier is formed with an n-type and no barrier (ohmic contact) with a p-type - the floating bubble (hole) and falling water drop (electron) analogy. If the metal's work function is less than the semiconductor's (its Fermi level is greater), an ohmic contact is formed with an n-type and a barrier is formed with a p-type. The work function being the thermodynamic (or perhaps photon energy) necessary to 'boil' an electron out of the material altogether, the difference between the material's vacuum energy and its Fermi level, such as happens with the electron guns of TV and other vacuum tubes that rely on electron flow in vacuum. The barrier is the discontinuous (depletion region) difference between the lowest energy of the semiconductor's conduction band which rises sharply at the point of contact and the conductor's Fermi level energy. Do you agree?

 

[eq1]

Seems right. You can also say because there are electrons in the conduction level of the semiconductor which can move to the empty energy states above the Fermi level of the metal a positive charge on the semiconductor forms due to the excess electrons, a negative charge on the metal side, leading to a contact potential.