# How do transistors work?

1. Sep 1, 2006

### gulsen

Yup.
I've been looking around about it for a while, I delved into electronics books, they mention things like hole/electron flow, energy bands, but I haven't understood how transistors actually work. I've heard that diodes, transistors and other semiconductor materials involve tunneling, so about diodes, I had this conclusion:

when you apply an external electric field to the ends of diodes, energy levels are altered, so does the tunneling probability. (of course, I haven't tried to compute it actually, I don't know how Schockley wrote that exponential thing). I hope this's something close to reality.

Well, I'm a physics undergrad who knows some quantum mechanics and tunneling. So, can someone explain what's going on within BJT and FET in terms of quantum mechnics? I'm aching for a real explanation!

Thanks!

2. Sep 1, 2006

### data1217

Diode is a PN junction
BJT is a PNP or NPN type junctions

Their working principle is similar. They are based on the different concentration of charge carriers (doping).

3. Sep 1, 2006

### Gokul43201

Staff Emeritus
Sadly,there is no tunneling in the regular PN junction, or a BJT, or in the Shockley Equation. The exponential term in the Shockley equation comes from the equilibrium charge carrier concentrations (which, is related to the band structure and temperature through the Fermi Dirac distribution ~ $[1 + exp(\Detal E/kT)]^{-1}$).

Now, there are things called Tunnel Devices (tunnel diodes, transistors, etc.) which are the limiting cases of their regular equivalent devices under conditions of very high doping levels. It is under these conditions that you have a large number of unoccupied carrier states in both bands (because the Fermi level no longer lies in the bandgap), that permit tunneling. The tunnel current then depends on the shape of the potential barrier and is calculated by making a suitable approximation.

But this is unrelated to the Shockley equation, which applies to regular devices, where the currents are diffusive (i.e, determined by a Fick's Law relation).

You'll find derivations in any of the standard device textbooks (e.g, Streetman, Bhattacharya or Sze).

4. Sep 1, 2006

### rbj

oh boy, semiconductor device physics! (warning, i'm not super good at it, but most electrical engineers had to learn some of it. but i take no responsibility for inaccuracies.) here is a little hand-waving explanation:

first a little p-chem (from an EE's POV): for some physical reason that i have long forgotten (having to do with the quantum mechanics of the hydrogen atom and then hand-waving some results of that analysis to bigger atoms), the electrons in atoms exist in somewhat stable "shells" of discrete energy levels. kinda like a staircase as opposed to a ramp. the nature of these discrete energy shells, what potential energy electrons possess in each shell and how many electrons each shell will hold essentially determine how these atoms will react chemically. i can only recall the first (or "bottom") two or three shells, the bottom shell holds at most 2 electrons, the next two holds at most 8 electrons each. given the number of electrons an atom has (it's "atomic number"), the bottom shell fills up first, then the next shell, and so on. this is what determines the form of the Periodic Table of Elements.

an atom that has these shells exactly filled is a very "happy" or "satisfied" atom and cannot be expected to react chemically with anything under normal circumstances. these are the inert gasses like Helium (He) or Neon (Ne) or Argon (Ar) at the extreme right of the Periodic Table. these are the only gasses or substances (that i know of) that are "monatomic", that is one atom per molecule. these atoms need no other atoms to grab electrons to fill an empty "hole" or to dump excess electrons on.

contrast this with Sodium (Na) and Clorine (Cl). Sodium, at the extreme left of the periodic table, has a complete shell with one extra electron and Clorine, at the extreme right (except for the inert gasses), has almost a complete shell but is missing one electron to complete the shell. call that a "hole" in the Cl atom. so suppose these two atoms happen to meet in a singles bar and that Na atom would like very much to plug that Cl atom's hole with his throbbing electron, and the Cl atom wouldn't mind it a bit. so they get together, the Na atom's extra electron jump over to the Cl shell, filling it, and spends nearly all of its time orbiting the Cl atom. fine, but these atoms were electrically neutral before doing the dirty deed (because the number of protons were equal to the number of electrons) so when that electron jumps over from Na to Cl, that leaves the Na atom positively charged (one more proton than electrons) and the Cl atom negatively charged and those two atoms are gonna stick together really tight because of electrostatic forces. think of that as the psychic bonding that happens to people (and many animals) after doing the horizontal bop.

(sorry for the anthropomorphizing, but it's the best way for me to imagine what is going on.)

okay, so you got the inert gasses (which are comparable to the celibates or those with removed hormones: "Who needs sex! I sure don't!") at the extreme right of the Periodic Table. then there are the elements on the extreme left and extreme right (just to the left of the inert gasses) of the Periodic Table who you might consider to be the young, nubile, (and horny) heterosexuals. think of the diatomic gasses ($\mbox{Cl}_2$ or $\mbox{O}_2$) as hot b1tches locked up away from the men and resigned to lesbian relationships (but you better watch out if they get loose) in the meantime. $\mbox{N}_2$ isn't so bad but if $\mbox{Cl}_2$ gets out, you better turn around and run!

now another interesting group in this microcosm are the atoms, in the Group IV cloumn of the Periodic Table, with outside shell exactly half filled. that is 4 extra electrons (or is it 4 missing electrons??) in the outside shell. this would be Carbon (C) or Silicon (Si) or Germanium (Ge). they are the hermaphrodites. they don't know if they be the girls or if they be the boys, but they ain't celibate. they just kinda hook up the way you might imagine two hermaphrodites hooking up. the 4 "extra" electrons of one Si atom sorta fills the need of 4 "missing" electrons in the adjacent Si atom. that is what a pure, undoped, semiconductor is and, except for the occasional electron that thermal energy kicks up out of their satisfied shells, they're not much of a conductor of electricity as such (pure silicon, say).

so now what happens is that your local neighborhood semiconductor factory infuses into this lattice of hermaphrodite atoms, some slightly less hermaphrodite atoms such as Boron (B) or Aluminum (Al) or Gallium (Ga) with 3 electrons in the outer shell (Group III in the periodic table) or Phosphorus (P) or Arsenic (As) with 5 electrons in the outer shell (Group V in the periodic table but better thought of as missing 3 electrons in the outer shell). the material doped with B or Ga (called "P" type silicon) would be missing an electron here or there (wherever there is the occasional B or Ga atom in the Si lattice) and that missing electron would be called a "hole". the material doped with P or As (called "N" type silicon) would have an extra electron here or there (wherever there is the occasional P or As atom in the Si lattice) and that extra electron would be called an "electron". both holes and electrons act as particles. an electron acts as a particle of positive mass and negative charge. a hole acts as a particle of positive mass and positive charge. a hole and electron combine to be a satisfied shell.

now, by themselves, both the P type silicon and the N type silicon are electrically neutral (same number of protons as electrons, even if there are some extra electrons or missing electrons in the outer shells), but what do you think might happen if you stick some P type silicon next to some N type silicon with some suitable glue? (this is a semiconductor diode.) at least around the "PN junction", the boundary or contact surface between the two, some of those extra electrons will really want to jump over and fill those holes on the other side. and some of those holes will really wanna jump over and satisfy those free electrons on their other side. when they do that, there will be a small voltage (called the "contact potential") because the P doped silicon will be more positively charged and the N doped silicon will be more negatively charged.

there is an equilibrium of forces because the positively charged P type material is trying to draw those electrons back, but the occasional empty shell "hole" is beckoning it to stay. or you can say the negatively charged N type material is trying to draw those holes back, but the occasional excess shell electron is beckoning it to stay. so if we did nothing, that situation, with an electrostatic charge pulling charges in one direction being opposed by the quantum mechanical physical chemsitry pulling the charges the opposite direction, could remain forever, if there were no other forces brought in from the outside. the number of electron/hole charges that have jumped over remains roughly constant. but it is a very precarious equilibrium.

so now we add another force, suppose we hook up a battery or some electrical source with the "+" terminal connected to the P silicon and the "-" terminal connected to the N silicon. then electrons in the N silicon will be given even more energy to jump over and combine with a hole in the P silicon (connected to the "+" terminal where that electron will eventually drain to). this resulting movement of charge is current and that diode is acting like a conductor of electrical current. this is called a "forward biased junction". those electrons and holes need time to find each other and recombine. the higher hole and electron density (which is what you have immediately beside the PN junction), the more rapid the rate of recombination. this means that there on the P side of the junction, there is a higher free electron density right beside the junction and, as you move away from the junction (toward the P terminal), more electrons have recombined with holes, fewer are left and the rate of recombination is proportionately less. if you set this up into a (first-order) differential equation, you will see that the free electron density declines in an exponential fashion $e^{-\alpha x}$ as you move away from the junction. and the same is what happens with the positive holes on the N side of the junction. keep this in mind.

now, suppose you hook the battery up the other way. those extra electrons will immediately drain into the "+" terminal (and the "holes" will drain into the "-" terminal) and what will be left is silicon with satisfied shells. no extra electrons or holes in the shells. like a piece of silicon rock (doesn't conduct very well). it won't be electrically neutral since the N type silicon will have extra protons in each Phosphorus atom and the P type material will have one missing proton in each Al atom, but that's what you would expect with the N connected to "+" and the P connected to "-". there will be no conduction of current that way. this is called a "reverse biased junction".

this is why they are called "semiconductors" sometimes they conduct electricity and sometimes they don't. it depends on what you're doing to them.

now that's diodes which are two-terminal devices that act like a one-way valve for current.

for bipolar transistors (BJT) they construct a sandwich of NPN material where the middle P section (called the "base") is very thin. when the transistor is used as an amplifier, the collector-base junction is reverse biased by a significant voltage (the voltage on the the collector N material is much more positive than the voltage of the base P material). so you might expect that no current flows (but wait and see). the base-emitter junction is forward biased by a teeny voltage (since it's forward bias, a large voltage would result in an enormous current). since it is forward biased, some amount of positive current flows from the base (P) to the emitter (N).

now that would be the end of the story (current flows across the forward biased base-emitter junction and no current flows in the reverse biased collector-base junction) except for one salient fact: the base wafer in this NPN sandwich is very thin. in the (positive) current that flows from base to emitter, that is equivalent to electrons (negative particles) flowing in the opposite direction (from emitter to base). but, because the base is thin, not all of those free electrons (and you want it so that very few do) will recombine before they drift to the other side of the base where the collector-base junction is. but now that junction is not devoid of charge (as would the reverse biased diode) because of those free electrons and they will see that sexy high positive voltage connected to the collector and will easily jump the junction.

so, if the base layer is thin enough, you can imagine that 1% of the electron current crossing the emitter-base junction will recombine and 99% will go to the high voltage collector. that is a current gain of 99. as you slightly increase the base-emitter voltage, the current will increase, but the fraction of that current that goes to the collector will increase by the same proportion. this is then fundamentally, a current amplifier or a current-controlled current source.

the depletion-FET is a different story but easily conceptualized. the source and the drain are terminals connected to a single chunk of N type silicon (an "N-channel FET") there is a reverse-biased junction from the gate (P material) to this N material channel. the more that this junction is reverse-biased, the more that the channel is depleted of free charges (electrons in N material) and the more that the channel looks like a piece of non-conductive rock and less like a conducting piece of doped-up silicon. so you use the reverse biased gate-to-channel junction to pinch off current flow through the channel. that's about all's i can say about the FET.

PNP BJTs or P-channel FETs, just reverse all of the voltages and exchange every occurance of the word "electron" with "hole" and vise versa. then do the same song-and-dance.

Last edited: Sep 1, 2006
5. Sep 1, 2006

### Gokul43201

Staff Emeritus
I hope you don't mind if I fix a couple of errors.

Errata:

The third shell actually holds up to 18 electrons. In general, the N'th shell can hold 2N^2 electrons.

To make this clear, a higher shell does not begin filling only after the previous shell is fully filled. That is only observed in the first 20 or so elements. In bigger atoms, electron-electron interactions make the energy dependent on not just the principal quantum number, n (or the shell number), but also the angular momentum quantum number, l (or the subshell number). Specifically, you will find in larger atoms (Z > 20 or thereabouts) that the more compact subshells of higher shells fill before the less compact subshells of lower shells.

The requirement for "happiness" is not that shells be completely filled, but that the valence shell have 8 electrons (the Octet Rule), or 2 in the case of helium-like species (Duet Rule). Fully filled outer shells are impossible to have after the first few elements for the reason I've described in the previous bit.

As per the above correction, we'd say Cl is an electron short of an Octet.

So, that's what you mean by "physical chemistry"!!

The conservative, right-wingers?

See, it's exactly half-way to an Octet, not a full shell.

If this were true, they'd form diatomic molecules. Closer to the truth is that it takes 4 more atoms to satisfy a single silicon atom. Each of these 4 atoms shares one of its 4 electrons.

These electrons and holes are localized at the locations of the donor or acceptor atoms only at very low temperatures.

Ooooh!!

The mass of the particles actually depends on the curvature of the bands and can be either positive or negative.

The thing opposing the electrostatic attraction is called a chemical potential - and it is defined in classical physics; no QM needed.

This is as far as I'm going - there's no more chemistry to entertain me!

Last edited: Sep 1, 2006
6. Sep 2, 2006

### rbj

not at all. "shells" vs. "octets", dunno the difference (this is dreged up after 30 years), but the idea is the same. if the octet is filled, the atom is pretty satisfied and will not react with others nor will have free electrons as charge carriers to conduct electrical current.

given a particular electric field, the holes will always move in the opposite direction than the electrons in the same field in a semiconductor. always. if they have negative mass then they also have negative charge, but as it was taught to me 30 years ago, it was just as well to say that the holes (which really aren't particles at all) behave as if they have positive mass and positive charge.

i dunno how one uses classical physics to determine the forces that draw electrons against the prevalent electric field. where does such a classical force come from?

7. Sep 2, 2006

### Gokul43201

Staff Emeritus
Yes, it doesn't change the big picture.

True, if you are refering to the effective electric field, rather than the applied electric field. Remember that electrons and holes also see the fields from the periodic lattice.

From the electrostatic interactions between the charged particles, which are too hard to model microscopically, and are hence treated macroscopically. It's the reason that particles diffuse outward from a source - even in the absence of an applied field. The force that causes the diffusion is the interparticle interaction.

Last edited: Sep 2, 2006
8. Sep 2, 2006

### rbj

whatever, but they always move in the opposite direction that the electrons do when an electrostatic field is applied. the only way for that to happen if they have negative mass is for them to also have negative charge.

now maybe it's only the meatball engineering-oriented semiconductor physics class we had 30 years ago, but there it was stated that the behavior (that is the motion under an electric field) that the holes have is identical if it is a negative mass with negative charge or positive mass with positive charge (it makes sense to me). other than the motion and the recombination, i see no salient net behavior of either the holes or free electrons in a semiconductor. if the salient behavior is the same for negative mass with negative charge or positive mass with positive charge, then what's the big deal? it makes much more sense to understand holes to be particles of opposite charge as the electrons, and then in that case, they have to be viewed as having positive mass. since holes really aren't particles at all, it's a conceptual vehicle to treat them as particles of some mass. i see no advantage to conceptualizing them as negatively charged (same as electrons).

9. Sep 2, 2006

### Gokul43201

Staff Emeritus
This is true in a vacuum, but not in a solid. Not only is it possible for an electron and a hole (in a crystal) to accelerate in the same direction, it is possible for two different electrons (with different momenta) to accelerate in opposite directions at the same instant. When averaged over sufficiently long times, however, an e and h will drift in opposite directions (I know not of any material where this last statement is untrue).

Yes, this is true of the average behavior (not the instantaneous behavior).

And for your subsequent discussion, it is the average behavior that's important, and there's nothing wrong in treating them as positively charged particles with a positive average effective mass. I was being a little too pedantic. My bad.

Last edited: Sep 2, 2006
10. Sep 2, 2006

### leright

wrong. With this contact potential, the P-type material is negatively charged and the N-type material is positively charged. Initially, both materials are electrically neutral, but then the charge gradient causes electrons in the N-type material to diffuse to the P-type material (as you said), but the "lack" of electrons on the N side (4 valence electrons instead of 5 on the phosphorus) will produce a positive charge, and the "excess" of electrons on the on the P side (4 valence electrons instead of 3 on the Boron) will produce a negative charge. So, the built in potential is positive at the N side and a negative at the P side.

As this charge seperation occurs, and electrons move over to the holes and this positive charge is produced at the N side and negative charge is produced at the P side an electric field is produced that points from the N side to the P-side, which means it wants to drive electrons in the opposite direction of the electron density gradient. So, eventually, as electons are moving from N to P an E-field is build up so that this charge flow is inhibited. The E-field that builds up for the charge flow to reach equilibrium is known as the built-in potential, and is generally around .65 to .7 volts.

Now, if we apply a voltage to the diode using a battery of some kind, for instance, and the positive terminal is at the P side of the diode and the negative at the N side, this built in voltage is reduced and the equilibrium condition is disturbed. Current is once again allowed to flow through the diode.

If we apply a voltage to the diode using a battery and the negative terminal is at the P side and the positive terminal is at the N side, the built in voltage is not decreased, but it is increased, which means no current will flow. However, is the external voltage gets high enough, current will in fact flow.

So, diodes allow current to flow when a voltage is applied positive at the P-side, but diodes don't allow current to flow when voltage is applied negative at the p-side. Also, look at the graph of the diode equation. Notice that the graph looks a lot like a battery when about 0.7V is applied across it. This is how complicated non-linear diode circuit problems can be turned into simple linear circuits. We can model the diode a battery!

I hope this helps. This may allow you to begin trying to understand transistors.

11. Sep 2, 2006

### leright

Gokul, yes, the 3rd shell holds 18 electrons, but there is always a valence of 8. when the 3rd energy level fills, the 3s and 3p orbitals fill, and then the 3d orbitals don't fill until the 4s orbital fills.

Regardless of the number of electrons the energy levels hold, the valence is still always 8

12. Sep 2, 2006

### leright

Ah, you already pointed out my above point.

13. Sep 2, 2006

### leright

BTW, is there an explanation as to why atoms are "happy" when they have valences of 8? I have always wondered this.

14. Sep 2, 2006

### rbj

i think you're correct. i switched the two around. N-type material has an extra free electron flying around when it's electrically neutral. P-type has a hole instead. when the free electrons leave the N-type material for the P, what remains are extra protons and the material is positively charged. and the P material gets those extra electrons and is negatively charged. this is what i meant but i was careless in my accounting.

one other thing, i believe that practical turn-on voltage for a silicon PN junction of 0.65v to 0.7v is significantly larger than the theoretical contact potential. it comes from a more emperically determined "corner" in the V-I characteristics of a practical diode with some series resistance. imagine an ideal diode in series with a small resistance. you will see two linear segments of the V-I curve that, empirically meet somewhere around 0.65 v. i don't think you can derive that voltage from the theory of semiconductor physics as you can with the contact potential.

Last edited: Sep 2, 2006
15. Sep 2, 2006

### rbj

i have no idea what the meaning of a "hole" in vacuum is. a hole is not really a particle. it's a void or absence of an electron where there might otherwise be one. i can only conceptualize the concept of a "hole" in the solid state context.

is this a QM thing?

it's what they do in my old Millman and Halkias electronics book.

once the "basic" ensemble properties of these particles or pseudo-particles (electrons and holes) are established, this hole-electron ballistics is done with classical physics. E field causes forces to act on the charged particles (both with positive mass) and they move. when they get into the others' territory, recombination starts happening and, because of that diffusion model described by a simple first-order diff. eq., the density of these holes or electrons that crossed over the junction decreases exponentially as one moves away from the junction. i am not sure, but i believe that this is the source of the exponential form of the volt-amp characteristics of both the diode and BJT (they're called the "Ebers-Moll equations" in the BJT).

no, now that i'm looking it up, i think the root of the exponential V-I characteristics is the Fermi-Dirac statistics. dunno how they are derived. it's been 30 years. i do Digital Signal Processing for a living, not semiconductor device design. most EEs that are not into this level chip design and fabrication don't remember this stuff.

but i wanted to give the OP an idea for how or why this NPN sandwich can conceivably be used to make a current-controlled current source or current amplifier (when attached to a decently high voltage at the collector). and it has to do with the finite rate of recombination and the thin-ness of the base wafer. not all of those charge carriers (in fact, a very small fraction) recombine in the base and the rest get "collected" by the collector. that is the salient picture i wanted to draw.

Last edited: Sep 2, 2006
16. Sep 2, 2006

### Gokul43201

Staff Emeritus
Oops. Being a little incoherent, aren't I? Ignore that sentence.

Yes. Here's the wiki on effective mass - http://en.wikipedia.org/wiki/Effective_mass

You'll find more in any recent solid state or device theory text.

17. Sep 2, 2006

### rbj

looks to me that all of those particles (electrons or holes) had positive effective mass listed in the article.

anyway, thanks for the error correction. as i said, all of this comes from my 30 year old memory of the topic which is not my specialty in EE at all.

but i thought that the OP needed a more intuitive explanation of what's going on.

Last edited: Sep 3, 2006
18. Sep 3, 2006

### inha

it can be negative aswell. and as it depends on the band structure it can vary in different parts on the energy band it resides in.

19. Apr 18, 2007

### oqan501

PN junction is a potential barrier, in terms of quantum mechanics.

20. Jan 20, 2008

### LydiaAC

The explanation of Leright of the built-in potential as a consequence of an equilibrium between diffusion and drift is better than that of Rbj of an increased rate of recombination because of the closeness of P and N materials. This last explanation make sense when the electric field begin to raise but after, when it is at middle way to its equilibrium value, there is already a depletion zone which separate P and N zones. On the contrary, the concentration gradient are always present, still when the equilibrium has been reached. The recombination explanations raises the answer that what happens with energy liberated by those events. The diffusion explanation can be done separately for valence band and conduction band and does not require liberation or absorption of energy.
Actually, intrinsic silicon have an appreciable conductivity (one of my colleagues, a Chemical Engineer more than eighty years old, insist on that silicon is a metal). The conductivity of a reverse biased diode is not the same as that of intrinsic silicon, but smaller by several orders of magnitude. And not always can be considered negligible. The built-in potential in PN junction is not problem for electrons in P side or holes in N side. In fact, they are helped for it. The reason of the low conductivity of the reverse biased diode is not the built-in voltage but the scarcity of these minority carriers. With abundance of minority carriers a reverse biased junction can have a very large conductivity.
I do not understand why you say that the reverse biased junction base-colector is in any way different from a reverse biased diode. It is not necessary for it to be different. Any electron in a P base is a minority carrier and as such, can pass without problem through a reverse biased junction. The thin base is a requisite for electrons can travel from emitter-base junction to base-colector junction without recombining but this thinness does not help them in any way to pass the reverse biased junction. They do not need help, they could do it in the most natural way: electric field push them in that direction.

Semiconductors are not called that name because sometimes conduct and sometimes do not. The name is from its intermediate conductivity more than isolators but less than metals.

The percentage of electrons recombined in base has nothing to do with the amplification of the BJT. This is controlled by the ratio of the doping between emitter and base. For a BJT can amplify, the dopage in emitter must be at least two orders of magnitude than that of the base. The forward biased emitter-base junction is a diode the current of which are due to two kind of majority carriers: electrons and holes. If dopage of N-emitter is 100 times dopage of P-base, the current of electrons will be 100 times that of holes. If you could separate both currents you will have a current amplifier. And that is what the inverse biased junction is for. Hole current from base to emitter cannot be fed by base-colector junction because holes in collector are very few, so this current must be fed externally by the emitter contact. Electron current from emitter to base has no problem crossing the base collector junction and go forward from emitter to base to collector. So, you have both proportional currents separated and using the proper circuit you can build an amplifier.
Actually the amplification could not reach 100 because of a) recombination b) reverse minority current in inverse biased junction c) other precisions. But that is the idea behind the BJT.

That is an important precision since both electrons and holes are really electrons but they are called "holes" if they are from valence band and "electrons" if they are from conduction band. Sometimes people ask if effective mass of holes are of the same magnitude as effective mass of electrons. The answer is that if you are referring to electrons in valence band, this is true, but if you are referring to electrons in conduction band ("electrons") then is not. Electrons and holes are alternative ways to treat electrons in a band and when people count both holes and electrons in the same band, they are counting the same carrier two times.

Actually, both are right. Effective mass depend on curvature of E-k diagram. Effective mass of holes are the negative of the effective mass of electrons "in the same band". However, in practical semiconductors, conduction band is almost empty and at the bottom of the E-k diagram, curvature is positive and so effective mass of electrons. Also, valence band is almost filled and in the top of the E-k diagram, curvature is negative and thus effective mass of electrons is negative. We decide to work this electrons as holes, and these holes have a positive effective mass. Really, if the hole does not happen to have a positive mass, the concept would be less useful. However, I do not know if in other fields people find that they could gain something by working with negative mass carriers.
Lydia Alvarez

Last edited: Jan 21, 2008