# Silicon PIN photodiodes

1. Apr 21, 2005

### student654321

i am writing a lab report into an experiment i did using a photodiode to detect alpha particles. I am curious however as to how a photodiode actually works. The alpha particles are emitted from the source, they hit the diode and then what happens? Its somehow changed into an electrical signal that comes up on the screen of my computer at a particular energy but how does it get there? Also theres a pre-amp and an amplifier being used which i am not sure about either.. Any help would be greatly appreciated.

2. Apr 21, 2005

### 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 if 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 who you might consider to be the young, nubile, (and horny) heterosexuals. think of the diatomic gasses ($\operatorname{Cl}_2$ or $\operatorname{O}_2$) as hot b_itches locked up away from the men and resigned to lesbian relationships (but you better watch out if they get loose) in the meantime. $\operatorname{N}_2$ isn't so bad but if $\operatorname{Cl}_2$ 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. now, suppose you hook the battery up the other way. those extra electrons will immediately drain into the "+" terminal (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. (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 why they are called "semiconductors" sometimes they conduct electricity and sometimes they don't. it depends on what you're doing to them.

now what you're doing is a little different. you have these diodes hooked up in a little circuit where you are likely not driving them with any external voltage. so you have that precarious equilibrium condition mentioned above. some electrons from the P material have jumped over to the N material and some holes from the N material have jumped over to the P (leaving the P material slightly positively charged and the N material slightly negatively charged and that charge keeps more electrons or holes from doing the same).

so some positively charge alpha particle come in to the N material and gives one of those holes (also positively charged but with a lot less mass) that would like to jump over to the P side but won't because the P side is positively charged, a little boot (positively charged particles repel each other and the alpha particle has a lot more mass/momentum than the hole, so guess who's gonna have to move) and that hole now has enough energy to make the jump. i dunno what happens to the alpha particle, but i don't think it combines chemically with the silicon or phosphorus, so i think it continues on its merry way. that leaves the positive side even more positively charged and the negative side even more negatively charged than what it was at equilibrium. then some charge is gonna want to move back to put that junction back to equilibrium. that is an electromotive force (or a voltage) and can be detected with an electronic circuit, just like the electric voltage coming out of a microphone (or your electric guitar) can be. that is what your pre-amp or amp is for.

(something similar might happen if a photon comes barrelling in and gives one of those extra electrons a kick up to a higher energy state.)

the electrical voltage generated by the photodiode is so small that, before the electronics used by your computer to read it (called an "analog-to-digital converter" or "A/D" or "ADC") can read it, that voltage has to be boosted. i can't explain how amplifiers work without explaining how the transistor (a PNP or NPN pair of junctions) works.

just pretend the pre-amp or amplifier is a magic box that makes little voltage fluctuations into bigger voltage fluctuations.

r b-j

Last edited: Apr 22, 2005
3. Apr 22, 2005

### student654321

So are you saying that the alpha particle is interfering with the movement of electrons from the negative to positive causing the holes to progress towards the positive? That seems to make sense to me. But.. what happens if the alpha particle hits the positive side instead wouldn't it have the exact opposite affect? I suppose it depends on how the diode is positioned but if it were to have its negative face turned downwards then wouldn't the alpha particle have to be stopped or deflected before entering the positive side, does its depletion in energy (transferred to the electrons presumably) cause it to have a different influence on the other side?

That was a very well written post and i thankyou because it helped me picture the diode structure and the influence of alpha particles but i still think i need a better understanding of whats going on here to write my report. Any takers?

4. Apr 22, 2005

### student654321

Ah i think its more that the alpha particle enters the diode which is placed so its n and p sides are laying on top of one another and not side by side. The alpha particle effectively creates turmoil pulling off electrons and making electron-hole pairs all the way through which are then forced to migrate to their respective side increasing the voltage in the circuit which when detected can be equated to the alpha particles energy. I don't know if this makes much sense?

5. Apr 22, 2005

### chroot

Staff Emeritus
This can all be explained with just a couple of sentences.

Essentially, a reverse-biased pn junction does not normally carry any appreciable current, because the pn junction is surrounded by a region depleted of all charge carriers.

The passing of charged particles like alphas ionizes (perhaps many) atoms in the semiconductor. When the alphas pass through the depletion region surrounding the reverse-biased pn junction, the newly-freed electrons are rapidly swept away toward the p-type semiconductor. (Additionally, the hole thus created is swept away toward the n-type semiconductor.) Thus, you'll see a blip of current when an alpha passes.

This small signal is then amplified so that it can be measured more easily, and then digitized for analysis by your computer.

- Warren

6. Apr 22, 2005

### jdavel

chroot,

"the newly-freed electrons are rapidly swept away toward the p-type semiconductor."

Isn't that backwards? The p side of the depletion region is negatively charged, so it repels the electrons.

I know this is picky, but the original poster seems to be learning this for the first time, and polarity issues are confusing enough even when the explanations get them the right way.

Or I could have it backwards!

7. Apr 22, 2005

### rbj

i think you're right. for forward-biased PN diode, the P is connected (perhaps through a resistor) to the "+" terminal and the N-type silicon to the "-" terminal. so reverse-biased it's the other way. i think "the newly-freed electrons are rapidly swept away toward the N-type semiconductor" is right.

i didn't know for either a photo-diode or an alpha particle detector, that the diode is reverse-biased, but it makes sense. i just thought that they were zero-biased and they were hooked up so that a small voltage was created by the incoming particle.

r b-j

8. Apr 22, 2005

### chroot

Staff Emeritus
Er, yes, good catch jdavel. I mixed up the polarity. The freed electrons migrate to the n-type semiconductor, while the holes migrate to the p-type semiconductor.

- Warren

9. Apr 23, 2005

### student654321

Ah i see so the P type is negatively charged because some of the electrons from the N type have filled the holes on that side.

But why does the alpha particle only have to interact with the silicon atoms in the depletion region and not everywhere else in the diode? I read somewhere that there needs to be a forward voltage to get rid of the region so that any current can pass through. Or can the alpha particle stimulate a current in this region?

Last edited: Apr 23, 2005
10. Apr 23, 2005

### rbj

no, i think the electrons that fill the holes in the P type silicon have come from the - terminal of the battery or power supply. if they came from the N type, the diode would be forward-biased (or not biased at all and this is simply the contact potential). that is if Warren is correct that the PN junction is reverse-biased in the particle detector. (he very well may be right. i hadn't seen the circuit so i was guessing that the diode was niether forward-biased nor reverse-biased.)

i don't think that is what Warren meant (i'll let him speak for himself). in that reverse biased setup, electrons from the - terminal fill the holes of the P material and electons in the N material drain into the + terminal. that creates a depletion region. the alpha particle will interact with a hole (a psuedo-particle, i'll admit, but they tell us it behaves as a positive charged and positive mass particle) in the P material, giving it a kick and enough energy, that along with the potential difference (the applied voltage that attracts that hole to the - terminal), it has enough to traverse the depletion region and go to the - terminal. (really, electrons are moving in the other direction.) that is a current that can be detected.

i haven't seen the circuit. Warren says the diode is reverse-biased and he may be right. i would think that a forward-bias diode would have so much charge moving that you wouldn't know the difference if an additional small amount was also moving due to detection of an alpha particle.

r b-j

11. Apr 23, 2005

### jdavel

rbj,

The depletion region isn't created by the power supply. It forms at the p/n junction with no voltage applied. It gets wider with a reverse bias, and narrower with a forward bias. But it's there with no bias.

12. Apr 23, 2005

### rbj

you're right. it's just easier for a non-physicist EE to think of it as not there during forward bias and there for reverse bias (no bias is a sorta case that isn't considered). leaving out the applications such as photo-detectors, we EE's think about diodes in one of two binary states, either conducting with very low resistance and about 0.6 volts drop, or not conducting. but it is a continuous function, ideally:

$$I = I_0 (e^{\frac{eV}{kT}}-1)$$

and not a discrete function. so, strictly speaking, qualitatively it is the same for all applied voltages. but it's hard to remember that (for a circuit guy, anyway).

r b-j