A very silly circuit

Inspired by your attempt I managed to breakdown the C-B junction. Used a MPSH-10 Rf transistor on the hunch it has thin base and low breakdown voltage. Full experiment is as follows:

Device beta at start 131

1. Pease style breakdown Veb, Vcb into 11meg DMM, Icb into short Ibreakdown Vcb Icb

500uA -513mV -0.54uA 1mA -528mV -1.1uA 2mA -539mV -2.09uA 4mA -542mV -3.76uA

Device beta after several mins Veb avalanche 61

1. Herold/Hobbs breakdown Vcb, Veb into 11meg DMM, Ieb into short I breakdown Veb Ieb

250uA +3.8mV 0uA 500uA +4.1mV 0uA 1mA +4.3mV 0uA

Device beta after 1-2mins Vcb breakdown 61

Despite the MPSH10 30V Vcb rating it took nearly 80V to breakdown.

The obvious difference was the in C-B breakdown case there was no opposite sign emf at the other junction, possibly just a very slight leakage at nA levels.

If the explanation is photo-electric then it looks like the C-B breakdown does not emit light or that no photo detection takes place in the E-B junction. I got the feeling that during the Veb breakdown the MPSH10 was producing a heftier output (-540mV) than the jelly-bean device I tried in the past (300-400mV) perhaps the thin-base is a big factor.

When I next get some time for bench messing around I'll try Phil's suggestion of trying various temperatures.

piglet

I was actually talking about reverse-biasing CB and looking at the Early effect. I think that if it's photocarriers, the Early effect will be small, and if it's diffusion, it'll be much bigger (bigger than in the same device in normal bias).

The reason that the collector can pull below base potential is that there's a frozen-in field. At a PN junction, you have a _huge_ chemical potential gradient, which pulls electrons into the P type and holes into the N type until the resulting E field balances out the chemical potential gradient. That E field pulls in any available free carriers, producing a current in the external circuit. The open-circuit voltage across the junction is in the forward-bias direction.

I doubt that the base is thin enough for the collector to see the emitter voltage. The avalanche process will be generating carriers of both polarities in the base, ISTM.

Fun though.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Nice--a linear 'beta'.

Interesting. Most of the avalanching will be occurring further into the collector region, due to its lighter doping, so there'll be fewer carriers that manage to diffuse upstream against the E field. It looks as though maybe the base is fully depleted at this bias, so you're seeing a little bit of the collector potential.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

I like your Vcb data more than mine. I have one DMM with high input impedance on the 200 mV scale. And that gave some strange numbers when looking at Vcb at low currents. I worry some about local RF getting in and screwing up the measurement... all these high impedances, it should be done in a shielded box. (too much work!) You are also getting ~x10 the current I say. (for the same zener current)

That's interesting. OK I got out another 50V supply and zenered the Vcb junction of the 2n3904. (~120V seemed to do it.) I got the same results you did. No Vbe voltage. (maybe a bit positive)

Now one thing that confuses me, (one of many), is that when I looked at the

3904 spec sheet I noticed that V_br_CEO < V_br_CBO. A spot check of other transistors showed the same behavior.

So this is most likely something silly that I should know about. But how can I even zener the cb junction without doing the same thing to the ce? And if I raise the emitter up some.. then it will breakdown to the base.

OK some hand-wavy explanations of your/our data. (This will be in the B. Pease photodiode model.. I don't understand Phil's model enough to even wave my hands at it. :^)

So how far a photon can travel in Si will depend of the abs. coef.

We don't know the wavelength, but I think the McKay article implied visible to UV. so maybe 10^5 cm-1 or a depth of 10^-7 meters, 100 nm. I'm not sure what a typical base thickness is, but a thicker base will mean fewer photons get through the base and are absorbed in the c-b junction. This could explain your higher currents and open circuit voltages. I'm not sure what to make of the lack of a negative voltage when zenering the c-b. That junction will be wider, (less doping in the collector.) but there should be a lot more photons too. (Because of the much higher break down voltage... I'd guess number of photons would go as the energy I*V, but that is a guess.)

George H.

Scratch scratch... (why is communication so hard without a white board?)

So you mean zenering E-B and looking at the C-B current as a function of the C-B (reverse) voltage. If it's the photodiode thing, then not much change in the current, (except for small leakage current) And if it's a diffusion thing, then the current will be bigger... because..(?).... because the depletion region will then reach (a bit) further into the base where the carriers are. (As some ratio of the doping densities in the B and C) Hmmm.. well depending on how important the absorption coefficient is, a bigger depletion region (in CB) might mean more photons captured there too.

Grin, you're always throwing these big words around to confuse we bear's of little brain. The chemical potential gradient is just due to the different doping densities (and signs) on the p and n side.

Say can I ask a silly and mostly unrelated question. Why isn't the built in potential measurable on the lab bench? If I put an electrometer across a diode I measure zero volts. (Where is the potential drop in the outside circuit?)

Yeah, Thanks for playing with piglet and I.

George (pooh bear) Herold.

Yeah, I noticed that too. I think the thinner base lets more light through (if the effect is optic). Also the beta degradation was much greater, about 50% whereas the jelly-bean part just degraded 10-15%.

There is no C-E junction to breakdown because the base is between them, I think the V_br_CEO and CBO have the third terminal (base or emitter resp) open circuited. The C-B junction has a high breakdown but when the base is floating as in the V_br_CEO test even the slightest C-B leakage will begin transistor action aka gain and the C-E conductance will rapidly increase.

I still haven't cut open a transistor and seen the light, I always imagine it as blueish-white. Did Jeroen Belleman say what he color saw?

Could be that a only heavily doped emitter junction makes photons?

piglet

I'll second that! Thanks Phil, I'm learning more about transistor innards.

piglet

With the base open, collector-base leakage gets multiplied by beta. There are four commonly-quoted CE breakdown voltages:

V_CEO (base open) V_CES (base-emitter short) V_CER (resistor from base to emitter) V_CEX (emitter-base reverse biased).

V_CEO < V_CER < V_CES < V_CEX, due to transistor action steadily decreasing down the list.

Transistor action is also absent in the VCBO measurement, of course.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Right. If you imagine Maxwell's Demon positioned between the N and P side, and suddenly opening the door, you get a Niagara of electrons going one way and holes going the other. (Energy-momentum diagrams and all that.)

Macroscopically, because current flows in the external circuit (even just air ions) so that the E field outside the die is zero. Interestingly this isn't true microscopically or even mesoscopically. If you fly a capacitive tip a few tens of microns over a silicon surface, you can measure changes in the surface potential due to work function variations. It turns out that there aren't enough states available to shield out E in less than ~0.1 mm. (The things you learn from customers.) ;)

Living under the name of Saunders. ;)

Besides, I go downstairs bump, bump, bump on the back of my head sometimes too.

(Two book references that parents will recognize and probably nobody else.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

There's a built-in field in a junction (as long as you're away from zero temperature) because of diffusion of high-density (majority) carriers in both the N and P regions. Photovoltaic devices use the generation of photocurrent and the built-in field to generate energy.

If you avalanche the BE junction, E minority carriers will go into the base. But, the base is THIN, so E minority carriers diffuse also into the C region, if they aren't recombined in the base (and E minority carriers are B majority carriers, they won't combine readily...). This current of electrons is the same as a current of photoelectrons, it recombines to generate energy.

I'm not seeing any compelling reason to assume photons are an intermediary. Prediction: terminal voltage will drop if you cool the transistor. Prediction 2: black-epoxy transistors, clear-epoxy phototransistors, and metal-can (light would stay in the silicon die, by internal rflection) transistors won't be markedly different.

I don't suppose any SPICE models really capture this effect, do they?

Well, for Phil's model, a thinner base means the CB junction is closer, and more carriers will diffuse in and fall down.

I haven't measured beta. Do you have some meter?

I think a thinner base/ lower breakdown voltage means higher doping. I picture this electron (or hole) in the depletion region getting accelerated and then crashing into defects and stuff in the lattice, sometimes it crashes into a dopant atom, and makes more carriers. With more dopant atoms around the chances of ionization increase... so it happens at a lower voltage.

Me either, (and no plans.) I'd want to stick the light into a spectrometer... and suddenly it's a project, and not bench DMM fun.

If anything, I could only make an argument the other way. (At some point there's only e*V of energy... The energy will go as charge*E-Field*mean free path.) with lower doping there are less defects and charge carriers may travel further before crashing into something. At some point the dopant atoms become the major scatters. (I think, ... I'm not really a semiconductor guy.... but I like to play one on SED :^)

Re: zenering the C-B junction... where are the photons? I have no idea what the width of the depletion region is when we're zenering the cb at ~100V. If we knew the dopant level I think we could estimate it, or measure the capacitance, or ask... If it's deep enough and the photons are blue enough, then very few may make it out.

I'm not sure how Phil will explain the lack of negative Vbe in this case. (The BE jucntion should be steeper, and be more in the base region.... I think.)

Thanks, and is the Icb vs Vcb, in the presence of Vbe breakdown, the Early voltage thing you were talking about.

Saunders? Sanders? or is that only in the US? You Canadian's and Brit's put extra 'u''s everywhere...:^)

I still play "Pooh" sticks in the creek with my kids. (My daughter was off looking at college #1 the other day. Junior year. Hardly kids anymore...)

George H.

OK can you explain why we see nothing when we zener the BC jucntion at ~100V and 1-2 mA? The steeper BE junction should pull in more of the base carriers. No? (This is fun! :^) George H. I may be wrong, but in the end I learn something.

Oh I really want to zener one of these $50 avalanche PD on digikey... huh, now only $35. next DK order!

Awesome, thanks. If I short b-e the collector breaks down sooner than C-B with the emitter open.. no worries that's easy to measure. George H.

On 2015-10-13 23:10, piglet wrote: [...]

Yes, blueish white it was. Surprised me. I'd expected something more monochromatic.

Jeroen Belleman

I've just tried with freezer spray and got an increase in C-B voltage. Device MPSH-10 in Veb breakdown Ie=1mA, at room temp Vcb -529mV, Icb

-1.0uA, after freezer spray Vcb -633mV, Icb -1.2uA (Ie was constant).

Perhaps when I get more time I'll try a more scientific plot vs temp.

Does the direction of the change help you nail down the effect to photo-voltaic or carrier diffusion? Pease vs Hobbs?

"Ring if an anwser is required, knock if an answer is not required"

piglet

Hmm, well the forward voltage of diodes goes up when the temperature goes down, so Vcb is not surprising. More current is more interesting.

For the photo-voltaic model there could be a number of reasons.

1.) More photons, perhaps at lower temps the avalanche makes more photons. did the Vbe reverse voltage (at 1 mA current) go up too? That would mean more energy has to be lost... more light. 2.) Well other hand wavy ideas.. but #1 looks best. It would be sweet if Vbe went up by 20%.

I still think that the lack of a negative voltage when we zener the bc junction still has to be explained in the Hobbs diffusion model.

"Think it over, think it under"

George H.

Aw shucks, I neglected to monitor the Veb, just kept eye on the current Ie. The emitter was fed via 10kohm so near a constant current, I did not have to adjust the psu voltage to keep Ie constant and it did not waver around more than 5-10uA (out of 1000uA).

I know PV solar cell output is higher at low temp which may support the optic explanation. But I do not understand the thermal diffusion hypothesis enough to know if that is proved or dis-proved by these observations.

piglet of little brain.

Not 100% sure. The increased Vcb is expected, the increased I_CB less so--naively, photodetection should leave it constant, and diffusion should (I think) reduce it at low temperature. Do you know the burden voltage of your meter? If it's anything significant at 1 uA, it might explain the change. (It probably isn't.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net