A very silly circuit

On Friday, October 9, 2015 at 8:48:43 PM UTC-7, George Herold wrote:
> On Friday, October 9, 2015 at 9:06:33 PM UTC-4, Phil Hobbs wrote:

> > Sure.  But the quantum efficiency of hot carriers generating visible 
> > photons is super low--it's happening in an indirect-bandgap semiconductor.

> > ... Everything ... in transistors is 
> > all about carriers diffusing around under the influence of electric 
> > field and doping density.  Why should this be any different?  On the 
> > face of it, it's very nearly the same physics as a normal BJT in saturation.

> OK, I really have no idea. 
> I was thinking maybe some diffusion /transport
> type thing,

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?

On Tuesday, October 13, 2015 at 5:10:50 PM UTC-4, piglet wrote:
> On 13/10/2015 20:10, George Herold wrote:
> > You are also getting ~x10 the current I say.  (for the same zener current)
> 
> 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%.

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.
 
  
> 
> >
> > 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.
> > https://www.fairchildsemi.com/datasheets/MM/MMBT3904.pdf
> > 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.
> 
> 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.
> 
> > 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 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?

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.   

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

      

> 
> piglet

On Tuesday, October 13, 2015 at 6:00:51 PM UTC-4, Phil Hobbs wrote:
> On 10/13/2015 03:41 PM, George Herold wrote:
> > On Tuesday, October 13, 2015 at 11:55:20 AM UTC-4, Phil Hobbs wrote:
> >> On 10/13/2015 10:06 AM, George Herold wrote:
> >>> On Monday, October 12, 2015 at 1:04:25 PM UTC-4, Phil Hobbs wrote:
> >>>> On 10/12/2015 11:41 AM, George Herold wrote:
> >>>>> On Saturday, October 10, 2015 at 5:17:58 AM UTC-4, Phil Hobbs
> >>>>> wrote:
> >>>>>> On 10/09/2015 11:48 PM, George Herold wrote:
> >>>>>>> On Friday, October 9, 2015 at 9:06:33 PM UTC-4, Phil Hobbs
> >>>>>>> wrote:
> >>>>>>>> On 10/09/2015 08:52 PM, George Herold wrote:
> >>>>>>>>> On Friday, October 9, 2015 at 6:47:21 PM UTC-4, Phil
> >>>>>>>>> Hobbs wrote:
> >>>>>>>>>> On 10/09/2015 06:13 PM, piglet wrote:
> >>>>>>>>>>> On 09/10/2015 20:31, Phil Hobbs wrote:
> >>>>>>>>>>>> I think Pease was completely wrong about the
> >>>>>>>>>>>> mechanism here.
> >>>>>>>>>>>>
> >>>>>>>>>>>> There's nothing special about photogeneration for
> >>>>>>>>>>>> this process--all you need is free carriers; you
> >>>>>>>>>>>> don't care where they originate.  The built-in E
> >>>>>>>>>>>> field of the collector depletion zone will make the
> >>>>>>>>>>>> collector go negative with respect to the base.
> >>>>>>>>>>>>
> >>>>>>>>>>>> It works in forward bias too--if you forward-bias
> >>>>>>>>>>>> the BE junction, the collector spontaneously pulls
> >>>>>>>>>>>> below base potential.  It's still above ground, of
> >>>>>>>>>>>> course, because of V_BE.  (I think this was one of
> >>>>>>>>>>>> JL's interview questions.)
> >>>>>>>>>>>>
> >>>>>>>>>>>> When you make the carriers by zenering B-E, the
> >>>>>>>>>>>> situation changes.  The free carriers make the
> >>>>>>>>>>>> collector pull negative with respect to the base
> >>>>>>>>>>>> just as before, except that the base is now at
> >>>>>>>>>>>> ground, so the collector pulls below ground.
> >>>>>>>>>>>>
> >>>>>>>>>>>> Cheers
> >>>>>>>>>>>>
> >>>>>>>>>>>> Phil Hobbs
> >>>>>>>>>>>>
> >>>>>>>>>>>
> >>>>>>>>>>> Very interesting. Since we know light is produced and
> >>>>>>>>>>> C-B junction can act as photo-diode then what kind of
> >>>>>>>>>>> experiment could we do to establish which mechanism,
> >>>>>>>>>>> photo-electric versus holes (or is it electrons)
> >>>>>>>>>>> shooting through the base region, is at work here?
> >>>>>>>>>>>
> >>>>>>>>>>> Would trying a thin-base transistor versus a
> >>>>>>>>>>> thick-base transistor reveal anything useful? Just
> >>>>>>>>>>> wondering aloud, I am a piglet of little brain on
> >>>>>>>>>>> solid-state physics.
> >>>>>>>>>>>
> >>>>>>>>>>> piglet
> >>>>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> One approach would be to zener the base-collector
> >>>>>>>>>> junction and see if you get more current from the
> >>>>>>>>>> emitter than the other way up.
> >>>>>>>>> Dang that would be fun.  Zener the CB at the same current
> >>>>>>>>> (?) 40- 60 V?  What do you expect?  More photons from the
> >>>>>>>>> CB, maybe? but is the BE junction optically thick? Say I
> >>>>>>>>> get a more negative voltage?  (Vbe and Vcb at the same
> >>>>>>>>> current are different.)
> >>>>>>>>
> >>>>>>>> I'd be more interested in the short-circuit current.
> >>>>>>>>
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> The quantum efficiency of the photogeneration process
> >>>>>>>>>> is pretty low, not to mention the (crappy)**2
> >>>>>>>>>> performance of phototransistors used as diodes, whereas
> >>>>>>>>>> George seemed to be getting hundreds of nanoamps out
> >>>>>>>>>> of the collector (37 mV @ 100kohm load).
> >>>>>>>>> Grin, mind you that was just a TH resistor held by hand
> >>>>>>>>> across the leads of a to-92. (to-92 in power supply
> >>>>>>>>> terminals) Still, 370 mV @ 10 Meg is 37nA... I had
> >>>>>>>>> several mA going the other way.
> >>>>>>>>
> >>>>>>>> Sure.  But the quantum efficiency of hot carriers
> >>>>>>>> generating visible photons is super low--it's happening in
> >>>>>>>> an indirect-bandgap semiconductor.
> >>>>>>>>
> >>>>>>>>>
> >>>>>>>>> Re: how to tell.  I'm not sure if this works, but if I
> >>>>>>>>> look at the shot noise from a photodiode and a forward
> >>>>>>>>> biased (FB) pn junction. Both shorted with some
> >>>>>>>>> resistance R... (I think this is really using noise to
> >>>>>>>>> measure impedance.. as a function of voltage/ current.)
> >>>>>>>>> Then the PD has more noise, a higher impedance, up to
> >>>>>>>>> ~50-100mV or so.
> >>>>>>>>
> >>>>>>>> I owe a lot to Bob Pease.  Professionally, his writing was
> >>>>>>>> a fairly significant part of my early education in circuit
> >>>>>>>> design, besides being fun.  Personally, he gave me very
> >>>>>>>> generous encouragement, both in print and face to face.  (I
> >>>>>>>> met him a couple of times in the early 2000s.)
> >>>>>>>>
> >>>>>>>> I just think he's wrong on this one.  Everything else in
> >>>>>>>> transistors is all about carriers diffusing around under
> >>>>>>>> the influence of electric field and doping density.  Why
> >>>>>>>> should this be any different?  On the face of it, it's very
> >>>>>>>> nearly the same physics as a normal BJT in saturation.
> >>>>>>>
> >>>>>>> OK, I really have no idea. I was thinking maybe some
> >>>>>>> diffusion /transport type thing, James said maybe no, you
> >>>>>>> said maybe yes. Data would help*.  Maybe you have an easier
> >>>>>>> way to measure the source impedance?  Is the DC impedance the
> >>>>>>> same? That would be a lot easier!
> >>>>>>
> >>>>>> Yup, data is good, I agree.  One possibility is speed.
> >>>>>> Photons propagate a lot faster than electrons diffuse.  That
> >>>>>> might be a pain to measure at these current levels.
> >>>>>>
> >>>>>> Another is temperature.  You'd expect normal diffusion current
> >>>>>> to go up with temperature, whereas photocurrent should be
> >>>>>> nearly constant.
> >>>>>>
> >>>>>> Another one is Early voltage.  There'll be a huge B-E depletion
> >>>>>> zone, so the effect of base narrowing should be a lot bigger in
> >>>>>> the diffusion case.
> >>>>>>
> >>>>>> Cheers
> >>>>>>
> >>>>>> Phil "Not a real solid state guy" Hobbs
> >>>>>
> >>>>> Hi Phil, Well my thinking was much simpler, but maybe wrong.
> >>>>> (Please correct me if I'm in error.) So if it's light onto a
> >>>>> diode I get a current source, with a fairly high compliance
> >>>>> (source resistance) like 100's of millivolts.
> >>>>>
> >>>>> And if it's a thermally activated diode (your idea) then it only
> >>>>> has 25mV of compliance, source resistance. (I'm not sure if it
> >>>>> makes sense to talk about compliance in this way.)
> >>>>>
> >>>>> So if I just look at the I-V from this as a function of loading
> >>>>> it with different R's, then some clever people should be to
> >>>>> figure out if it's thermally or optically excited. (I'm finishing
> >>>>> a schematic and then can squeeze in some quick measurements.)
> >>>>
> >>>> Hi, George,
> >>>>
> >>>> I think what's happening is that you're forcing electrons into the
> >>>> base region, which then fall into the collector's potential well--a
> >>>> lot like an ordinary saturated transistor.
> >>>>
> >>>> If the photocarriers are generated in the same location as the
> >>>> avalanche current, there should be no difference in the compliance
> >>>> behaviour, ISTM.   I'm thinking that if the base is thin,
> >>>> photocarriers might be generated further into the collector
> >>>> region.
> >>>>
> >>>> I'd expect the temperature dependence to be the easiest thing to
> >>>> nmeasure.
> >>>>
> >>>> I'd also suspect that zenering the CB junction would give a bit
> >>>> more current due to the higher doping gradient in the BE junction.
> >>
> >>>
> >>> Hi Phil,  I hope you saw my data. (I tried to zener the CB junction,
> >>> I got the reverse bias voltage up to 102 V and ran out of power
> >>> supply, with no current.)
> >>>
> >>>  From your comments you seem to think that this ~300 mV of compliance
> >>> is expected.  I must admit that when you compare this to a saturated
> >>> transistor it just confuses me.  (I don't understand saturated
> >>> transistors all that well either.)  For a saturated transistor I have
> >>> this model (picture) of electrons diffusing across the (lowered
> >>> potential) base and into the collector, such that the collector
> >>> (voltage) eventually is below the base.
> >>
> >> 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).
> >
> > 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.
> >
> >
> >>
> >> 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.
> >
> > 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.
> 
> 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.)

Thanks, and is the Icb vs Vcb, in the presence of Vbe breakdown,
the Early voltage thing you were talking about.  
> 
> >
> > 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?)
> 
> 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.) ;)
> 
> >
> >>
> >>>
> >>> For the zenering EB condition, it seems like I'm pulling holes out of
> >>> the base, raising the voltage barrier between the emitter and
> >>> collector. And I don't see how this allows more electrons to fall
> >>> into the collector. How are the current's linked?
> >>
> >> 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.
> >>
> >>>
> >>> I kinda like the two diode model of the transistor here... That seems
> >>> to work for me. (Oh and it's OK to just ignore this... it's not all
> >>> that important.)
> >>
> >> Fun though.
> >
> > Yeah, Thanks for playing with piglet and I.
> >
> 
> 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.)
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.
> 
> 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

On Tuesday, October 13, 2015 at 7:21:19 PM UTC-4, whit3rd wrote:
> On Friday, October 9, 2015 at 8:48:43 PM UTC-7, George Herold wrote:
> > On Friday, October 9, 2015 at 9:06:33 PM UTC-4, Phil Hobbs wrote:
> 
> > > Sure.  But the quantum efficiency of hot carriers generating visible 
> > > photons is super low--it's happening in an indirect-bandgap semiconductor.
> 
> > > ... Everything ... in transistors is 
> > > all about carriers diffusing around under the influence of electric 
> > > field and doping density.  Why should this be any different?  On the 
> > > face of it, it's very nearly the same physics as a normal BJT in saturation.
> 
> > OK, I really have no idea. 
> > I was thinking maybe some diffusion /transport
> > type thing,
> 
> 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?

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!
  

On Tuesday, October 13, 2015 at 5:52:05 PM UTC-4, Phil Hobbs wrote:
> On 10/13/2015 03:10 PM, George Herold wrote:
> > On Tuesday, October 13, 2015 at 11:43:47 AM UTC-4, piglet wrote:
> >> On 13/10/2015 15:06, George Herold wrote:
> >>>
> >>> Hi Phil,  I hope you saw my data.
> >>> (I tried to zener the CB junction, I got the reverse bias
> >>> voltage up to 102 V and ran out of power supply, with no current.)
> >>>
> >>>   From your comments you seem to think that this ~300 mV of compliance is
> >>> expected.  I must admit that when you compare this to a saturated transistor
> >>> it just confuses me.  (I don't understand saturated transistors
> >>> all that well either.)  For a saturated transistor I have this model (picture)
> >>> of electrons diffusing across the (lowered potential) base and into the collector,
> >>> such that the collector (voltage) eventually is below the base.
> >>>
> >>> For the zenering EB condition, it seems like I'm pulling holes
> >>> out of the base, raising the voltage barrier between the emitter and collector.
> >>> And I don't see how this allows more electrons to fall into the collector.
> >>> How are the current's linked?
> >>>
> >>> I kinda like the two diode model of the transistor here...
> >>> That seems to work for me.
> >>> (Oh and it's OK to just ignore this... it's not all that important.)
> >>>
> >>> George H.
> >>>
> >>
> >> 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
> >
> > 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)
> >>
> >> Device beta after several mins Veb avalanche 61
> >> 2.
> >> 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
> >
> > 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.
> > https://www.fairchildsemi.com/datasheets/MM/MMBT3904.pdf
> > 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.
> 
> 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.
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. 
> 
> 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

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

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

Jeroen Belleman

On 13/10/2015 16:55, Phil Hobbs wrote:
>>> I'd expect the temperature dependence to be the easiest thing to
>>> nmeasure.

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

On Thursday, October 15, 2015 at 12:49:34 PM UTC-4, piglet wrote:
> On 13/10/2015 16:55, Phil Hobbs wrote:
> >>> I'd expect the temperature dependence to be the easiest thing to
> >>> nmeasure.
> 
> 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?

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

"Think it over, think it under"

George H. 
> 
> piglet

On 15/10/2015 20:33, George Herold wrote:
> did the Vbe reverse voltage (at 1 mA current) go up too?  That would mean
> more energy has to be lost... more light.

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.

On 10/15/2015 12:49 PM, piglet wrote:
> On 13/10/2015 16:55, Phil Hobbs wrote:
>>>> I'd expect the temperature dependence to be the easiest thing to
>>>> nmeasure.
>
> 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 answer is required, knock if an answer is not required"

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
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