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. > > 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.netHi 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.
A very silly circuit
Started by ●October 3, 2015
Reply by ●October 13, 20152015-10-13
Reply by ●October 13, 20152015-10-13
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 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 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
Reply by ●October 13, 20152015-10-13
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). 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.> > 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. 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
Reply by ●October 13, 20152015-10-13
On 10/13/2015 11:43 AM, 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 -3.76uANice--a linear 'beta'.> > 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 >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.> 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 > >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
Reply by ●October 13, 20152015-10-13
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 -3.76uAI 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 0uAThat'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.> > 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. > > pigletOK 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. http://www.pveducation.org/pvcdrom/pn-junction/absorption-coefficient 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.
Reply by ●October 13, 20152015-10-13
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. 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?)> > > > > 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. George (pooh bear) Herold.> > 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
Reply by ●October 13, 20152015-10-13
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%.> > 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?> 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? piglet
Reply by ●October 13, 20152015-10-13
On 13/10/2015 20:41, George Herold wrote:> On Tuesday, October 13, 2015 at 11:55:20 AM UTC-4, Phil Hobbs wrote: >> >> Fun though. > > Yeah, Thanks for playing with piglet and I. > > George (pooh bear) Herold. >>I'll second that! Thanks Phil, I'm learning more about transistor innards. piglet
Reply by ●October 13, 20152015-10-13
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. 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
Reply by ●October 13, 20152015-10-13
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.)> > 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.) 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
