On 10/09/2015 08:00 AM, dagmargoodboat@yahoo.com wrote:> On Thursday, October 8, 2015 at 3:17:27 PM UTC-4, George Herold wrote: >> On Thursday, October 8, 2015 at 2:29:39 PM UTC-4, John Larkin wrote: >>> On Thu, 8 Oct 2015 11:02:52 -0700 (PDT), George Herold >>> <gherold@teachspin.com> wrote: >>> >>>> On Thursday, October 8, 2015 at 1:43:08 PM UTC-4, dagmarg...@yahoo.com wrote: >>>>> On Thursday, October 8, 2015 at 1:19:50 PM UTC-4, John Larkin wrote: >>>>>> On Thu, 8 Oct 2015 09:59:55 -0700 (PDT), dagmargoodboat@yahoo.com >>>>>> wrote: >>>>>> >>>>>>> On Thursday, October 8, 2015 at 11:59:16 AM UTC-4, piglet wrote: >>>>>>>> On 08/10/2015 16:46, Piotr Wyderski wrote: >>>>>>>>> piglet wrote: >>>>>>>>> >>>>>>>>>> You don't even need an opto-coupler for that. As Pease wrote: just use a >>>>>>>>>> 2N2222 (or similar) with base grounded, zener the emitter junction with >>>>>>>>>> plenty mA and see a small negative voltage at the collecter. The B-E >>>>>>>>>> zener emits light that excites the B-C junction as a photodiode. >>>>>>>>> >>>>>>>>> Since it's not April, the 1st, so.. please explain. :-) >>>>>>>>> >>>>>>>>> Best regards, Piotr >>>>>>>>> >>>>>>>> >>>>>>>> I thought my description did explain it. Go on, try it: >>>>>>>> >>>>>>>> <https://www.dropbox.com/s/g9zwy6wyb4lhx4a/peasecon.pdf> >>>>>>>> >>>>>>>> Have fun. >>>>>>>> >>>>>>>> piglet >>>>>>> >>>>>>> http://electronicdesign.com/site-files/electronicdesign.com/files/archive/electronicdesign.com/files/29/6362/figure_01.gif >>>>>>> >>>>>>> Applied for bias cancellation... >>>>>>> http://electronicdesign.com/analog/single-supply-op-amp-input-bias-current-cancellation >>>>>>> >>>>>>> >>>>>> >>>>>> Probably noisy. And maybe drifty. >>>>> >>>>> Dunno. Zeners are still lower noise than band-gaps, right? >>>>> >>>>> Woodward's one of the best, but it has been eighteen years since he >>>>> wrote that. And he was souping up what is basically a high-test LM324, >>>>> so it just might not matter that much. >>>>> >>>>> www.linear.com/product/LT1013 >>>>> >>>>> Cheers, >>>>> James Arthur >>>> >>>> A quick glance says the noise is not outrageous. >>>> ('scope, AC couple, 2mV 20 MHz BW limit, about 2mVp-p.. >>>> And most of that is still the 100 MHz that my lab is bathed >>>> in. I had to turn the room light off also. High impedance and >>>> all that. >>> >>> The source is high impedance so any capacitance (scope and such) >>> lowpass filters the voltage noise. The ED application depends on the >>> current into a summing point. Being optical, the current will have at >>> least shot noise, plus anything the zener-optical effect adds. >>> >>> Pease conjectured that the effect is optical. Has that been verified? >> How would you verify it? (Crack it open and look for light?) >> Do you get light out of 7 volt zeners? > > Supposedly you can do exactly that--crack open a transistor can and see flashes > of blue light. At 7 volts, you wonder why it isn't mostly U.V.Hot carrier emission is a commonly-used diagnostic tool in failure analysis labs. A couple of old colleagues of mine used it for picosecond time-resolved studies of gate performance. (Google for PICA--Jimmy Tsang and Jeff Kash.)> >> Could you have some higher energy electron >> from the avalanche make it all the way across the base and then fall >> down into the collector? (I think that gives the right sign.) > > For an NPN the electrons would be avalanching the wrong direction, from > base to emitter, headed away from the collector, right?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 -- 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
A very silly circuit
Started by ●October 3, 2015
Reply by ●October 9, 20152015-10-09
Reply by ●October 9, 20152015-10-09
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
Reply by ●October 9, 20152015-10-09
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. 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). Thicker base transistors would make much less efficient phototransducers. 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 9, 20152015-10-09
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.)> > 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. 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. George H.> > Thicker base transistors would make much less efficient phototransducers. > > 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 9, 20152015-10-09
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. There's a fatal allure in that ground lead that makes people think that there's new physics going on, when there 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
Reply by ●October 10, 20152015-10-10
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! George H. *I'm always reminded of a line by B. Pippard, Physics of Vibration Vol I, paraphrasing, "being not as smart as other's, I need to have some data, to guide my thinking." (Which is science in a nutshell.)> > There's a fatal allure in that ground lead that makes people think that > there's new physics going on, when there 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
Reply by ●October 10, 20152015-10-10
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 -- 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 12, 20152015-10-12
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" HobbsHi 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.) George H.> > > -- > 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 12, 20152015-10-12
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.net
Reply by ●October 12, 20152015-10-12
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.netOK quick data... that I haven't thought about. So I measured short circuit current (Iss) and open circuit voltage (Voc) at different reverse bias currents. And then the voltage with different load resistors. I rev Iss VoC 10M 1M 0.33M 0.1M (mA) (uA) (mV) (mV) (mV) (mV) (mV) 1.0 0-0.01 329 324 87 32 8 2.0 0.09 329 338 166 54 16.7 4.0 0.26 330 351 295 108 32.3 8.0 0.57 332 350 331 212 65 The open circuit voltage was a bit wonkie. I changed slowly when I loaded it. To me it looks more like a photodiode. George H. (I'm heading home early, my son is home alone and it's a beautiful day, we will do something outside!)
