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.
I tired a big fat transistor too (TIP31) it didn't zener till a higher voltage the collector/ base gave less voltage ~150 mV with (I'm guessing) a higher impedance.
George H.
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?
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
How would you verify it? (Crack it open and look for light?) Do you get light out of 7 volt zeners?
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.)
George H.
Don't know. I think zeners and even CMOS gates make a tiny amount of light. In CMOS, the light has been used for logic diagnostics, with massive signal averaging.
Something like that maybe.
I think Phil has found PHEMTs that can bias their own gates negative a little, sort of an electron venturi effect.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
I haven't tried cutting the top off a TO-18 part (yet) but photon emission by avalanche breakdown in silicon has been known a long time:
piglet
Great thanks, I think McKay was "Mr." avalanche breakdown at bell labs, I've read some of his other papers.
George H.
I did and yes, there is light.
Jeroen Belleman
It explained the construction, but are you really sure Pease presents the correct explanation? Photons generated in silicon?! My guess would be some kind of thermoelectric effect.
Thank you.
Best regards, Piotr
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.
For an NPN the electrons would be avalanching the wrong direction, from base to emitter, headed away from the collector, right?
An LED-to-LED coupler would be cool. You can make several volts without needing +12v for Vbe(br), and with much better efficiency.
Ordinary photocouplers should work too, as long as the base terminal is brought out. Saves all that optical shielding hassle too.
Cheers, James Arthur
Well (as I assume you know) the breakdown voltage depends on the doping level, and how far electrons have to travel (mean free path and all) till they get enough energy to ionize impurity atoms. (piglet's article link sounds nice.)
Oh, yeah... never mind. Since there is light the led->PD idea sounds fine.
There are these (mosfet drivers???) where one led couples to a whole string of PD's to make a high voltage (at low current). I was stringing a few together and getting ~100V or so. George H.
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.)
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
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
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.)
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.
I'd be more interested in the short-circuit current.
Sure. But the quantum efficiency of hot carriers generating visible photons is super low--it's happening in an indirect-bandgap semiconductor.
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
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.)
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
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.)
George H.
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
OK 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 65The 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!)
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