A very silly circuit

OK so no 20% changes.

No no, that's not your part. piglet is the quite guy in the corner, who mumbles the right answer. :^) :^) George H.

Well if not more photons from Vbe, maybe less carriers bumping into thermal stuff, (phonons.. mumble..) and hence more available for radiation producing interactions.

George H.

OK my hand wavy model looks like this... (I haven't tried to stick in numbers... It's likely wrong.)

(I'm not sure about the small positive voltage, this is only the lack of negative.)

If the photons are blue-ish then the absorption coef, penetration length can be short, ~100nm. And if the collector is lightly doped, the depletion layer is wide, and almost all of the energy loss (photon generation) is happening more than a penetration length away. Only a few photons make it out of the collector. (and are lost in whatever makes the small positive voltage. :^)

George H.

Hmm, I was thinking the other night that Bob was a smart guy, and he may have tried zenering the cb. Explaining the lack of a negative signal there seems key. I must admit I don't understand the fully depleted idea. I thought the much higher Vcb reverse voltage meant most depletion is in the collector... Hmm, but how much charge is involved, is there more one way or the other? is there any way to tell if you don't know the doping?

You can measure capacitance. C*V gives some charge. (measuring the C at 100V bias is not something I can do, without some work.)

Hey for the photon idea, Take off the "top" and see how much light there is when you zener cb. (Well that only checks my no photons get through idea.) Can you grind a to-92, dissolve the plastic?

On 15/10/2015 23:04, Phil Hobbs wrote: Do you know the burden

When reading 1uA the p.d. across it is 1mV. Or in other words the meter is 1kohm.

piglet

Here are two photos:

My setup of an old TO-18 npn with lid snipped open.

Shows a spot of light from the die when E-B broken down. Note the camera gets the colour badly wrong - the image redish/orange is really blueish-white just as Jeroen described. I had to use 20mA to get it bright enough to photograph and then I still had to enhance the image.

I tried breaking down the C-B junction but saw no light, but lack of high enough voltage meant avalanche current was 400-500uA (which when E-B breaks down does make enough light for the eye to see).

However I can confirm that the E-B junction is photo-sensitive to room lighting like the C-B junction. So if C-B avalanche were to emit photons then I think the E-B junction ought to respond. My guess is therefore that only E-B breakdown makes light.

piglet

Hi piglet, I thought I might try this today.

Measure the current as a function of Vcb, which I think was one of Phil's ideas... though I'm not sure. Hopefully he'll respond. In theory this should show no change in the current if it's a photo thing. (Well at very small voltages maybe a small increase.) And I think the diffusion idea would show more current with higher ~10's of volts on Vcb. How much more I'm not sure.

Anyway I'm going to buckle down, get my work done and may have some time in the afternoon.

Later, George H.

Finished work at ~3:00 and by 3:40 my lab bench looked like this,

So I've got a nice TIA to measure the base-collector current. First results for zero bias on Vbc.

Ibe(mA) Ibc(nA)

0.303 32 0.605 64 1.02 108 2.003 210 4.01 407 8.01 760

The 3904 is linear now too...

I then tried to take data at a fixed Ibe and change Vcb. But it didn't work so well, the emitter base current would drift around. (probably not the most stable power supplies.) So then I looked at changes in Icb as I switched Vcb from 0 to 50 V. In which case I say about a 1% increase in the Icb

Ibe(mA) Icb(Vcb=0) Icb(Vcb=50V)

1.003 99.6 100.7 2.003 198.1 199.8 4.01 387.7 391.1

Phil is this the magnitude of the expected change for your diffusion idea?

As some final data I was trying to see if the change is linear in Vbc. The problem is things are not at all stable at the 1% level. I set Ibe for 2.0 mA measured Icb, increased Vcb to some voltage, record number, reduce back to zero, record number.

So the nominal current was ~190nA, the change at 50 V was 1.85 nA 25 V was 1.2 nA 11 V was 0.9 nA

For whatever it's worth it doesn't look to be linear. (maybe squarerootish)

George H.

[and charged particles are accelerate and and generate energy}

It might be geometry; the CB breakdown generates carriers over the full area of the base, but you only harvest current in the smaller area of the BE depletion region. Any charge that recombines before it gets to the depletion zone doesn't get any acceleration, and no net energy is generated. It might be amusing to experiment on CMPT404A or other chopper transistor with symmetric beta (which implies some geometry symmetry).

Another issue is the thinner BE depletion region, due to higher doping. That charge cloud has to separate while traversing the depletion region, or no net generation occurs. If it spends little time in the region, the separation (0.5 * a *t**2) is small regardless of the high field intensity, the "a" factor.

Hmm OK, My understanding of the diffusion idea, is that all it takes is charge in the base, then some diffuse into the other junction. I was thinking that the same base current should cause similar amounts of diffusion current.

George H.

I have a final piece of temperature data, I don't have freeze spray, but I hit the transistor with a heat gun (on low) and warmed it up till Icb decreased by ~10%, At the same time the

2 mA of eb current (monitored with 1 k ohm, driven from a voltage source) decreased by ~5%. (The numbers are a bit loose, holding a heat gun, looking at meters and writing down numbers.)

The measurements were disappointing in that the Icb (@constant Ibe) decreased by ~10% during the ~1/2 hour I was taking data. Zenering eb at 1-2 mA is not so good. (at least for 2N3904's)

piglet, I think I'm done playing with this. I'm not sure of the mechanism, I'd lean more towards the photo idea, mostly because of the temperature data, but I wouldn't be surprised if Phil was right.

"When you wake up in the morning, Pooh," said Piglet at last, "what's the first thing you say to yourself?"

"What's for breakfast?" said Pooh. "What do you say, Piglet?"

"I say, I wonder what's going to happen exciting today?" said Piglet.

Pooh nodded thoughtfully. "It's the same thing," he said."

-- A.A. Milne

We'll find some exciting thing tomorrow. :^) George H.

Gosh, George it is exciting to read another post of yours! That is very good work. C-B does look to me like a photodiode.

It does look (pun accidental) as if the photoelectric idea is at play here. I can see (pun intended) light coming from an avalanched E-B jct and I can see (pun intended) the C-B jct is photo-sensitive to external room light so placing the two next to each other looks (wow-another pun) kinda obvious. However I am also happy to accept diffision could also be happening, solid-state physics is pretty much a closed book to me.

piglet

BTT for the Bob Pease e-b zener negative bias generator discussion (for Gerhard, over in the Two-transistor circuits thread)

Nice measurements by piglet and George Herold.

Cheers, James Arthur

There are photovoltaic optocouplers that produce small but useful amounts of output power, hundreds of microwatts, ballpark 10 volts and

20 uA. Most small SSRs use them internally. I've used them as linear gate drivers in high-voltage amps.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

Sometime in the past I tried stacking up a bunch of the PV opto-couplers to make a 5V- 100V DC source. The output impedance was the pits.

George H.

ADI has magnetic couplers that transfer useful power, too. Ain't cheap, though.

I use a lot of those small dual-winding inductors (DRQxxx types) as switcher power transformers.

The ADI things are practically broadcast transmitters.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

I didn't have any trouble getting one past the FCC.