A very silly circuit

On 10/15/2015 03:33 PM, George Herold wrote:
> On Thursday, October 15, 2015 at 12:49:34 PM UTC-4, piglet wrote:
>> On 13/10/2015 16:55, Phil Hobbs wrote:
>>>>> I'd expect the temperature dependence to be the easiest thing to
>>>>> nmeasure.
>>
>> I've just tried with freezer spray and got an increase in C-B voltage.
>> Device MPSH-10 in Veb breakdown Ie=1mA, at room temp Vcb -529mV, Icb
>> -1.0uA, after freezer spray Vcb -633mV, Icb -1.2uA (Ie was constant).
>>
>> Perhaps when I get more time I'll try a more scientific plot vs temp.
>>
>> Does the direction of the change help you nail down the effect to
>> photo-voltaic or carrier diffusion? Pease vs Hobbs?
>
> Hmm, well the forward voltage of diodes goes up when the temperature goes
> down, so Vcb is not surprising.  More current is more interesting.
>
> For the photo-voltaic model there could be a number of reasons.
> 1.) More photons, perhaps at lower temps the avalanche makes more photons.
> did the Vbe reverse voltage (at 1 mA current) go up too?  That would mean
> more energy has to be lost... more light.
> 2.) Well other hand wavy ideas.. but #1 looks best.  It would be sweet if
> Vbe went up by 20%.
>
> I still think that the lack of a negative voltage when we zener the bc junction
> still has to be explained in the Hobbs diffusion model.

It has to be explained in either model, ISTM.  The _positive_ BE voltage 
looks to me as though the base is fully depleted, so that the emitter is 
effectively shorted to the base.

(Given that I was quite fond of Bob, I'd be happy if he was right about 
this one.)

Cheers

Phil Hobbs

-- 
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

On Thursday, October 15, 2015 at 5:51:01 PM UTC-4, piglet wrote:
> On 15/10/2015 20:33, George Herold wrote:
> > did the Vbe reverse voltage (at 1 mA current) go up too?  That would mean
> > more energy has to be lost... more light.
> 
> Aw shucks, I neglected to monitor the Veb, just kept eye on the current 
> Ie. The emitter was fed via 10kohm so near a constant current, I did not 
> have to adjust the psu voltage to keep Ie constant and it did not waver 
> around more than 5-10uA (out of 1000uA).
OK so no 20% changes.  
> 
> I know PV solar cell output is higher at low temp which may support the 
> optic explanation. But I do not understand the thermal diffusion 
> hypothesis enough to know if that is proved or dis-proved by these 
> observations.
> 

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

On Thursday, October 15, 2015 at 6:04:34 PM UTC-4, Phil Hobbs wrote:
> On 10/15/2015 12:49 PM, piglet wrote:
> > On 13/10/2015 16:55, Phil Hobbs wrote:
> >>>> I'd expect the temperature dependence to be the easiest thing to
> >>>> nmeasure.
> >
> > I've just tried with freezer spray and got an increase in C-B voltage.
> > Device MPSH-10 in Veb breakdown Ie=1mA, at room temp Vcb -529mV, Icb
> > -1.0uA, after freezer spray Vcb -633mV, Icb -1.2uA (Ie was constant).
> >
> > Perhaps when I get more time I'll try a more scientific plot vs temp.
> >
> > Does the direction of the change help you nail down the effect to
> > photo-voltaic or carrier diffusion? Pease vs Hobbs?
> >
> > "Ring if an answer is required, knock if an answer is not required"
> 
> Not 100% sure.  The increased Vcb is expected, the increased I_CB less 
> so--naively, photodetection should leave it constant, and diffusion 
> should (I think) reduce it at low temperature.  Do you know the burden 
> voltage of your meter?  If it's anything significant at 1 uA, it might 
> explain the change. (It probably isn't.)
> 
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. 

 
> Cheers
> 
> Phil Hobbs
> 
> 
> -- 
> Dr Philip C D Hobbs
> Principal Consultant
> ElectroOptical Innovations LLC
> Optics, Electro-optics, Photonics, Analog Electronics
> 
> 160 North State Road #203
> Briarcliff Manor NY 10510
> 
> hobbs at electrooptical dot net
> http://electrooptical.net

On Thursday, October 15, 2015 at 6:06:34 PM UTC-4, Phil Hobbs wrote:
> On 10/15/2015 03:33 PM, George Herold wrote:
> > On Thursday, October 15, 2015 at 12:49:34 PM UTC-4, piglet wrote:
> >> On 13/10/2015 16:55, Phil Hobbs wrote:
> >>>>> I'd expect the temperature dependence to be the easiest thing to
> >>>>> nmeasure.
> >>
> >> I've just tried with freezer spray and got an increase in C-B voltage.
> >> Device MPSH-10 in Veb breakdown Ie=1mA, at room temp Vcb -529mV, Icb
> >> -1.0uA, after freezer spray Vcb -633mV, Icb -1.2uA (Ie was constant).
> >>
> >> Perhaps when I get more time I'll try a more scientific plot vs temp.
> >>
> >> Does the direction of the change help you nail down the effect to
> >> photo-voltaic or carrier diffusion? Pease vs Hobbs?
> >
> > Hmm, well the forward voltage of diodes goes up when the temperature goes
> > down, so Vcb is not surprising.  More current is more interesting.
> >
> > For the photo-voltaic model there could be a number of reasons.
> > 1.) More photons, perhaps at lower temps the avalanche makes more photons.
> > did the Vbe reverse voltage (at 1 mA current) go up too?  That would mean
> > more energy has to be lost... more light.
> > 2.) Well other hand wavy ideas.. but #1 looks best.  It would be sweet if
> > Vbe went up by 20%.
> >
> > I still think that the lack of a negative voltage when we zener the bc junction
> > still has to be explained in the Hobbs diffusion model.
> 
> It has to be explained in either model, ISTM.  The _positive_ BE voltage 
> looks to me as though the base is fully depleted, so that the emitter is 
> effectively shorted to the base.

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.  



> 
> (Given that I was quite fond of Bob, I'd be happy if he was right about 
> this one.)

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?



> 
> Cheers
> 
> Phil Hobbs
> 
> -- 
> Dr Philip C D Hobbs
> Principal Consultant
> ElectroOptical Innovations LLC
> Optics, Electro-optics, Photonics, Analog Electronics
> 
> 160 North State Road #203
> Briarcliff Manor NY 10510
> 
> hobbs at electrooptical dot net
> http://electrooptical.net

On 15/10/2015 23:04, Phil Hobbs wrote:
   Do you know the burden
> voltage of your meter?  If it's anything significant at 1 uA, it might
> explain the change. (It probably isn't.)
>

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

piglet

On 16/10/2015 03:51, George Herold wrote:
> 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?
>

Here are two photos:

<https://www.dropbox.com/s/6rkc80amj2aybuk/Veb_Setup.jpg>
My setup of an old TO-18 npn with lid snipped open.

<https://www.dropbox.com/s/2fvi5i6tsphp0yg/Veb_AvalancheLEQ.jpg>
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

On Friday, October 16, 2015 at 8:48:30 AM UTC-4, piglet wrote:
> On 16/10/2015 03:51, George Herold wrote:
> > 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?
> >
> 
> Here are two photos:
> 
> <https://www.dropbox.com/s/6rkc80amj2aybuk/Veb_Setup.jpg>
> My setup of an old TO-18 npn with lid snipped open.
> 
> <https://www.dropbox.com/s/2fvi5i6tsphp0yg/Veb_AvalancheLEQ.jpg>
> 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.
https://www.dropbox.com/s/ja9ce4cgvtt8zz3/Icb_vs_Vcb.JPG?dl=0

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. 

On Friday, October 16, 2015 at 9:27:23 AM UTC-4, George Herold wrote:
> On Friday, October 16, 2015 at 8:48:30 AM UTC-4, piglet wrote:
> > On 16/10/2015 03:51, George Herold wrote:
> > > 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?
> > >
> > 
> > Here are two photos:
> > 
> > <https://www.dropbox.com/s/6rkc80amj2aybuk/Veb_Setup.jpg>
> > My setup of an old TO-18 npn with lid snipped open.
> > 
> > <https://www.dropbox.com/s/2fvi5i6tsphp0yg/Veb_AvalancheLEQ.jpg>
> > 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.
> https://www.dropbox.com/s/ja9ce4cgvtt8zz3/Icb_vs_Vcb.JPG?dl=0
> 
> 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,
https://www.dropbox.com/s/d8y8hpwyy5jbzw0/DSCN0181.JPG?dl=0

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. 
 

On Tuesday, October 13, 2015 at 5:45:58 PM UTC-7, George Herold wrote:
> On Tuesday, October 13, 2015 at 7:21:19 PM UTC-4, whit3rd wrote:

> > There's a built-in field in a junction (as long as you're away from zero temperature)
> > because of diffusion of high-density (majority) carriers in both the N and P regions.
 [and charged particles are accelerate and and generate energy}
> > I don't suppose any SPICE models really capture this effect, do they?

> OK can you explain why we see nothing when we zener the BC jucntion 
> at ~100V and 1-2 mA? The steeper BE junction should 
> pull in more of the base carriers. No?  
> (This is fun! :^)

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.

On Friday, October 16, 2015 at 5:50:54 PM UTC-4, whit3rd wrote:
> On Tuesday, October 13, 2015 at 5:45:58 PM UTC-7, George Herold wrote:
> > On Tuesday, October 13, 2015 at 7:21:19 PM UTC-4, whit3rd wrote:
> 
> > > There's a built-in field in a junction (as long as you're away from zero temperature)
> > > because of diffusion of high-density (majority) carriers in both the N and P regions.
>  [and charged particles are accelerate and and generate energy}
> > > I don't suppose any SPICE models really capture this effect, do they?
> 
> > OK can you explain why we see nothing when we zener the BC jucntion 
> > at ~100V and 1-2 mA? The steeper BE junction should 
> > pull in more of the base carriers. No?  
> > (This is fun! :^)
> 
> 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.