The Cryogenic P-N Junction

[CaptnYellowShirt]

In the best tradition of volt-nutting, I've recently been spending a good chunk of my free time dreaming up ways to generate a more stable voltage reference.  So, I've been reading up on zener-based voltage references, checked out a few books on semiconductor physics, meditated under a waterfall on the nature of the p-n junction, and (would you believe it?) watched a few you-tube videos.

All of this has left me with some serious questions about some of the explanations I've read. In trying to sort the wheat from the chaff here, I've conducted the following experiment. I'd like your help with it, because I'm at a bit of a loss...

Here's my understanding of how the p-n junction works: Take a slab of silicon and sprinkle in some p-type and n-type dopants on either side. The unequal concentration of ions causes a diffusion of electrons and holes in opposite directions across the p-n interface. At some unspecified period after this diffusion starts, the diffusion pressure comes into balance with the electric field of the unequal charge being developed on the two sections of silicon. At room temperature, for the geometries and concentrations of dopants normally used, this electric field generates a voltage of around 0.7V  -- your typical 'diode voltage drop'.

So it would make sense if one could measure this voltage? Sure, why not? It's called the "built-in voltage", and here's what I found* for 3 zener-style diodes:

Model	BZX79C18	1N5232B	MTZJT-774.3
Zener Voltage	4.3v	5.6v	18v
Built-In Voltage	0.25279v	0.21200v	0.27350v
Stdev	0.02021	0.00823	0.00080

* HP 3457A - 100 PLC - 10 samples


I don't know about the voltages being different that the 0.6-0.7v that one might have expected here. I'm guessing that has to do with the zener-style doping?

I'm not sure how one might model the 'source' of a diode.... uber high impedance voltage source? uber low current source (diffusion current)? Either way, it should be driven by diffusion, which is fundamentally driven by temperature. So it would follow that if one were to cool the diodes, one should observe a drop in built-in voltage -- and maybe a lower spread?

Well you'd be wrong. And by "you", I mean "I".

At first I tried this with a freezer, and got weird results. Then I tried it with dry ice... and got even weirder results. So today (as evidenced by by freeze-dried beer post an hour or so ago --  https://www.eevblog.com/forum/projects/liquid-nitrogen-chilled-beer/), I bought some liquid nitrogen and tried this experiment at 77K. Here are those results....

Model	BZX79C18	1N5232B	MTZJT-774.3
Zener Voltage	4.3v	5.6v	18v
Built-In Voltage	0.71267v	0.64509v	0.82416v
Stdev	0.00579	0.00296	0.00132

Anybody see anything weird here? Because unless I blacked out from nitrogen asphyxiation, this looks wrong. How could something driven by diffusion strengthen at such a low temperature? And not just strenghten, but increase by a factor of nearly 3?  No, really... I'm asking...

[T3sl4co1l]

How are you measuring the "built in potential"?

I don't recall that you can measure it independent of the junction as well...

(You can also have a doping gradient without a junction; of course, as soon as you connect it with something, it just ends up canceling out, just like with thermocouples.)

If you're measuring the junction potential V = Vth * exp(I/Is), remember that Is varies with temperature as well.  IIRC, the physical manifestation is the density of states around the bandgap.  I recall there's a 3/2 power in there somewhere, but I forget what.  Should review my textbook...

Tim

[CaptnYellowShirt]

How are you measuring the "built in potential"?

That's a good question. It crossed my mind that maybe I'm measuring gibberish...

As seen in the pictures, the diodes are attached to a short (~12") twisted pair of belden 24awg hook-up wire (FR-1 insulation - pvc). The leads run back to my HP 3457a in its DC voltmeter mode. Regular 60/40 solder was used.

First, I wondered if I was measuring anything at all... maybe its just a float voltage? But these numbers were oddly specific (and with such little spread) to be a simple float charge. To double check,  I shorted the leads a few times on a few tests and found that over the next 30-60secs, the voltage would rise back up to the previous level and stay there.

Second, I thought maybe it was a thermal issue. But since all the junction pairs are at the same temperature... I'm not sure if it could generate 1/2v ?

What's left? Photoelectric effect? Charge injection via air currents? Maybe I'm missing something simple.

[Rerouter]

to remove your measuring setup from the equation, reverse the direction of the diode, if the voltage stabilises at the same voltage but reveresed polarity your onto something, otherwise it could be leakage current and the zeners leakage is reducing as it cools causing a greater drop,

[muvideo]

Hello,
I'm ignorant here, I havent studied semiconductors thoroughtly (yet),
but thinking to energy conservation one questions crosses my mind:
a normal voltmeter can measure a static potential stealing a small
current from the measurement points. The highest the input impedance,
the lowest the current needed to obtain a measure* .
So for a p-n junction (in thermal equilibrium and in total darkness)
from where does come the electon flow needed to take the measurement?

Another question is: how is the behaviour of the metal contacts
of the leads to semiconductor? Dont do they become junctions as well?

*I dont know your meter data, but probably it's input "resistance" will
be equivalent to tens of GOhm, that means it will require something around
10^-10 to 10^-12 A to take a measurement at 0.7V
The best electrometers still require flows of thousands electrons per second

[ConKbot]

Off the top of my head, I'd guess the diode and lead wires were picking up ambient EM fields (or if you were holding the wires, 60 Hz), and partially rectifying them.  The leakage of the diode would drop as you chill it, so may be leaking less, and the voltage you measure increasing.  Being zener diodes, I could see their reverse leakage not being optimized like it would on a silicon diode.

[T3sl4co1l]

Another question is: how is the behaviour of the metal contacts
of the leads to semiconductor? Dont do they become junctions as well?

Yes, Schottky junctions in fact.  There are two kinds: ohmic and rectifying.  Just about any material making a junction to normally-doped semiconductor results in a rectifying junction -- though particular materials have the most useful Schottky barrier characteristics, I think usually platinum silicide or something like that.

The standard technique to make ohmic contacts in silicon is to heavily dope the surface, then apply a similar metal: for example, n+ doping with aluminum on top (if any aluminum diffuses in, it only becomes more n+).  The n+ doping level is similar to that used in silicon "wires" -- poly-Si, etc., and is fairly conductive already.  Basically, making the silicon look as much like a metal as possible before actually connecting it to a suitable metal.  I think this works just fine on p+ doping as well, but I'm not sure if there's an extra step there.

Tim

[free_electron]

tempco of a diode is -2mV/degree kelvin...
drop it 200 degrees below ambient (77K ) and you gain 400millivolts.

[CaptnYellowShirt]

tempco of a diode is -2mV/degree kelvin...
drop it 200 degrees below ambient (77K ) and you gain 400millivolts.

What mechanism is responsible for the tempco?

[CaptnYellowShirt]

Here's some more data for you all to chew on.  I also measured the Zenver voltage before, during, and after their plunge into the icy depths of my 1/2gal igloo thermos. ~20mA reverse bias current (very stable source). Here's what I found...


Before...

Model	BZX79C18	1N5232B	MTZJT-774.3
Zener Voltage	4.3v	5.6v	18v
Zener Voltage	-4.6154v	-5.6967v	-19.4390v
Stdev	0.00024	0.002587	0.0481

During....

Model	BZX79C18	1N5232B	MTZJT-774.3
Zener Voltage	4.3v	5.6v	18v
Zener Voltage	-4.6991v	-5.3469v	-15.5830v
Stdev	0.0000068	0.0000248	0.000394

After....

Model	BZX79C18	1N5232B	MTZJT-774.3
Zener Voltage	4.3v	5.6v	18v
Zener Voltage	-4.6187v	-5.7021v	-19.5302v
Stdev	0.0001092	0.0005127	0.01489

Its interesting to note that after the diodes were chilled and warmed back up again, they seem to have 'calmed down' a bit in their reverse bias voltage -- the spread is much tighter.

[robrenz]

What about all the thermal EMF's going on in that setup? are they exactly cancelling or not?

[CaptnYellowShirt]

What about all the thermal EMF's going on in that setup? are they exactly cancelling or not?

I'm thinking so. The two solder junctions in the LN2 are at the same temp and the two lugs on the front of the HP DMM are (basically) at the same temp.

[T3sl4co1l]

'Sides, even with such a large temperature difference, it'll only amount to a few milivolts?
Tim

[uwezi]

It's getting late and later tonight 

T3sl4co1l is right: you cannot measure the built-in voltage! So in order to analyze your data you need to tell us, how you measured...

When you measured the unloaded potential difference of the diodes with an instrument with high enough input impedance you might have measured the open-circuit voltage of the diodes, while these were working as tiny, tiny solar cells.

If you measured the forward voltage drop at an arbitrary forward current through the diodes, then it just this what you measured, together with the associated temperature coefficient of this voltage - as stated above, around -2 mV/K.

So, how can one determine the built-in voltage? There are some ways to do it, either using the variation of the junction capacitance with applied voltage - running the diodes as varactors,
http://arxiv.org/pdf/1011.3463

or from opto-electronic measurements, preferably at different temperatures and light-intensities...

[ejeffrey]

Hello,
I'm ignorant here, I havent studied semiconductors thoroughtly (yet),
but thinking to energy conservation one questions crosses my mind:
a normal voltmeter can measure a static potential stealing a small
current from the measurement points. The highest the input impedance,
the lowest the current needed to obtain a measure* .
So for a p-n junction (in thermal equilibrium and in total darkness)
from where does come the electon flow needed to take the measurement?

This is *exactly* the right question to be asking.

If the PN junction is at the same temperature as the volt meter, in the dark, and the volt meter is a perfect meter with a finite input impedance, and there are no external noise sources, the measured DC voltage will be zero.  There will be some fluctuations around zero (e.g., johnson noise), but the DC component has to be zero.

So, if you measure something else, one of those assumptions has to be violated.  That means that the (room temperature) results are meaningless because they are caused by some uncontrolled error source.  Possible sources of error:

1) The input bias current of the meter is biasing the junction.  Unless you have a crappy meter or a very low leakage diode, this shouldn't be a problem.
2) You are picking up some electromagnetic noise in your leads, and rectifying it with the diode.  I think this is the most likely.
3) Thermocouple effects -- your system is not in thermal equilibrium.  This could happen at any junction between dissimilar metals, but the seebeck coefficient of PN junctions is particularly high.  Still, I don't think this can account for the 200 mV difference at room temperature.
4) Light.  If you use glass body diodes, they act as photodetectors and will self-bias from the photocurrent.

Of course, the change in voltage when you cryogenically cool it is due primarily to the thermocouple effect.  It won't be an accurate measurement, since whatever sources of error are giving you the 200-300 mV error at room temperature probably won't stay constant.

If you want to know how to model the small signal impedance of an unbiased diode, you can estimate it by taking the derivative of the shockley diode equation and evaluating it at zero voltage.  You end up with dV/dI = n*Vt/Is. n is the ideality parameter (estimate 2 for most diodes), Vt is the thermal voltage (25 mV), and Is is the saturation current of the diode.  Typical values will be a couple of megaohms at room temperature, much less at elevated temperature.

[CaptnYellowShirt]

Here's what I'm looking at:



The caption reads: "A p–n junction in thermal equilibrium with zero-bias voltage applied. Under the junction, plots for the charge density, the electric field, and the voltage are reported."

[T3sl4co1l]

The diagram is missing thermally generated charges, which diffuse across the depletion region and deliver current.  You have to deliver the same current in order to achieve "zero bias" as in the diagram.  Otherwise, with no external current flow, the voltages balance out, so that no terminal voltage is generated, which is as should be expected.

Tim

[jpb]

The temperature variation in forward characteristics is normally derived from the Einstein relation (NO not E=mc^2!!)

http://en.wikipedia.org/wiki/Einstein_relation_(kinetic_theory)

this gives the balance between diffusion and drift currents (diffusion is just random movement of electrons/holes because there are more in one place than another and drift due to the electric field).

The built-in voltage should be proportional to T but for the zeners you seem to have found the opposite! Edit: VT is proportional to T but you need to also take into account the thermal dependence of the reverse saturation current as pointed out by uwezi below

The reason for this is probably that the conductivity drops with a dropping temperature in a semiconductor. Less electrons are able to escape into the conduction band so what you are probably measuring is an increase in the resistance of the ohmic contacts or of the part of the diode that is just conducting.

Note I'm not an expert on Zeners so may have this completely wrong. The thing is Zeners are designed to be reverse-biased and break-down not be forward biased.

[uwezi]

The built-in voltage should be proportional to T but for the zeners you seem to have found the opposite!

No, the voltage over a diode under forward conduction at a constant current drops with increasing temperature! For silicon diodes (as probably all zener diodes on the market) with about -2 mV per Kelvin (or centigrade).

The reason for this is probably that the conductivity drops with temperature in a semiconductor. Less electrons are able to escape into the conduction band so what you are probably measuring is an increase in the resistance of the ohmic contacts or of the part of the diode that is just conducting.

The conductivity of a semiconductor increases with increasing temperature, because you get more and more electrons which are thermally excited from the valence band to the conduction band.

In metals the conductivity drops with increasing temperature, because the increasing thermal vibration of the metal atoms disturbs the path of the electrons.

[jpb]

The built-in voltage should be proportional to T but for the zeners you seem to have found the opposite!

No, the voltage over a diode under forward conduction at a constant current drops with increasing temperature! For silicon diodes (as probably all zener diodes on the market) with about -2 mV per Kelvin (or centigrade).

The reason for this is probably that the conductivity drops with temperature in a semiconductor. Less electrons are able to escape into the conduction band so what you are probably measuring is an increase in the resistance of the ohmic contacts or of the part of the diode that is just conducting.

The conductivity of a semiconductor increases with increasing temperature, because you get more and more electrons which are thermally excited from the valence band to the conduction band.

In metals the conductivity drops with increasing temperature, because the increasing thermal vibration of the metal atoms disturbs the path of the electrons.

Yes, I meant to say the conductivity drops as the temperature drops (which is what was happening in the experiment).

V_T is proportional to temperature (i.e. equal to kT/q) so for a given voltage the current should increase, so for a constant current the voltage across the junction should drop if temperature is decreased (as in the experiment). The general conductivity, as discussed, increases as T increases which goes the other way I think.

Assuming the ideality factor n is 1 (for ease of maths) then the diode equation is I₀(e^V/V_T-1) so given V_T = kT/q if T goes to half T then to keep the exponential term constant V also has to go to half its previous value.

Edit:
BUT this doesn't take into account the temperature dependence of I₀ , see uwezi's post below.

[CaptnYellowShirt]

I'm going to re-try the experiment in the next few days (maybe earily next week). My LN2 dewar was an ebay special, and as such I think the vacuum seal is broken -- the LN2 only sticks around for about 10 hours after I buy it. So, I'd like to get all my experimentation setup, have a clear test procedure outlined, and then go buy the stuff.

I'd like to get everybody's input for what I should try next. The main goal of the first experiment was to see if I could measure the "Built-In Voltage" to test my understanding of the quasi-steady, zero current bias state of the diodes. What I'm really struggling with here is the notion that V_bi (or *anything* for that matter) is increasing as I cool the p-n junction. At some point it has to go down? What happens at 2K?

So remeasuring V_bi is the first thing I'd like to retry. From the comments I've read already there have been a few suggestions: I may be measuring something else other than the built-in voltage -- this may be due to a) thermal issues, b) current bias of the voltmeter, c) the metal-semiconductor junctions in the diode, d) photo-electric effects from the two diodes that were glass. Anymore?

The thermal issues are easy to check -- I can just reverse the hookup and re-measure. The bias current of the HP 3457a is hard to measure. It has a input impedance of something like 10 GOhms, and its voltmeter-mode current measures in (with another 3457a) at +/-0.1nA. Which, to my mind is zero for this experiment. Any more comments on this would he helpful -- I have access to an electrometer and semiconductor characterization object if these would be more helpful.  I don't know much about this metal-semiconductor junction idea. I can see where every diode must be some combination of three diodes in one -- (M-S(p)-S(n)-M configuration). I can't think of anyway to test the built-in voltage without invoking this other diode idea? For d) I'm going to black out the test chamber next time.

But I have an opportunity here to test a set of things that are beyond typical range of most hobbyist's labs. So if anyone would like to suggest an experiment they'd like to see the results of, let me know. Forward/reverse bias voltages on any type of diode? I have a set of MOV's that I'm going to test too... Etc etc.  There also exists the possibility of obtaining liquid helium, if the experiments show promise.

[CaptnYellowShirt]

<eating my hat>

So I said the input impedance to the HP347a was really high. But I decided to actually go test it.... the actual input impedance of the HP 3457a in the range it's operating in is a little less than 10MOhms. (9.9946 MOhms on mine)

Breakout the electrometer....

[uwezi]

Yes, I meant to say the conductivity drops as the temperature drops (which is what was happening in the experiment).

Which would be normal for a semiconductor - but the conductivity was never measured. In order to measure the conductivity you need a piece of semiconductor without a junction! Then you will find that the conductivity is

\sigma = q ( µ_n * n + µ_p * p)     // let's see if the forum can keep a \mu in place

where \mu is the mobility of either electrons (n) or holes (p), where n and p are the density of free electrons and holes. The mobility might have a slight temperature dependence as well, but the main part here is that the density of free electrons and holes increases exponentially with increasing temperature.

V_T is proportional to temperature (i.e. equal to kT/q) so for a given voltage the current should increase, so for a constant current the voltage across the junction should drop if temperature is decreased (as in the experiment).

V_T is not the built-in potential, it is the thermal voltage, or rather the mean thermal energy of an electron divided by its elementary charge.

Assuming the ideality factor n is 1 (for ease of maths) then the diode equation is I₀(e^V/V_T-1) so given V_T = kT/q if T goes to half T then to keep the exponential term constant V also has to go to half its previous value.

You got the diode equation alright, but you are neglecting the temperature dependence of I₀ !

I₀ = I₀₀ e^-Eg/kT

For the analysis let's assume that the applied voltage is much higher than V_T which is 26 mV at room temperature, then we can neglect the -1 in the parenthesis and the diode equation becomes (assuming ideality factor n=1)

I = I₀₀ e^-Eg/kT e^qV/kT =  I₀₀ e^(qV-Eg)/kT

In order to see in which direction the temperature coefficient goes, we have to look into (qV-Eg)/kT. Your interpretation is completely correct: to keep the exponent constant with changing temperature. I.e. for half T we need a voltage V which makes (qV-Eg) half the value as well - for all reasonable voltages over a diode, qV < Eg.

log I = log I₀₀ + (qV-Eg)/kT

V = kT/q * log I/I₀₀ + Eg/q

This seems to still indicate a positive temperature coefficient, but wait a second. How big is this strange I₀₀?

Let us take a standard 1N4148: NXPs datasheet states about 0.67 V at 5 mA, it's a silicon diode with a band gap of 1.1 eV

0.67 V = 0.026 V * log (5 mA/I₀₀) + 1.1 V

we see that the logarithm must be negative!

-16.54 = log (5 mA/I₀₀)  or then at 5 mA:  V = 1.1 V - 16.54 * kT/q

which gives an I₀₀ =  76200 A

There is of course no current of 76200 A anywhere inside a small 1N4148, it is just a parameter in the equation. But it is the reason why the diode equation gives us a dropping forward voltage over a diode with increasing temperature.

Look again at the equations above. For T=0 K at absolute zero the diode equation predicts that the forward voltage drop independent of current will reach the value of the band gap in eV. But even before we reach these low temperatures with a real diode, other things will happen...

[sync]

The bias current of the HP 3457a is hard to measure.

You can simplly measure it with a resistor. Connect a high value resistor directly to the multimeter, e.g. 1Mohm. Short the resistor (alligator leads, etc). Put the meter into DCV and wait until the readings have stabilized (Thermal EMF, etc). NULL the meter and remove the short. Now the readings are the voltage across the resistor caused by the bias current.

[CaptnYellowShirt]

The bias current of the HP 3457a is hard to measure.

You can simplly measure it with a resistor. Connect a high value resistor directly to the multimeter, e.g. 1Mohm. Short the resistor (alligator leads, etc). Put the meter into DCV and wait until the readings have stabilized (Thermal EMF, etc). NULL the meter and remove the short. Now the readings are the voltage across the resistor caused by the bias current.

A 1MOhm resistor directly across the terminals gives about 2uV... 2*10^-12A

The electrometer reports about 1.82*10^-12A.


This is interesting though... The other HP 3457a gives values around 8.5*^-12A. The only difference (beyond something that happened at the factory?) is the one reporting the higher leakage current has rear/external terminals -- which are shoved up against wall -- while the other one doesn't.... Something for me to investigate tomorrow.

Beyond that, playing around with the electrometer and the diodes in an unscientific way shows three things: 1) Light does have an effect on the glass diodes,  2) the room temperature current range is around 10^-12A, and 3) scratching my beard near the thing wildly effects its readings.

So if I'm going to get good measurements on the second try, I'm going to need to be mindful of these picoamp sources.