The Cryogenic P-N Junction - Page 1
Forum section: electronics / projects

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 25, 2014, 05:04:19 am
content: 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:ModelBZX79C181N5232BMTZJT-774.3Zener Voltage4.3v5.6v18vBuilt-In Voltage0.25279v0.21200v0.27350vStdev0.020210.008230.00080* HP 3457A - 100 PLC - 10 samplesI 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....ModelBZX79C181N5232BMTZJT-774.3Zener Voltage4.3v5.6v18vBuilt-In Voltage0.71267v0.64509v0.82416vStdev0.005790.002960.00132Anybody 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...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 25, 2014, 05:52:59 am
content: 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

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 25, 2014, 06:19:20 am
content: Quote from: T3sl4co1l on February 25, 2014, 05:52:59 amThat'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.

author: Rerouter
author_rank: Super Contributor
timestamp: February 25, 2014, 06:49:23 am
content: 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,

author: muvideo
author_rank: Frequent Contributor
timestamp: February 25, 2014, 07:53:25 am
content: 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 smallcurrent 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 contactsof the leads to semiconductor? Dont do they become junctions as well?*I dont know your meter data, but probably it's input "resistance" willbe equivalent to tens of GOhm, that means it will require something around10^-10 to 10^-12 A to take a measurement at 0.7VThe best electrometers still require flows of thousands electrons per second

author: ConKbot
author_rank: Super Contributor
timestamp: February 25, 2014, 11:42:26 am
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 25, 2014, 06:09:51 pm
content: Quote from: muvideo on February 25, 2014, 07:53:25 amYes, 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

author: free_electron
author_rank: Super Contributor
timestamp: February 25, 2014, 07:12:29 pm
content: tempco of a diode is -2mV/degree kelvin...drop it 200 degrees below ambient (77K ) and you gain 400millivolts.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 25, 2014, 07:17:13 pm
content: Quote from: free_electron on February 25, 2014, 07:12:29 pmWhat mechanism is responsible for the tempco?

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 25, 2014, 07:52:33 pm
content: 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...ModelBZX79C181N5232BMTZJT-774.3Zener Voltage4.3v5.6v18vZener Voltage-4.6154v-5.6967v-19.4390vStdev0.000240.0025870.0481During....ModelBZX79C181N5232BMTZJT-774.3Zener Voltage4.3v5.6v18vZener Voltage-4.6991v-5.3469v-15.5830vStdev0.00000680.00002480.000394After....ModelBZX79C181N5232BMTZJT-774.3Zener Voltage4.3v5.6v18vZener Voltage-4.6187v-5.7021v-19.5302vStdev0.00010920.00051270.01489Its 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.

author: robrenz
author_rank: Super Contributor
timestamp: February 25, 2014, 09:06:01 pm
content: What about all the thermal EMF's going on in that setup? are they exactly cancelling or not?

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 25, 2014, 10:25:36 pm
content: Quote from: robrenz on February 25, 2014, 09:06:01 pmI'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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 25, 2014, 10:38:22 pm
content: 'Sides, even with such a large temperature difference, it'll only amount to a few milivolts?Tim

author: uwezi
author_rank: Unknown
timestamp: February 26, 2014, 12:34:56 am
content: It's getting late and later tonightT3sl4co1l 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/1011.3463or from opto-electronic measurements, preferably at different temperatures and light-intensities...

author: ejeffrey
author_rank: Super Contributor
timestamp: February 26, 2014, 12:36:22 am
content: Quote from: muvideo on February 25, 2014, 07:53:25 amThis 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.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 26, 2014, 02:25:28 am
content: 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."

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2014, 03:29:27 am
content: 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

author: jpb
author_rank: Super Contributor
timestamp: February 27, 2014, 01:57:07 pm
content: 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 belowThe 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.

author: uwezi
author_rank: Unknown
timestamp: February 27, 2014, 07:37:53 pm
content: Quote from: jpb on February 27, 2014, 01:57:07 pmNo, 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).QuoteThe 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.

author: jpb
author_rank: Super Contributor
timestamp: February 27, 2014, 08:31:24 pm
content: Quote from: uwezi on February 27, 2014, 07:37:53 pmYes, I meant to say the conductivity drops as the temperature drops (which is what was happening in the experiment).VTis 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(eV/VT-1) so given VT= 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, seeuwezi'spost below.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 27, 2014, 09:32:01 pm
content: 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.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 27, 2014, 09:57:55 pm
content: <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....

author: uwezi
author_rank: Unknown
timestamp: February 27, 2014, 10:12:01 pm
content: Quote from: jpb on February 27, 2014, 08:31:24 pmWhich 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 ( µn* n + µp* p) // let's see if the forum can keep a \mu in placewhere \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.QuoteVTis not the built-in potential, it is the thermal voltage, or rather the mean thermal energy of an electron divided by its elementary charge.QuoteYou got the diode equation alright, but you are neglecting the temperature dependence of I0!I0= I00e-Eg/kTFor the analysis let's assume that the applied voltage is much higher than VTwhich 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 = I00e-Eg/kTeqV/kT= I00e(qV-Eg)/kTIn 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)/kTV = kT/q * log I/I00+ Eg/qThis 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 eV0.67 V = 0.026 V * log (5 mA/I00) + 1.1 Vwe see that the logarithmmustbe negative!-16.54 = log (5 mA/I00) or then at 5 mA: V = 1.1 V - 16.54 * kT/qwhich gives an I00= 76200 AThere 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...

author: sync
author_rank: Frequent Contributor
timestamp: February 27, 2014, 10:35:00 pm
content: Quote from: CaptnYellowShirt on February 27, 2014, 09:32:01 pmYou 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.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 28, 2014, 12:42:04 am
content: Quote from: sync on February 27, 2014, 10:35:00 pmA 1MOhm resistor directly across the terminals gives about 2uV... 2*10^-12AThe 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.

author: BravoV
author_rank: Super Contributor
timestamp: February 28, 2014, 02:34:23 am
content: Quote from: CaptnYellowShirt on February 28, 2014, 12:42:04 amShaving included ?Nice thread and thanks for sharing this.

author: uwezi
author_rank: Unknown
timestamp: March 01, 2014, 12:31:41 am
content: Whether you trust my derivation or not - it's been quite silent here - I do not only want to give negative feedback.If you want to do interesting stuff at low temperatures consider the following:take a yellow-green LED (GaP-type, not the "modern" InGaN ones), cool it down while running it at a modest forward current (5-10 mA). Look at the color of the light emission and describe it here!if you are in AVR µcontrollers: try to characterize the internal reference voltage and oscillator frequency over temperature. It would be interesting information, which sadly is missing in otherwise brilliant datasheets.These are experiments which I really would like to do myself, if I had the time...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 01, 2014, 03:08:03 am
content: So, back in Inorganic Chemistry, we had a lab about solid state physics... band gap etc. So, we dug out a box of LEDs and lit them up at room temperature and under LN2.The stated premise was that BG is inverse with temperature, so the color changes "up" on cooling (red (GaAsP) towards yellow; blue (GaN) towards violet; etc.).So, naturally, I grabbed a green LED (GaP). And it went red, not blue. Asked the instructor... he said "oh, well this one'sdifferent, go pick a different color"...Chemistry is a lot like English: they like to have their rules and all (compare Lewis Octet Rule to "I before E except after C"), but... they have an awful lot of them. And a far more massive list of exceptions to those rules. So, they aren't really *rules*, are they?I suppose this is true of all fields. Certainly, textbook (HS to undergrad level) electronics has the same problem, where only the simplest, least useful approximations are taught as fact. Thevenin's theorem, maximum power transfer theorem: yes all well and good, but I don't suggest applying both at the same time to *my* audio amp. But go ahead, do try it with *your* amp.Tim

author: uwezi
author_rank: Unknown
timestamp: March 01, 2014, 07:29:57 am
content: Quote from: T3sl4co1l on March 01, 2014, 03:08:03 amThat's actually what I was after! The explanation can be found in the textbook by E. Fred Schubert,http://www.ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/- but it really is not straight forward!

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 04, 2014, 03:40:51 am
content: Sorry its been a few days. Work and all getting in the way of the fun stuff...So in interest of investigating the big difference in leakage currents from my two HP 3457a DMM's as mentioned previously in this post, I cracked both of them open tonight and took a look-see. The things I found were interesting (at least to me).To review, the problem was the two meters were generating wildly different leakage currents (2e-12 A vs 9e-12 A) when operating in their DCV mode. My initial thought was about the rear terminals of the devices -- the one with the large leakage current had them, the other didn't. Perhaps the wall the meters were shoved up against were causing a source effect?The first attached picture shows the leads of the front and rear terminals joining up to the DMM's mainboard. The top leads are from the rear terminals, the bottom are from the front terminals. Clearly they are isolated and switched. However, the spacing there is only a few mm. So I decided to unplug the wires from the rear terminal face... That is the wires are still attached to the mainboard, but the molex-like connector that joins up with the rear face place was unplugged and left resting on top of one of the aluminum shields in the box (local ground - not mains earth).I hooked the meters back up, and what I saw surprised me. Rather than quickly (5sec) floating up to around the +2.5v that was common in both meters before. The modified one *slowly* (50sec) floated up to around 0.6v. The non-modified one continued to quickly float at around 2.5v.So I thought, what's going on here? There's no rear terminals.... oh but there is a switching card! So I opened 2nd meter up and found a similar situation. Here, I did the same thing -- unhooked the molex-like connector from the switching card, wrapped it back up and found that the 2nd meter acted like the 1st one.... taking about 50sec to float up to around 0.6v.Now, with a 1MOhm resistor across the terminals, the first meter measures 1.121 uV and the second measures (1.319 uV)... compared to the 1.82 and 8.5 uV before.Case and point though.... even after the modifications, the hp 3457a would be loading diode with something like 1pA

author: ejeffrey
author_rank: Super Contributor
timestamp: March 04, 2014, 07:10:52 am
content: Quote from: CaptnYellowShirt on February 27, 2014, 09:32:01 pmAs I explained before, the zero current bias state of a diode is zero voltage (if the diode is at the same temperature as the meter). Any non-zero value you measure is a measurement error, of which there are several possible sources.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 04, 2014, 07:44:15 am
content: How do you explain the drawing I posted showing the "Built In Voltage" for a p-n junction in thermal equilibrium? Is that non-sense? I'm really asking.The point of these experiments is to drill down on this built in voltage idea... its critical to the way many texts explain the inner workings of a p-n junction (not just how to mathematically model them in a "small-signal" sense). So if built in voltage and carrier diffusion stuff is non-sense I'd like to conduct an experiment to demonstrate that to myself.So I'm addressing your #1 concern about the meter biasing the junction. And I've found its a real possibility... so I'm looking into another meter. My thought here is to use a differential null meter.As far as #2 goes -- that I'm picking up some EMI and rectifying it -- also a possibility. I didn't do it during the test, but an easy check here would be to switch modes on the DMM to AC and see if there's an AC signal?For #3, thermal emfs... this is a common misunderstanding. Thermal emf's are not created *at* junctions. They are a result of thermal differences throughout an entire circuit. But either way, I agree with you in noting the measured voltages seem to be a few orders of magnitude off to be explained by thermal emf.Lastly, #4 - glass diodes and the photo-electric effect. Two of the three diodes were glass, the 3rd was a metal can. I think the glass ones were effected by light (and the shadow of my hand moving over the dewar to reach the multimeter), but in broad strokes, all three diodes showed the same voltage effects.Are there any more? I'm never been that great at double-checking myself on things I don't yet know.

author: muvideo
author_rank: Frequent Contributor
timestamp: March 04, 2014, 08:06:30 am
content: Quote from: CaptnYellowShirt on March 04, 2014, 03:40:51 amSorry, I'm not following you here, in voltmeter mode?If so 7-8uA is huge, perhaps you meant 7-8pA, if so AFAIK it's pretty good.Did you measure a 1MOhm shorting the inputs or a 1MOhm in series with a V-source?Do you have a pAmmeter or a stable enough V-source to take the above measurements?

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 04, 2014, 08:17:08 am
content: Quote from: muvideo on March 04, 2014, 08:06:30 amMy mistake! I'll correct that.My finger slipped when using iPython to do the math. Big difference between 10*6 and 10**6!Should just be 1 pA.

author: uwezi
author_rank: Unknown
timestamp: March 04, 2014, 09:34:01 am
content: Quote from: ejeffrey on March 04, 2014, 07:10:52 amExactly.Just look at the diode equation, which describes so nicely how current and voltage in a diode behave (neglecting series and shunt resistances) in the darkI = I00e-Eg/kT(eqV/kT- 1)The only point where current I is zero is at V=0 and vice versa at I=0 you will only measure V=0 give or take some additional factors like light-induced currents, thermal EMF, offset currents,...The built-in potential is not a voltage which can be measured at the terminals of a diode.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 04, 2014, 04:09:37 pm
content: Isn't this a little like saying a real-world capacitor won't exhibit a self recharge effect or an ESR because the equation V = 1/C int{0}{t} i(tau) d_tau doesn't predict it?All electrical components exhibit odd, non-modeled-for behaviors if you look closely enough.I'm still not saying I *can* measure a diode's built in voltage, but if I can't, it's surely not because of an exponential-law modeling equation of an in-circuit-diode says I can't.Data, data, data, I cannot make bricks without clay.

author: uwezi
author_rank: Unknown
timestamp: March 04, 2014, 06:10:31 pm
content: Quote from: CaptnYellowShirt on March 04, 2014, 04:09:37 pmYou have a good point here, but you are also talking to someone who has written aPhD thesison diodes and spends half of his job time working with solar cells (essentially pn junctions in the light) and the other half teaching electronics.Of course the diode equation is only a model on how a diode should behave, it is not a law of nature but a description of an observation (first) and the result of a model (later). Furthermore the form I quoted is neglecting the effects of shunt- and series-resistance which you will find in every real-world device. But from measurements under controlled conditions I can tell you that the diode equation holds. At zero current the voltage over a diode is zero. Actually we regularly measure the current-voltage characteristics of solar cells between 100 K and room temperature, both in the light and under dark conditions. From the open-circuit voltage under illumination of a well-behaved pn-junction at different temperatures you can extract the bandgap of the semiconductor, but whatever you do, the built-in potential is nothing which will show up on the terminals of a diode.Coming back to the diode equation: what you very often will see is that the current-voltage characteristic deviates from a pure exponential behavior at very low applied voltages, much lower than the value of the built-in voltage (still better: built-in potential). But you will always find that the direction of the current changes at an applied voltage of zero volts - showing zero current at zero volts.So then the model of a built-in voltage must be wrong?No, it is not, it is just not the full truth. I just tried to find suitable material on the net, perhaps the following will help to make things a bit more clear:http://www.ee.sc.edu/personal/faculty/simin/ELCT563/04%20P-n%20junction%20basics.pdf

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 03:56:46 am
content: I performed another test today.In this experiment, I hooked up several of the 5.6v zener diodes (Motorola 1N5232B) I used in the first experiment in different configurations. I measured their output with my electrometer. Here's what I found:Diode ConfigurationCathode-to-PositiveCathode-to-NegativeOne-2e-9 A+2e-9 ATwo-parallel-4e-9 A+4e-9 AThree-parallel-6e-9 A+6e-9 ATwo-series-0.6e-9 A+0.6-9 AThree-series-0.8e-9 A+0.8-9 ATwo(+)/One(-)-2e-9 A+2e-9 AThe last entry there is measuring three diodes in parallel -- one turned against the other two.Room temp: 22.4CElectrometer input impedance is 100MOhms in this setup. So a current of 2e-9A translates into about 0.2V. The fact they seem to add in parallel suggests its less a 'built-in voltage' and more of a built-in charge-rate source. Next I'll need to do a set of tests to form a load-line to get a better idea of the equivalent circuit of the diode.Attached are pictures.The rod is teflon.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 05:01:39 am
content: After the tests with the zener diodes, I wanted to know if these tests would hold for a regular diode.I stopped off at radioshack and grabbed one of their 'assorted bags'. I mainly got stuff from the 1N400x series.I ended up using a group of a few 1N4001's. At first nothing happened. But then I realized that after handling them I had some kind of goo building up on my finger tips. From the way they look, I wouldn't be surprised if Tandy swept them up off a factory floor.Anway, I put the diodes though a wash of alcohol and this is what I got:Diode ConfigurationCathode-to-PositiveCathode-to-NegativeOne4e-13 A-3e-13 ATwo-parallel5e-13 A-5e-13 ATwo-Series1.6e-12 A-1.8e-12 AFour-series4e-12 A-3.2e-12 AThese lower values there were somewhat unrepeatable. They were measured with the electrometer in its 100G Ohm mode, so a hair or slightly moist air current (or a bit of goo on the diode body) could have thrown the experiment off.So maybe what I'm seeing here isn't the built in voltage after all? *Something* is happening with these zeners. And to some extent its happening with a 'regular' diode, but its certainly not as strong of an effect.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 05, 2014, 06:19:23 am
content: The obvious absurdity that should be screaming at you is...If there IS a terminal voltage difference...How much current is behind it? And where does the V*I =powercome from?Doing the experiment in light vs. dark conditions (or with glass vs. plastic packages) may provide additional clues.Incidentally, vacuum tube diodes exhibit short circuit leakage current (typically on the order of uA), and O/C voltages around 2V (the current in this region transitions from V^(3/2) law to an exponential tail, so the available power is considerably less than SCC * OCV; the tail, BTW, is exponential for the same reason semiconductor diodes are exponential over most of their range -- statistics!). This is thermodynamically acceptable because the cathode is hotter than the plate; indeed, a vacuum tube is a [ridiculously inefficient] electric heat engine! Likewise, applying a thermal gradient to a semiconductor diode results in nonuniform mobility and built-in potentials and such, but not an isothermal diode.Tim

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 06:47:14 am
content: Quote from: T3sl4co1l on March 05, 2014, 06:19:23 amA 10M Ohm resistor in the place of the diodes in both tests yields a reading of zero.Quote from: T3sl4co1l on March 05, 2014, 06:19:23 amIf you look closely at the photographs, you'll see the diode chosen for the experiment is in a silvered case. However, I'll look at it under a microscope tomorrow. Maybe there's a window to the semiconductor I didn't see?Quote from: T3sl4co1l on March 05, 2014, 06:19:23 amGood question. Its something I've been wondering too. In the 'built-in voltage' model I'm exploring with the experiment, charge carriers are forced across the p-n junction by a diffusion pressure until the electric field of the separated charges arrests any further diffusion. I'm not sure if the diffused or undiffused state contains less entropy? I'd have to think about this first before making any further guesses. But you're right, its gotta come from somewhere... or does it?! Naw, just joshin' ya. Seriously though, if this ends up being a time machine or perpetual motion machine, I call dibs...

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 07:04:18 am
content: Quote from: uwezi on March 04, 2014, 06:10:31 pmCould be. It is rather simplistic.I've been reading this book:http://ecee.colorado.edu/~bart/book/book/title.htmAny sections you think I should focus on?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 05, 2014, 07:55:58 am
content: Quote from: CaptnYellowShirt on March 05, 2014, 06:47:14 amRelative to an electrometer (or an ice cold junction at an arbitrarily low voltage), 10M is a short circuit.I'd be curious to see what a polystyrene cap (if you can find one) of about 100pF does over time (if the electrometer is on the order of 10^12 ohms, we're talking minutes of time constant). That should be much more representative of a junction near zero bias.Remember, an experiment doesn't prove anything if it's self-validating. Test every counter-case you can, especially the simple ones, and you have a much more convincing proof!Down at some level, there should still be a residual current, due to cosmic rays. It may be very difficult to detect.Tim

author: uwezi
author_rank: Unknown
timestamp: March 05, 2014, 07:56:57 am
content: Quote from: CaptnYellowShirt on March 05, 2014, 07:04:18 amWell from just quickly scrolling down the long page, Ch.4 seems to be quite ok.Back to your recent experiments: now do them in a shielded Faraday cage. As a kid I had a quartz wristwatch running on the rectified power from a nearby AM radio station (AFN in northern Germany). I don't expect that you are living near a strong AM transmitter like I was, but as pointed out byT3sl4co1l, now that you see voltage and current you either have tapped a source ofcompletely free energywhich I missed when studying the current-voltage curves of hundreds of diodes close to V=0 and I=0ORyou are getting the power from somewhere else. Any diode (Schottky and pn) has the ability to rectify stray RF and mains signals...

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 04:04:06 pm
content: Quote from: uwezi on March 05, 2014, 07:56:57 amWell I thought about this too. Lets see what you think. I used the Hp 3457a dmm to measure the DC and AC voltages on one zener diode that was used in this experiment. At room temperature, with the diode wired directly to the screw terminals on the dmm, and left alone for 10mins, I found a DC voltage of around 0.21 volts and two AC voltages -- 3mv / 24kHz (with my florescent lights on) and 1mv / 60hz (with the lights off).And I actually already am in a bit of a faraday cage. My workshop is sheet metal clad on all 5 sides, and the bench where I am taking these measurements is in-set into a hill (so its like 5-6 feet below ground). Nothing works in the shop -- no cell phones, no FM radio, no AM radio. But there's always setting the experiment up inside of another cage. The electrometer is battery powered, so I'll take a look at that.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 05, 2014, 04:06:12 pm
content: Quote from: T3sl4co1l on March 05, 2014, 07:55:58 amGood idea. I'll see if I can find something in my parts bin.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 06, 2014, 08:02:50 am
content: Quote from: CaptnYellowShirt on March 05, 2014, 06:47:14 amIn thinking about the problem from an entropy perspective, I may have found what is really at issue here -- terminology. And it is like you all have been saying: you can't measure the 'built-in voltage' with a voltmeter because the term 'voltage' in 'built-in voltage' and 'voltage' with respect to a voltmeter are two entirely different things.Check me on this explanation ...When I say voltage as in something my voltmeter can measure I'm actually talking about galvanic potential -- or the *net* force motivating the flow of electronics in a circuit. The source of this really could be anything. Classically its generated by a chemical potential, but any system that wants to dump/accept electrons has a galvanic potential. This is the voltage that is measured by a voltmeter.When I say 'built-in voltage', I'm talking about a simple electric field -- which, if one could probe using magic-physics-text-book probes, it would have a scalar potential of X volts. But since no such device actually exists, I do the next best thing which is to rely on the flow of electrons to divine where the field is and how strong it is. Again, the net force causing this flow is the galvanic potential. Its measured with a voltmeter. And for most cases it is taken to be equivalent to the potential developed by the electric field in question.However in the case of the p-n junction the built-involtageelectric field is balanced by the diffusion force. These forces (which also include thermal and photonic) are only acting on a narrow slice of the p-n junction. Sure, there's a galvanic potential to be measured, but it only exists over this narrow window because there's no motivation for electrons to leave the p-n junction. Since its way easier for the electrons to balance the competing forces of diffusion and built-in eletric fields by simply drifting across the p-n junction (rather than around a larger circuit), "built-in voltage" is not something you can measure with a voltmeter.Really the diagram that I posted earlier should show the voltage potential dying off as distance from the junction is increased --reflecting the non-divergence of the electric field and the effective nullification of the electric dipole in the p-n junction as distance from the diode increases to infinity.

author: uwezi
author_rank: Unknown
timestamp: March 06, 2014, 09:01:07 am
content: Hi CaptnYellowShirt,I guess we are approaching a common denominator here.However, your last paragraphs would imply that - if you only would make the slab of silicon thin enough - you still would expect to see a flow of current and an external voltage over the diode. But this is not the case. It would still defy the other basic law of physics, the conservation of energy.Any even so slight flow of electrons driven by an even so small potential difference would mean that your diode would deliver electric power to the outside world. This power would have to come from somewhere - possible source being an external flow of heat through the diode, light hitting the diode, external electromagnetic fields,...Just a junction in thermal equilibrium with its surrounding, either at room temperature or at cryogenic temperatures is not good enough.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 06, 2014, 11:04:55 am
content: Like I said, you can get a difference in potential from a thermal gradient. Under isothermal conditions? This is silly.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 06, 2014, 04:37:02 pm
content: Quote from: uwezi on March 06, 2014, 09:01:07 amThe paragraph that starts with "However..." or "Really..." ?

author: uwezi
author_rank: Unknown
timestamp: March 06, 2014, 04:54:17 pm
content: Quote from: CaptnYellowShirt on March 06, 2014, 04:37:02 pm"However..."

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 06, 2014, 05:01:55 pm
content: Quote from: uwezi on March 06, 2014, 09:01:07 amAh, I see what you mean. I'm going to put that one on in the back of my head and let it simmer for a while.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 06, 2014, 08:09:53 pm
content: I've been walking around my lab for the last 20 mins with my electrometer in hand performing what to an outsider might look like a ghost-hunt.I've hooked up the various diode configurations I've tried over the past week and seen if location makes a difference. Long story short, I've been plagued by a combination of two problems.1. The 'silvered' zener diodes (5.6v) are light sensitive. It's not something which is readily notable until you hit it with a *really* bright light. In this case, the application of light becomes notable when using one of those premo 3w LED flashlights up close. But it does look like what I was measuring in the case of the zener tests was photoelectric current -- which would explane why they add in parallel.2. A CFL bulb that is positioned over my shoulder on the ceiling has been emitting a pretty horrible RF noise. It's not something I checked in the earlier tests because it was like 3m away from my work bench. I thought, hey if none of the lights ON my desk or within arm's reach is showing an effect, how could this one 3x the distance away do anything? Well it does. And I think this explains the 'regular' diodes adding in series. It would seem that the pick-up loop formed by the diode ring needs a characteristic diameter to pick up this bulb's RF. One or two diodes doesn't really cut it, but 3 or 4 is big enough.So taking the Faraday cage advice, I put the electrometer in a steel bucket and put a steel plate over most of the top of that bucket. And wouldn't you know it... no current.I couldn't seem to produce any voltage or current by way of a thermometric effect. I heated either side of the diode's leads using a soldering iron. The only response I seem to get there is from the pointed tip of the iron "looking at" the lead. It might be that there's is a thermal response, but it would seem to be overwhelmed by (what I'm guessing is) a electrostatic effect.So It would seem I need to do a few things before moving onto my next stage of cold research: 1. whatever experiment I perform needs to be properly shielded and performed in the dark, and 2) in general I need to think about replacing a few CFL's with incandescents... oh wait... I can't buy them any more.... curse you obama!

author: uwezi
author_rank: Unknown
timestamp: March 06, 2014, 08:41:27 pm
content: Quote from: CaptnYellowShirt on March 06, 2014, 08:09:53 pmWhat a detective story! Congratulations!But it's not your president's fault that incandescent ligth bulbs are so inefficient as they are. Perhaps you should jump the step and go directly to DC powered LEDs with the power supply either far away or powered from batteries.Colleagues of mine recently had problems with 12 V halogen light bulbs in their measurement setups. The measurements (on wafer-level MOSFETs) would be performed in the dark inside a Faraday cage, but in order to position the sample there were halogen lights installed. Never before had this been a problem, but then my colleagues complained that something was wrong with the measurement setup: for every single transistor the gate would blow after the first measurement.Then they consulted me.By accident and with the involuntary help of another colleague we tracked down the problem to the halogen lights!In one setup the 240 V/12 V 50 W transformer for the halogen light bulb, including the power cord had been installed inside the Faraday cageBut it was not enough to get the transformer out of the cage - no, even when the transformer (made in the 1980s in Eastern Germany) was connected to a wall outlet on the other side of the room we could consistently blow the MOSFET gates inside the Faraday cage just by switching on or off the transformer - no lamp attachedIt proved to be spikes which through power line and protective earth and through the semiconductor parameter analyzer found their way to the gate of the MOSFETs.The other setup guided the light through a fiber-optic cable into the Faraday cage - but the problem was similar and most probably the same cause.Since two months now both setups are equipped with 3 W LED lights, powered with linear regulated constant current sources from old-style transformer wall warts. The transformer is never switched - switching of the light is done by slowly ramping the current down and up.As far as I know not a single MOSFET gate has been destroyed since...

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 06, 2014, 08:51:12 pm
content: Quote from: uwezi on March 06, 2014, 08:41:27 pmOh no, it's totally his fault -- that there fox news said so. Basically every problem any american currently faces (including, but not limited to jock itch) is obama's fault.

author: SeanB
author_rank: Super Contributor
timestamp: March 06, 2014, 09:11:44 pm
content: You do know most 3W LED units have a rather horrid SMPS inside them? If you want low noise you rip it out and make a constant current supply using a LM317 and a high enough supply voltage, or use the 3 LED dies in parallel with a resistor and drive from 12V. When the power goes out here ( Yay for load shedding, thanks Eishkom for having wet coal and no other contingency plans as usual SNAFU) the RF noise level drops dramatically, and it is not me radiating, I went hunting as well looking for FRI sources.

author: uwezi
author_rank: Unknown
timestamp: March 06, 2014, 10:17:50 pm
content: Quote from: SeanB on March 06, 2014, 09:11:44 pmI was talking about the "naked" 3 W LED chips - definitely no SMPS inside...

author: uwezi
author_rank: Unknown
timestamp: March 06, 2014, 10:44:14 pm
content: absolutely off-topic:this is the circuit I used. RC sets the soft-fading, Rsettogether with the NPN transistor Q2 determines the LED current. Primitive? Sure, but it works!

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 07, 2014, 05:36:30 am
content: For a the experiments I have in mind, I'm going to need a few containers to store the DUT and the LN2.To take extended measurements, the components will need to stay cold for a long period of time. The containers will have to be well insulated so the LN2 will stick around for at least 12 hours.I have two candidate containers -- both constructed out of materials from the thrift store. The first (on the left) is a coffee carafe which has been slimmed down to fit inside of a 64oz Igloo-brand drink cooler. The carafe's insulation method is a double-walled, fully silvered, quartz vacuum dewar. The drink cooler on the outside is (hopefully) to hold in some of the 78K nitrogen gas that will be enveloping the carafe once it boils off.The second (on the right), is a double walled stainless steel vacuum thermos -- same kind you'd put your soup in for the day. In a similar way to the carafe, its been slimmed down to fit inside of a slightly smaller drink cooler.The caps for both devices are similar in nature -- air filled plastic plugs about 2" in depth. The caps for both drink coolers can also be installed. Even if they are just thin single walled plastic, I figure they will help hold in the cold nitrogen gas.Lastly, to give them both the best fighting chance they can get (I really don't want to be filling these things up constantly), I placed them in a small bucket that also has a cap.To test them, I put 0.5 L of LN2 in each one. My money's on the coffee carafe outlasting the thermos, but I'll let you know how it goes.Stay frosty!

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 10, 2014, 02:33:06 pm
content: Well, the stainless steel thermos sucks.It killed the the 1/2 L of LN2 in just under 3 hours. The outside of the insulated drink cooler (red) was cold to the touch and collecting frost when I inspected it at the 2 hour mark -- a sure sign something's wrong. I'm not sure if this is an issue with the vacuum (or lack thereof) of the thermos flask, or if the stainless steel simply conducts that kind of heat. Either way, its right out.The silvered glass vacuum flask fared much better. It kept the 1/2L of LN2 frosty for 24hours (+). When I checked on it again in 24hrs, there was still around 0.1L in the flask.I repeated the experiment with a full 1L of LN2 in the vacuum flask and that lasted 48hrs (+) -- holding about 1/4L after the 2 days.I am considering an improvement of making a deeper insulated stopper/cork for the flask out of styrofoam. Any other suggestions?

author: SeanB
author_rank: Super Contributor
timestamp: March 10, 2014, 05:58:58 pm
content: The more styrofoam the better, just leave a small gap for expanding gas to escape. As thick as possible walls, floor and lid, with a long path so heat has to travel a long way into it. Find a foam supplier in your area and get a block about a half meter on each side and cut the hole out of it.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 14, 2014, 02:05:48 pm
content: I have built two more glass vacuum dewars in the same style as the (blue) one in the previous post. One is a similar 1L size, the second is larger measuring in at around 2L of storage space.I've also found a cheap supply of LN2 (e.g. free) but its a self-serve kind of situation. Which means I need to bring my own transfer hose, etc, etc. Not having properly researched cryogenic hose construction, I pulled a teflon-core braided stainless steel hose out of my junk bin and fitted it up to accept the 1/2" male flare that is standard on the 180L pressurized LN2 dewars.So I was aware going into this that the teflon wouldn't really be very flexible after the LN2 hit it. But I thought, that's no big deal i'll be able to flex the hose into the correct position, turn the LN2 on, and when the teflon freezes I'll just need to uncouple the flare fitting and walk a stiff 4ft hose back to my truck. BTW, cryogenic transfer hoses are typically made of corrugated stainless steel with a stainless steel outer braid. And there's a reason for this...As soon as the LN2 hit the hose I made, I started to hear this crackling. No biggy right? Well, yeah... until the crackling became just a cracking. What I was left with was like one of those weeping garden hoses that waters plants over its entire length... more LN2 was getting into my dewar by flowing on the outside of the hose than on the inside.Back to the drawing board... I actually have a set of corrugated stainless hoses in my junk bin, but they have a weird swagelok type of connector on them that I haven't been able to find an adapter for. Going to pay a visit to the hose guy later today. Wish me luck.

author: SeanB
author_rank: Super Contributor
timestamp: March 14, 2014, 07:46:42 pm
content: I know those too well........ Went in one day to fill a LOX unit, and found a little 1m snake busy in the room in the cradle where my hands were. Went and unscrewed the vent line and played a little game of cryo snake, and had a stiff scaled stick after 5 minutes. As I was using cryo gloves when picking it up I dropped it and it shattered. That room was a perfect place to go and get cold, as it had a 1kl cryo tank in it that was always venting a trickle of vapour. Nasty thing to do to the smokers was to quietly "borrow" the pack and go and fill it with LOX, then leave it back without them noticing. Next cigarette would light and burn down to a stompie in about 10 seconds. Even nicer, we were prohibited from polishing our work shoes....Apparently wax polish and LOX do not mix well.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 16, 2014, 07:00:10 am
content: As mentioned in the previous post, I had a set of corrugated stainless hoses laying around in my junk pile. But because they had an unknown swagelok style fitting on their ends, I couldn't connect them to anything -- even themselves.So I decided to make some creative modifications. To connect the two hoses, I turned a coupling from a 3/4" stainless steel bolt that could fit over top of the swagelock threads and TIG'd the entire thing together. For the end of the hose, I turned a 90deg flare fitting out of the remaining stock of 3/4" stainless bolt, and also TIG'd that on.I went back to the LN2 storage area last night and tried my new hose on it. It worked like a champ.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: March 30, 2014, 08:24:14 pm
content: Here's an update on the 'home-brew' cryo dewars:I've settled on a general design which encapsulates a glass coffee carafe inside of an Igloo-brand-style thermos. The empty space is stuffed with fiberglass insulation and a cap/plug is made from stryofoam.I've made a few versions from whatever was available at my local thrift stores. Its interesting to note that there is a HUGE variation on the quality of these carafes that seems to align well with the county of origin. Anything from China or Taiwan hasn't worked at all -- they've just cracked when pouring in the LN2. Ones from Japan, Germany, or the USA have all worked reasonably well, but inside of those groupings there's still a wide variation on the heat gain rate.The one pictured below came from a Bunn 2L pump-style coffee carafe -- like you might find in a coffee shop. Its done the best so far with a heat gain of 0.584 watts (mass loss of -10.6 g / hr), which blows most commercial cryo dewars out of the water (the large 30L dewar that I transport the LN2 specs in at -67.2 g / hr).Variations thereof seem to deal with the depth of the 'neck/mouth' section (better performing ones have deeper necks) and the quality of the silvering. Some have a mirror-like finish all the way around. Others have spots where the silvering is thin or missing. The worst performing one comes from a 1L pour-style carafe meant for home use which specs in at 0.832 watts (-15.1 g/hr).For an up coming experiment I am planing, I've created a 'cold box' to keep long term variations in temp to a minimum. You can see pictures of this below, also. Its made from 1.5" of stryofoam packed inside of a steel cabinet (~1/16" reinforced sheet metal) so it should also offer good rejection of E&M noise.Currently, I've added 10 wires from a 30awg(?) 3M ribbon cable to each dewar. Its just slung in the mouth of the dewar and out the top of the outer 'igloo' case. I'm trying to get a rough idea of how much heat several copper kelvin leads will pipe into the dewar. I'll report that when I get some data.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 21, 2014, 03:38:03 pm
content: I've built two new cryo dewars using Bunn's "Airpot" 2L style carafes -- so far I've found these to be the highest quality both in terms of resisting cracks and thermal insulation.I've also changed the style of the caps on all of the dewars from a simple cylinder of styrofoam to a thick block of styrofoam that's been 'carved' to fit down into the neck of the dewar, around the outside of the neck, and then above it by 1.5". I've also glued a disc of aluminum foil at the bottom of the neck plug to reduce radiative heat transfer. These improvements have reduced the heat gain to around 0.52 watts (~ 10% reduction).The goal of this work is to place a set of components down into the LN2 for long term study. So the next step is to figure out how to 'wire' the dewar. Obviously I'm going to need something that has a fairly low overall thermal conductivity. The experiment in the last post (the ribbon cable) failed because the cable tended to curl up and pull away from the LN2 (weird). So next I tried a set of 29awg magnet wires. I could probably go thinner, but this is the thinnest I had laying around.I tried several configurations of the wire ... wrapping it a few times around the neck of the dewar to isolate the outside and inside temperatures, etc etc. But there was very little variation between tests. In the end I decided to simply run the wires straight into the dewar under the sytrofoam cap (no need to waste wire, right?). The tests showed an addition of 0.028 Watts / wire. In other words, 18 wires has about the same thermal conductivity as the entire dewar.To put that in perspective... these 18 wires have a cross section of 1.16 mm^2, while the surface area of the dewar is something like 76,600 mm^2 !!2402