Measuring 10kV with a 34401A - Page 1
Forum section: electronics / general

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 24, 2016, 09:45:02 pm
content: Hi,I've build a drift measurement box for voltages up to +/-10kV. Calibration is done by switching some resistors from series in parallel configuration. This enables a calibration with 625.000V instead of 10kV.To verify this I looked at the calibration uncertainties of some cal labs here in northern germany. The best I could find (except the PTB) has an uncertainty of 100ppm at 10kV.I wonder if I can do it by my own with an uncertainty of 100ppm. Therefore, I took some salvaged Caddock USF371 (20Meg 0,1% 5ppm/K) out of my treasure chest and connected pairs of two to get 10Meg and then I connected 9 pairs in series to get 90Meg. I connected this 90Meg in series with the input of an Agilent 34401A. Afterwards, I connected my Fluke 5440B and set it to 1000V. The 34401A was switched to the 1000V range, I wrote the reading down and calculated the gain.My uncertainty assumption:- 34401A 20ppm of reading + 6ppm of range (24h spec 1000V)- USF371 voltage coefficient: 0.02ppm/V * 1000V = 20ppm- USF371 self heating: 1000V² / 20Meg = 50mW -> 50mW * 40K/W = 2K -> 5ppm/K * 2K = 10ppm20ppm + 6ppm + 20ppm + 10ppm = 56ppmI think in worth case my reading could be of by 56ppm if I calculate the voltage with the calculated gain (relative to the Fluke 5440B -> additional 6ppm to the PTB)Is this right or did I miss anything?Best Philipp

author: acbern
author_rank: Frequent Contributor
timestamp: November 25, 2016, 07:42:18 am
content: I think you missed the abs. uncertainty of the 10M input resistance (and its voltage / load and temperature dependency; these are not necessarily part of the 34401 uncertainty spec; essentially only the divider factor is) and the uncertainty of the 5440A used for calibration. Also, if you want to improve the design, each of the nine 10M resistors can be placed in a shielded case, which is connected to a ladder divider, so the shields see the average voltage of each resistor. See Fluke 752A manual for details.

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 25, 2016, 08:12:59 am
content: Quote from: acbern on November 25, 2016, 07:42:18 amOk, I missed that. The absolute value is included in the calculated gain, but the change in value is a real problem here... Thanks!!Looks like I have to build a complete divider and use the high impedance 10V Range.Quote from: acbern on November 25, 2016, 07:42:18 amI meant this with "relative to the Fluke 5440B". For absolute uncertainty I would add 6ppm (5440B 1year spec for 1000V)Quote from: acbern on November 25, 2016, 07:42:18 amThanks for this hints. However, this is only a quick & dirty solution to verify the "real" instrument.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: November 25, 2016, 08:56:52 am
content: Quote from: e61_phil on November 25, 2016, 08:12:59 amWith high impedance dividers you need to be careful with the uncertainty from the 34401A input current. With 10M impedance every picoamp will add 1ppm error at 10V range, and the input current also depends on the input voltage... . The 34401A spec is <30pA at 25C (so potentially up to 30ppm error from that only, and variable with volage and temperature), however you better check if it actually meets the spec!CheersAlex

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 25, 2016, 09:09:24 am
content: I added two Caddock USF371 to get 100Meg and I added a 100k S102K (treasure chest..).new uncertainty assumptions:- 10G input impedance in parallel to 100k gives 10ppm of error, but I think this should be included in the gain factor (therefore ignored)- 30ppm uncertainty from USF371 stays the same- S102K 0,1ppm/V adds 1ppm- S102K max 2,5ppm/K adds less than 0,2ppm- 34401A 24h spec 10V range: 15ppm + 4ppm30ppm + 1.2ppm + 19ppm ~ 50ppm maxIf I add 3°C temperature margin it will result in 3K * 2.5ppm/K + 3K * 5ppm/K ~ 23ppmI think I should easily stay within 75ppm of uncertainty or did I miss anything again?Edit:@Alex: Thanks for this hint. If I take the input current instead of the input resistance into account and assume it isn't included in the gain factor cause of volatge dependencies I have to add 3ppm uncertainty I think.

author: Dr. Frank
author_rank: Super Contributor
timestamp: November 25, 2016, 11:28:16 am
content: Quote from: e61_phil on November 24, 2016, 09:45:02 pmIt's a problem, that the 'voltage coefficient' is not clearly defined. Am I right, that no power dissipation effect is specified for these Caddocks?I assume, that this parameter simply accounts for the self heating effect, i.e. for its T.C. and thermal resistance, although it's a linear parameter, not a quadratic one.I further assume, that any deviation due to electric fields can be neglected, so no linear effect should be visible.You may simply measure this dissipation/voltage effect by building a 1kV voltage divider (to < 10V output voltage), using one Caddock 50MOhm, and an appropriate S102k, and using the very good linearity of your 5440B, and maybe you have a 3458A available. 34401A linearity is also ok, but maybe not sufficient.I expect that you find a strictly quadratic deviation function, when you measure the linearity between 50...1kV.That would prove, that the 'voltage coefficient' gives an idea about the power dissipation effect as an upper limit value.Frank

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 25, 2016, 11:54:39 am
content: Caddock defines a Voltage Coefficient and a separate TC. In my case here the effect of the voltage coefficient is greater than the effect of self heating. Therefore, I assume it is a "real" voltage coefficient. The voltage coefficient is defined as 0,02ppm/V max. I think this would be the worst case. Caddocks specifications are usually very conservative.I bought some matched pairs from caddock a while ago for another project and the application engineer calculates the effect of the voltage coefficient in the same way.Nevertheless, a VC measurement would be a nice one..By the way I tried the divider again and now the readings look like:Drift Measurement Unit: 9962,5334470A with divider: 9962,54I shifted from the 34401A to the 34470A mainly because of the very nice scaling functionThe reading with the 90Meg in series to the 34401A showed about 280ppm difference to the other unit.

author: Dr. Frank
author_rank: Super Contributor
timestamp: November 25, 2016, 12:19:16 pm
content: Quote from: e61_phil on November 25, 2016, 11:54:39 amYes, I know. For other precision resistors, (low voltage ones, I have to admit), like the BMF from AE, they instead define the T.C. and additionally the dissipation / temperature rise, i.e. 0.14°C/mW, but no voltage coefficient.Quote from: e61_phil on November 25, 2016, 11:54:39 amDid you measure that, or do you assume that only, from the datasheet?Quote from: e61_phil on November 25, 2016, 11:54:39 amWell, you estimated this p.d. effect, and it was lower than this assumed pure voltage coefficient effect. So my assumption could be true as well, as Caddock may really specify very conservatively, so that the quadratic over linear effects are generously included.The quadratic p.d. effect is well understood, as the mechanism can be easily calculated.But I'd really like to learn about the physics of the linear voltage coefficient, as I did not find any explanation yet.FrankP.S.: I just downloaded their datasheet..http://www.caddock.com/Online_catalog/Mrktg_Lit/TypeUSF.pdfCaddock spends a whole paragraph about a 'power coefficent', describing in detail the p.d. effect.The boo-boo is, that this parameter later on is NOT specified at all!They only specify the 'voltage coefficient' of 0.02ppm/V, without explaining what this parameters means, in contradiction to the 'power coefficient'They also missed to specify the heat transfer resistance, which is necessary for calculation of the p.d. effect, or the 'power coefficient'.Where did you get these 40K/W from?That's the point in Caddocks specifications, that always disturbed me.They are not better than Vishay, concerning fat lettered promotion w/o content:"Zero Nominal TC""Extremely Low Power Coefficient"

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 25, 2016, 01:39:06 pm
content: Quote from: Dr. Frank on November 25, 2016, 12:19:16 pmHere my assumption is from the datasheet. But I also measured some USF271 a while ago (see attachment). They are specified with 2ppm/K and I measured a temperature rise of 1.6K (Pt100 glued to the resistor) and a drift of about 2ppm (5Meg USF271 in series with VHP101 120k @500V). I assumed the TK was better than specified (1.25ppm/K vs. 2ppm/K) and the thermal resistance seems to be lower too.Quote from: Dr. Frank on November 25, 2016, 12:19:16 pmI read something about semi-conductor effects in high ohmic resistors. These effects aren't linear. On very high resistor values (few 100G or T Ohms) this can cause many 100ppm of error which aren't relatet to thermal effects.Quote from: Dr. Frank on November 25, 2016, 12:19:16 pmI asked Caddock, because I can't find it in the datasheetPS: Quoting in this forum is really a pain.

author: Dr. Frank
author_rank: Super Contributor
timestamp: November 25, 2016, 02:07:56 pm
content: Quote from: e61_phil on November 25, 2016, 01:39:06 pmYes indeed, it is.... a pain in the ... headThank you for all the information, and about the HV - effect, although it's a bit diffuse, as a Thin Film alloy is not necessarily a semi-conductor..I thought in the direction of a piezo-effect, as the substrate is ceramic, and might be distorted by applying a high voltage... but the electric field would be in parallel to the ceramic instead of perpendicular to it, so I would not expect a big effect.So anyhow, I suggest to make a linearity measurement and solve for linear and quadratic terms in the deviation from a linear fit.Frank

author: Kleinstein
author_rank: Super Contributor
timestamp: November 25, 2016, 02:36:31 pm
content: The surface of the thin film of wire is often a kind of insulating oxide. There is no hard line from insulator to semiconductor - so this surface layer like a semiconductor may show nonlinear conduction effects. If everything is nice symmetric, there should be no linear effect, but one never knows how perfect symmetry is. Having a few resistors in series should reduce any linear effect, as it can compensate in parts.The ceramic substrate should not be piezoelectric - but much weaker electrostriction is of cause possible. For the direction of the field, there should not be that much difference. But for the long form the field will not be that large.

author: Marcus_S
author_rank: Contributor
timestamp: November 25, 2016, 05:24:45 pm
content: Hi,I think it might be interesting to have a careful look at the technology of the high-ohmic resistors. As far as I know a lot of them are made as thick-film resistors. And thick-film resistors are reported to have a significant voltage coefficient.Unfortunately it is a little complicated to measure the voltage effect as it is connected with the self-heating of the resistor. As the voltage coefficient is said to be a non-linear effect it could be characterized by distortion measurements (1).Long, long time ago I read some papers about thick film inks... Perhaps I should try to remember... The very much simplified model of the conduction mechanism in the 1970ies/1980ies was:The conducting material in thick film resistors (mostly and mainly ruthenium oxide) was converted into extremely small particles (by different processes) which are dispersed in a glass matrix. Depending on the concentation and the distribution of the particles they form a dispersion of single particles being separated by the glass or a network of chains of particles. These ruthenium oxide particles were reported to be semiconducting.But in generall the glass matrix is not inert. Depending on the chemistry of the glass and the conditions during resistor manufacturing the glass dissolves a smaller or larger part of the ruthenium oxide and form different compounds as lead ruthenates and ruthenium plumbates and a lot of compounds in between. But all these compounds were reported to have a low (but ohmic) conductivity. This low-conducting glass separates the ruthenium oxide particles but if the distance of the particles is small enough conduction by tunneling of electrons through this glass-barriers occurs.And as the simple ruthenium oxide/lead containing glass system is not satisfying regarding resistor properties a lot of additives are mixed into the thick film inks and into the glass (e.g. bismut and this is reported to form different conducting bismut ruthenates...) to improve resistance value and t.c. and pulse loading capability and matching of thermal expansion coefficient to the substrate and wetting of the substrate and ...And this nice combination of semiconducting and ohmic and tunneling conduction mechanisms leads to interesting HV-behaviour of the resistors.(End of memories...)Therefore I would not dare to predict the high-voltage behaviour of thick film resistors from the datasheet. If I would build a high-precision (thick-film-resistor)-HV-divider I would try to characterize each resistor separately.Best regardsMarcus(1) Holmes and Loasby: Handbook of Thick Film Technology, Electrochemical Publications Limited, 1976.

author: Dr. Frank
author_rank: Super Contributor
timestamp: November 25, 2016, 05:56:18 pm
content: Well, both from Kleinstein and Marcus_S is still speculation, isn't it?Usually, Thick Film resistors have T.C.s as high as 100ppm/K, whereas Thin Film is 50ppm/K maximum, down to 5ppm/K or less.It's nowhere said, that Caddock uses a Thick Film process (printed, metal-oxyde in glass matrix), or that their material is more like a Thin Film (sputtered alloy) system.Caddocks calls their resistive material / system Tetrinox Film resistors, so obviously they use some metal Oxydes, but the T.C. is as low as 2..5ppm/K.So maybe somebody finds their Tetrinox patent from about the 1970ties?Frank

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 25, 2016, 08:41:36 pm
content: Quote from: Marcus_S on November 25, 2016, 05:24:45 pmMy thought was: Is it worth to go a whole day into a calibration lab with at least 2h of driving to get an uncertainty of 100ppm or is it possible to achieve this uncertainty with parts out of my treasure chest.I needed a "second opinion" of some high voltage measurements to validate the calibration procedure with only 625V. The resistors used in the drift measurement box are matched Caddock sets with guaranteed TC below 1ppm/K sitting on a teflon bar and connected with small aluminium "eggs" to avoid sharp edges. Everything is in a big aluminium block and temperature stabilized with a stability better than 50mK (I measured 10mKpp over a few hours).For stability measurements the VC should be less important. But it is very interesting.

author: Marcus_S
author_rank: Contributor
timestamp: November 26, 2016, 10:51:49 am
content: Frank,yes, there are some speculations. And Caddock does not indicate a resistor technology in their USFXXX-datasheets. And I did not claim that the USF2XX/USF3XX-types are standard thick film resistors.But I was curious thereafter and did a short search but did not find any usable information regarding Tetrinox. Then I had a look into other datasheets. Caddock indicates the use of their Tetrinox-process (at least) for their (flat) USFS (= selected USF370/371) and (cylindrical) USG and TG. And I found an old patent. Having a look into US3,858,147 which claims a non-inductive pattern for cylindrical resistors (= USG, TG?) and which describes a printing and firing process for making the resistive layer I still assume (assume!) that they use some highly specialized and highly developed type of thick-film process (calling it Tetrinox) for the USFXXX-types also. But this deduction will certainly be too far-fetched.All I wanted to point out is that from my very little experience from the past it is not very easy (but will be possible with a lot of effort) to make a thick film ink which gives such good resistors and that it is not very easy to control a thick film manufacturing process giving a high yield of such good resistors. Therefore I would be careful and would try to characterize each specimen of thick film resistors for demanding high-end applications.Philipp,I hope to understand your experimental setup. You think about a hamon-type resistor with four resistors (10000 V / 625 V = 16)? I would be a little anxious (but I have no experience or proofs or data!) if the resistors see 2500 V during measurement and being calibrated at 625 V.Best regardsMarcus

author: Alex Nikitin
author_rank: Super Contributor
timestamp: November 26, 2016, 11:29:35 am
content: One of possible ways to evaluate/measure the voltage coefficient is to build two dividers from same type and value resistors, the second twice as long, and compare the outputs over the working range - up to 10kV. The voltage stress will be different and the temperature difference due to a different power dissipation can be accounted for (and will be visible as a drift with a change in voltage).CheersAlex

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 26, 2016, 01:51:31 pm
content: Quote from: Marcus_S on November 26, 2016, 10:51:49 amYes, you're right. It is (already done) a chain of 20 resistors divided in 4 chains which can switched from series to parallel. Perhaps I'm wrong but I thought this is the best compromise to bring the resulting voltage into the region of normal voltage calibrators (1100V max). 2 chains will result in 5kV, 3 chains in 1111.11..V and 4 chains in 625V. But even at 10kV every resistor will only see 500V and this will result in 10ppm of VC in worst case (7.5ppm if you assume a linear behavior). Other configurations aren't even possible, because I couldn't find high voltage relays with sufficient insulation at open state.Quote from: Alex Nikitin on November 26, 2016, 11:29:35 amI think it is still not easy to account for the different power dissipations. Perhaps one could measure the step response for different voltages (like my USF271 measurement).

author: VintageNut
author_rank: Frequent Contributor
timestamp: November 26, 2016, 04:42:12 pm
content: If you want to characterize the voltage coefficient of a resistor, why not just characterize one resistor at a time by sweeping voltage and measuring current?

author: e61_phil
author_rank: Frequent Contributor
timestamp: November 26, 2016, 07:15:47 pm
content: Quote from: VintageNut on November 26, 2016, 04:42:12 pmMy USF271 measurement was only one resistor. And if you replace the internal current shunt with a better resistor like a Vishay S102, Z201 or VHP101 you will end up with a voltage divider.It is very hard to judge if the drift is temperature related oder VC related.

author: VintageNut
author_rank: Frequent Contributor
timestamp: November 26, 2016, 07:26:59 pm
content: Quote from: e61_phil on November 26, 2016, 07:15:47 pmYes that makes perfect sense. You should hold constant voltage and change temperature to characterize temperature coefficient. Then sweep voltage which will cause a temperature dependent change as well as a voltage coefficient change.If you can measure temperature of the device, then you can subtract the change in resistance due to temperature and the remaining change in resistance is due to voltage coefficient.This requires having a good temperature chamber.