Measuring capacitance (R+C as voltage divider confusion). - Page 1
Forum section: electronics / beginners

author: grumpydoc
author_rank: Super Contributor
timestamp: June 12, 2014, 08:23:01 pm
content: OK, the background to this is that I have recently acquired a Seaward Primetest 100 PAT tester1. I have a Primetest 50 which I use for electrical safety testing of stuff I sell on ebay2but I haven't been totally satisfied with it as it only has a "pass/fail" indication and doesn't do leakage current3. The Primetest 100 displays the insulation resistance and the earth bond resistance and adds a leakage current measurement.There was no cal certificate with the unit so I want to make sure it works properly, to check the insulation resistance/leakage measurements - I came up with this:The test voltage for insulation resistance is 500V hence two 510k resistors rather than a single 1M resistor and the high voltage cap, I checked these on the Keithley and they came to just over 1.03M - the Primetest read 1.04M so that's OK.Now, caps tend to have wide tolerances and only relatively few values are available - I wanted to have a total AC leakage current of 0.75mA as that's the regulatory limit so I thought I could trim the current by having a resistor in series.I have a couple of multimeters capable of measuring caps, none terribly accurately, but they seem to agree that the 10nF cap is, indeed, 10nF give-or-take a few % That gives an impedance at 50Hz of 318310 ohms - add the 130k (actually 130300 ohms) plus the current through the two 510k resistors and that comes to 0.766mA.The Primetest reads 0.8mA which although 4% high is actually in spec which is within 5%Unfortunately this doesn't validate the PAT tester as well as I'd like - knowing the test values to only a similar tolerance to the meter spec isn't really good enough - I'd like to aim at something closer to 0.5%.So I had the clever idea of measuring the impedance and working out the capacitance from that. So I took just the 10nF cap and 130k resistor and drove them with a 50Hz sine from my sig gen.The SG goes up to 20V p-p (into a high impedance load, 10Vp-p into 50 ohms) which is 7.071V RMS, I measured 7.079 on the Keithley 2015.BUTI get 5.866V rms across the cap and 2.387V rms across the resistor - I sort-of expected these to add up to 7.079V, not 8.25VMy brain does start to hurt a bit when we get to AC circuit analysis - can anyone explain the measurements without getting too mathematical? I thought I could ignore voltage/current phase differences with this set-up but perhaps not. I'm reasonably certain the readings are correct. At £30 "not working" I couldn't resist - it just took cleaning the battery contacts and the PCB pads for the front panel buttons and it was fine. Sadly while the other £30 Primetest 100 that I got at the same time responded to a similar approach it also turned out to have arealfault as well. You win some, you lose some. Partly because I think it's a good idea and partly because I think that the Electrical Safety Regulations 1994 say you have to - I'm probably the only ebay seller who bothers though. This now gives me the conundrum that the Tek 2035 that I'm working on "failed" the leakage test. Lots of discussion on the Seaward forums about similar problems with PCs - probably due to the mains EMC filters.

author: Andy Watson
author_rank: Super Contributor
timestamp: June 12, 2014, 08:39:56 pm
content: How much capacitance does the input to your meter have?

author: grumpydoc
author_rank: Super Contributor
timestamp: June 12, 2014, 08:42:56 pm
content: QuoteSpec says < 100pF

author: Andy Watson
author_rank: Super Contributor
timestamp: June 12, 2014, 08:50:06 pm
content: I think you should be reading 2.677 Vrms across your resistor and 6.554 Vrms across the capacitor. Something in your measurement system is imposing a not insignificant impedance in series with the components that you are measuring.

author: grumpydoc
author_rank: Super Contributor
timestamp: June 12, 2014, 09:04:16 pm
content: Can't see it putting anything significant in series, it's just a resistor and cap soldered in the middle, croc clips to the other two leads and measurements directly at the components.Of course, duh, the input is 1Mohm plus <100pF and the 1M is going to significantly affect the readings.In fact double duh because Dave did a video on the subject.However it doesn't actually help my confusion - that will go in parallel with the component and should tend to reduce the reading.

author: Andy Watson
author_rank: Super Contributor
timestamp: June 12, 2014, 09:31:08 pm
content: The 1M is probably making the difference. I had assumed that, being a DVM it would have a resistive input component much greater then 1M - one too many assumptionsWhen you add capacitive (or inductive) components to resistive components you need to use vector addition, i.e. Zrc = sqrt(R^2 + Xc^2). Workout Zrc, then use V/Zrc to find the current (the currentmustbe common to both the capacitor and the resistor in this series circuit). Find Vr by simply multiplying, I x R. Ditto, find Vc by multiplying I x Xc. If you've done it right you should be able to add the voltages togetherusing vector additionand get the voltage you started with, i.e. Vosc = sqrt(Vr^2 + Vc^2).

author: grumpydoc
author_rank: Super Contributor
timestamp: June 12, 2014, 10:06:04 pm
content: QuoteYeah, I was thinking of the >1Gohm spec for the DCV range, completely missing the fact that the AC input is totally different.QuoteAllright, with the 10nF as an example:The impedance of the cap at 50Hz is 318310ohms, so the combined impedance is sqrt(1303002+3183102) or 343947ohmscurrent = 7.079V/343947 = 20.58uAvoltage across resistor = 130300x 20.58uA = 2.682Vvoltage across cap = 6.55VOK, that's more-or-less what you gotChecking - sqrt(2.6822+6.552) = 7.078 (close enough given rounding).TheWikipedia pagesays that for an RC divider (or low pass filter) would have a Vin/Voutratio ofPlugging the values in would give 6.54V across the cap so that agrees.My next problem - apart from how to set up the measurement, is this:If the impedance of the series RC network is 343947ohms, not 448610ohms then the current through it at 240V1is 0.698mA - add the 0.23mA going through the two 510k resistors and that's 0.929mASo either the Primetest is reading woefully under or I still don't understand the calculations the Primetest doesn't actually measure at 240V - it does a "substitute test" where it measures at 40V and scales the answer by 6.

author: Andy Watson
author_rank: Super Contributor
timestamp: June 12, 2014, 10:41:12 pm
content: Quote from: grumpydoc on June 12, 2014, 10:06:04 pmAh but, the current through the RC path is no longer in-phase with the current through the two 510k resistors - so you can't just add them.QuoteGoing beyond one R and one C does start to get mathematical! I've not done this sort of analysis like this for some 30 years and I would have to have a good think about expressing the problem with phasor diagrams. - that's as simple as the maths gets. On the other hand, if you don't mind delving into complex arithmetic (i.e. imaginary numbers), the way I would tackle this problem is to express each component in its complex form and then use the standard series/parallel impedance formulas to arrive at an impedance for the whole network.QuoteDo you know how it arrives at the voltage reduction? If it is using a transformer then you can probably ignore it, but if it is using a resistive or capacitive divider you will need to factor this "burden" into your calculations - you will no longer be applying 240V across your phantom load.

author: grumpydoc
author_rank: Super Contributor
timestamp: June 13, 2014, 07:48:07 pm
content: QuoteI'm actually pleased to hear that since I'd prefer the conclusion to be the Primetest is so-many% accurate not "it's broken"If I can't work it (current through the network), can I measure it by sticking a sense resistor at the top of the network and looking at the voltage across that?QuoteI've started to realise that!QuotePhasor diagrams are probably the way to go - especially as I found some notes athttp://www.kwantlen.ca/science/physics/faculty/mcoombes/P2421_Notes/Phasors/Phasors.htmlwhich include a worked example - a sub network of which is the network I want to solve.Even so it will stretch my maths to understand it - I can sort of see what's happening geometrically but it's a long time since I did any vector math.QuoteIt doesn't start with 240 and reduce down, it starts with 9v (6xAA) and generates 40V rms @ 50Hz - not sure how it current senses.One thing this has done is provided a concrete reason for buying an LCR meter - I think I might be able to figure out the maths for the network if I take it in small steps (and possibly with some hints) then figure out how to measure things to confirmifI know the values. I doubt my maths is going to be up to re-arranging the stuff to come up with the cap value from the measurements.

author: grumpydoc
author_rank: Super Contributor
timestamp: June 13, 2014, 09:00:07 pm
content: QuoteOK, well it's more plugging the numbers in than properly understanding what is going on but I get an impedance of 294947ohms for the combined network - that gives 0.813mA leakage at 240V which is pretty much spot on what the Primetest 100 is reading, phew.I understand AC networks a lot better as well - not nearly enough to fly solo but definitely enough to know what to work on to improve my knowledge further.Andy, if we ever meet, I owe you 1x the beverage of your choice.