Anyone have Current probe circuit or high bandwidth current sensing circuit - Page 1
Forum section: electronics / projects

author: Allen_Du
author_rank: Contributor
timestamp: June 06, 2022, 05:57:38 am
content: Hi All,I have a application is to measure the current of high voltage circuit, the current is 0~6A, voltage is 400V DC.The sensing circuit cannot be same ground with high voltaage circuit. bandwidth at least 10MHz. sensing circuit output will be sent to ADC and FPGA. Accurate need to be 0.1%~1%.I am try to look for Current probe circuit but didn't find useful information.Can anyone help advice if have similiar experience, thanks in advance!

author: jonpaul
author_rank: Super Contributor
timestamp: June 06, 2022, 06:08:41 am
content: Bonjour 400 V is not "high voltage" just mains or medium.You need 10 A (peak or continuous?)DC or AC coupled?AC coupled: Pearson 411, 410 current transformerhttps://pearsonelectronics.com/pdf/411.pdfCost ~ $400DC coupledA6303 current probehttps://w140.com/tekwiki/wiki/A6303Use with Tektronix AM503, A B current probe amplifiers and mainframe TM50xhttps://w140.com/tekwiki/wiki/AM503Combination cost (used) ~ $1000Bon chanceJon

author: moffy
author_rank: Super Contributor
timestamp: June 06, 2022, 06:16:18 am
content: I have been looking for a reasonably high bandwidth current probe, AC only, and have run across this:https://www.ebay.com.au/itm/133992914189?_trkparms=amclksrc%3DITM%26aid%3D1110018%26algo%3DHOMESPLICE.COMPLISTINGS%26ao%3D1%26asc%3D20200818141841%26meid%3Df48bb38132b343d3b5abc46caa1d1c57%26pid%3D101111%26rk%3D2%26rkt%3D12%26sd%3D114841170268%26itm%3D133992914189%26pmt%3D1%26noa%3D0%26pg%3D2563228%26algv%3DThematicAlgo%26brand%3DUnbranded&_trksid=p2563228.c101111.m2109&amdata=cksum%3A133992914189f48bb38132b343d3b5abc46caa1d1c57%7Cenc%3AAQAHAAABALwVoSbxFsxNTl7u%252B0ZoqBb6AwCdyaTU5W%252BLvnW2toVQeGKuHPKYIkWYN41IQRr9lTggTWajWKmlFBfvkllV7yVUED%252BvuKtl4IihStfeUzaRKYuORAbEJpjgWlLq9w0BvTm%252F4NEOzj7aPTglLxB3Dso3cAEtlKGQ9AvWfUeZLzDl0r6tubmDTVfzus7riDedOBkmJRzaK4VW%252FmxOdYZ5%252B8W5mGLMo5CDScDjAHBzWiM0Wcp2uOj5osq6%252FAj2TfSNADDLxFQDnkr8RegeyPNOndOZvlFlu9AjQzHnAruoyTVusC3O1P1Qorq2yYnc0n%252BkslJcZY2w9xBFKrZmiKYcCSM%253D%7Campid%3APL_CLK%7Cclp%3A2563228A Rogowski coil that with a suitable integrator is good up to 1MHz. I assume a smaller coil could have higher bandwidth.Also there is a TI app note: tidubv4a.pdf or TIDA-01063 where they design and build a PCB Rogowski coil sensor with signal conditioning circuitry that goes up to 20MHz, which they feed directly into an ADC.

author: Allen_Du
author_rank: Contributor
timestamp: June 06, 2022, 08:48:01 am
content: Hi Jon and moffy,Appreciate your reply and the information.The application is for DC 400V. TIDA-01063 seems for AC.The Current probe cost is too high, I have to integrate the current sensing circuit in the product, cost cannot be much high.

author: inse
author_rank: Super Contributor
timestamp: June 06, 2022, 09:09:31 am
content: 10Mhz and 0.1~1% accuracy and isolation is tuff specs.No need to wonder why Tek & co take 1000‘s for such equipment.If you can do with less bandwidth, maybe something like Allegros ACS37002 might work for you?

author: moffy
author_rank: Super Contributor
timestamp: June 06, 2022, 09:13:37 am
content: the following isolated 16 bit ADC, STMs ISOSD61, with 25MSps:https://www.st.com/en/data-converters/isosd61.html#documentation

author: jonpaul
author_rank: Super Contributor
timestamp: June 06, 2022, 10:19:02 am
content: See hall effects sensors OEM,LEM Switzerlandexpect 1..5% best accuracy of any coil, CT, HE.jon

author: Allen_Du
author_rank: Contributor
timestamp: June 06, 2022, 01:24:37 pm
content: Wonder if the attached circuit work? The op amp are using isolated power supply.

author: eugene
author_rank: Frequent Contributor
timestamp: June 06, 2022, 01:50:04 pm
content: ACS730promises 1MHz bandwidth and 4% maximum error. About $5 (if you can find them.) You might need to find a way to work with that.

author: Zero999
author_rank: Super Contributor
timestamp: June 06, 2022, 01:58:21 pm
content: What's the lowest frequency of interest?When I needed to measure a high current pulse, I just used a current transformer. There are off the shelf units, or make your own with a toroidal ferrite core. Of course this will only work with AC.

author: coromonadalix
author_rank: Super Contributor
timestamp: June 06, 2022, 02:15:02 pm
content: Quote from: Allen_Du on June 06, 2022, 01:24:37 pmyou dont have any isolationAn acs 7xx hall effect sensor will be your best choice vs accuracy and pricing, tons of arduino projects around it .... you'll get isolation and a small foot print ....

author: Marco
author_rank: Super Contributor
timestamp: June 06, 2022, 02:52:42 pm
content: The cheapest in volume is an isolation amplifier like that one from TI with a normal resistive shunt and calibration. The shunt will have inductance, but the dimensions of the shunt and the PCB are relatively constant over temperature, so the inductance is relatively constant, so can be calibrated out with a little math.Requires a bit of engineering obviously, so you might prefer just spending the money for a drop in solution. Such as a SPT-414-01 coaxial shunt, with an isolation amplifier. Probably costs in the same range as that Pearson current sense transformer.PS. up to a MHz or socombining a Hall sensor with a cheap current transformerworks, but above that you'll either need a very expensive current transformer like that Pearson mentioned above orroll your own.

author: jonpaul
author_rank: Super Contributor
timestamp: June 06, 2022, 03:59:39 pm
content: Rebonjour, ISO amps have limitations re CMRR and max V, perhaps risky on 400VAC lines unless safety approved eg VDE/TUV/UL/ETL.The Pearson current transformers are capable of very low LF cutoff and 50 MHz, but costly $400.DIY and Chinese clones are not nearly as good.Hall effect accuracy and drift cannot achieve the 0.5% the OP asked for (?)As always a tradeoff cost><accuracy><BW, etc.Jon

author: ajb
author_rank: Super Contributor
timestamp: June 06, 2022, 04:18:49 pm
content: Digitizing on the sense side of the isolation barrier offers some additional options. At that point, take your pick of single-ended ADCs to convert the shunt voltage at the required bandwidth and precision. Pay attention to the noise figures of the isolated power supply you'll need, since a lot of cheap isolated DC-DC converters exhibit poor regulation--you might need a secondary regulator to get the required stability. Getting the digital signals across the isolation barrier is as simple as choosing a suitable digital isolator IC, which gives you built-in level shifting as well.

author: Marco
author_rank: Super Contributor
timestamp: June 06, 2022, 08:45:31 pm
content: Forgetting about DC for a moment, how about a circuit like the following for AC? R1+L1 are a 0603 shunt soldered film side down. VR1 is a tiny SMD pot, which adjusts the low pass to compensate for the inductive high pass of the shunt. C2 needs to be a film capacitor, for the low self inductance. The potentiometer is near ground level simply for safety, having at rails voltage would just allow C1 to be a lower voltage capacitor, which is hardly relevant. The AC rails voltage is also coupled down to get a differential voltage near ground, then you use a AD8129 to get a single ended output.Capacitive differential coupling seems a relatively cheap way to get an AC current shunt signal to ground with very high bandwidth ... then again, I've never seen it actually donePS. the 400V rail has to be slightly well behaved so as not to saturate the AD8129 inputs.PPS. C2/C3 could be a little smaller/cheaper, 1 uF would be plenty.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 01:46:44 am
content: I looked ACS7XX, but out of stock, I tried AKM CQ3201 instead, also 1MHz.it has around 0.2us delay comparing with current probe, settling time around 1.5us, I guess ACS or AKM will be the choice if no better options.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 01:53:44 am
content: TI isolated amplifier bandwidth just 200-300kHz, I tried AMC3301, it has around 2.5us delay, might not be the good choise for this application.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 02:04:20 am
content: Quote from: Marco on June 06, 2022, 08:45:31 pmThe application is to measure DC 10us pulse, not sure if this circuit works for it.

author: Marco
author_rank: Super Contributor
timestamp: June 07, 2022, 02:15:28 am
content: Quote from: Allen_Du on June 07, 2022, 02:04:20 amI think it can be build to substantially high bandwidths. Bandwidth wise the limiting factor will be the inductances in the low pass filter but I don't think those will be too bad. If necessary you can make C1 ~10x smaller and the potentiometer ~10x larger, which will reduce effect of inductance even more without having too much impact on noise though it will reduce CMRR a bit. AD8129 has plenty of bandwidth and very good CMRR. The question is more how stable the low pass filter compensation for the shunt inductance is ... too much work for me to calculate how thermal drift influences itYou still have to combine the AC measurement with a DC measurement, but that's true for most approaches. Any old high side current measuring IC can use the same shunt to do a DC measurement.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 02:41:02 am
content: Quote from: ajb on June 06, 2022, 04:18:49 pmThe isolation power supply as the picture, I am using TI SI6501 for isolation, and TPS65133 provides +/-5V to op-amps , I think should be good.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 02:54:19 am
content: Quote from: Marco on June 07, 2022, 02:15:28 amLet me try if can find pspice model of AD8219 and simulate the circuit.

author: Allen_Du
author_rank: Contributor
timestamp: June 07, 2022, 06:02:23 am
content: I built a prototype board as the attached circuit, when I use arduino generates a 1V PWM pulse(op-amp is using isolated power supply, arduino using none-isolated power), put the PWM pulse as input of the sensing circuit, the output waveform looks good.But when I connect it to 400VDC circuit, the output is not as expected...

author: Jeroen3
author_rank: Super Contributor
timestamp: June 07, 2022, 06:21:51 am
content: Sensitec used to have the CMS3015 which is a magnetoresistive current sensor with 2 Mhz bandwidth on a ceramic substrate.It is end of life, probably didn't sell enough, but perhaps they can still offer something similar.

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2022, 09:55:10 am
content: Quote from: Allen_Du on June 07, 2022, 02:04:20 amIf it's just a short pulse, a current transformer, such as the Murata 54050C or 54100C will work. The good thing about a 1:50 ratio (the 54100C has a 1:1:100 ratio, so connect the primary windings in series for 50:1) is if your oscilloscope has a 50 Ohm input, no external components are required, to give 1V/A out. The DC will of course be lost, but that doesn't matter, if you just want the pulse shape.https://uk.rs-online.com/web/p/current-transformers/7157055https://www.mouser.co.uk/ProductDetail/Murata-Power-Solutions/54050C?qs=RJ0JL0kt8YwnDD8xm0%252BNVw%3D%3Dhttps://docs.rs-online.com/0455/0900766b80eb1ad9.pdf

author: Marco
author_rank: Super Contributor
timestamp: June 07, 2022, 10:51:48 am
content: Quote from: Zero999 on June 07, 2022, 09:55:10 amThat's what the open source current probe does, only with P8208NL, they only manage a couple MHz -3db bandwidth.The researchers who rolled their own current transformer with a powder iron core got up to 30 MHz -3db, near the Pearson. They do have an important reminder in their paper, due to phase distortion you need a lot more bandwidth than you think with a simple 6 dB per octave fall off. Phase will distort faster than response falls off.

author: Marco
author_rank: Super Contributor
timestamp: June 07, 2022, 10:57:13 am
content: Quote from: Allen_Du on June 07, 2022, 06:02:23 amHow the hell didn't it just blow up? You're shorting the 400/-400V rails through the clamping diodes. It needs to be AC coupled with high voltage film capacitors.Also a three opamp instrumentation amplifier, discrete or integrated, won't remotely reach the high frequency CMRR of an AD8129.Also a parallel capacitor isn't a correct compensation for shunt inductance, it needs to be a RC filter with the voltage across the capacitor measured (likehere). Which has the added benefit you can use a potentiometer for the resistance and adjust the knee of the filter to match that of the shunt.

author: Marco
author_rank: Super Contributor
timestamp: June 07, 2022, 01:41:54 pm
content: In your configuration it should look something like this before the differential to single ended amplifier (VR2 adjusted to same resistance as VR1, will improve supply voltage rejection).PS. the diode clamps are a good idea since those 1u capacitors will pass a fair bit of current before they are biased, the diodes should be low capacitance though, 1n4148 will do. They could still get damaged if the -400V rail ramps too fast, 1u is a lot of capacitance to fill ... maybe the potentiometer should be at -400V, then the coupling capacitor could be smaller ... hmmm, I'm not the one getting electrocuted. With the potentiometer(s) at high voltage it would like the bottom schematic.

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2022, 09:52:24 pm
content: Quote from: Marco on June 07, 2022, 10:51:48 amThe Murata part I tested had a higher bandwidth, than the data sheet stated. I also made my own with a ferrite bead and got a bandwidth in the 10s of MHz.

author: Allen_Du
author_rank: Contributor
timestamp: June 08, 2022, 03:31:05 am
content: Quote from: Marco on June 07, 2022, 10:51:48 amCould you share me the link of the open source current probe, I only found this one, it uses hall sensor and current transformer at the same time, bandwidth 5Mhz.

author: Allen_Du
author_rank: Contributor
timestamp: June 08, 2022, 05:22:23 am
content: Quote from: Marco on June 07, 2022, 01:41:54 pmHi Marco, thanks for the suggestion, may I ask what tool are you using for simulation. I tried use LT spice do transient simulation for the AD8129 circuit, but didn't work,BTW, I cannot get access to the ground of 400V, I attached a picture of the application, I want to measure the current on the load.

author: Allen_Du
author_rank: Contributor
timestamp: June 08, 2022, 05:46:50 am
content: Quote from: Zero999 on June 07, 2022, 09:52:24 pmHi Zero, did you use it for DC current measurement, may I ask which part did you use?

author: EugenioN
author_rank: Contributor
timestamp: June 08, 2022, 09:08:04 am
content: Because those Sensitec sensor didn't work. Ask me why I know!They work fine until the temperature changes, then you get unpredictable drift in gain and offset. Wild drift! Also self heating is a major issue when frequency get high enough to enter the skin and proximity effects territory.A complete waste of time, money and reputation.My two cents.

author: Marco
author_rank: Super Contributor
timestamp: June 08, 2022, 04:50:38 pm
content: Quote from: Allen_Du on June 08, 2022, 05:22:23 amI use MPLAB Mindi, which is Microchip licensed version of Simetrix. It can only handle limited nodes though, Microchip models don't count against the node count but other models do. That's why I don't add the AD8129 itself.QuoteI can't help there, I can't stand the GUI of other simulators and the numerical convergence is just god awful.QuoteI thought your load was causing the current pulse itself and was sitting between 400v and -400v rails ... but you have a low side MOSFET pulling a current pulse through a load from a 400V rails to circuit ground. In that case putting the current sensor on the return line is really bad, that MOSFET drain has huge voltage swings. My circuit certainly won't work on the return rails and even a current transformer isn't entirely immune to such a large common mode voltage swing on the primary, it can swamp your signal on the secondary. Only a digitally/optically isolated sensor should reasonably be put on the return line and still give decent results, even some digital isolators would struggle with that common mode swing.You are supposed to put the current sensor on the relatively steady rails, the 400V line, a high side current sensor. That way the most the sensor has to deal with is some supply rails noise instead of 400V common mode swings. So for my circuit that means back to the original, the graph shows the coupled supply rail noise on VOUT- and the differential output voltage.Even if you think you have a really good reason to put the current sensor on the return line, you probably don't have a good enough reason. Even if you think it's impossible to do it differently, it probably isn't as impossible as you think. It's really incredibly bad to put it there.PS. as you see, for a 20 us pulse you don't really need DC at all. A pulse train will slowly shift the bias of the output, but you could correct for that (you know the bottom is zero). The pulse shape comes through pretty much undistorted with just AC coupling, you can see a tiny bit of droop on the pulse top but it's around 0.15%. With 100nF coupling capacitors instead of 10nF, the droop becomes pretty much invisible. The Pearson current transformer has a low enough low frequency cut off that it would not distort the pulse shape either, but cheaper SMD current transformers would, they just don't go low enough.

author: Marco
author_rank: Super Contributor
timestamp: June 08, 2022, 07:06:48 pm
content: Quote from: Allen_Du on June 08, 2022, 05:22:23 amI might have misinterpreted this, I thought you meant you couldn't put the sensor below the MOSFET. Do you mean the entire circuit is flapping in the wind and you're not allowed to ground any part of it, nor have a ground which is at least within a couple of V of the circuit ground?My circuit does need the supply rail of the circuit to be at a relatively steady voltage relative to the measurement side ground. It can fluctuate a couple V, but no more than that (no voltage fluctuation relative to measurement ground beyond the supply voltage of the AD8129). If the whole current pulse circuit is just randomly floating around by 100s of volt on every point relative to measurement side ground you need galvanic isolation.

author: Allen_Du
author_rank: Contributor
timestamp: June 09, 2022, 02:06:22 am
content: Quote from: Marco on June 08, 2022, 07:06:48 pmYes, I cannot connect the sensing circuit to any point of inside of PSU, what I can do is connect it to either 2 sides of load.

author: Marco
author_rank: Super Contributor
timestamp: June 09, 2022, 03:04:49 am
content: Quote from: Allen_Du on June 09, 2022, 02:06:22 amYou are missing my point. Is the PSU grounded/earthed?The circuit I posted is capacitively coupled to the 400V rail, but it needs the 400V rail to remain steady relative to the measurement circuit ground. The PSU needs to share a ground with the measurement circuit ground.A little bit of ground loop disturbance doesn't matter, but if the PSU is floating it won't work.

author: Allen_Du
author_rank: Contributor
timestamp: June 09, 2022, 03:33:11 am
content: Quote from: Marco on June 09, 2022, 03:04:49 amPSU housing is connected to earth ground, but 400V not, 400V is floating.

author: Marco
author_rank: Super Contributor
timestamp: June 09, 2022, 04:07:44 am
content: Quote from: Allen_Du on June 09, 2022, 03:33:11 amDoes it need to float? If you ground the 400V rails you can have the high side shunt at ground level ... and that makes things a whole lot easier.

author: Allen_Du
author_rank: Contributor
timestamp: June 09, 2022, 04:53:35 am
content: Quote from: Marco on June 09, 2022, 04:07:44 amThe problem is PSU is ordered from supplier, cannot change it.

author: Marco
author_rank: Super Contributor
timestamp: June 09, 2022, 05:00:06 am
content: If it's really floating, you don't have to change anything ... just connect ground to the 400V output on the outside.Same as you can take a lab supply and plug a ground into any of the outputs.PS. if it's not really floating don't do that

author: Allen_Du
author_rank: Contributor
timestamp: June 09, 2022, 06:39:38 am
content: Quote from: Marco on June 09, 2022, 05:00:06 amThanks Marco, might check with supplier first.

author: Marco
author_rank: Super Contributor
timestamp: June 09, 2022, 05:54:28 pm
content: To give an example of what you can try if it's floating, you can directly amplify a shunt voltage and put it on coax. For the common mode chokes get two ferrite toroids and wind the power cable and coax on them. That's why the source termination, to deal with all the reflected EMI. I don't know if this will be sufficient to deal with conducted EMI, I have no intuition for the magnitudes of the effects involved.