Isolated current sense instrumentation - Page 1
Forum section: electronics / beginners

author: eecook
author_rank: Regular Contributor
timestamp: May 23, 2018, 02:49:25 pm
content: Hi All,I am designing a current sense circuit for educational (my own education) purposes and can't seem to move forward in one minor aspect. I have gone through many application notes and online info but the information seems contradictory to me.In particular the circuit I want to design is depicted in the following image:A current transformer with a termination resistor generates a voltage proportional to the current on the primary. This goes through CM and DM filtering and into the InAmp.The question is, do I need R4 and R5 (in red) to provide a current path to bias the InAmp input stage?I have read repeatedly that indeed that is what I should be doing, the canonical example being thermocouple instrumentation. On the other hand however, I stumbled upon this (http://www.ti.com/lit/ug/tiducy7/tiducy7.pdf) appnote, where all the measurements seem to neglect this fact (see Figure 18 for instance).Of course I am going to build it and test it. I just want to make sure I understand what is going on behind the scenes.Thanks,

author: Dave
author_rank: Super Contributor
timestamp: May 23, 2018, 05:21:34 pm
content: You need to set some DC bias on the input pins, otherwise nothing is keeping the potentials on the inputs from floating around, likely out of the common mode input range of the amplifier (because of high impedance inputs).The application note is completely irrelevant here, because that circuit DOES have a DC bias due to the very fact that you're measuring voltages on resistors, which are referenced to the ground of the circuit and not an isolated source like a transformer.

author: David Hess
author_rank: Super Contributor
timestamp: May 23, 2018, 08:34:00 pm
content: Quote from: eecook on May 23, 2018, 02:49:25 pmThe transformer provides galvanic isolation unlike figure 18 so a return path to ground for the amplifier's bias current is necessary.An instrumentation amplifier and balanced filtering is not needed with the output of a current transformer. The secondary of the current transformer is galvanically isolated so one side may be referenced to ground (or any other reference) and single ended circuits used.

author: JS
author_rank: Frequent Contributor
timestamp: May 23, 2018, 09:44:36 pm
content: You don't need them, as you don't need CM filter. Just tie one end of the transformer to your circuit's ground and the other to the input of the opamp with a proper shunt resistor, note the power disippation of the shunt as it depends on the measured current and the transformer ratio. NM (your DM) filtering could be still used, the transformer would already be filtering something but as it's a transformer working in current mode, heavily loaded, it might filter very little.Then the DC bias would be provided throw the secondary of your transformer, this could be a problem if you are trying to measure very low currents with a tini tiny transformer and a high DC bias opamp, but for most cases it won't.Note that the amp in your figure is configured as a differential amplifier block, not an operational amplifier, you will need to add some resistors and probably compensation for your opamp to work as a gain block.JS

author: eecook
author_rank: Regular Contributor
timestamp: May 24, 2018, 03:59:02 pm
content: Quote from: Dave on May 23, 2018, 05:21:34 pmThank you for the input Dave. The application note has GND for the power electronics and AGND for control and instrumentation. I was assuming that this two grounds were not connected. So I guess that judging by the fact that the circuit in the app note presumably works and sharing grounds is the only way in which the InAmp gets properly biased, that's how you conclude the two grounds are electrically connected. Is this an accurate assessment?

author: eecook
author_rank: Regular Contributor
timestamp: May 24, 2018, 04:02:14 pm
content: Hi David thank you for your commentQuote from: David Hess on May 23, 2018, 08:34:00 pmWhy? What prevents CM currents from flowing into the circuit?

author: eecook
author_rank: Regular Contributor
timestamp: May 24, 2018, 04:04:05 pm
content: Quote from: JS on May 23, 2018, 09:44:36 pmThank you JS.Same comment as Dave here. I cannot see why it is not needed.As for the InAmp block, it is just a placeholder to avoid drawing the rest of the circuit, but you get the idea

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 25, 2018, 05:17:32 am
content: Quote from: xxxThe idea is that opamps have bias currents, and these need to be managed.But these bias currens can have widely different orders of magnitude, depending on the opamp used.Providing a DC path to GND is not always the best option.Alterternatively, you can split R1 in 2 equal resistors, and use that as the grounding point for your amplifier.Have you thought about WCS1800?Its an Ali / Ebay special. HALL sensor, upto 35A and the PCB often has an on-board amplifier.

author: eecook
author_rank: Regular Contributor
timestamp: May 28, 2018, 03:12:16 pm
content: Quote from: Doctorandus_P on May 25, 2018, 05:17:32 amHow would you be able to tell in advance, that a path to GND is not the best option? and what would the alternative be?QuoteThat would still be providing a path to ground.QuoteI didn't know that one in particular, so thank you for bringing it to my attention. I have considered other hall effect sensors like one in the ACS722 series from Allegro. The problem is that I need to be able to measure a minimum 25mArms and these devices seem to be too noisy in that range.

author: JS
author_rank: Frequent Contributor
timestamp: May 30, 2018, 04:20:26 am
content: You have a coupling transformer, it should be pretty little CM signal if done properly with a good transformer. You might have some capacitive coupling between the windings you need to consider but it shouldn't be hard to deal with. For starter you will tie one end of the transformer to ground so no CM there, and pretty hard to deal with the other side of the coupling, it will make a CL HP filter with the interwinding capacitance closer to the side not tied to ground and the secondary inductance. If you tie to ground the side of the sec that's closer to the pri you already solved most of the problem.If your opamp has a high input bias current placing the DC bias path to ground (with split rails, or reference with single rail) and you use high resistance would create an offset, so you might want to offset the point where the bias current is being fed so you don't have that offset. If you use a low input bias current this is less of a problem.Another way of compensating that is to use equal resistance path to both inputs of the opamp, but now you have to consider the input bias current offset and drift and input current noise, as you might not need such high resistance on both inputs this might not be optimal.A TL072 with about 100pA of input bias and a 1MΩ to ground you will have 100µV offset. Not much of a problem considering the offset of that opamp is an order of magnitude higher than that.If you have an NE5534 with about 1µA you now have 1V of offset. That's 2 or 3 orders of magnitude over the offset. Also you will have a lot of noise and drift using 1M with such opamp.JS

author: eecook
author_rank: Regular Contributor
timestamp: June 12, 2018, 09:23:46 pm
content: Hey Guys,Following JS's advice, I'll go for the single-ended version. I've been playing with NL5 and decided to go with the circuit depicted below. Some background first...The goal of the design is to measure AC mains load currents with the following spec:Current range: 2.5mA to 1A.Freq Range: 50Hz to 2kHz (to capture harmonic content).Signal conditioning to be fed to an ADC with 2.5V full-scale and a sampling rate of 40ksps.Some of the relevant parts are, the current transformer (PN: FIS155NL) and the op-amp (PN: MCP6L1T-E/OT).I would love to read some thoughts.This is the schematic:This is the frequency response (from I1 to VX):

author: David Hess
author_rank: Super Contributor
timestamp: June 12, 2018, 10:44:10 pm
content: That looks workable but isn't R2 awfully low in value? I get 12.4 millivolts with a 1 amp input and the transformer datasheet shows that values up to 87.5 ohms total secondary resistance allowing an output up to 175 millivolts total should be acceptable for 50 Hz operation.

author: eecook
author_rank: Regular Contributor
timestamp: June 13, 2018, 12:23:22 am
content: Quote from: David Hess on June 12, 2018, 10:44:10 pmIt is horribly low, the problem is the magnetizing inductance of the current transformer. If I increment R2, I lose flatness (and increase phase distortion) in the frequency response.

author: JS
author_rank: Frequent Contributor
timestamp: June 13, 2018, 01:55:08 am
content: The R loading the transformer should be low, as it work as the shunt multiplied by the transformer ratio, if too high the burden voltage is high and the core could saturate and loose precision. Current transformers are rated for that resistor, so datasheet should express the maximum value or power of that reaistor. The lower the better for the transformer. It also depends on frequency, as higher frequency would allow for higher resistance/coupled power.The topology seems rigt and the bode plot as well, I cant see the values as I can only download a low q pict right now.JS

author: David Hess
author_rank: Super Contributor
timestamp: June 13, 2018, 02:39:03 am
content: I am just saying that it is awfully low based on the transformer specifications which I checked for operation down to even 50 Hz. Of course there is lots of leeway given the operational amplifier's low input noise and AC coupled operation and maybe you wanted to operate down below 5 Hz.If you are looking for operation down to the lowest possible frequency, then no load resistor is required at all. Instead use the transformer to drive the summing node of an inverting amplifier so the load resistance becomes "zero" and only the winding resistance remains. The feedback resistor for the inverting amplifier then sets the gain in volts/amp or transimpedance.

author: JS
author_rank: Frequent Contributor
timestamp: June 13, 2018, 02:51:21 am
content: The peoblem seems the choice of the transformer, you should look for a low turns ratio designed for low freq, not the case of the FIS155 which is 1:500 for 500kHz.You could wind your own transformer, small core, M6 lamination should do. Not so many turns, easy transformer math applies here, use your desired output voltage and input current to pick a turns ratio and shunt.JS

author: capt bullshot
author_rank: Super Contributor
timestamp: June 13, 2018, 06:43:31 am
content: There are current transformers specially designed for your purpose. They are specially designed to give accurate phase and amplitude results at the power line frequency including harmonics to some kHz.Typically, they have a 1:1000 (ballpark) ratio and a very high main (magnetizing) inductance (some 10H). In terms of accuracy, noise, linearity and immunity to external magnetic fields, they outperform any open loop Hall type sensor.An example would be the datasheet I've attached.

author: eecook
author_rank: Regular Contributor
timestamp: June 13, 2018, 11:15:11 am
content: Quote from: capt bullshot on June 13, 2018, 06:43:31 amI don't see how that is better than the one I am using (also designed for current sensing). It has an even worse turns ratio and inductance.

author: eecook
author_rank: Regular Contributor
timestamp: June 13, 2018, 11:48:29 am
content: Quote from: JS on June 13, 2018, 02:51:21 amI will indeed have to seriously consider designing my own xfmr, but I will leave that for a 2nd iteration. At the moment it is not a possibility as the current sensing is part of a larger system and I have a time constraint for a working prototype.In terms of the bandwidth I need, the FIS155 is the the only off-the-shelf part I found that met my spec. The price to pay is the turns ratio.In any case, while the circuit could certainly be improved, it doesn't seem all that bad, does it?Thanks again for your insight JS.Cheers.

author: capt bullshot
author_rank: Super Contributor
timestamp: June 13, 2018, 01:06:36 pm
content: Quote from: eecook on June 13, 2018, 11:15:11 amThe CT's main inductance and the resistive part (sense resistor plus CT's internal copper resistance) form an L/C high pass. Its corner frequency is determined by (total R)/2*pi*L. As inductance increases by n² but resistance increases by n (assuming using the same wire, which is not true), it's easier to achieve low corner frequencies by increasing n (at first approach). Additionally choosing the appropriate core material gives a large increase in L without affecting R.So this is why (by providing a much larger L) these line frequency CTs give better results than yours though they have higher R.Edit: The CT whose datasheet I attached, has a main inductance of about 650H @100Hz and a resistance of 460 Ohm, put these values into your simulated circuit to see the results. For this particular CT and your chosen CT (which has 470mH only), the core material makes for the large difference in main inductance.

author: JS
author_rank: Frequent Contributor
timestamp: July 02, 2018, 01:28:36 am
content: Quote from: eecook on June 13, 2018, 11:48:29 amDesigning a transformer shouldn't chew much time, as it's working with very little voltage they are easy to work with, build and test. Also, once you have a proper transformer, the next step is much easier, building an amplifier for a proper transformer could be much easier than working with a wrong transformer to start with.As they work in current mode they are usually much more linear with less effort than voltage transformers, where core saturation can happen and makes things awful.I've used quite a few measurement transformers, usually 5A output, for the desired input, 100A, 500A, whatever, as 5A-10A is usually an easy current to deal with and many instruments support that, as any half decent DMM or using a shunt to probe around it. 5A are nominal, but the specs usually hold tight up to 120% and they can stand over that for a short period. When transients are huge and you need the lower range is common to short the transformer during start up, and then start measuring the steady state.1:1000 is also common, but you need to wind 1000 turns on the sec, kind of a pain compared to 100 turns, as the primary is usually 1 turn.The circuit for this is quite straight forward, µCurrent could do a great job if one of the ranges is fine for your needs, if not adjust that design, you don't really need to care for the burden voltage so you can have a much higher one and let your transformer make it small enough, that eases up the opamp choise, as voltage errors become much smaller, still the topology remains, shunt, non inverting amplifier with gain to get the desired output, dual rails makes it easy. Then there are the frequency considerations, to get up to a few kHz isn't hard at all, filter over the needed bandwidth so you don't have unnecessary noise.If you want to make it more complex there are several things you can add, for a current mode transformer you could compensate for the winding resistance and get a much wider frequency range and improved linearity.JS