Question : Circuit to measure low current without resistive insertion lost - Page 1
Forum section: electronics / beginners

author: BravoV
author_rank: Super Contributor
timestamp: November 20, 2012, 02:49:02 am
content: While reading EDN, found this interesing circuit ->EDN : Measuresmallcurrents without adding resistive insertion lossQuestions :- It says measuring current at "almost" zero drop, how low is that ?- How this circuit's accuracy say using a precision auto-zero op-amp ?- As an EE wannabe/hobbyist, for me this circuit looks too good to be true, whats the catch if any ?Please, any other comments or critiques regarding this technique if you have any, tia.

author: IanB
author_rank: Super Contributor
timestamp: November 20, 2012, 03:19:31 am
content: From reading the comments under the article the circuit shown is a simplified illustration of a practical circuit that may be found in a textbook. The version illustrated appears to need a few extra details added before it can be used in a real application.It looks rather like the active version of a bridge circuit, where you adjust a known value to exactly balance an unknown value by nulling out the offset between them. Having done that you can read off the known value as giving the measure of the unknown value.

author: Psi
author_rank: Super Contributor
timestamp: November 20, 2012, 09:16:11 am
content: Ya don't always need to add a resistor to sense the current using the Vdrop resistive method.It's possible to use the existing wiring as your sense resistor. You just need a really sensitive amplifier to pick it up as its so small.

author: TriodeTiger
author_rank: Regular Contributor
timestamp: November 20, 2012, 09:27:19 am
content: Quote from: Psi on November 20, 2012, 09:16:11 amWouldn't that require a milliohm meter? You'll find V but have no tangible R any longer. I think current sensing (guessing V instead of R, or already knowing V) and knowing I through magnetic fields makes more sense. I suppose that is a current transformer.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 20, 2012, 09:43:48 am
content: maybe not what you need but FWIW... transimpedance buffer amp (current to voltage converter)http://pdfserv.maximintegrated.com/en/an/AN3428.pdf

author: Psi
author_rank: Super Contributor
timestamp: November 20, 2012, 09:45:51 am
content: Quote from: MmCoffee on November 20, 2012, 09:27:19 amYeah, you need to calibrate to get R.It's not too bad if you're using a PCB track though.

author: BravoV
author_rank: Super Contributor
timestamp: November 20, 2012, 10:05:11 am
content: @IanBI'm afraid I'm lost, bridge circuit ? Elaborate further please.@PsiThat method is not practical, as MmCoffee's concern and that will be only for one off project. What interested me is the "almost" zero drop feature and yes, I have Dave's uCurrent and it works fine so far, and thinking if this circuit is good enough, why not build it as measurement gadget to complement my uCurrent.Also this circuit is quite intriguing (at least for me) and really like to hear analog guys to chime in more about it.@MechatrommerThanks though, btw that circuit already mentioned at that EDN page, check it out, and I'm aware of this technique too. Fyi, that current to voltage converter circuit has that 1 M Ohm resistor drop/burden voltage which is the weakpoint that this circuit doesn't have. Hint .. "zero drop" ...

author: Harvs
author_rank: Super Contributor
timestamp: November 20, 2012, 12:47:42 pm
content: As for the first question, how low is almost zero drop? If you were living in an ideal world where the gain of the opamps were infinate, had zero input offset and infinate input imedance (and hence zero bias current) then there would be zero voltage drop. As we have to use real components, those three things are going to work against you. Suggest you grab a datasheet for a commonly available op-amp and have a go at some circuit analysis. Wouldn't be hard and could be a telling exercise. Also throw into that the resistor tolerances (i.e. the top and bottom resistors not matching by say 1%) The other thing to consider is differing gain, offset and input bias current between the two amps.If you haven't totally lost interest by this stage, try calculating the output accuracy you're going to get keeping all those factors in mind. It's probably going to be quite telling just how expensive this will get by the time you buy enough precision parts to keep the whole circuit at a level of acuracy you're after.The other thing to consider is the power consumption of the circuit...

author: robrenz
author_rank: Super Contributor
timestamp: November 20, 2012, 01:05:09 pm
content: Quote from: MmCoffee on November 20, 2012, 09:27:19 amYou would only need the micro-ohm meter to calibrate your trace or wire resistance between the sense lead connection points once during construction. Then its just the voltage drop over that resistance (as long as the input impedance of the voltage measuring device or circuit is very high)

author: IanB
author_rank: Super Contributor
timestamp: November 23, 2012, 11:36:23 am
content: Quote from: BravoV on November 20, 2012, 10:05:11 amBasically, like this:If the three fixed resistors are equal in value then the voltage at point B is zero. In the example shown our current to be measured is returning to ground. We adjust VR1 until the galvanometer needle reads zero, meaning that the voltages at points A and B are equal. At this point VAand VBare both at zero volts, so that the current to be measured is returning to ground with no additional voltage drop imposed by the current measurement. Having made this adjustment, it is now possible to determine the current by measuring VR1 (perhaps via a calibrated dial) and applying Ohm's law.The op amp circuit for measuring current is doing something essentially similar, by setting up a virtual ground and adjusting currents until the voltages balance, only it is doing it automatically.A further modification of the basic op amp circuit as described in the referenced article is to relax the constraint that VAand VBare at zero volts and instead allow them to float at the circuit voltage by making the lower right R also variable (VR2) and adjusted in tandem with VR1. The left side of the bridge will now sink current and the right side of the bridge will source an exactly equal current as long as VR1 and VR2 are equal in value.On reflection this last bit isn't right and doesn't work as I have described it. I'm sure there's a passive circuit that would work but I can't think of it immediately.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 23, 2012, 02:55:00 pm
content: Quote from: IanB on November 23, 2012, 11:36:23 amnow there's the problem. where can you find two "equal" resistors? let alone three. btw i'm still confused by what "resistive insertion lost" means is. i can understand by adding series shunt resistor that can be a lost, but transimpedance opamp i posted earlier? where the lost can be? the +ve input of the opamp is just a virtual ground, basically you can remove the opamp circuit from the picture of your analysis, you only care with the feedback resistor, one only that you need to measure its resistance to some degree of precision. is it worth it to match or exact the value of 2 or more precision resistors? you may as well complicate things further unnecesarily. btw what the amperage we talking about? in OP its "small current" how small? power engineer will consider 1A is small

author: BravoV
author_rank: Super Contributor
timestamp: November 24, 2012, 08:34:23 am
content: @IanB, thanks for the explanation & illustration, I'm somehow get "part of it", pardon for noobness. I guess I will try this out my self once I had the chance, circuit wise should be easy enough looking at the component counts.Quote from: Mechatrommer on November 23, 2012, 02:55:00 pmGoogle for resistor pairing manually, its relatively easy to sort a pair out from a bunch of resistors from the same batch down to 0.1% accuracy or even better, as long you have the required and good enough resolution, like bench-top dmm class.Quote from: Mechatrommer on November 23, 2012, 02:55:00 pmAgain, find the articles on resistor pairing method manually, its a piece of cake Cik Shafri.Quote from: Mechatrommer on November 23, 2012, 02:55:00 pmThis thread is created for learning purpose, hence its in Beginner section, even its not practical. Just really curious of the concept of low drop out current measurement.About the current scale, it depends heavily on the op-amp's current capability of course, looking at common precision op-amps spec, I'd say this is only suitable for sub mA measurement, CMIIW.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 24, 2012, 10:23:23 am
content: Quote from: BravoV on November 24, 2012, 08:34:23 am"Encik" not "Cik". "Cik" is for ladyi havent read "resistor pairing" carefully but i currently doing on something require "very close" resistor matching, i just sorted out to using multiturn trimpot, i dont have time to DMM and sort all resistors batch on the bench and match which number is closest to which.the more precision you require, the catch is always... "Time". i guess. at mA maybe still forgiving, but when ones gets down to uA range, things can get exponentially complicated i believe, i never too good at % accuracy calculation though, just trying and error. maybe you can ask people who have built precision resistor decade box down to mOhm or uOhm accuracy how difficult, costly or "timely" it can be.

author: SeanB
author_rank: Super Contributor
timestamp: November 24, 2012, 10:55:33 am
content: Pairing takes time. Lat time I was doing that was to bin assorted 3V3, 3V6 and 3V9 1W zener diodes to make some 6V7 zeners to upgrade cheap emergency lights. I used a resistor fed from 5v to do the sorting, and sorted them into 50mv bins for mixing and matching to do the final selection. Out of 10 of each value zener diodes I was able to get 6 sets that were close to 6.7V, the other sets came to between 6.5 and 7.1V in value, so they were just used as straight diodes to get the extra 0.7V drop to compensate for the voltage drop across the isolation diode in the light.These were the cheapest and quickest method to fix the charger, which just used a series resistor fed from 9-11V to charge a 6V SLA battery, I got tired of doing yearly battery replacements, and they were getting expensive. Was not going to polish these bull specimens further, any more sophisticated a charger would have been overkill, all I did as a final upgrade was to add a 470uF 50V radial electrolytic across the power rails so the Royer oscillator would start reliably when the mains failed. They came with a 10uF cap fitted, and the 470/50 units were part of a big grab bag i had bought recently. They now do not kill the battery so fast, and now at least will start.

author: Colfaxmingo
author_rank: Contributor
timestamp: November 27, 2012, 02:45:37 pm
content: Matching resistors can be a real nuisance. This reminded me of this article I read by Howard Johnson (Handbook of Black Magic guy)http://www.edn.com/electronics-blogs/signal-integrity/4363408/7-solution