Need to calibrate 200A shunt to read correct for amp scale on meter across shunt - Page 1
Forum section: electronics / beginners

author: DonJuane
author_rank: Newbie
timestamp: August 22, 2013, 04:33:33 am
content: My last physics course was 40 years ago so be gentle ;-) I am purchasing a shunt and meter pair that works in combo with each other, reading out DC amps in the range of 0-200A, shipped with the meter and a shunt marked 200A. The objective is to read out current draw on a 12VDC battery. The meter is wired across the shunt and the meter reads in an amps scale (while measuring volts across shunt) and there is a screw on the back of the meter that must be used to calibrate it. How do I calibrate the meter to read the correct value of amps across the shunt while calibrating it using the voltage sensitivity adjustment screw on the back of the meter?I would assume all this. I would want to calibrate the shunt maybe at mid range, say at 100A load and this would be done in line with my 12V automotive battery and the 200A shunt all three in series. Now the supplied 200A shunt I don't know the resistance of and I know that my cheap VOM would not give me a reliable resistance value going that low. The shunt is labeled as simply "200A". From my electronics background, I know this setup I purchased is in reality a digital meter (needs calibrating) and is only reading the parallel voltage drop across the shunt (high amp resistor) but from my memory I believe I would need a known true value of resistance, a true known current draw or true known power draw to calculate E=IR or P=EI to use some absolutely true value as my load in the circuit, so I could accurately tune the meter to read 100A or the amp draw that is truly occurring.So if I am on board up to here, it seems I just need to find something of an automotive nature with a current requirement somewhere near the middle of the shunt capacity (200A) to make my meter read correctly by perhaps tuning it at the halfway point of the 200A capacity shunt, which would be 100A applied as a median load. As a load an automotive starter comes to mind or maybe 4 sets of full headlamps but I don't have those and I think the starter draws current all over the scale depending on if the cylinder is firing at the moment or the wear of the mechanics, in other words, not constant enough to use for calibration. At 12DC, to get in the mid range of the shunt's 200A capacity, which is around 100A I am thinking I would need way too many light bulbs and possibly have to go to a copper wire coiled or something as a load, maybe in suspended in oil in a glass jar to dissipate heat or to devise something that would not build up head and dramatically change the resistance, but then I would need to do all kinds of power calculations and heat dissipation calculations to ensure my coiled copper wire load did not melt. Or more simply, how do I come up with a known load in the desired current range for a good calculation?And the funnest part of my dilemma, I want to figure this out without buying anything, of course not minding winding some copper wire resistors. There would not be a big point in purchasing an accurate dummy load for my apparatus if I would be using it only once. I have a cheap digital VOM and a medium precision old large faced analog VOM to use as instruments. I can't think of a single thing that I would trust as a load and using its labeled "power/amps pulled" to use in the calculation, i.e. like a bunch of headlights in parallel, etc.Any help is appreciated.

author: Rerouter
author_rank: Super Contributor
timestamp: August 22, 2013, 07:19:24 am
content: if you want to play it smart, feed in a low mV signal to the meter first and work out its range, then possibly break into your high beam circuit with the shunt to work out the resistance of the shunt, (voltage across it)a well made shunt should not change resistance drastically as it approaches its upper range due to heating, so it would be safe to say once you work out your resistance you can calibrate it at your bench with a millivolt source,

author: poorchava
author_rank: Super Contributor
timestamp: August 22, 2013, 07:45:10 am
content: I think the shunt will be in the order of .1 miliohm. Anything higher, let's say 1mR and at 200A there will be 40W dissipated in the shunt. Kind of hard to measure such thing without very precise equipment.For determining the exact resistance of the shunt I would pass like 10A through it and measure the voltage with a precision low offset opamp (first shot: OP07) in differential configuration or maybe an instrumentation amplifier if you have one at hand. As for calibrating the voltmeter itself, maybe a stable voltage source and a precision divider?Btw, why don't you use a current transformer or a hall based sensor? They are much easier to use and waste much less energy. Look herehttp://tnij.org/farnell(The link is just shorted, no adware here).LEM is a reputable company and IMO under 20€ for a 200A CT is a reasonable price.

author: M. András
author_rank: Super Contributor
timestamp: August 22, 2013, 08:48:07 am
content: i bet if that shunt for a panel meter it will be either 50mv output at full current or 75mv

author: Unknown
author_rank: Unknown
timestamp: August 22, 2013, 10:11:03 am
content: The resistance at 10 A, let alone the 10 mA that a multimeter will use, is going to be different from the resistance at 200 A due to heating. The only way to accurately calibrate a 200 A current shunt is to feed a known current of about 200 A through it.

author: madworm
author_rank: Frequent Contributor
timestamp: August 22, 2013, 11:55:54 am
content: Feed the shunt with the current you will most likely measure most of the time and use a clamp-meter as a reference (e.g. Fluke 365 and the likes)?

author: Paul Price
author_rank: Super Contributor
timestamp: August 22, 2013, 12:27:55 pm
content: Assume that the shunt when purchased is already calibrated to accuracy guaranteed and it is designed to deliver precisely 50 or 75mV for 200A.Now by applying one of these known voltages to your meter, you will get it to read very close to full scale. Now, all you need to do is just calibrate the meter. Apply exactly 50 (or if it takes 75mV or 100mV) to move the indicator to full scale. Now adj. the calibration pot on the meter to read exactly full scale and you are done!

author: Dr. Frank
author_rank: Super Contributor
timestamp: August 22, 2013, 12:32:51 pm
content: Quote from: DonJuane on August 22, 2013, 04:33:33 amMuch too complicated.If you buy a meterin combowith this 200A shunt, then they should simply play together without "calibrating" anything. The meter will display the current in the correct format, I assume.Otherwise, shunts normally have a defined resistance, and meters were calibrated on delivery. So I do not see the necessity to calibrate, but perhaps to determine the correct scale factor, which should then be an even value.This can be determined from I (full scale) = U (meter full scale) / R (shunt)It's impossible to give correct hints without knowing anything about the components you have bougt.If you publish here the link to the purchased items, we could check that.If nothing of the above mentioned applies, then you will not have an easy possibility to trim the meter. You have recognized the problem by yourself, that you do not have a consumer in the car that would draw a current sufficiently constant and well defined.The only possibility is to switch on the headlamps of your car. As those light bulb have a standardized power consumption of say 55W or 60W each, they will draw 10A, relatively accurate.This can be used to check, if the shunt and the meter display the correct order of current.As your last electricity courses took place 40 years ago, I have to to warn you, fumbling around with such high currents.Wrong handling or mistreatment could either damage the battery and other devices in your car, but also the shunt.Do not - under no circumstances - measure the current directly across the battery, from one pole to the other!Solid cables and clamps are necessary for all measurements, if you have a regular load (head lamps, starter, etc.) in the measurement loop.Otherwise, you will create electric sparks, or overheat the cables and might hurt yourself severely.Try to avoid opening the battery current loop of your car and measure the current indirectly with a current clamp, as proposed by others already.Frank

author: johansen
author_rank: Super Contributor
timestamp: August 22, 2013, 06:32:17 pm
content: Quote from: alm on August 22, 2013, 10:11:03 amthis hasn't been true for any professional shunt built in the last 100 years since resistance alloys have been developed for 1% change in resistance across reasonable temperatures.

author: Unknown
author_rank: Unknown
timestamp: August 22, 2013, 06:55:51 pm
content: Quote from: johansen on August 22, 2013, 06:32:17 pmProfessional maganin shunts indeed have a fairly low tempco, so drift should be minimal as long as you don't get them crazy hot. I'm not sure if you can say the same for cheap shunts, however.And even for high quality shunts, the keyword is accurately. Ask any metrologist how to calibrate a shunt. Ask the manufacturers how they calibrate their shunts. They're not using a DMM or low Ohms meter.

author: robrenz
author_rank: Super Contributor
timestamp: August 22, 2013, 07:13:34 pm
content: That's why some micro ohm meters have 200A drive currents likehere

author: Unknown
author_rank: Unknown
timestamp: August 22, 2013, 07:17:05 pm
content: Fair enough. I was half expecting that the link would be a picture of one you acquired for $100 sitting on your bench.

author: robrenz
author_rank: Super Contributor
timestamp: August 22, 2013, 07:39:50 pm
content: I have 10 to 310A CC DC power supply that I can use. (My tig welder, old saturable reactor style)

author: crisr
author_rank: Contributor
timestamp: August 22, 2013, 07:50:59 pm
content: I think it would be very tricky to adjust such meter without some sort of reference test equipment such as a DC clamp meter. One thing I can think of is that you could make a bunch of equal wire-wound resistors that when connected in parallel would yeld around 0.100 - 0.136 ohms (100A @ 13.6V); connect them all in series and measure the resistance (much more precision that way); then connect them all in parallel, calculate the precise resistance, connect in series with your shunt and battery and measure the voltage drop, adjusting your meter according to the calculated current. At around 100A, remember safety, use beefy (really beefy) wires and connectors.

author: Paul Price
author_rank: Super Contributor
timestamp: August 22, 2013, 09:32:20 pm
content: Crisr: I don't quite get the logic of your method.Suppose I bought quan. 10 of .01ohm sense resistors 1-watt 1% (33cents US ea. quan 10 Farnell.com). If I just connect them in parallel with a Kelvin connection for measurement, OK, so then I have a 10 watt array that will measure quite close to.001 ohms, probably well within 1% because their value errors will likely average out. Now if I apply 100-amps to these resistors all in parallel this way but also in series with the shunt you have, then the .001 resistor array will develop .1volts and yet not exceed their power rating and I will have an accurate current reading to also calibrate the meter/shunt hooked up at the same time.So what is the advantage in connecting them in series to get a result inconsistent with their parallel connection in final use and get a resistance reading that will undoubtedly not be their actual parallel resistance value due to soldering and lead resistance etc, and so just add error to upset the accuracy of your measurement and calibration goal?

author: Unknown
author_rank: Unknown
timestamp: August 22, 2013, 10:09:18 pm
content: Measuring 1 mOhm accurately is quite tricky, even with a decent bench meter. You can shove a large current through there and measure the voltage drop, but then you're back to needing a large, known current. Putting them in series gives you 0.1 Ohm, something that is much easier to measure. Of course at these resistances leads and connections may start to become an issue.Just buying resistance with a sufficiently low tolerance would also work, but probably more expensive.

author: robrenz
author_rank: Super Contributor
timestamp: August 22, 2013, 10:24:37 pm
content: From my experience your conductors connecting many paralleled resistors need to be about 10000x less resistance than the resistors if you want the voltage drop to be anywhere near the calculated value and for the load carried by the resistors to be evenly distributed. I posted an example inThis threadAlso at 1 ohm and below thermal emfs are a large part of the errors and you need to use current reversal or offset compensated ohms techniques and obviously 4 wire measurements.

author: Paul Price
author_rank: Super Contributor
timestamp: August 23, 2013, 12:30:40 am
content: Robenz, I don't think you quite got it right at needing 10000x less resistance to parallel sense resistors. It is all a question of correct layout and symmetry.For example, imagine 10 units .01 ohms to be paralleled:Lay them out on a breadboard symmetrically in a circle. You connect a 4-in long #12 wire to each side of the .01 ohm resistor to a single common circular feed point, all wires are tied together to the power feed points, which are two circular pieces of PCB stock about 2-in in diameter so as to create power in /out points and all having the same wire length from the center of the common center point to each shunt from this circular feed. These individual spaghetti leads to the resistors swamp out the resistance of each accurately known value .01 ohm shunt, so their exact length should be matched but not to a micro-inch.What you are doing next is connecting all the resistor Kelvin points at the shunts together with common length flying wires and to a second pair of two circular common Kelvin connection measuring points on each side of the shunt array, same way, using wires going up from the PCB or breadboard into the air to a common circular piece of PCB stock and you've then will see you have perfect symmetry and resistance balance and no need for some probably unattainable 10000x less ohms connection theoretical setup for paralleling the shunt resistors to attain any kinda high accuracy.Since you then know the resistance of the shunt just made as very very close to .001 ohm you only need to calibrate the unknown 200-amp shunt and meter.

author: Paul Price
author_rank: Super Contributor
timestamp: August 23, 2013, 12:34:50 am
content: alm. I don't get it!If you use a .1 ohm resistor as a shunt and feed it 100 amps you are dissipating I*I*R watts or 10000 x.1 watts!!!!Watch out for the fireworks.

author: The Electrician
author_rank: Frequent Contributor
timestamp: August 23, 2013, 01:53:08 am
content: Quote from: DonJuane on August 22, 2013, 04:33:33 amYou say what you're buying is a digital meter; are you buying something like this:http://www.ebay.com/itm/ON-OFF-DC-0-200A-White-Lcd-Digital-Ammeter-Current-Panel-Amp-Meter-Shunt-/330929864739?pt=LH_DefaultDomain_0&hash=item4d0cf50023The shunt itself is trimmed at the factory to have exactly (within the accuracy class) 75 millivolts across the measurement terminals when 200 amps passes through the shunt. You don't need to measure the shunt resistance.All you have to do is make sure your meter reads 200 amps when 75 millivolts is connected to the input. Follow the instructions in post #6.The shunt only dissipates 15 watts when carrying the full 200 amp current. It's made of manganin and 15 watts of heating will not push the resistance of the shunt out of its 0.5 accuracy class.Alternatively, you could buy one of these:http://www.ebay.com/itm/Analog-Amp-Panel-Meter-Current-Ammeter-DC-0-200A-Shunt-/110735392312?pt=LH_DefaultDomain_0&hash=item19c857fe38Then the little analog millivoltmeter will already be calibrated for use with the shunt. I notice the analog meter is accuracy class 2.5, but the individual Chinese shunts also on eBay are accuracy class 0.5.Or, you could buy a clamp on ammeter:http://www.ebay.com/itm/Digital-Clamp-Multimeters-DC-Ammeter-Professional-Tester-Meter-Clamp-Meter-/281090944569?pt=LH_DefaultDomain_0&hash=item4172536e39It has a 40 amp and a 400 amp DC range. With a clamp on, you don't have to make up cables capable of carrying 200 amps safely. When I was involved in the inverter business, we used a lot of 4/0 welding cable for this purpose, and it's expensive!

author: crisr
author_rank: Contributor
timestamp: August 23, 2013, 02:25:31 am
content: Quote from: Paul Price on August 22, 2013, 09:32:20 pmWell, you said you didn't want to buy anything and wouldn't mind winding some resistors, so this is one way I thought it could be done. If you just buy 1% power resistors of course it could be much simpler. The reason for measuring the (hand-made) resistors in series before wiring them in parallel is that you can get a much greater accuracy. Supposing each resistor (that you make yourself each using exactly the same length of wire) is around 1 ohm, measuring 10 ohms with an ordinary multimeter would yeld a smaller error than if you tried to measure each one individually and less still than if you tried to measure all of them in parallel. Oversize the wires and solder junctions and you shouldn't have problems.Once you calculate the resistance of the parallel array, and put that in series with your load, whatever current flows in the circuit you can calculate by measuring the voltage drop in your array and calibrate your meter to that.Edit: And yes, if you build a .1 ohm array that is used as both shunt and load, you will be dissipating 1000W in it. Beefy wires, and make the measuring and calibration fast, so that temperature does not affect the resistance by a whole lot and that you do not set anything on fire!Edit 2: It's probably not practical anyway, because if you wound 10x 1 ohm resistors out of 1mm2 (17AWG) wire, you would need about 625m of it.

author: robrenz
author_rank: Super Contributor
timestamp: August 23, 2013, 02:26:36 am
content: Quote from: Paul Price on August 23, 2013, 12:30:40 amI think I have it right for layouts that people would typically use. Resistors in a ladder arrangement and just two sense connections for the whole paralleled group. In the post I linked above I state that I intentionally make the spice layout poor to show the effects. My real point is many don't realize that the ratio of conductor R to resistor R becomes very important with low R resistors. It is not noticed with higher values because the 10000:1 ratio is typically met or exceeded with normal traces or wires. Wattage balance of the group is also very influenced by this.I think your solution does address the issues but I think you would agree that it is not very practical to implement.EDIT: From my spice experimentation and real world testing it seemed like the 10000:1 was a good general rule of thumb. I have never seen this addressed in any literature, its just the number that worked out to have minimal shunt voltage error and good wattage balance.

author: Unknown
author_rank: Unknown
timestamp: August 23, 2013, 11:28:38 am
content: Quote from: Paul Price on August 23, 2013, 12:34:50 amAs crisr describes it: make resistors with unknown value, put in series, measure accurately, then put in parallel and use as shunt. This should give you the average resistance of the resistors, and allows you to calculate the resistance of your very low value shunt without any special equipment.

author: eevblogfan
author_rank: Frequent Contributor
timestamp: August 23, 2013, 12:18:10 pm
content: HeyQuoteyou need to aim to perhaps the halfway point of the power being dissipated on the shunt , in that case , you meant 141.42135623730950488016887242097A , witch is going to dissipate 7.4999999999999999999999999999926W on your sunt ( halfway point of 15W ) ( LOL 0.00001yA and 0.0000001yW res)Don't accept to have better accuracy on that stupid shunt , you need bigass 30W+ shunt in order to measure say 3W safely at say 4~8 degrees change , that's going to change 0.008% on 10PPM metal , but what is the type this shunt utilizes ? , perhaps one that with zero error trimming at say 1W load , will claim up to 0.5% when 15W passed on the tinny shunt , even if you are going to trim ( adjust) the system , you need to know that measuring 75mV at better then 1% absolute accuracy , is hard , hence it may be 1~2% of the goal , I really have to wait until I know what purpose that serves ..now before any judgment , please shear with us the need for calibrating ( actually perhaps you meant to adjusting ... ) ?0.5% on 200A is okay for basically any application normally guy would have imagine ..I can measure up to 54A to the 0.02% accuracy , but most of the stuff you are perhaps going to measure , will accept 1% error , and even more ..I don't know the res you got there , but I think it is going to be only 200 count meter , in that case the uncertainty is +- 1 count *on good meter * , in that case , you need the shun't value not to change at 200A (15W) load , and .... that's not going to work with that small shunt ... not even close ..Cheers

author: johansen
author_rank: Super Contributor
timestamp: August 23, 2013, 08:17:35 pm
content: some of you guys are going way overboard with the standards you expect of things.the expensive equipment posted that delivers 200 amps to measure .01 micro ohms...That current is required so that the manufacturer can guarantee .2% accuracy at a resolution of .01 micro ohms..01micro ohms times 200 amps is 2 microvolts.The chopper stabilized opamp inside can only guarantee 0.1uV of offset voltage? so that's 5% right there you can never get rid of.and then you have the voltages generated by the temperature differences in the dissimilar metals, between the kevin clips and the shunt.and the shunt is warming up during the measurement...I can measure micro ohms with a 7$ 1KG drug scale off ebay, a D cell battery, and my VC99 with #12 wires sticking out the back of the meter, to about 1% accuracy. the resolution at the lcd is 0.1 gram. the max is 1000 grams. the ADC inside it is probably 12 bits, but it seems to have noise at the 11th bit, which means the readings are always plus or minus .2 grams, which is about .48 microvolts of offset error in the amplifier.this means i need at least 10 amps flowing to measure 1 microohm to 5% accuracy.not bad for 40$ in equipment. the accuracy is the limits of the kelvin varley bridge used to determine that full scale 1000 grams is 2.4 millivolts..and that depends on the battery voltage inside the scale.1% accuracy for the amp meter, 1% for variation in the scale, 5% for the limitation of .48 micro volts.i could use my kelvin varley bridge to calibrate an external shunt and then make a much more accurate measurement of the current if i borrow a standard OHM from someone, and i have a weston cell available that i know is 1.018 or 1.019 volts..you can get a 4.5 or a 5 decade kelvin varley bridge off ebay for about 50-200$ in the form of a differential vacuum tube volt meter. you might get lucky and find the saturated weston cell in those things still good. which is where i got mine.

author: DonJuane
author_rank: Newbie
timestamp: September 25, 2013, 05:45:20 am
content: Thanks for all the replies! Have been away for a while and just now getting back home and looking into this. I was very surprised to see all of this activity since I had to drop this project and am now returning to finish it as a colder weather one. Thanks for all the debate!Maybe I should be more specific. The problem I face is that I do not have any elaborate test equipment and this meter must be adjusted to a variety of supported shunts:http://www.ebay.com/itm/261203869239?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1423.l2649The meter is the same regardless of the shunt (I believe) and one therefore has to set the adjustment screw to match the shunt to make it read correctly. Also and unfortunately this meter and shunt setup as I can see only measures the current through the negative terminal of the load, so I have to isolate the negative terminal of a recreational vehicle 12V battery ground terminal and attach the shunt between the battery (-) and the car metallic frame in order to use this setup. Isolating the positive would be much easier because dealing with the ground of course, means the gen-set starter motor is grounded to the steel vehicle frame and the part I am not really interested in cannot be isolated or eliminated. I could supply a separate ground bus for all the inside accessories which are my true objectives to measure, but that could be expensive and tedious. I am looking for something simple.Also a much smaller shunt would do for non-starter-time use, but isolating all the grounds of the house wiring is not practical. So with this limitation and worried about a start of the gen-set smoking any smaller shunt I am just guessing that a 200A shunt would handle the start cycle but don't know that for certain. I am also afraid the more common loads in the vehicle will read less accurately due to such a large shunt and a smaller one might be better if I could isolate. The only way I can think of to use a smaller shunt with this meter (negative terminal load) is to perhaps use a solenoid that bridges the shunt during the start cycle of the gen-set and instead of the 200A shunt use (perhaps) a 100A shunt instead.Thanks again!

author: M. András
author_rank: Super Contributor
timestamp: September 25, 2013, 06:50:26 pm
content: Quote from: DonJuane on September 25, 2013, 05:45:20 amthe starter current would be much higher then 200 amps thats for sure

author: DonJuane
author_rank: Newbie
timestamp: September 25, 2013, 07:48:33 pm
content: http://www.amazon.com/STARTER-MARINE-KOHLER-GENERATOR-GM23910/dp/B007AMC5AYThe generator(genset) starter here is advertized as a 1.2KW device. At 12 volts, would that not be I=P/E or amps = 1200W/12V or around 100 amps?

author: DonJuane
author_rank: Newbie
timestamp: September 25, 2013, 08:05:27 pm
content: After looking at all the various shunt combinations offered with this same style meter, I notice that all the shunts are rated at xxxAMPS/75mv. At an aha moment I am now assuming this means that there is 75mv voltage across the shunt when it is at full load. If that is true, then all I have to do is come up with a 75mv source, put it across the yellow and black leads of the panel meter and turn the adjustment screw until the meter reads what the shunt value is rated as (in amps).Then if that's true, the challenge will be to find a 75mv source. A search for "75mv voltage regulator" doesn't turn up much. I guess I could come up with it through a couple of known resistors across a known low voltage source, and that's assuming there is negligible current drain across the meter leads.