How to measure >10A with DMM ?

I was trying to take an amp reading for JimJones in this thread with the 19.2V (16 x 2/3A) light, Jim was getting readings of 11.5 Amps, which I cannot measure with my new Triplett that like most Flukes I looked at has a 10A max measurement.

I was looking at this electroplating forum where they were talking about setting up an external shunt to be able to take higher amp readings, but don't know if it is reliable, fairly straightforward to hook up, etc.

Any feedback?

You can use either a shunt or a current clamp. A (properly calibrated) shunt is very accurate and less expensive, but also less flexible and more cumbersome to use (since it will be useful over a relatively narrow range, you'll need to do some calculations to use it, and has to be installed in line in the circuit). A clamp is safer, easier to use (you clamp it around a wire and take a relatviely direct reading with your DMM), and more flexible for general use (it doesn't interrupt the circuit), but more expensive.

If this is for AC, you can use a (cheaper) transformer-type clamp, which plugs in to the mA jacks on your DMM and usually reads 1mA per Amp. If this is for DC, you'll need a "Hall effect" type clamp (they also work for AC), which contains some circuitry and a battery of its own and usually reads 1mV per Amp (you plug it in to the Volts or mV jack on your DMM).

If you go the shunt method, be sure you really understand how it works, what's going on, and how to calculate your results. Note that a high current shunt has to have very low resistance. If you want to make your own you'll need the ability to accurately measure very low resistance in order to calibrate it. Handheld DMMs (even very good ones), generally can't do this. For proper results, you need to use the 4-wire method, which is generally only available on bench/lab meters.

H. Caul

Well since it is probably only going to be for this one 19.2V light, it sounds like it is over my head at the current time. Maybe after I start reading The Art of Electronics !

In this case, you want a very small voltage drop across the current sensing resistor so as not to interfere too much with the bulb's drive voltage and therefore current draw. I would try to keep the voltage drop across the resistor below 0.1 V for sure, and this is where you start to reduce accuracy since your meter is only going to measure mV. If you select a 0.01 Ohm resistor, then your voltage drop at 11.5 Amps = 11.5*0.01=0.115V, already reducing the bulb voltage by that much. A 0.01 Ohm resistor would need to be able to dissipate 11.5*11.5*0.01=1.3225 Watts. To get closer to the desired drive voltage you really need an even smaller resistance value. There are 0.001 Ohm resistors that would only drop 11.5*.001=0.015V, but then your voltage reading differing by 0.001V affects your current accuracy by 0.001/0.001=1 Amp, where current=voltage/resistance from V=I*R. A "compromise" would be if you used a 0.005 Ohm resistor so that your reading would be about 11.5*.005=0.057 V, with a 0.001 V deviation on your meter resulting in an error of .001/.005=0.2 Amps.

Edit: Of course, there is always another way (and another...). You could get a good estimate of the current by using a larger value resistor like the 0.01 Ohm, and let the voltage drop at the bulb but get more accuracy on the current reading. Also measure the voltage at the bulb pins while the resistor is in series. Then, the usual bulb voltage would be approximately equal to the sum of the measured bulb voltage with the resistor in circuit plus the voltage dropped by the series resistor. This allows you to use incandescent bulb re-rating formulas (once found on CPFWiki - though it seems to be down for some time now) to estimate what the current draw would be at the usual drive voltage. These formulas are out there somewhere... If you are interested in this approach I could search for the formulas and help you apply them.

I think this method would be more accurate than using a DMM capable of measuring this current (>10A) because simply placing the DMM in series with the circuit will drop a few tenths of a volt away from the bulb due to resistances in the DMM and the lead wires.

HCaul said:

If you want to make your own you'll need the ability to accurately measure very low resistance in order to calibrate it. Handheld DMMs (even very good ones), generally can't do this. For proper results, you need to use the 4-wire method, which is generally only available on bench/lab meters.
H. Caul

Click to expand...

The typical DMM has a constant current source of 1mA at the lowest resistance range and measures resistance by measuring the voltage across the leads on principle of V = I*R.

To raise the resolution as R gets small, you increase I and that's what ready made low resistance meters do. You can easily measure a very low resistance using two ordinary DMMs, even Harbor Freight cheapies. Connect an ammeter and shunt to be calibrated in series with a 2A or so load and apply power. Measure the voltage right across the shunt under load. This is a four wire method, since the voltage measurements are taken using a separate set of wires from the current source feeds.

Let's say you get 0.015v, 1.9A.

0.015v = 2 * R

0.015v/2A = 0.00789 ohms, so the calibration is 7.89mV/A

Yes, and that's a good way to get accurate readings in a pinch. But while ou can certainly do a 4-wire resistance measurement with two different meters, but I wouldn't recommend that someone try to do that to build a current shunt until they were fully conversant with basic electronics, e.g., Ohms law, power and heat dissipation, etc.

H. Caul

While I agree that one should be mindful of power requirements for circuit components and be able to calculate these before using them, I think of this method as better than a typical low current four point method.

I know there are expensive four point meters that can be outfitted with a power booster to do a fantastic job, but two cheapie meters can do the same thing. The reason a typical low current four point method is sometimes inaccurate stems from the heat derating characteristic of resistors.

As a resistor is heated it will increase in resistance. Low currents do not heat the resistor and therefore do not include this effect. For accurate current measurements you need to measure the resistor at the same power level (current) so it is at the same temperature in both cases. One can use power resistors mounted to heat sinks to reduce this effect.

Edit: Perhaps I came off like I was attacking another poster's position. Sorry, but I was meaning to add to it. In some cases the power dissipated in the current sense resistor is too low to be of consequence and it can be ignored, so it largely depends on what you are measuring and how you design the current sense "circuit".

Geez, there's a lot of opinion flying around. Use a calibrated shunt. You can get one pretty cheap. It's basically a known resistance (about .01) ohm bar of metal. Measure the voltage drop across it and you have your current measurement.

so a v-drop of 1.0 volts is 100.0 amps of current. They work well when the v drop is less than about 1.0 volts.

adnj said:

Geez, there's a lot of opinion flying around. Use a calibrated shunt. You can get one pretty cheap. It's basically a known resistance (about .01) ohm bar of metal. Measure the voltage drop across it and you have your current measurement.

so a v-drop of 1.0 volts is 100.0 amps of current. They work well when the v drop is less than about 1.0 volts.

Click to expand...

A lot of opinion flying around? I don't see any disagreement here, or room for disagreement, for that matter (I=V/R, not just a good idea, it's the law). What "opinions" are you talking about?

I think the main question is which of a varity of options best suits the original poster's equipment, knowledge, experience, and budget. That probably means either a Hall effect clamp (more expensive but easiest and safest to use) or a pre-calibrated shunt of the appropriate size (less expensive but requires a bit more confidence about what's going on). I'd not recommend a DIY shunt, although that's certainly cheaper, to someone who doensn't understand fully how to make low resistance measurements and calculate power dissipation, etc.

H. Caul

Until I get my "Art of Electronics" book which is coming, I agree with HC. I confess that even the solutions posted here are beyond my confidence/understanding level allowing me to proceed safely.

Maybe if someone could give me a practical source to get one of these solutions, and exactly how to interface it with my DMM....it would be very much appreciated. HC, you did such a WONDERFUL job in explaining things in that other thread....maybe you can bring it down to my level again?

I found this listing of calibrated shunts, but $145 seems like a lot of money to just be able to figure out how many amps are flowing in a 19.2V flashlight....nor do they explain how to use it for my intended purpose.

I also wondered why my Triplett DMM did not make the connection when I put the leads between the NEG terminal of the battery pack (while in the light) and the side edge of the light when turned on. That works in all my other lights <10A, and turns on the light when the leads complete the circuit. I would have thought it should have given me an "OL" error when the current was above its 10A limit. (It is reported by JimJones to be about 11.2A)

Thanks again very much for all your help figuring this out.

LuxLuthor said:

Until I get my "Art of Electronics" book which is coming, I agree with HC. I confess that even the solutions posted here are beyond my confidence/understanding level allowing me to proceed safely.

Maybe if someone could give me a practical source to get one of these solutions, and exactly how to interface it with my DMM....it would be very much appreciated. HC, you did such a WONDERFUL job in explaining things in that other thread....maybe you can bring it down to my level again?

I found this listing of calibrated shunts, but $145 seems like a lot of money to just be able to figure out how many amps are flowing in a 19.2V flashlight....nor do they explain how to use it for my intended purpose.

I also wondered why my Triplett DMM did not make the connection when I put the leads between the NEG terminal of the battery pack (while in the light) and the side edge of the light when turned on. That works in all my other lights <10A, and turns on the light when the leads complete the circuit. I would have thought it should have given me an "OL" error when the current was above its 10A limit. (It is reported by JimJones to be about 11.2A)

Thanks again very much for all your help figuring this out.

Click to expand...

A current shunt is just a very low value resistor with a known value (typically much less than 1 ohm for higher current applications) that you put in series with the load you want to test (e.g. cut one of the wires going between the battery and the load and attach one lead to one side of the shunt and the other lead to the other side). The resistance of the shunt has to be very low relative to the resistance of the load (otherwise it will reduce the power going to the load).

Because of Ohm's law, the voltage ACROSS the shunt (as measured with your DMM on the VOLTAGE / milliVolts setting, with the leads hooked up to either side of the shunt) will be directly proportional to the current flowing though the load. For example, if the shunt has a resistance of .001 Ohms, you will measure 1mV per Amp flowing through the circuit. You'll need to use a meter with < 1mV accuracy with a shunt of this value.

If you have the ability to measure very low resistance values (but with your equipment, you can't do this directly), you can make your own shunt out of wire. Otherwise, you can buy a commercial shunt of the appropriate size. If you're seeing prices like $145 those are probably either high precision or very high power shunts, complete overkill for anything you'd want to do. Check ebay and other suppliers.

I found this web page, which may help explain things in more detail: http://www.rc-electronics-usa.com/current-shunt.html
They seem to sell a suitable one for about twenty bucks, but you should be able to do even better than that on ebay or at surplus electronics places.

As for measuring current with your DMM: I don't know if your particular meter is has a fuse on the 10A leads, but if so you may have blown it when you took that measurement.

Even if you were measuring less than that, you might STILL blow the fuse in your DMM when you use the built-in current shunt. Many circuits (including most flashlights) draw a much higher "surge" current when they first start up. It quickly settles down to the steady value, but can last long enough to blow a fuse. The solution is to short the current leads of your DMM together when you start the device up, and separate them only after its actually on and running. (This applies ONLY to series current measurements through the DMM, not to voltage or any measurements with an external shunt).

H. Caul

LuxLuthor;

One easy way that I use is a 50 Amp shunt.

It measures 50mv @ 50 amps. Each mv on your digital meter = 1 amp.

Actually, it has less effect on the circuit that way also.

Larry Cobb

HC, I REALLY REALLY appreciate your help on all these occasions. Giving me that link now has me understand exactly what is happening, and I don't mind spending $25 bucks on something like this, so I ordered that one. Thank you very much again.

I did not know that the initial surge of the 23V (off the charger) 16 x 1.2V 2/3A NiMH Elite 1500 mAh cells in the startup of that Osram 90W light was a lot higher than 10A....and I expected to see an "OL" message.

I took out the DMM 10A fuse and having known that a smaller NiMH flashlight has about a 2A bulb draw, I put a wire place of the 10A fuse slot to test it on this smaller light, and it did show the correct current....so that fuse is blown. Crap!

It's weird with this 10A 600V fuse that has a white ceramic center rod...and there is no way to visibly tell that the fuse is blown. Good to know also that I can cross the DMM leads until the bulb is lit, then separate them to take a reading without the surge. Good Lesson.

LEDite said:

LuxLuthor;

One easy way that I use is a 50 Amp shunt.

It measures 50mv @ 50 amps. Each mv on your digital meter = 1 amp.

Actually, it has less effect on the circuit that way also.

Larry Cobb

Click to expand...

Oh that would be a nice option also. Do you know where those are sold?

Edit: Like this one? Is there any reason to consider the setup that HC mentioned instead of this one. I had just caught your post before ordering the one at HC's link.

Lex,

That shunt looks like a better deal for your purposes. Use as described above, in series with the circuit you're measuring, and hook your mV meter across it (DMM on mV setting). 1mV = 1A, up to 50mV=50A.

H. Caul

OK, sweet.

One last question..I just want to make sure this DMM will be accurate enough to interpret the mV range. This Triplett 9045 "True RMS" DMM has one selection for Volt-DC (defaults to AUTO with a range to select decimal places manually) and the spec for DC Voltage are:

Range -----Resolution
600mV ------ 0.1mV
6V ------------1mV
60V ----------10mV

Accuracy is +/- (0.5% rdg + 5 digits)

LuxLuthor said:

OK, sweet.

One last question..I just want to make sure this DMM will be accurate enough to interpret the mV range. This Triplett 9045 "True RMS" DMM has one selection for Volt-DC (defaults to AUTO with a range to select decimal places manually) and the spec for DC Voltage are:

Range -----Resolution
600mV ------ 0.1mV
6V ------------1mV
60V ----------10mV

Accuracy is +/- (0.5% rdg + 5 digits)

Click to expand...

You'll always be using 600mV range and the accuracy will be 0.5% the read value + 5 counts. So, 0.10000mv could read anything from -0.1mV to 0.6mV. That's just the way digital meters work.

You could also get something like this:
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=320108437758 , although these tend to ship from Hong Kong / China

Handlobraesing said:

You'll always be using 600mV range and the accuracy will be 0.5% the read value + 5 counts. So, 0.10000mv could read anything from -0.1mV to 0.6mV. That's just the way digital meters work.

You could also get something like this:
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=320108437758 , although these tend to ship from Hong Kong / China

Click to expand...

That's an interesting thing on EBay, but it seems hard to beat the simplicity of a shunt....this unit looks like it has more things that could go wrong.

I'm not understanding what your description of the accuracy means. My level of understanding is that if I get a reading of something that is say... 500 mV then the accuracy is +/- 0.5% or in this example the accuracy is +/- 2.5mV ?

Reason I'm asking is if the shunt is dealing with 1mV=1A then a reading of that light I mentioned and using the shunt, and getting a reading of say.... 12mV would = 12A but the range would then be +/- (12mV x 0.005) or 11.94A to 12.06A correct?

LuxLuthor said:

That's an interesting thing on EBay, but it seems hard to beat the simplicity of a shunt....this unit looks like it has more things that could go wrong.

Click to expand...

That thing is a a bundle of simple shut + digital panel meter.

I'm not understanding what your description of the accuracy means. My level of understanding is that if I get a reading of something that is say... 500 mV then the accuracy is +/- 0.5% or in this example the accuracy is +/- 2.5mV ?

Click to expand...

Let's say you have 0.1mV resolution +/- 0.5% + 5 digits
You feed it 100.00000mv from a calibrator.

Your meter would have +/- 0.1mV in the internal reading PLUS +/- five counts in the least significant digit (the tenth of mV place), so acceptable readings are 99.4mV to 100.6mV

The percentage is relative to input, the +/- 5 digits is an allowed fixed error margin.

Reason I'm asking is if the shunt is dealing with 1mV=1A there could be quite a wide innacuracy. For example, if I have a correct understanding of the accuracy, then a reading of that light with the shunt of 12mV would = 12A but the range would then be +/- (12mV x 0.005) or 11.94A to 12.06A correct?

Click to expand...

If there is 12A going through the shunt, the output is 12mV. Your DMM can only resolve to the tenth of a mV, so its acceptable reading would be 12.0mV +/- .5% + 5 LSD. Since 0.5% would be around 0.01 here, it's not a factor. When you add the 5 LSD error, the acceptable reading would be 11.5 to 12.5mV or 11.5 to 12.5A. You should expect DMMs to have 50% error in the least significant digit (the far far right digit).

Common laboratory procedure is to use an instrument that can resolve ten times beyond the accuracy sought.

You can get 1% current sensing resistors from Digi-Key for a couple dollars. EG: http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=1787&Row=32479&Site=US Which is a 0.01 ohm, 5w resistor. This would work great for measurements around 10-20 amps (yielding a resistor drop of 0.1-0.2v, and acceptable power dissipation).