200mA limit on my Amprobe 5XP-A - Page 1
Forum section: electronics / beginners

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 12:55:00 pm
content: I bought this one a few years back for some hobby learning. After some basic experiments I kind of let the electronics thing dropNow that I'm back at it I and have a better understanding of DMM it was kind of disappointing to read the current limitations. I'm not sure how much it will effect me and suppose that it depends what I'm working on. For my R/C stuff , testing amp draw is certainly out. Right now I'm just putting little circuits together like, resistors in parallel vs serial. Again pretty low amps.Searching I read somewhere that in general measuring current is not a typical thing. Can that be true or just a poor misinformation site? I have seen some DMM's in the same price range that go up to 10A readings. I wonder how trustworthy and even safe those can be?Anyway sorry to start another thread on multimeters. After thinking about it some I believe this one can probably suffice for the time being while I learn but am not 100% sure. Outside of that the meter is okay but no capacitance, no autorange (to me not a big deal), pretty accurate.

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 01:11:34 pm
content: Hi"Searching I read somewhere that in general measuring current is not a typical thing. Can that be true or just a poor misinformation site?"It is slightly incorrect. In order to make a measurement of current, the meter must be placed in line with the voltage source, most of the time this can be inconvenient which puts people off making this measurement. So, yes (I would say) a good proportion of people do not make current measurements due to the inconvenience factor but some people do."I have seen some DMM's in the same price range that go up to 10A readings. I wonder how trustworthy and even safe those can be? "A 10A (or other multiple Amps) range is very common.In general (so this is not always true for every DMM) the high current range actually has the lowest value shunt (resistor) so actually produces the lowest burden voltage and therefore the most true representative current measurement (simple answer is - it is pretty trustworthy). However, most meters only has a resolution of 0.1 (100mA) in the 10A range, so if you are trying to measure less than 100mA, the 10A range is not going to be of any use.Safe ?When dealing with large currents, there is always a risk, usually of starting fires. The 10A shunt (resistor) is typically around 0.1ohm - pretty much a short circuit. If you short out a power supply that is not going to current limit or shut down (eg a battery) the current just goes up to what every it can, the wires get hot and may start a fire.

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 02:29:57 pm
content: Thanks Mosher! I suppose if the need arises something like thishttp://www.ebay.com/itm/like/281514343083?lpid=82&chn=ps&ul_noapp=true0-100V 200A DC Digital LED Voltmeter Ammeter Amp Volt Meter + 200A shunt 12v 24v might be a solution. Maybe it is a better solution then in a low end DMM?

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 02:52:32 pm
content: "I suppose if the need arises something like thishttp://www.ebay.com/itm/like/281514343083?lpid=82&chn=ps&ul_noapp=true0-100V 200A DC Digital LED Voltmeter Ammeter Amp Volt Meter + 200A shunt 12v 24v might be a solution. Maybe it is a better solution then in a low end DMM?"I would worry about the accuracy and safety of something like that.A Current Shut is all that is needed to measure current, and all that is, is a resistor.The resistor need to be a fixed known values, not too high so as to not cause too much of a 'burden voltage'and high enough power as to be able to cope with the power that it will dissipate.A (cheap) meter can then just read the volts across the shunt and the current can be calculated.eg 1ohm gives 1V per Amphttp://uk.farnell.com/current-sense-through-hole-resistorsThey tend to be lower than 1ohm to avoid a) power dissipation and b) high burden voltage.The meter needs to have a good mV range to be able to read the low voltageseg 0.1ohm gives 0.1mV per AmpNote, you need to know to 0.1% or better precision the value of the resistor or the results will be too inaccurate.So your DMM can be used, just get yourself a high precision, high wattage, low value power resistor and use the mV (uV if you got it) range.

author: rstofer
author_rank: Super Contributor
timestamp: July 04, 2016, 04:36:03 pm
content: A common designation for shunts is 50 mV. That is, they drop 50 mV at full scale regardless of the scale. That way the panel meter movements are all the same - 50 mV full scale.Google for '50 mV shunt' - they are all over the place.You can put the shunt in the high side (between the battery + and the load) and there are high side current amplifiers just made for this application. The output will be a low level voltage suitable for uC ADC gadgets.http://www.ti.com/lit/ml/slyb194a/slyb194a.pdf

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 04:43:33 pm
content: Thanks for the information guys! I'll be reading up on the shunt. It sounds like the way to go.

author: retiredcaps
author_rank: Super Contributor
timestamp: July 04, 2016, 05:17:49 pm
content: Have a look at Martin's video as well re current shunts.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 04, 2016, 05:33:41 pm
content: Quote from: MosherIV on July 04, 2016, 02:52:32 pmEr....No. A typo, maybe?A 0.1 ohm Current Sense Resistor gives a voltage drop across the resistor of 0.1V (100mV) per amp.V=IRVdrop=I x (0.1 ohm)So if I = 1 amp then Vdrop= 0.1 V or 100 mVPower dissipated in the resistor will beP=I2RP = 12x 0.1 = 0.1 W or 100 mWIt is also important, if measuring any kind of higher frequency current, to use a non-inductive CSR.If you use high-side sensing you must realize that the meter and CSR are floating up at the full power supply voltage. This is why low-side current sensing is often preferred-- especially when using an oscilloscope.

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 06:04:01 pm
content: And the prize for deliberate mistakes goes to...................alsetalokin4017Well spotted. My mistakeSorry for any confussion.

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 07:23:20 pm
content: Quote from: rstofer on July 04, 2016, 04:36:03 pmMade it through the pdf but will need to let it sink in a bit. From what I can understand, the monitor uses the shunt and then outputs to i2c?

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 07:57:26 pm
content: "Made it through the pdf but will need to let it sink in a bit. From what I can understand, the monitor uses the shunt and then outputs to i2c?"Probably, but do not worry it is not relevant to you. You should look at shunts, which as I siad is just a resistor. I have no idea how accurate the 50mV shunts are.Do not worry about 'high side' or 'low side' measurment. This is irrelevant to the shunt or measuring it with a dmm, hand held dmm are battery powered and completely isolated so should be able to measure high or low side shunts.

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 08:36:59 pm
content: Not entirely sure how to choose the CSR. I'm thinking up to 12V and the ability to get 30A.R = V/IR = 12/30R = 0.4mOhm is the CSR I'd need.Does this make sense?

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 04, 2016, 09:09:40 pm
content: Quote from: Stuartambient on July 04, 2016, 08:36:59 pmNo. 12v/30a is 0.4 Ohm, not 0.4 milliOhm. And the voltage you are concerned with is the voltage across the CSR, not the total supply voltage.But if you are measuring up to 30 amps you have to be concerned about the power dissipation in the resistor. For example, at 30 amps a 0.01 ohm resistor will dissipate 30 x 30 x 0.01 = 9 Watts. The voltage drop across this resistor will be 30 x 0.01 = 0.3 V or 300 mV for a 30 amp current.For such large currents you really do need a proper "shunt" of very low resistance. If you can find one of 0.4 mOhm ( = 0.0004 Ohm) then 30 amps will give you 30 x 0.0004 = 12 mV drop and it will dissipate 30x30x0.0004= about 360 milliWatts.Note that shunts are specified with voltage drop AND amps. A "50 mV shunt" could be 50 mV/500 A, or 50 mV/50 A, etc. meaning for example a 500 amp current will give a 50 mV drop across that shunt (which therefore has a resistance of R=V/I = 0.050/500 = 0.0001 ohm).Or this one which is 50 mV/100A:http://www.newark.com/productimages/standard/en_US/4854561.jpg

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 09:14:58 pm
content: "Not entirely sure how to choose the CSR. I'm thinking up to 12V and the ability to get 30A.R = V/IR = 12/30R = 0.4mOhm is the CSR I'd need."The choice of shunt will be a compromise of a number of parameters :Max current (which you have specified)Resolution (volts per Amp you want or probably more what can you measure with your meter, what is the smallest value you want to measure and what is the smallest volt your meter can measure)What resistors can you actually get. Remeber it has to be accurate, you probably will not be able to measure sub ohm resistors accurately so you will depend on the initial accuracy. The power rating, it needs to handle the power, P=I*I*RThe resistor also needs to have a low temperature coefficient, which means it does not change value too much as it gets hotter

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 04, 2016, 09:33:14 pm
content: Now... we have the issue of measuring current in R/C stuff. This means you will likely want to be measuring pulsed currents, like from an ESC controlling a 3-wire brushless DC motor, yes?Nothing beats an oscilloscope for that kind of work. But the good news is that most DMMs do a fairly good job of _averaging_ current or voltage readings from pulsed circuits. So if you choose your shunt properly and measure a voltage drop value you should get a reasonable average value for a pulsed current. The peaks of course will be higher and to see what they are you'll need a scope.

author: rstofer
author_rank: Super Contributor
timestamp: July 04, 2016, 10:14:28 pm
content: Quote from: MosherIV on July 04, 2016, 09:14:58 pmOne thing for sure, you don't want your sense resistor to drop 12V at 30A. You won't have much voltage left for the load. A 50 mV shunt drops 50 mV at full load and that might be a pretty high value for robotics. You just don't want to give up much voltage to the motors by dropping it in a shunt.A 50 mV shunt for 30A would be a 0.001666... Ohm resistor. The power rating would be 0.05V * 30A or 1.5 Watts

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 04, 2016, 11:35:35 pm
content: Quote from: alsetalokin4017 on July 04, 2016, 09:09:40 pmI found this Murata, though I haven't completely finished searching.http://www.digikey.com/product-detail/en/murata-power-solutions-inc/3020-01099-0/811-1095-ND/1926138.0005 so at 30A would be a .015mV drop. Though these say are to be used with their ammeters, but these would be considered a proper shunt from what you are saying.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 05, 2016, 04:23:26 am
content: Quote from: Stuartambient on July 04, 2016, 11:35:35 pmV=IRVdrop=I x Rshuntso Vdrop= 30 x 0.000530 amps across 0.0005 ohms = 0.015 V, not mV. That is, 15 mV.P=I2x Rshuntso Power dissipation will be a little under half a Watt.Be careful about those decimal places and abbreviations! 1 mV = 0.001 VYes, this shunt seems like a good one for your purposes. Expensive though! (but now you will be concerned with the accuracy and precision of your DMM. What if the current you are reading is, say, 2 amps? 2 x 0.0005 = 1 mV. )

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 05, 2016, 02:30:39 pm
content: Accuracy is probably why these shunts are matched up with the manufacturer's ammeters. As for my DMM, I guess it will be an experiment plus trial and error.? I see two products though to consider. One is the type I linked above and the other is current sensing resistors and amplifiers. Seems the later would offer more accuracy even if I had to bypass my multimeter and use a microcontroller. Might be a bit more work as well putting the circuit together. Either one should work.The talk about high side and low side though has me confused. What started me on this is creating a simple parallel circuit with a resistors. I want to measure across each resistor to get a current value. Seems like this would be neither low or high side. Same if I put the shunt between a controller and motor. I think Mosher was saying a few posts up that it wouldn't matter.

author: MosherIV
author_rank: Super Contributor
timestamp: July 05, 2016, 04:20:05 pm
content: Hi"Seems like this would be neither low or high side. Same if I put the shunt between a controller and motor. I think Mosher was saying a few posts up that it wouldn't matter."It dos not matter if you use a shunt and your dmm.Not sure about the current sense device.

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 05, 2016, 05:12:10 pm
content: Quote from: MosherIV on July 05, 2016, 04:20:05 pmThat is what I thought. I'll have to investigate the CSR's though it seems the CSR sitting between the dmm and circuit would work the same way.The other thing is I put some random resistors in a series with the lowest being a 410 Ohm. This was to test the accuracy of the meter at the lowest voltage setting of 200mV and I got 4.3mV across the resistor (@9V). The other resistors were a 1M and a 2.2K. Also tried a 100 Ohm in the series and got .9mV. Seems on mV there is only one digit to the right of the decimal so I'm not sure about the accuracy.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 05, 2016, 06:13:54 pm
content: If you put resistors in series, then the current through the series stack of resistors is determined by the total value of the resistors (and of course the supplied voltage and any other load.)I = Vsource/ (Rcsr1+ Rcsr2+ ... + Rcsrn+ (Rload))Now, if you measure voltage drop across each resistor, or the total stack, and apply Ohm's Law as before, you will read _different voltage drops_ if the resistor values are different, but you should see the _same current_ value in every case (after doing the math.)I = Vdrop/ RcsrxLooking at your series example, we have a 9 Volt supply and three resistors in series, 1 meg + 2.2k + 410 ohm = 1,002,610 ohms (more or less depending on tolerance values.) So the current through the circuit will be I = V/R = 9/1,002,610 = 0.000008977 A or about 9 microAmps.Now, if you measured across the 410 ohm resistor and got a Vdrop of 4.3 mV, this equates to I = V/R = 0.0043/410 = 0.000010488 A or about 10 microAmps -- a result that is in agreement with the first calculation (due to the tolerance values of the resistors and the accuracy of your DMM.)QuoteParallel resistors are different. You cannot get readings from individual resistors in a parallel array. The entire array will have a resistance according to the following formula:1/Rtotal= (1/R1) + (1/R2) + ... + (1/Rn)And when you measure across any one you are measuring across the entire stack. So your current will be found, as usual, byI = Vdrop/ Rtotal

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 05, 2016, 06:42:56 pm
content: Quote from: alsetalokin4017 on July 05, 2016, 06:13:54 pmYes, that is what I calculated.Quote from: alsetalokin4017 on July 05, 2016, 06:13:54 pmMissed that partSo it is off some but if the current was 5A would that make the difference significant?Quote from: alsetalokin4017 on July 05, 2016, 06:13:54 pmThat makes sense. Thank you!

author: alsetalokin4017
author_rank: Super Contributor
timestamp: July 05, 2016, 07:16:43 pm
content: You're welcome.QuoteMost of the calculated difference is probably due to not knowing the exact value of the 1meg resistor. The resistors you used in the series example were probably 5 percent tolerance. Even a one percent tolerance on a 1 meg resistor would give you +/- 10K ohms on the value. You could plug in some values to the series equation and see how much that affects your final result. This is why CSR resistor values need to be accurately known, with close tolerances.IIRC, the current shunt you linked to up above has a 0.25 % tolerance (one part in 400), which is pretty good. And the DMM will probably be more accurate on higher ranges. You can check the spec sheet for your DMM to find out its accuracy parameters. So if you are calculating 5A from your shunt and DMM Vdropreadings, you could vary by 1/4 of one percent due to the shunt's tolerance level, or +/- 12.5 mA . If your DMM has, say, +/- 1 percent accuracy in the range you are using, then a similar calculation gives a possible variation of +/- 50 mA in your result due to the DMM's accuracy level.

author: Stuartambient
author_rank: Regular Contributor
timestamp: July 05, 2016, 08:02:10 pm
content: Okay, think I should be alright from here. Thank you for all the help!