Multimeter burden voltage correction

Discussion in 'General Electronics Chat' started by Russel, Mar 7, 2011.

  1. Russel

    Thread Starter New Member

    Mar 6, 2011
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    Am I understanding multimeter burden voltage correctly?


    I have a Fluke 85III multimeter with a burden voltage of 1.8mV/mA on the 40mA and 400mA ranges. So, if I am taking measurements at lower current levels, I can correct for burden voltage mathematically?

    For example, measuring 300mA on a 12V cirucut with a Fluke 85III:

    Burden voltage of the meter is 1.8mV/mA.
    So 1.8mV * 300 = 540mV, the voltage drop in the circuit caused by the meter while measuring current.

    0.540V / 0.3A = 1.8Ω meter resistance

    12V circuit, drawing (measured with Fluke 85III) 300mA.
    So 12V / .3A = 40Ω circuit resistance including the meter.

    Corrected circuit resistance:
    40Ω (circuit and meter resistance) - 1.8Ω (meter resistance) = 38.2Ω

    So, the calculated current without the meter in series:
    12V / 38.2Ω = 0.314mA


    Let me try that again with a 3V circuit and 100mA measured with a Fluke 85III.

    1.8mV * 100 = 180mV meter voltage drop.

    0.180V / 0.1A = 1.8Ω meter resistance

    3V / 0.1A = 30Ω (3V circuit with measured 100mA current)

    30Ω (circuit and meter resistance) - 1.8Ω (meter resistance) = 28.2Ω

    So, current without the meter in series should be:
    3V / 28.2Ω = 106mA

    Does this make sense?
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Here you are assuming that the load circuit behaves like a pure resistance, so that the circuit current is directly proportional to the applied voltage.

    This will by no means always be the case. Electronic circuits will often have non-linear voltage / current relationships, and even such things as motors and incandescent lamps do not behave exactly like resistors.
     
  3. Russel

    Thread Starter New Member

    Mar 6, 2011
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    Yes, I'm assuming no inductive or capacitive reactance. A theoretical circuit to help me understand if I get the jist of multimeter voltage burden. I guess I've misinterpreted voltage burden...time for more reading.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Many nonreactive loads still don't obey Ohm's law. For example, most op amps, if in an application where they are actively biased in a DC circuit, will appear as a nearly constant current sink on the positive supply pin, and as a nearly constant current source on the negative supply pin.
     
  5. someonesdad

    Senior Member

    Jul 7, 2009
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    It's a good question, Russel. I've never worried about it too much, but if you are, it could be worth characterizing. I agree with Adjuster's comments, so the task is to figure out how the multimeter responds. This shouldn't be hard to do experimentally: include a known resistance in the circuit and characterize the currents with and without the multimeter in the circuit. Use a second multimeter to measure the voltage across the known resistance and translate it into a current (just stay well below the resistor's power rating). You can also use the second multimeter to measure the first multimeter's voltage burden and you'll want to think about whether you need to correct for the input impedances.

    A second experiment could do the same thing for AC current at different frequencies; this would be illuminating to see where reactances or nonlinearities become important. I'd be interested in seeing you post your measurements and conclusions.
     
  6. Russel

    Thread Starter New Member

    Mar 6, 2011
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    OK, so my understanding of meter voltage burden isn't entirely off base. It's just that any given circuit may not react as if the meter was purely resistive. I think this is sinking in...

    Someonesdad, I tried a simple experiment. I soldered two 100Ω 10W resistors in parallel and measured the resistance as 49.8Ω. Then I put the 49.8 Ω resistor in series with the Fluke 85 III set to mA range and connected this to an adjustable linear power supply. The current limiting was turned all the way up on the power supply (3A max.) and the voltage limit all the way down. Then I turned the voltage limit on the power supply up until the Fluke meter in the circuit with the resistor indicated a stable 200.0mA. I measured 9.981V across the resistor with a second multimeter (being careful to only contact the resistor leads.) Then I measured the voltage drop across the Fluke 85 III meter and it's two test leads at 382.4mV, and across the negative lead and the meter at 378.2mV.

    So, the current calculated by the voltage drop at the resistor should be:
    9.981V / 49.8Ω = 200.4mA

    Calculating the meter resistance (including both test leads):
    382.4mV / 200.4mV = 1.9082Ω

    The meter resistance with the negative lead only:
    378.2mV / 200.4mA = 1.887Ω

    Just for fun, the resistance of the positive test lead:
    382.4mV (meter and both leads) - 378.2mV (meter and negative lead only) = 4.2mV
    4.2mV / 200.4mA = 21mΩ

    It wasn't until I put everything away and sat down to calculate that I realized I measured the resistance of the two 100Ω resistors in parallel while they were at room temperature, but adjusted the voltage at the power supply with the resistors warmed up for stability. I never checked if the resistance changed with temperature. That kind of throws my numbers out the window. I'm going to have to check the resistor when it is up to temperature.

    I guess I should have measured the voltage drop across the resistor with and without the Fluke meter in the circuit, as I just realized Someonsdad suggested. Well, I'm learning...more to try tomorrow.

    Suggestions are welcome.
     
  7. eblc1388

    Senior Member

    Nov 28, 2008
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    You are talking of 100mA through a 10W 100Ω resistor, that's only 1W.

    I doubt that this small amount of heat would heat up the resistor that much to have change the resistance of the resistor to seriously affect your measurement result.

    I have absolutely no idea of what your are trying to prove or investigate, besides verifying Ohm's law.
     
  8. Russel

    Thread Starter New Member

    Mar 6, 2011
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    Correction:
    Calculating the meter resistance (including both test leads):
    382.4mV / 200.4mA = 1.9082Ω

    I agree that isn't much for a pair of 100Ω 10W resistors in parallel. The resistors got quite warm to the touch, but not hot enough that you had to pull your hand away. I just don't know from experience if the resistance would change with temperature so I figured that I should confirm that before considering the current calculated from the voltage drop across the resistor to be accurate.

    While I'm spending a lot of time with Ohm's law, my purpose is to determine if I can correct for voltage burden mathematically. If I measure 300mA on a 3V circuit with my Fluke meter (1.8mV/mA burden voltage) I can expect the measurement to be off by about 23%. (Assuming the circuit is reacting in a purely resistive manner.)
    1.8mW * 300 = 540mV voltage burden
    540mV / 300mA = 1.8Ω meter resistance
    10Ω - 1.8Ω = 8.2Ω circuit resistance without the meter
    3V / 8.2Ω = 366mA current without the meter

    Apparently, the way the meter affects the circuit depends upon what is in the circuit. So, I tried a simple experiment to see if mathematically correction agreed with empirical measurements.

    To be honest, I'm just curious about it. I think it is always good to know the accuracy of your test equipment. In the example above, measuring 300mA on the 400mA range with a result that is off by about 23% seem excessive. I was really surprised to find out that the Fluke multimeter (as well as other multimeters) would measure so far off under certain circumstances.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    They are probably wirewound resistors. If they are nichrome, which I believe is fairly common, they will have a TCR of about 0.04%/°C. Lower TCRs are available, though, probably at higher prices than nichrome.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    As a side note, you can calculate the current sense resistor directly from this:
    Rsense=1.8mV/mA=1.8Ω.
     
  11. eblc1388

    Senior Member

    Nov 28, 2008
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    Not easily.

    The inclusion of the ammeter would create burden and thus reduce the voltage seen by the load. Unless the load is a linear resistor, then maybe you can correct for the drop.

    Most loads are not linear and could even take more current at reduced input voltage to complete a task. e.g. motor driving a load....
     
  12. someonesdad

    Senior Member

    Jul 7, 2009
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    Russel, I think your curiosity is a good thing. There's nothing like making a measurement and knowing, rather than just assuming that things work a particular way.

    Years ago I blew out the current-measuring fuses for my Fluke multimeter and I'm too cheap to go find and buy new fuses. Thus, I use an HP 3435 multimeter I bought in the 1970's as my current meter on my bench (I have a nice clamp-on AC/DC meter for out in the field). It has the nice feature of using cheap AGC fuses for the current measurement, although it doesn't provide the industrial-type protection that the Fluke fuses do. Anyway, I used a DC power supply to put current through a GR resistance box, measured the current with the 3435 in ammeter mode, and measured the burden voltage with my Fluke multimeter (which I know is an accurate voltage measuring device). I've attached the plot.

    The ammeter behaved as a linear resistance except for the 2000 mA range (its highest range), where it was nonlinear; a quadratic with a positive coefficient gave a better fit. This is consistent with Joule heating and a positive temperature coefficient of resistivity, which I'd assume is what is going on.

    Truthfully, though, it's rare that this burden voltage is important. Usually, I measure the current with one meter and measure the voltage of interest (e.g., across the load) with another meter.
     
  13. Russel

    Thread Starter New Member

    Mar 6, 2011
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    I certainly need to spend more time learning the fundamentals of electronics. I feel like I'm in a little over my head on this forum, but that is good, there is plenty here for me to learn.

    I was taking some measurements today on a NiMH battery charger for AA and AAA cells made by Sanyo. This is an example of why I would like to correct for meter voltage burden. The charger label indicates a charge rate of 300mA with 2 or 4 AA cells. (It's a four bay charger that charges two or four cells at a time.) I measured 195mA maximum with my multimeter while a cell was charging. At 1.25 to 1.35V, I wonder how much influence the meter resistance has on the circuit. Of course, in this case a simple calculation with Ohms law isn't going to do the trick.

    The charger is a Sanyo model NC-MQN06U. The charge waveform is interesting in that the charge current pauses for about 12.5ms every second. I assume this is so the device can measure the cell voltage without the charge current affecting the measurement.

    Note: The oscilloscope was set to DC coupling at the time of the waveform capture. The scope happened to be set to AC coupling when I retrieved the the waveform from memory, that's why you see the AC symbol next to CH1 on the bottom left.
     
  14. someonesdad

    Senior Member

    Jul 7, 2009
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    The easiest way to measure current is to stick a resistance in the circuit and measure the voltage drop across the resistance; this is how most ammeters on digital multimeters work. That inserted resistance can influence the circuit's behavior.

    Another approach is to measure the magnetic field surrounding the inductor. You've no doubt seen AC clamp-on meters that do this and there are clamp-on meters that also measure DC current, usually by using the Hall Effect. Here's one I've used that does a reasonable job down to around 10 mA (it resolves to 1 mA, but I don't often trust it to that level); another is the HP428, an old vacuum tube instrument that is still doing work in a number of labs.
     
  15. Russel

    Thread Starter New Member

    Mar 6, 2011
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    I can't find any inductive meters that aren't more than I want to spend. The one you pointed a link to, Someonesdad, does appear to be a good value for the cost. So, that leaves measuring voltage drop across a resistor. I ran across a lithium Ion battery charger review where the gentleman that wrote the review used a simple method with resistors that looks like something I would like to try. He used a strip of circuit board, plain with copper on both sides, that could be placed between one of the terminals of a common cell (AA, 18650, etc.) and the terminal of the battery charger being tested for current output. Test leads were then soldered to each side of the strip of circuit board. Then, rather than measuring the current by connecting the test leads to an ammeter, 10 - 0.1Ω 1% resistors were soldered in parallel from one side of the circuit board to the other (leaving enough bare circuit board to slip between the cell and the charger terminals) to create a resistance of 10mΩ, the test leads are then connected to a voltmeter to measure voltage drop while the charger operates. Using a 10mΩ resistor, that works out to about 10mV/A. So, it looks like measuring voltage drop across a low value resistance may be the most inexpensive method with a reasonable degree of accuracy.
     
  16. someonesdad

    Senior Member

    Jul 7, 2009
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    I understand about not wanting to spend money. You can make your own shunts to measure the current. I made relatively high current shunt from a chunk of 3 conductor 12 gauge Romex and it worked well (I haven't the foggiest idea what it was for now, but it worked :p and it's still in my shop). Since 12 gauge copper wire has a resistance of about 5.2 mΩ/meter, you could get 10 mΩ from slightly less than 2 m. I've also used 40 mil diameter stainless steel safety wire (get a spool of it from Harbor Freight) because it has a higher resistivity. Chromel from type K thermocouples works too; it's essentially a Nichrome-type alloy.
     
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