I recently suggested a plan of removing one end of a resistor from a breadboard to remove it from the circuit in order to measure its resistance, but my mentor suggested that this won't work for high precision measurements (0.025% error) because of leakage current entering the DMM. I haven't heard of leakage current before. What do I need to know, and how might leakage current affect my designs?
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3\$\begingroup\$ Generally you were misinformed if your DVM was a decent device and/or was battery powered. \$\endgroup\$– Andy akaMay 16, 2013 at 21:32
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\$\begingroup\$ You are on a breadboard, the only way that you can measure without leakage affecting you if you truly have the instrumentation available is by removing the resistor. \$\endgroup\$– placeholderMay 16, 2013 at 21:41
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\$\begingroup\$ What did your mentor say when you asked him this? \$\endgroup\$– Dave TweedMay 16, 2013 at 22:07
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2\$\begingroup\$ Breadboard, DMM, 0.025% error (=250ppm) ... eh!? How many decimals does your DMM show? Please say 6 or more, otherwise it is useless to worry about a 250ppm error. And what kind of resistor do you use that has such low tolerance (and thermal drift) that it justifies such accurate measurments? \$\endgroup\$– jippieMay 17, 2013 at 6:46
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1\$\begingroup\$ @jippie Turns out I misremembered the spec sheet of the DMM and added and extra 0, its really 0.25% error. It does have 5 decimal places though. The resistor is being used as a one-off to calibrate an in-circuit current sensing system, so I don't really care about thermal drift. Similarly, I don't really care what value the resistor has, as long as I know what it is. The DMM and the circuit that I suggested leaving one end of the resistor attached to do share a common ground at the wall socket, does that change anything? \$\endgroup\$– DrewMay 17, 2013 at 13:52
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2 Answers
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TL;DR - You are probably fine disconnecting one end and just measuring resistance like that. Other thing such as DMM lead resistance needs to be considered however.
You are measuring resistance and not voltage so your mentor is not correct in assuming the voltage measurement input model of the multimeter. When measuring voltage leakage current can be a big problem. Like this:
In the above image the series resistance is a low impedance and so the voltage readout on the DMM is very close to the true 9V.
VM = (9 V * 10 M Ω)/(50 Ω + 10 M Ω) VM = 8.99996 V
Here we can see how the input impedance of 10Mohms can skew voltage measurements on high impedance nodes. ie like so:
VM = (9 V * 10 M Ω)/(1 M Ω + 10 M Ω) VM = 8.18182 V
The DMM introduced a 10% error in our measurement! Here a leakage current through the 10Mohms caused an unexpected voltage drop and introduced error in our measurement. Since we never know the series resistance of our circuit beforehand when making a voltage measurement we can never calibrate it out.
Now for resistance measurements the mental image of the DMM should be like so:
(for now ignore the HIsense and LOsense nodes). Here a DMM measures resistance by sending a constant current and measuring the voltage developed over the resistance as a result. Here the 10Mohm input resistance of the DMM is irrelevant as it won't affect the constant current being sent by the DMM.
What is important however are the resistances of the wire especially if you are measuring a very low value resistor. As such you may need to do 4-wire sensing to calibrate those resistances out:
"Current (amps) is forced through the source leads (HI, LO). As a result, a voltage develops across the resistance (ohms) under test. By measuring the voltage directly across the resistor using the sense leads, the voltage drop of the interconnects (RLEAD) is ignored." (http://www.ni.com/white-paper/3981/en/). You can google more on this.
Measuring in circuit resistance is also sometimes possible. A 6-wire technique must be employed like so that guards the voltage of the developed across the resistor being measured (here a 30kOhm). To use this technique however more needs to be known about your full circuit. You can google more on this as well.
Resources:
http://www.ni.com/white-paper/3981/en/
http://www.ni.com/white-paper/3296/en/
http://wiki.pickeringtest.net/Resistance+Measurements+with+a+DMM
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If you're measuring resistance, a DMM typically forces a current and measures the resulting voltage. So leakage current in this context makes no sense. If you're measuring a voltage, then maybe he's referring to the input resistance of the DMM which typ is 10Mohm. This will definitely affect your reading if youre probing a high impedance node. That is, if a circuit is operating at a microamp and the node is at 1V, the meter will draw 100nA which is 10% of operating current. Think of the meter as a 10M resistor you're applying to your circuit.
