Measuring very low resistance

Hey folks,

Once, don't remember where, I saw something on how to measure a very low resistance (I need about 0.1 Ohm) with DMM. Any pointers? I have a Fluke 87 III.

draw a known current through it (say 1A), and measure the voltage drop?

I've been reading Bob Pease's "Troubleshooting analog circuits" and he definitely recommends Jakob;s approach. Measuring voltage drop with a known current is the best way to measure low resistance.

Cheers,

Kris

Way back in yester-year I worked at a company that used really cheap pc boards which were notorious for having shorts between the traces. To find the shorts, we used a Fluke meter with 4-wire, low ohm measurement capabilities. I don?t remember the model #, though (20 years ago so I?ve slept approximately 7200 times since then, not including naps! :shock: ).

It was probably <$500 back then, but if you could find one of these cheap on evil-bay, I believe it?d serve you well.

Yes,

If you've got a PSU with a variable current-limit (and a good output, i.e. greater than 1A continuous!) connect the "unknown resistance" to this via a DC ammeter. Start with the current-limit set to a low value. Set the PSU to a low voltage, and then bring up the current limit so that 1A flows through the "unknown resistor". Then measure the voltage across the sample. R=V/I , so 1V=1R. The PSU is acting as a chunky "constant current source".

BTW, I once did this and got screwy results- make sure that you place the voltmeter across the resistor only. I had it connected from the +V and 0V of the PSU, forgetting about the slight (but significant) resistance of the leads and ammeter :roll:

A good method is to get a pair of nice big all-metal earth terminal blocks. Connect heavy-gauge leads to the PSU and ammeter, and then connect the voltmeter leads directly to the blocks, and place the specimen directly between the blocks. The screw terminals keep everything nice and tight, and the large area of the blocks equate to a low resistance, and ensure that the voltage across the resistor leads is what you're measuring (you'd be surprised by the resistance of a 4mm plug lead connected to a croc-clip lead connected to a resistor leg, it really starts having an effect at 1A!)

Mark

You need a Fluke 8842 or the newer 45 or equivalent bench meter with four wire ohms as mentioned above.
If you use the current method, you now need 4 place accuracy on your current source and your voltmeter.
Also, current will generate a little heat, which will mess things up unless you have like a Dale LVR series mil resistor with a very low heat coef.
:sam:
PS This is if you need accuracy.
For a ballpark easy to do method, do the current thing.

Thanks for the Fluke #, cj. Looks like they are going on evilbay for around $350. I didn't think they sold new for much more than that! There's got to be something cheaper.

Marik probably wants to measure a ribbon mic's ribbon resistance, so I wouldn't recommend any method that puts a sizable current through it.

[quote author="Flatpicker"]
Marik probably wants to measure a ribbon mic's ribbon resistance, so I wouldn't recommend any method that puts a sizable current through it.[/quote]

How did you guess :?: :shock: :green:

Yeah, I need to measure ribbon once or twice, so it doesn't pay to get something expensive. I think idea of measuring current would be the best--did not think about it.

Thanks!

[quote author="Marik"]...How did you guess :?: :shock: :green:[/quote]Oh, probably just dumb luck. :wink:

Hey, did you notice you used your 100th post quoting me? :green:

> I need to measure ribbon

OH!

So you don't want to run 10 Amps through it.

You are going to need a transformer anyway. Wire the ribbon and the transformer. If possible, leave out the magnet so the ribbon won't vibrate.

Set your audio signal generator to midband, 1V output. Put a 10K and a 100 ohm resistor in series across the output. Be ready to wire the transformer to the 100 ohm resistor.

You need an audio millivoltmeter. Some cheap DVMs will do that. Otherwise use a line-amp and an audio voltmeter. Know the gain of the line amp.

Check the voltage across the 100 ohm resistor. Should be about 10mV, adjust so it is.

Now bridge the transformer secondary across the 100 ohm resistor. The 10mV will be less. If it drops to 5mV, then the total transformer+ribbon impedance at the transformer secondary is 100 ohms. It probably won't be that neat, you will have to do math.

Now short the ribbon with the fattest piece of copper you can get in there. Measure again. That is the impedance of just the transformer.

Might be interesting to try with the ribbon disconnected.

And sweep the audio band.

The reason I am taking you the long way: if the ribbon isn't the main part of the total ribbon+transformer impedance, then you are wasting valuable ribbon. You may want a lower-R transformer, or you may want a higher-R (thinner, lower mass) ribbon. Testing with the transformer tells you a lot about ribbon transformer matching, and also transforms the awkward 0.01-0.1 ohm impedance up to like 100 ohms. And anyway you will never "see" the ribbon without a transformer: the untransformed output is too low-voltage for the real world. So measuring through the transformer is realistic.

Thanks PRR, that's great!

I am working with 0.8um ribbon, so I needed to know if output impedance will be too high with Lundahl 1:37 LL2911.

Great stuff PRR!

:thumb:

Mark

Pardon an ignorant question.

Why wouldn't a simple Wheatstone bridge be able to measure the resistance? It seems that if the ratio resistors were, say, 1 ohm and 10k ohms, then the rheostat resistance would be on the order of 1k, assuming the ribbon were about 0.1 ohm. With 10 volts across the bridge this would pass a current of about 10 ma through the ribbon, which I'd think it would be easily able to handle. If it were nulled with a ua meter, I'd think this would give reasonable precision.

Of couse, all leads should be kept very short. The meter and power supply leads shouldn't introduce any imbalance if they were of equal length and gauge.