Measuring tiny unknown resistance?

You would have some very unusual meters if they can't measure resistances of less than 100 ohms. As digital multimeters go they almost universally (from the very cheapest to the most expensive) have a minimum resistance scale of

200 ohms. On the 200 ohm scale the maximum resistance that can be measured is 199.9 ohms. The minimum resistance that can be measured is limited by the accuracy of that scale. On the 200 ohm scale, this means the minimum resistance that can be measured is 0.1 ohm. The last digit is not very accurate, so it could be off by a little. Measuring a resistance between 0.5 ohm and 1 ohm will not be too especially accurate, but it will likely be accurate to about as good as within 0.1 ohm. Sometimes the meter will not be calibrated to zero and will thus produce a fixed offset to your measurement. You can manually compensate for this by shorting the meter leads together on the 200 ohm resistance scale and observing the result. It may for instance measure 0.2 ohms, but for a hard short circuit it should be 0.0 ohms. So, take you measurement of your 0.5ohm to 1 ohm resistance and then subtract 0.2 ohms from the displayed value.

To get better accuracy and to further extend the measurement range down to lower values you need another technique. One way is to build a 100mA constant current source using an LM317 linear regulator. Then apply the

100mA constant current to the unknown low value resistance and measure the voltage across it. Ohms law predicts V=IR, so for I=100mA, the voltage that appears will be one tenth of the resistance. IE: if the meter reads 50.0mV, then the resistance is 0.500 ohms.

If you have a low value, high power resistor you could put it in series with your unknown resistance. Apply a voltage then measure the voltage across the know resistance and across the unknown resistance. Since the current through both is the same, the ratio of the two resistances will be equal to the ratio of the two voltages. Therefore:

R1 = R2*V1/V2

Where V1 = voltage across R1, V2 = voltage across R2

To get an accurate measurement the known resistor value should be a similar order of magnitude to the unknown resistance, then you will get sensible voltages across both resistances.

Obviously, your known resistor must be capable of handling the current.

Gareth.

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To add to what others have posted, note that the trick to accurate low-resistance measurements is to measure the voltage drop across the unknown resistance with

*different leads* than those driving the current through it. (Unlike the the conventional ohmmeter arrangement.) So, if you use the suggestion of a series power resistor, you can measure the voltage across that to determine the current, then measure the voltage across the unknown and compute its resistance.

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Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

try to measure current thru it & voltage across it & calculate the resistance than ...

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You could use a 12 ohm resistor is series with the 12 volt battery so you get about 1 amp of current. It will heat up with about 12 watts of heat, so you need a large resistor. Or, use a tail lamp bulb which is a about 2 amps. Then use your multimeter to measure the exact current. Also measure the voltage across the unknown resistance and divide this by the known current.

So, if you read 0.6 volts across the resistance, and the current is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

-Bill

Stick a 100ohm resistor in series with the defroster lead. Switch on and measure the DC voltage across the heater. A 1ohm defroster would give 100mV. A 0.5ohm defroster would give 50mV etc. regards john

"john jardine" skrev i melding news:ctdl62$g86$ snipped-for-privacy@news7.svr.pol.co.uk...

100mV.

Or, like this:

o-------| R1 |--------| R2 |-----------|!' Gnd Ucc

Ur2 (voltage across R2) = (Ucc * R2) / (R1 + R2) // The basic formula.

Turning the formula to get the value of R2 is quite easy: UR2( R1 + R2 ) = Ucc * R2 ( (UR2 * R1) / R2 ) + UR2 = Ucc

// This is the important stuff So finding R2 is to calculate this formula: R2 = (UR2 * R1) / (Ucc - UR2)

Hope this helps : )

"Bill Bowden" skrev i melding news: snipped-for-privacy@z14g2000cwz.googlegroups.com...

You could use a 12 ohm resistor is series with the 12 volt battery so you get about 1 amp of current. It will heat up with about 12 watts of heat, so you need a large resistor. Or, use a tail lamp bulb which is a about 2 amps. Then use your multimeter to measure the exact current. Also measure the voltage across the unknown resistance and divide this by the known current.

So, if you read 0.6 volts across the resistance, and the current is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

Disagree. When he measures 0.6 volts across the resistance, there is no longer 1 amp through the circuit any more because the total resistance will be more then 12 ohms. The higher the "unknown" resistance is, the higher error on the measurment.

If using my formula (erlier post, same thread) the resul will be: "unknown resistance" = (0,6V * 12ohm) / (12V - 0.6V) = 0.63 ohm

I don't quite follow the logic. It doesn't really matter what the total resistance is, or the battery voltage. We only need to know the current through the circuit, and the voltage across the unknown resistance. If we measure 1 amp through the circuit, and 0.6 volts across the unknown resistance, the value of the resistance will be *exactly* 0.6 ohms, unless you are considering the error introduced by the meter resistance, in which case you need to know the meter resistance, which wasn't given.

-Bill

Yes, very good formula. But the meter resistance measuring voltage is quite high, probably several megohms, so it has little effect in parallel with 1 ohm or less. Probably less than .001%

I would imagine the series meter resistance measuring current would be of more concern, since the resistance in series is very low and the meter resistance could be a significant part of the total.

I think only one known resistor is needed to figure this out, but I haven't worked it out.

-Bill

"Bill Bowden" skrev i melding news: snipped-for-privacy@z14g2000cwz.googlegroups.com...

That's also possible to find.

If you couple two resistors in series and measures the voltage over one of them, say f.ex that you have R1 and R2 in series and measure the voltage over R2. If you know the exact resistance for both R1 and R2, then the formula for resistance in the meter is as follow:

Rmeter = (UR1 * R1 * R2) / ( Ucc*R2 - UR1*R1 - UR1*R2 )

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50.0mV,

Ohmmeters are notoriously lousy for measuring low resistances. While the nominal range may allow readings to 0.1 ohms, it is likely that at this level the error may be, in fact, much higher - esesentually making the reading no more than a continuity test. Such things as a thin oxide coating on the terminals . can blow the reading to hell as the ohmmeter will not have the "oomph" to break down this layer as happens in normal use. You need to be able to supply enough voltage (something near nominal.) to drive a reasonable current - and measure V and I. Someone mentioned use of a known resistance in series (check its power rating ) and measuring the voltage drop across this (getting, by the way, the current fromV/R) and that across the unknown. This works (and in fact is related to the way most multimeters handle current measurements) Use of a such a shunt (low resistance) in series with the load, and measuring the voltage across the shunt is a lot cheaper than making a constant current source in the 5+ A range. 100ma won't do it for proper measurement of the resistance of a 12V device drawing 12 to 24A.

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