My muiltimeter does not measure resistance lower than .1 ohm. If I wanted to find lenght of a certain size wire, I want to measure the resistance of the conductor. Example would be the resistance of #2 awg is .1987 ohm per 1000 feet. does anyone know of a way to measure ohm below .1 ohm?
Less than .1 ohm
- Thread starter EEC
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zog
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EEC said:
Hmmm, a micro ohmeter?
Actually there is no way to measure any resistance, it is just a calulated value and you need more current to measue that small of a voltage drop or your meter will read Outside Limits (O.L), most micrometers have a 10A output.
Bob NH
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wireguru said:
Put that precision 10 Ohm resistor in series with the wire.
Connect a DC voltage source, such as a Nickel/metal hydride or NiCd cell, to the wires. You will get something like 120 milliamps through the circuit.
Measure the DC voltage across the 10 Ohm resistor, and the voltage across the wire.
The resistance of the wire is Resistor Ohms x voltage across the wire divided by the voltage across the resistor.
Example:
Voltage across the resistor is 1.22 Volts. Voltage across the wire is 0.049 Volts.
Resistance of wire is 10 x 0.049 / 1.22 = 0.40 Ohm
LarryFine
Master Electrician Electric Contractor Richmond VA
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Wheatstone bridge?
quogueelectric
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Cobble stone bridge.
wptski
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You'd need a high quality DMM that has the REL or Zero function to null the resistance of the test leads at least.
I don't believe that zeroing out the resistance of the leads of a standard DMM will give sufficiently accurate measurements of the low resistances in question. Four-wire (Kelvin) measurement is needed in this case, either by using an expensive micro-ohmmeter, or by using the technique suggested by Bob_NH.
The Low Level Measurements Handbook by Keithley Instruments may be worth a look.
The Low Level Measurements Handbook by Keithley Instruments may be worth a look.
gar
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080904-2231 EST
EEC:
Create a moderately high current source. Use a Variac at the input to a transformer with a 1 turn secondary that can produce maybe 50 A. A toroidal transformer would be easy to use. You need an ammeter for the current level you plan to use. A very low resistance connection at the far end of the cable is required to join the two wires. Also a millivoltmeter is needed. A Fluke 27 is probably satisfactory. You will do a 4 terminal measurement. This means you use two terminals to supply current to the resistance under test, and two more terminals that contact the resistance inside of the points where the current is injected.
Make a series loop of the transformer secondary, ammeter, the current connections to the wire to be measured, and the wire. The resistances of the connections to the wire under test from the current source are not particularly important because of the 4 terminal test. You just want moderate stability of current. Adjust the Variac starting from 0 to get your desired current. The current does not have to be any particular exact value, you just need it large enough, and know the value.
Put the voltmeter in the appropriate millivolt range, and put the meter probes in direct contact with the wire under test just inside of where the current is supplied. These are your third and forth terminals.
Calculate the resistance. For example 0.001 ohms * 50 amps = 0.05 V or 50 MV. This corresponds to a loop length of 1000/0.1897 = 5270 ft/ohm, and for 0.001 ohm this is 5.27 loop ft or 2.63 ft to the end of the cable.
Measure the voltage across the voltage points with no current to verify there is no induced voltage that might be an error.
This method eliminates including the contact resistance of the current source terminations. It does nothing to solve a problem at the far end joint connecting the two wires. However if there was a measurable voltage drop across that far end connection you could measure it and subtract the error out.
.
EEC:
Create a moderately high current source. Use a Variac at the input to a transformer with a 1 turn secondary that can produce maybe 50 A. A toroidal transformer would be easy to use. You need an ammeter for the current level you plan to use. A very low resistance connection at the far end of the cable is required to join the two wires. Also a millivoltmeter is needed. A Fluke 27 is probably satisfactory. You will do a 4 terminal measurement. This means you use two terminals to supply current to the resistance under test, and two more terminals that contact the resistance inside of the points where the current is injected.
Make a series loop of the transformer secondary, ammeter, the current connections to the wire to be measured, and the wire. The resistances of the connections to the wire under test from the current source are not particularly important because of the 4 terminal test. You just want moderate stability of current. Adjust the Variac starting from 0 to get your desired current. The current does not have to be any particular exact value, you just need it large enough, and know the value.
Put the voltmeter in the appropriate millivolt range, and put the meter probes in direct contact with the wire under test just inside of where the current is supplied. These are your third and forth terminals.
Calculate the resistance. For example 0.001 ohms * 50 amps = 0.05 V or 50 MV. This corresponds to a loop length of 1000/0.1897 = 5270 ft/ohm, and for 0.001 ohm this is 5.27 loop ft or 2.63 ft to the end of the cable.
Measure the voltage across the voltage points with no current to verify there is no induced voltage that might be an error.
This method eliminates including the contact resistance of the current source terminations. It does nothing to solve a problem at the far end joint connecting the two wires. However if there was a measurable voltage drop across that far end connection you could measure it and subtract the error out.
.
Bob NH said:
well i had 1/2 of it
made me think of something, i have a PCB i made with a microcontroller and current sense amplifier to measure DC current up to a few amps via a current sense resistor, and get the readings via a USB connection. Bet it could be used here......
One little point from all these equations that is missing: temperature.
Awhile back we had a thread (or two) here that discussed the issue of wire lengths left on reels...here is one such thread:
http://forums.mikeholt.com/showthread.php?t=81107&highlight=greenlee
Greenlee "had" a very accurate wire measuring tool....I cannot seem to locate the model # and my link to it in the thread linked to comes up "404".
I know there is a similar thread on here that *may* even have the tools model number. This other thread is older than the one linked to above.
iwire? (Bueller?... Bueller?... Bueller?) do you recall?
Awhile back we had a thread (or two) here that discussed the issue of wire lengths left on reels...here is one such thread:
http://forums.mikeholt.com/showthread.php?t=81107&highlight=greenlee
Greenlee "had" a very accurate wire measuring tool....I cannot seem to locate the model # and my link to it in the thread linked to comes up "404".
I know there is a similar thread on here that *may* even have the tools model number. This other thread is older than the one linked to above.
iwire? (Bueller?... Bueller?... Bueller?) do you recall?
gar
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080905-0814 EST
Near room temperature the resistance of copper varies about 0.393% per deg C. If excessive current is not used for the test current and the duration of the test current is short the wire won't deviate much from the ambient temperature. Even a 20 deg C rise only increases the resistance by the ratio 1.08 . What length accuracy is required? Also the drawing accuracy of the wire diameter and alloy need to be considered as error sources.
Your required wire length accuracy will determine what you need to do experimentally.
Another method of measurement is propagation velocity of an electric pulse along the cable. The velocity of light is about 982 ft per microseconds. The velocity of propagation of a pulse along a cable is in the range of 0.7 times the velocity of light, or about 687 ft per microsecond. An experiment I ran on CAT-5 cable showed about 0.2 microsecond for 150 ft, and for a different brand 1.5 microsecond for 1000 ft. These calculate to 0.2*986 = 196 ft, and 150/196 = 0.76 for the factor relative to the speed of light, and 1.5*986 = 1479 ft. and 1000/1479 = 0.68 for the ratio. There is a lot of variation here between different cables. I think the resistance method is probably a better approach. Photos of my experiment are at photos P1 and P3 at
http://beta-a2.com/cat-5e_photo.html
.
Near room temperature the resistance of copper varies about 0.393% per deg C. If excessive current is not used for the test current and the duration of the test current is short the wire won't deviate much from the ambient temperature. Even a 20 deg C rise only increases the resistance by the ratio 1.08 . What length accuracy is required? Also the drawing accuracy of the wire diameter and alloy need to be considered as error sources.
Your required wire length accuracy will determine what you need to do experimentally.
Another method of measurement is propagation velocity of an electric pulse along the cable. The velocity of light is about 982 ft per microseconds. The velocity of propagation of a pulse along a cable is in the range of 0.7 times the velocity of light, or about 687 ft per microsecond. An experiment I ran on CAT-5 cable showed about 0.2 microsecond for 150 ft, and for a different brand 1.5 microsecond for 1000 ft. These calculate to 0.2*986 = 196 ft, and 150/196 = 0.76 for the factor relative to the speed of light, and 1.5*986 = 1479 ft. and 1000/1479 = 0.68 for the ratio. There is a lot of variation here between different cables. I think the resistance method is probably a better approach. Photos of my experiment are at photos P1 and P3 at
http://beta-a2.com/cat-5e_photo.html
.
It's pretty easy to make a 4-wire ohmmeter that will likely be accurate enough for what you are trying to do. I got the idea from an aviation electrical website. It is similar to what gar is saying, and likely the same as Bob_NH (I didn't look for his post) Yeah, I know we are building ground structures, not airplanes - but the science still applies.
Probably you have everything you need is already in your toolbox.
You will need two $50 DVMs. One needs a 10A scale the other needs to measure millivolts.
The current source is pretty simple - use a D-cell. The article I've linked below says it will put out about 7A short circuited. The one I built only put out about 3A - maybe it was an old battery. Seven amps would be better.
Take a look at the article and see what you think. Considering the price, it is amazingly accurate.
http://www.aeroelectric.com/articles/grnding.pdf
carl
Probably you have everything you need is already in your toolbox.
You will need two $50 DVMs. One needs a 10A scale the other needs to measure millivolts.
The current source is pretty simple - use a D-cell. The article I've linked below says it will put out about 7A short circuited. The one I built only put out about 3A - maybe it was an old battery. Seven amps would be better.
Take a look at the article and see what you think. Considering the price, it is amazingly accurate.
http://www.aeroelectric.com/articles/grnding.pdf
carl
gar
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coulter:
I quickly scanned the article and the primary problem I see is that there are only two probes. This means the interface resistance(s) of the probe(s) to the item under measurement are part of the measured resistance. This is why the voltage probes should be independent of the current source.
.
coulter:
I quickly scanned the article and the primary problem I see is that there are only two probes. This means the interface resistance(s) of the probe(s) to the item under measurement are part of the measured resistance. This is why the voltage probes should be independent of the current source.
.
If one were trying to measure a micro-ohm, that would certainly matter. If the desired resolution is a milliohm - maybe not.gar said:
As you have said (alluded?), the idea is that one does not have to spend $M$ for a micro-ohmeter. Inexpensive equipment and a little applied physics will do fine for a lot of tasks. I would never suggest this for an exacting customer application - one could never get a traceable certificate.:roll:
The other point, is there are simpler current sources than using a transformer, rectifier, and a series resistance. A D-cell with soldered leads works pretty well and is dirt cheap.
You like doing experiments. Set it up and give it a try.
I build one of these a few years back. For the probes, I soldered the ends of the wires together and clamped the soldered ends on the test piece with vice grips. It appeared to work very well.
edit: Ah... finally saw Bob NH post. Yes, similar method.
carl
wptski
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I've used a milliohm meter with Kelvin clips and compared that to using a Fluke 1507 insulation tester which can be zeroed and it does help.steved said:
It is called a DLRO (Digital Low Resistance Ohmmeter). It is a 10 and 1 amp current source and uses 4-points, two to inject the current, and two to measure the voltage developed from the current. They will rad down to .1 micro-ohm. Otherwise you have to use a TDR.
Eitherway they are very expensive meters.
Eitherway they are very expensive meters.
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