Resistance amplification? 3

[timeline1968]

I have a project that I am working on that is used to show, very simply, the difference in resistance between different materials. I'm using wires of a relatively small diameter, however I'm seeing that the resistance is going to be so small that it will not be measureable using standard multimeters (~.04 ohm).

Is there some way that I can amplify the resistance for all these wires so that it will be measureable, and distnctly different by the same amount for every wire?

Any assistance would be greatly appreciated.

 

[Skogsgurra]

The standard way of doing such measurements is to let a known (and constant) DC current pass through the specimen and show the voltage drop across it. If you use 1 amp, your reading vill be 1 V = 1 ohm. If 1 A is too much current, you can use less current and then use a convenient constant (like .1 or .01) to read resistance from voltage drop.



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

If you connect your different samples in series and maintain a relatively constant current the voltage across each sample will be proportional to the resistance of that sample.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[IRstuff]

You can use either longer wires or a micro-ohmmeter:

http://www.ietlabs.com/IET/LOM-510.html

http://www.testequipmentdepot.com/AEMC/MicroOhmmeters/6250.htm

They're not cheap, since you're paying for a substantial reduction in electronics noise. You could probably rent one for a week or so. Don't forget that with low resistances, you need a Kelvin connection to eliminate the probe contact error.

TTFN

FAQ731-376

 

[OperaHouse]

A cheap way is to use a 7805 regulator in current mode producing .5 to 1 amp. That requires using a five or ten ohm resistor from the output pin to the common pin. Review regulator app notes for details. Requires a lot of heat sinking, but that will make the voltage readable by a 200mV digital meter. Then just calculate resistance.

 

[bogeyman]

Get hold of a 3.5 digit DPM module with differential reference and signal inputs.

Use a D ell as the power source with a series resistance of 1.2 ohms to limit the source to about 1A.

Connect a 0.1 ohm across the DPM ref input and the test sample across the signal input. Connect the DPM common to ref lo.

Power the ref hi from the +1.5v 1.2 ohm source and link the ref low to the sample resistance, return the other end of the sample resistance to -ve of the 1.5v source. Use 4 wire connections (one wire each end for the current and one each end for the DPM input)on the ref and sample resistances.

This will measure 0 to 0.1999 ohms. You can scale the values and test current for other ranges.

The display reads the ratio of the ref and sample resistances.

 

[timeline1968]

First off, thanks for all the replies. Second: I'm not an electrical guy and this is going to be used to show students the different resistances of materials, theoretically by mounting short lengths of wires on a board. I need the simpler methods such as running a current across the wires or maybe adding the regulator/resistor in order for this to work for me. This is not an industrial application where we can procure expensive equipment or use complicated methods.

I tried to just run a voltage across the wires (from a D-Cell battery) however I am seeing no drop at all (I'm using a Fluke 189 DMM to read the voltage). I suspect this is because the wires are so short (about 60cm long) and there just isn't enough resistance to matter. Will the regulator method gain me anything in this case?

 

[OperaHouse]

The 1 amp regulator will provide 40 mV not a lot of resolution. I have used a toroid (donut style) AC power transformer to create high currents. Even a small 40W transformer can create currents of 10-20 amps at under 1/2 volt for these tests. I break out the usual epoxy center so I can loop a couple turns of #12 wire. Down side is ac measurements with many voltmeters are very poor.

Adding a couple regulators in parallel can increase the DC current and make readings easier.

 

[sreid]

Perhaps very fine wire. 40 ga is about 1 Ohm/ft,

37 ga about 0.5 Ohm/ft, 34 ga about 0.25 Ohm/ft. Every three gage size decrement halves the resistance.

 

[melone]

Don't use short wires, use large rolls. For example, measure the resistance of the entire spool of wire.

 

[waross]

I assume you have the same gage wire for all of your samples.
I would use a board (1"x6", or 1" x 8") as long as you feel is reasonable in the class room. The length may be 3Ft., 4 ft., 6 ft., or even 8 ft.
Run your samples the length of the board. Connect alternate ends together so that all samples are in series.
Now run 1 amp through the series arrangement. The higher resistance of the series arrangement may make the current easier to stabilize.
Measure the voltage across each section with a mill-ammeter.
With 1 amp flowing, the voltage will equal the resistance. 40 ma will infer 40 milli-ohms.
A tip o measuring voltages;
There will probably be some non linear voltage drops in the connections between the different wires.
You may avoid this error by measuring the voltages just beside the connections.
For example, consider an iron wire connected to an aluminum wire connected to a copper wire. When measuring the voltage drop across the iron wire, position your meter probes so that they both contact the iron, just beside the splices. When measuring the aluminum wire, move the probe to the other side of the connection so that it contacts only the aluminum.
Good luck!!

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[timeline1968]

@ Operahouse: Too complex and costly

@ Sreid: I've thought of this, however since students will be using these themselves, I'm afraid they will break the wires. Maybe a possibility if I can somehow protect the wires.

@ melone: I can't use big spools of wires because not all types of wires come insulated and measuring the resistance across the ends of a big spool of uninsulated wire won't get me where I need to go

@ waross: Yes, all wires are the same gage, all around 24. I will experiment with this idea, however I still think that the the resistance will be too small. Thanks for the detailed explanation. I'll start futzing with this now.

Thanks to everyone. Please keep the brainstorming ideas coming!

 

[timeline1968]

The series option didn't work. Not enough resistance.

How valid would this be:
Connect a D-Cell battery in series with a multimeter. Set DMM to measure amps. Put one lead on far side of power circuit. Measure amperage at short end, then at long end. There seems to be enough to measure when it comes to amps (I get differences of at least 0.04).

Then have them use the V=IR equation at both ends of the wire, subtract the differences in R to get resistance of the wire and compare that value to the theoretical value?

Plausible?

 

[waross]

Well #24 copper wire has a resistance of 25 Ohm per 1000 ft. 6 ft would be 0.15 ohms. You won't see much with an analog meter, but you should get a good reading with a digital multi-meter.
Have I slipped a decimal place here? How long are your wires?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[timeline1968]

The wires are 60cm long.

Yes, you're right, the copper and silver wires in the set will have very little measurable resistance, however some of the other materials seem to work well using this method(iron, constantan, Ni-Chrome, stainless steel). I can just use the low resistance wires (copper, aluminum, silver) to show how little resistance there is and maybe make references as to why aluminum used to be used for wiring and no longer is standard.

 

[OperaHouse]

Too expensive!!! I can assure you that many here would like to have a low cost method to measure low ohms. Sometimes the best lesson learned is that it can't be done easily. I finally got a nice HYPATIA 309 that will easily calculate those resistances for any current from 1-100A. For years I used the methods I mentioned which cost less than $3. Clearly you should look for some other physical property to demonstrate. Calculating how a lamp goes through an nearly 10:1 resistance change as it warms up is more practical.

 

[Skogsgurra]

timeline1968,

I am somewhat puzzled to see how you reject all these solutions to your "selfinflicted" problem. My answer, for one, is a perfectly valid solution. All you need is a power supply with settable current limit and a cheapo digital voltmeter. Not anything you need to buy, but exists in plurality in every technical school worth the name.

What is the reason you make this simple (and, if I may express a personal meaning) rather uninteresting lab so difficult?

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[timeline1968]

I'm sorry, I didn't mean to offend anyone. Maybe a little deeper explanation regarding the use of this may be in order.

I work for a company that manufactures basic science equipment for physics labs (and classes) anywhere from elementary school up to the college level. We specialize in the middle school area though, and rarely are our products used at the college level.

When it comes to making demonstration equipment for use in public schools, cost is the number one driving force. Certainly I could make a demonstration unit that would cost $100+ or more, but no teachers would purchase it. We have 50+ years of experience showing this, especially nowadays. These demos must be inexpensive ($20-30) and simple since many of the teachers using them do not have expertise in every area of physics, and don't have access to equipment other than a d-cell battery and some wires.

All this is meant to accomplish is to show young kids that different metals have different resistances and maybe to expose them to some initial formulas regarding electricity. I highly doubt that a technical college or advanced university lab would pick this up. That's not our primary focus.

Thanks again for every-one's input, and I again apologize if I offended anyone with any of my remarks.

 

[davidbeach]

I think that if you had given the full story in your original post the whole thread might have taken a different direction, and may well now. People can only respond to the information given. I don't have a solution to offer, but I'm sure that the creative folks here will not take the new set of givens and come up with something creative and probably something that produces a far more interesting result than just different numbers on a meter. Besides a DMM doesn't meet your definition of inexpensive.

 

[IRstuff]

A D-cell will put out in excess of 10 A into a dead short. However, it's not a constant current, so it'll fluctuate while you're doing your experiment and give you variable results.

You're basically doing something wrong somewhere. A Fluke 189 has a 1 uV resolution. You should be able to use a constant current LED supply, ala

http://www.national.com/pf/LM/LM2755.html

which puts out 30 mA, and be able to measure the voltage drop across a 1 milliohm resistance with plenty of resolution. Or any of a variety of approaches. A DC wallwart with 5V output and 200-ohm series resistance will give you a relatively constant 25 mA current source, since all your test resistances are less than 1 ohm.

You should review Kelvin connections, which may be why you're having such difficulties in the measurement:

http://www.allaboutcircuits.com/vol_1/chpt_8/9.html

In fact, the whole subject of using Kelvin connections is probably worth a lecture alone.

TTFN

FAQ731-376