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Re: Resistance measurement in the 0.5 - 1.0 Ohm range

Subject: Re: Resistance measurement in the 0.5 - 1.0 Ohm range
From: "Chris"
Date: 12 Mar 2006 10:00:07 -0800
Newsgroups: sci.electronics.basics
Ed wrote:
>
> Thanks, Chris. I really appreciate the completeness of your reply.
> BTW, if
> a person wanted to get better accuracy, say +/- 5% what would he have
> use? A Wheatstone
> bridge and a precision decade box?
>
> Ed

Hi, Ed.  The kelvin method of measuring resistance is very accurate for
measuring low ohms.  Using the kelvin method, you put a known current
through the resistor using two leads, and then use two other leads to
measure the voltage across the resistor.

Even with a coil that is sensitive to too much power, you can make a
fairly accurate resistance measurement using a DMM with a 200mV range,
if you have a precision series resistor to give a known current.

You've got your 12V battery already.  See if you can find a 100 ohm
resistor that's 3 watts or more.  The better the precision and the
higher the wattage, the better.  I like to scrounge power resistors
when I find them, so I've got a collection of 10 watt and up 1%
resistors to do this type of thing.

Now, hook up your battery in series with the 100 ohm resistor and the
coil, like this (view in fixed font or M$ Notepad):


|         ___
|    .---|___|-----.
|    |  100 ohm    |
|    |             o
|    |             |
|    |             o------.
|    |             |      |
|   +|          |  |      |
|   ---    120mA|  C|    / \
|    -          |  C|   (DMM)
| 12V|          |  C|    \_/
|    |          |  |      |
|    |          V  |      |
|    |             o------'
|    |             |
|    |             o
|    |             |
|    '-------------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Your coil resistance is very small in relation to the 100 ohm resistor,
so for the sake of calculations let's assume it's basically zero.  Now
measure your battery voltage under load, then calculate the load
current using Ohm's Law:

I = V / R

Assuming your battery puts out 12.00V under load, and your resistor is
100.0 ohms, you'll have 120mA going through the circuit.  Now use your
DMM  on the 200mV range to measure voltage across the coil, and use
Ohm's law again to calculate coil resistance.  Let's assume you read
60mV across the coil, then you can calculate:

R = V / I = .060V / .12A = 0.5 ohms

With even the cheapest DMM, you can get a resistance reading which is
pretty much within the tolerance of your series resistor.  Even cheap
DMMs have 1% or better accuracy on the 200mVDC range.

Make sure you place the DMM leads on the device to be measured
carefully, so you don't measure anything but the coil.  Post back if
you're unsure on this.  It's the major source of error in kelvin
measurements.  The rule of thumb is, current on the outside, potential
on the inside.

By the way, your question about power is a good one.  But at this low
current, it shouldn't be an issue.  Power = voltage * current, so

P = .12A * .06V = a little over 7 milliwatts for a half ohm coil, which
shouldn't even get it warm.

Good luck
Chris


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