How do you measure and use a low value power resistor?

[technix]

One of my component suppliers gave me a power resistor for free for some reason. It is one of those ones with integrated aluminum heatsink and mechanical mounting holes. Whatever marking that was on there has long been scrubbed away.

When I measure it with the ohms range on my multimeter, the result seemed a bit inconsistent to me since the resistor is almost within the range of the resistance of the leads I am using. How should I measure it then? I don't have a multimeter capable of four-wire measurement, but will a ghetto four-wire measurement solution using a bench power supply, a Uni-T UT61D and a UT61E work? How should I arrange the two different meters for the maximum accuracy?

How do I estimate the power ratings of such a resistor?

Also what can I do with it? The leads on both ends of the resistor is millimeters thick and very hard to bend, so should I solder it to a PCB directly at all? If not how should I attach it to a PCB?

[Conrad Hoffman]

Put it in series with a lowish value resistor that you do know, say a 10 ohm 1% or similar. Attach to a power supply and put a volt or two across them. Measure the voltage drop across the 10 ohm to get the current, then measure the voltage drop across the unknown resistor and use ohms law to calculate the value. Done carefully, this will get you any accuracy you want, based on your knowledge of the reference resistor and the accuracy of your meter. Just watch your power dissipation as the resistors may change on heating.

[technix]

Put it in series with a lowish value resistor that you do know, say a 10 ohm 1% or similar. Attach to a power supply and put a volt or two across them. Measure the voltage drop across the 10 ohm to get the current, then measure the voltage drop across the unknown resistor and use ohms law to calculate the value. Done carefully, this will get you any accuracy you want, based on your knowledge of the reference resistor and the accuracy of your meter. Just watch your power dissipation as the resistors may change on heating.

The unknown resistor is about the 0.2 ohm range. Also I don't have low value reference resistors handy, but I do have two multimeters.

[Heartbreaker]

The unknown resistor is about the 0.2 ohm range. Also I don't have low value reference resistors handy, but I do have two multimeters.

You have a fine power resister inside both of your multimeters. Furthermore they are already hooked up to measure the voltage across this resistor and calculate the current for you, similar to what Conrad Hoffman suggested.

When dialing up the voltage, do the multiplication of the current and voltage in your head and assume that all the power i dissipated into your unknown resistor. From your description it sounds as if the resistor is of a 5-10W type.

[mjs]

Use a lab supply to push 1A (or more) through it and measure voltage over it on the mV scale of your multimeter. At 1A 1mV=1mΩ

[Insatman]

@ MJS    I built a precise 1A current supply box for this purpose some time ago because my bench supplies all time vary by as much as a few mA.   I can get stability of 10's of uA using this kind of current source.  So measurement is more precise and doesn't vary all over the place when you measure

[technix]

Use a lab supply to push 1A (or more) through it and measure voltage over it on the mV scale of your multimeter. At 1A 1mV=1mΩ

@ MJS    I built a precise 1A current supply box for this purpose some time ago because my bench supplies all time vary by as much as a few mA.   I can get stability of 10's of uA using this kind of current source.  So measurement is more precise and doesn't vary all over the place when you measure

That is why I am thinking about using two meters. One meter in amps range measuring the actual output current from the power supply, and the other measuring the voltage developed across the resistor.

[beanflying]

As the couple of posts above higher test currents give you much better results. My Quadtech millivoltmeter runs 1A on m range https://www.atecorp.com/atecorp/media/pdfs/data-sheets/quadtech-1880_datasheet.pdf It runs rings around my Agilent 34401a (1mA test current) at very low resistances even using 4W on both due to this increased test current. Thermal variations from self heating can also step in with prolonged tests so watch that too.

If you don't want to roll a dedicated constant current source then the option Conrad suggested works too.

[Terry01]

Maybe a decent picture of "said resistor" will help the guys help you to work out the value and power rating of your unknown resistor. 

[capt bullshot]

That is why I am thinking about using two meters. One meter in amps range measuring the actual output current from the power supply, and the other measuring the voltage developed across the resistor.

Yes, that would be the correct answer considering your equipment. Just take care to connect the voltmeter "Kelvin Style" to the resistors leads. Usually these resistors come in standard values, so "round" your result to the nearest standard value (e.g. 0.1 Ohm, 0.22 Ohm, whatever).

[Terry01]

That is why I am thinking about using two meters. One meter in amps range measuring the actual output current from the power supply, and the other measuring the voltage developed across the resistor.

Yes, that would be the correct answer considering your equipment. Just take care to connect the voltmeter "Kelvin Style" to the resistors leads. Usually these resistors come in standard values, so "round" your result to the nearest standard value (e.g. 0.1 Ohm, 0.22 Ohm, whatever).

When you say to make sure to connect it "Kelvin style" do you mean like this?

[beanflying]

Maybe a decent picture of "said resistor" will help the guys help you to work out the value and power rating of your unknown resistor. 

Until you have an accurate resistance the size and then likely wattage is a moot point.

[capt bullshot]

When you say to make sure to connect it "Kelvin style" do you mean like this?

Yes

[technix]

Here comes the measurements:

Resistance: 250mΩ.
Body length: 27mm.
Body width: 14.5mm.
Body height: 13mm.

The body I am talking about includes the aluminum heatsink it comes with Any estimate of its power rating? A similarly sized 33Ω resistor has a rating of 25W.

[beanflying]

Looking at the dimensions it is about the same size as a few 25W Ohmite resistors I have around here. Some useful data here https://www.ohmite.com/assets/docs/acl_hs10.pdf

Heat sinking is really key if you want to push it toward that or if you are considering using it as a current shunt you will want to keep the power way down to minimize any Temp Co issues.

[soldar]

The unknown resistor is about the 0.2 ohm range. Also I don't have low value reference resistors handy, but I do have two multimeters.

You can easily make low value resistors. Copper wire has a well defined resistance and can be easily found. Resistance can be easily calculated. 1000 meters of copper wire with a diameter of 1 mm will have a resistance of 22.323 Ohm. 

If you have nichrome wire you can easily measure the resistance of a length and then calculate what fraction you need for the resistance you want to obtain.

[mjs]

Copper has a resistance tempco of roughly 0.4%/K at room temperature, which means that +40°C temperature rise will result i 16% change in resistance. Not very accurate.

When I use a lab supply to measure small resistances, I usually just measure and set the current to 1.00 A, then connect resistor and measure voltage over it. Quite fast and no need to fiddle with two meters.

[technix]

Copper has a resistance tempco of roughly 0.4%/K at room temperature, which means that +40°C temperature rise will result i 16% change in resistance. Not very accurate.

When I use a lab supply to measure small resistances, I usually just measure and set the current to 1.00 A, then connect resistor and measure voltage over it. Quite fast and no need to fiddle with two meters.

My lab power supply don’t have the best accuracy, so an additional meter is thrown in there as a ammeter.