How do you measure and use a low value power resistor? - Page 1
Forum section: electronics / projects

author: technix
author_rank: Super Contributor
timestamp: December 16, 2018, 03:16:51 pm
content: 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?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: December 16, 2018, 03:20:41 pm
content: 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.

author: technix
author_rank: Super Contributor
timestamp: December 16, 2018, 03:22:41 pm
content: Quote from: Conrad Hoffman on December 16, 2018, 03:20:41 pmThe 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.

author: Heartbreaker
author_rank: Unknown
timestamp: December 16, 2018, 04:01:46 pm
content: Quote from: technix on December 16, 2018, 03:22:41 pmYou 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.

author: mjs
author_rank: Regular Contributor
timestamp: December 16, 2018, 07:54:57 pm
content: 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

author: Insatman
author_rank: Frequent Contributor
timestamp: December 17, 2018, 03:59:52 am
content: @ 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

author: technix
author_rank: Super Contributor
timestamp: December 17, 2018, 04:14:21 am
content: Quote from: mjs on December 16, 2018, 07:54:57 pmQuote from: Insatman on December 17, 2018, 03:59:52 amThat 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.

author: beanflying
author_rank: Super Contributor
timestamp: December 17, 2018, 04:18:18 am
content: As the couple of posts above higher test currents give you much better results. My Quadtech millivoltmeter runs 1A on mrangehttps://www.atecorp.com/atecorp/media/pdfs/data-sheets/quadtech-1880_datasheet.pdfIt 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.

author: Terry01
author_rank: Frequent Contributor
timestamp: December 17, 2018, 09:02:20 am
content: 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.

author: capt bullshot
author_rank: Super Contributor
timestamp: December 17, 2018, 09:19:27 am
content: Quote from: technix on December 17, 2018, 04:14:21 amYes, 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).

author: Terry01
author_rank: Frequent Contributor
timestamp: December 17, 2018, 10:02:29 am
content: Quote from: capt bullshot on December 17, 2018, 09:19:27 amWhen you say to make sure to connect it "Kelvin style" do you mean like this?

author: beanflying
author_rank: Super Contributor
timestamp: December 17, 2018, 10:20:17 am
content: Quote from: Terry01 on December 17, 2018, 09:02:20 amUntil you have an accurate resistance the size and then likely wattage is a moot point.

author: capt bullshot
author_rank: Super Contributor
timestamp: December 17, 2018, 10:22:40 am
content: Quote from: Terry01 on December 17, 2018, 10:02:29 amYes

author: technix
author_rank: Super Contributor
timestamp: December 25, 2018, 09:51:14 am
content: 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.

author: beanflying
author_rank: Super Contributor
timestamp: December 25, 2018, 10:12:18 am
content: Looking at the dimensions it is about the same size as a few 25W Ohmite resistors I have around here. Some useful data herehttps://www.ohmite.com/assets/docs/acl_hs10.pdfHeat 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.

author: soldar
author_rank: Super Contributor
timestamp: December 25, 2018, 10:51:47 am
content: Quote from: technix on December 16, 2018, 03:22:41 pmYou 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 havenichromewire you can easily measure the resistance of a length and then calculate what fraction you need for the resistance you want to obtain.

author: mjs
author_rank: Regular Contributor
timestamp: December 25, 2018, 02:33:17 pm
content: 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.

author: technix
author_rank: Super Contributor
timestamp: December 26, 2018, 03:29:21 pm
content: Quote from: mjs on December 25, 2018, 02:33:17 pmMy lab power supply don’t have the best accuracy, so an additional meter is thrown in there as a ammeter.