Need help identifying resistor. - Page 1
Forum section: electronics / beginners

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 17, 2024, 02:29:13 am
content: Attached to this Post is a resistor that is in question. I obtained this particular resistor from a donor board.According to the way that I interpret the color bands it should be .25 Ohm +/- 5%.Is my interpretation correct?I also measured with a multimeter and got .252 Ohm. I do not put a lot of stock into this as I know that low ohm values are not accurately measured with the multimeter and a 4 wire method should be used.Because of that fact I decided to test it using Ohm's law.I set a regulated linear lab supply to 12.011 V ( measured). I then put the multimeter in series with the resistor and the reading I got was .536A. Using Ohm's law the resistance value is 22.4 Ohm.I then verified the resistance of the leads by shorting them together and got .080 Ohm.So I am off by a factor of 100 from what is actually measured through Ohm's law and what I interpret the color codes to be and what the meter measured the resistance at.I was wondering if someone would please chime in to help clarify some of this.I would like to thank everyone that contributes in advance.

author: themadhippy
author_rank: Super Contributor
timestamp: January 17, 2024, 02:42:12 am
content: current limiting by your power supply? is it able to chuck out nearly 50A?

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 17, 2024, 02:48:03 am
content: Quote from: themadhippy on January 17, 2024, 02:42:12 amThat would be it. The power supply is rated for .5A. You live and you learn. Thank you so much.

author: Jwillis
author_rank: Super Contributor
timestamp: January 17, 2024, 09:43:37 am
content: The colour code does indicate a value of 0.25Ω. If your cold resistance at 20oC was measured at 0.25Ω then the active resistance becomes 22Ω then it sure sounds like that's a PTC resistor.Put together your test rig again and restart at a lower voltage at room temperature and see if the current drops at a steady rate as the resistor warms up. A lower voltage will slow down the temperature increase of the resistor making it easier to monitor.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 17, 2024, 02:08:56 pm
content: Quote from: Jwillis on January 17, 2024, 09:43:37 amThank you for the explanation as well as the suggestion. I know what a PTC resistor is I never knew they came in a package that resembled a regular resistor.Here is the problem I see with your suggestion. The lab power supply that I am using tops out at about .5A. Assuming that the resistor is .25 then the voltage for the experiment would need to be set at .125V for everything to function without the current regulation of the power supply being a factor.Would it not be easier to just apply heat from a heat gun to test to see if it is a PTC resistor?I am still going to try your suggestion but will also try a different experiment that I thought of which would entail starting at around 5 V. Then a multimeter would be connected in series to measure the current. I should theoretically still get the .536A because of the current limiting on this power supply. Then slowly start turning the voltage down and see if I can catch the point at which the current starts going down. At that point the current limiting on the power supply should not be an issue.Am I missing anything from my train of thought?

author: ebastler
author_rank: Super Contributor
timestamp: January 17, 2024, 07:22:01 pm
content: I don't see a reason to assume you have a PTC resistor. The 0.5A current limit of the power supply used for the test fully explains the observation. Occam's razor etc... Jwillis probably posted his suggestion before he had read the subsequent posts by themadhippy and yourself.

author: Jwillis
author_rank: Super Contributor
timestamp: January 17, 2024, 08:23:40 pm
content: Quote from: ebastler on January 17, 2024, 07:22:01 pmI did not notice the .5A limit on current as you said. That would certainly explain the strange behaviour and measurements. It's likely not a PTC given all the information.

author: DavidAlfa
author_rank: Super Contributor
timestamp: January 17, 2024, 09:14:13 pm
content: Doesn't seem a PTC, but a shunt (Current sensing resistor).Though the first band looks brown to me, red perfectly matches 0.25 Ohms, so I'd say it's all right!If you measured 0.5A, either the DMM ammeter had high internal resistance, other PSU limited the current beyond 0.5A.0.5A over 0.25 ohms would only drop 125mV and disspiate 62mW, this won't cause any heating nor resistance change in a PTC!

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 01:32:52 am
content: Thank you to everyone that has contributed so far. I agree with the post that was debating between Brown and Red for the color, I personally had a very hard time telling what it is and I have direct access to the item. It might be one or the other.Keep in mind that this resistor was pulled off of a board so it might actually be faulty.So I did the following experiment to further this exploration.I set a random voltage of 14.2 mV ( it's the best I could do with a fidgety knob). I verified this voltage with 3 separate meters. The cheapest being a $40 meter and the more expensive one being a $500 meter. All 3 were very close in line so let's assume that voltage correct.I then proceeded to take a current measurement. On the A scale i got .048 A. ( the multimeter is capable of 20 A on this range for 30 seconds). I then went to the mA and got 6.028ma.That would make the resistor .3087 Ohm on A range and 2.355 Ohm on the mA range.Then I set the voltage to .0027V got .008A which is .3375 Ohm then I switch to the milliamp range got 1mA which respectively gives a value of 2.7 Ohm.The meter is somewhat consistent in its discrepancies between A and mA range selection for the 2 voltagesNot sure why there is such a discrepancy.Any insight would be greatly appreciated.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 18, 2024, 01:47:35 am
content: Quote from: JJ_023 on January 18, 2024, 01:32:52 amSince you've got more than 2 meters, put an ammeter and a voltmeter in circuit at the same time. Have 4 attachment points to the resistor like this:

author: DavidAlfa
author_rank: Super Contributor
timestamp: January 18, 2024, 01:50:03 am
content: At such low voltage and resistance, wires will heavily affect the testing, easily adding those 50mOhms.Also the 14.2mV source voltage is just crazy.Just set the PSU to 1V, 1A CC limit, measure voltage at the resistor terminals, the reading in volts will be exactly the ohms.I'd assume the initial 0.25 Ohm reading was ok.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 02:12:27 am
content: Quote from: Kim Christensen on January 18, 2024, 01:47:35 amThank you for the suggestion. Where do you suggest I put the best meter. My guess would be to measure current?

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 02:18:24 am
content: Quote from: DavidAlfa on January 18, 2024, 01:50:03 amI get your suggestion and it makes a lot of sense the issue I am having is that the lab power supply will only putting out .5A before going into current limiting so I have those limitations that I am dealing with.I am trying to use a low-voltage to circumvent those issues. It's not digitally controlled but is a knob which is somewhat finicky to get an exact voltage that's why I am using these crazy figures because basically where it lands is where I measure, because once I start getting into an area where it is not finicky then the voltage is too high and I end up dealing with the current limitation of the power supply.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 02:39:02 am
content: Quote from: Kim Christensen on January 18, 2024, 01:47:35 amSo I followed your suggestion and my hypothesis on putting the more expensive meter to measure current.so here is what I get.On the course amperage setting. I get 22.1mV and .143A as a reading which would equate to .1545 Ohm.I then repeated everything utilizing the mV range on the meter. I get 2.8mv and a current of 18.395mA which would equate to .1522 Ohm.I don't understand why the meter is dropping the voltage reading by a factor of 10 simply by going to mV from V. The numbers in the end seem to come out awfully close together so it is seeming that the actual resistance might be 15 ohm. As someone previously said they could not tell the difference between brown and red (nor could I) and Brown would make it 15 ohm.If anybody is willing to explain some of these oddities (to me they are oddities) it would be greatly appreciated.Thank you to everyone that has contributed so far.

author: wraper
author_rank: Unknown
timestamp: January 18, 2024, 03:18:00 am
content: Unless OP and multiple people in comments are color blind, I fail to see how do you all get 0.25 ohm since first band is obviously brown, not red, and it should be 0.15 ohm.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 18, 2024, 03:22:38 am
content: I see your resistor as possibly being Brown Green Silver Gold which equals 0.15 ohm 5% which matches what your measuring.Your ammeter & leads in series has resistance. So when you swap out the ammeter with a different one, or change ranges on the one you have, there will be a different voltage drop across it. Thus the remaining voltage across the resistor will be different as will the current through it. But ohms law will always apply as you've seen.To minimize the differences when swapping ammeters, you can set your power supply open circuit voltage to apx 6V and rely on the fact that the power supply will limit the current to apx 0.5A (As long as the ammeter and resistor can handle 0.5A of current)

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:26:18 am
content: Quote from: wraper on January 18, 2024, 03:18:00 amI guess I am colorblind along with everybody else. lol

author: wraper
author_rank: Unknown
timestamp: January 18, 2024, 03:31:46 am
content: Quote from: DavidAlfa on January 17, 2024, 09:14:13 pmI would not trust 2 lead measurement, especially without ensuring reliable contact and that the way you attach resistor is the same as when you zero out lead resistance. Just connecting probes closer/further from the handle can affect the reading.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:36:43 am
content: Quote from: Kim Christensen on January 18, 2024, 03:22:38 amI think it is turning out to be .15 Ohm as well as a pretty effective test for colorblindness.I understand everything you are saying in the 1st paragraph. It's all basic Ohm's law stuff. And like I said I had a feeling the multimeter was influencing this discrepancy. I guess the part that I am having a hard time understanding is why when you switch from the course current setting to the mA setting why would there be a different voltage drop. I guess the explanation is the course and fine current settings differ internally by that difference. For some reason it's not clicking.I think I understand your 2nd suggestion. The idea is you would be creating a known fixed current to take the measurements off of. That is a pretty clever way to accomplish that by utilizing the power supplies current limiting capability ( although I did notice when it was happening in the beginning the current limiting would drift). Thanks for the knowledge. I think the multimeter and the resistor would fail at 40 A.

author: ArdWar
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:37:12 am
content: Quote from: JJ_023 on January 17, 2024, 02:29:13 amMeasuring low resistance with a "simple" two lead multimeter will not necessarily give accurate reading. The connector, cable and test point resistance will add a non-insignificant amount to the measured resistance.Quote from: JJ_023 on January 17, 2024, 02:29:13 amOh no...Just curious, what do you think will happen. What's your expected measurement values.Quote from: JJ_023 on January 18, 2024, 01:32:52 amYou're most likely approaching the measurement accuracy and resolution limit for your ammeter. Unless you got a good meter the 20A range is not expected to give accurate reading at low current (or any current at all in my experience)Quote from: wraper on January 18, 2024, 03:18:00 amDon't be so harsh lol. With the proliferation of cheap low CRI LED bulbs you can't ever sure what color in a photo are supposed to be in real life.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:39:04 am
content: Quote from: wraper on January 18, 2024, 03:31:46 amThanks for pointing that out, I was actually aware of the added resistance so I would always try to clip onto the resistor as close as possible to the body of the resistor. Using the body of the resistor as a stop would guarantee pretty decent repeatability.

author: ArdWar
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:42:56 am
content: Quote from: JJ_023 on January 18, 2024, 03:39:04 amNot really, the "lead" resistance here is mostly the metal-metal contact interface between your test pin and component lead. I don't think anyone can even get consistent resistance value since it's pretty much depends on how much you force them together, and what compound is actually contacting. A thin oxide layer for example will throw off your measurement.

author: wraper
author_rank: Unknown
timestamp: January 18, 2024, 03:45:44 am
content: Quote from: ArdWar on January 18, 2024, 03:37:12 amWell, even with monitor color reproduction variation, there is no way this would show red on a display. it's still one photo and at least OP should have seen if it does not match with real life. To get such a picture with red band you should have lighting with almost no red spectrum in it.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:49:57 am
content: Quote from: ArdWar on January 18, 2024, 03:37:12 amThank you for your contribution. I was aware that a 2 lead method of measuring resistance was not anywhere near ideal that is why I tried using Ohm's law to at least kind of guess where I am at on resistance.If there was no current limiting on the power supply (which actually saved my oversight) and the resistor was actually .25 Ohm the current would've been just over 48 A. I am aware of those implications.As far as the accuracy of the amp meter It's a $500 multimeter and supposedly has decent accuracy (+/- .10% +2). .01uA is its capability for resolution.I still don't quite understand the discrepancy between A readings and mA readings.I am glad I did make this post because it pointed out some oversights on my behalf as well as imparting some additional knowledge.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 03:54:18 am
content: Quote from: ArdWar on January 18, 2024, 03:42:56 amThanks for pointing that out. I tried to use micro clips because I figured it would be a lot more repeatable. I have a considerable amount of experience with metrology when it comes to measurements having to do with machining so I am always keeping in mind things like thin oxide Layers, repeatability, temperature drifts, etc. so I tried to carry some of that discipline over to EE. Even though I try sometimes my inexperience catches the best of me as illustrated in some of the examples in this post.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 18, 2024, 04:04:03 am
content: Quote from: JJ_023 on January 18, 2024, 03:36:43 amThat's exactly the case. Multimeters set to "current" are actually millivoltmeters reading the voltage drop across various shunt resistors internal to the meter. The lower current ranges have higher resistance shunts than the high current ranges.ie: When you change current ranges on your meter, you're inserting different value shunt resistors in series with the load.QuoteMost lab power supplies have adjustable current limiting. You would normally set this to a sane level such as 100mA or so.In the absence of a power supply with built in current limiting, you could put a resistor in series with the output to limit the short circuit current to a sane level.ie: If you had a known 100 ohm 2W resistor in the junkbox, you could put that in series and set the supply to 10V. Then the current would never go above 100mA.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 18, 2024, 04:20:45 am
content: Quote from: Kim Christensen on January 18, 2024, 04:04:03 amSo the way you explain the current settings makes sense to me now. I was aware of how multimeters function when measuring current. My understanding is through a shunt resistor which is basically a piece of thick wire of a known resistance (looks like a giant staple on the inside of a multimeter).My problem is I am accumulating knowledge about these topics but I still lack the experience to string that knowledge together to get a higher understanding of the subject matter. So it's like I am aware of and know all of the pieces individually but sometimes have a hard time fitting those pieces together. So your explanation ( as well as others on this forum) help my understanding of the subject matter and how all of these things come together.I appreciate your suggestion on the 100 ohm resistor. My gut feeling tells me it probably would not be as accurate as the 4 wire method that I implemented previously that was suggested to me by somebody on this post. Is my assumption correct?Again thank you for helping out.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 18, 2024, 04:45:22 am
content: Quote from: JJ_023 on January 18, 2024, 04:20:45 amFor the high current range, it's sometimes just a piece of wire. But for the lower ranges they'll be precision resistors.Set one of your meters to measure resistance. Set the other one to measure current on the mA range. Connect the red lead of one meter to the red of the other. Do the same for black. Now change current ranges and watch the ohms reading on the other meter change.QuoteThe 100 ohm resistor (For example) will not effect the accuracy of the measurement but does limit the resolution. It simply sets the max current through the entire circuit. If you changed it to 220 ohms, you'd still get the same result after working out ohms law on the DUT:

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 18, 2024, 05:35:21 am
content: Quote from: JJ_023 on January 17, 2024, 02:29:13 am1. It looks like a resistor. It is identified by you as 0.25 R resistor. It is measured by you as 0.25 R resistor.2. If something looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 19, 2024, 01:35:36 am
content: Quote from: Kim Christensen on January 18, 2024, 04:45:22 amThanks for the explanation as well as the experiment suggestion.Here it is what I got doing the range experiment.On the A setting I could barely read any resistance it was flickering between .1 and 0 Ohms.On the mA setting it showed 2 Ohms.On the uA setting it showed 101.2 Ohms.Now moving onto the resistor experiment again. I set up the experiment and started cranking the voltage until I was slightly above 100mA.Here are the results.15.5 mv (using a cheaper meter)101.15 mA ( using the best meter)This calculates to the resistor value of .15324 OhmsThis is in line with the previous experiment, as well as confirming my colorblindness.Thank you so much for walking me through this. Like I said I knew all of this knowledge previously in terms of the different bits and pieces I just never thought of it at the time to put it together.My follow-up question is how accurate do you think that reading is? If you had to guess a percentage.The meter is capable of .1 % accuracy on DC current. The meter that was used for the voltage is accurate to .8%My guess would be around 1%.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 19, 2024, 01:37:04 am
content: Quote from: Vovk_Z on January 18, 2024, 05:35:21 amIn this particular case the duck turned out to be a duckling that measured .15 Ohm thus confirming my and everyone else's colorblindness.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 19, 2024, 03:14:44 am
content: Quote from: JJ_023 on January 19, 2024, 01:35:36 amThat sounds about right... Apx 0.9% if I calculated that correctly. (Ignoring the tiny voltmeter circuit loading error)

author: wraper
author_rank: Unknown
timestamp: January 19, 2024, 06:02:55 pm
content: Quote from: JJ_023 on January 19, 2024, 01:35:36 amYou need to look into full spec for the range you are using. Usually it comes as percentage + number of counts of least significant digit. So if you're measuring at the bottom of the range, it can be way worse than percentage spec. Also you need to consider resistor temperature (temperature coefficient of resistance), as your measurement can be right for the conditions but not exactly what you think it is.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 19, 2024, 09:29:49 pm
content: Quote from: wraper on January 19, 2024, 06:02:55 pmThat is a valid point. The good multimeter is .10% +2 on current. The so-so multimeter is .8% +/-1.I am well aware of temperature drift associated with resistors. I do appreciate your statement and it would be valid for a particular instance where you are designing a circuit that would operate over different temperature ranges. I am wondering how that statement is valid when I am trying to ascertain the value of it at a particular moment in time. Wouldn't temperature already be considered in that equation?Thank you for your input and time.

author: wraper
author_rank: Unknown
timestamp: January 20, 2024, 02:21:19 am
content: Well, you measured the current and voltage drop across the resistor. Say you should be within 0.2% according to specs of your meters. The problem is that resistor could be at 120oF at the time, for example. So while you measured with 0.2% tolerance, it's only true at this certain resistor temperature and can be quite a bit more off from resistance at room temperature. In this particular case, ~1.5mW of power dissipation won't meaningfully rise the temperature but you need to be aware of this in general. The same goes for meter specs, usually tolerance spec is given for a quite narrow temperature range. And that's considered it had been timely calibrated.Although in general you get significantly more accurate measurements than the meter specs but you cannot be 100% sure without doing all proper procedures.

author: JJ_023
author_rank: Frequent Contributor
timestamp: January 20, 2024, 03:42:31 am
content: Quote from: wraper on January 20, 2024, 02:21:19 amThank you for such an elaborate explanation.One of the things I do enjoy is reading data sheets, especially ones from Texas Instruments and Analog. Those always have temperature charts indicating what you are talking about in a graphic. I also know that whatever the rating is typically you try to discount 1/3 from the value. So if a resistor might dissipate 3 W I would try to keep it around 2 W or less at whatever temperature that corresponds to.I was wondering where you got the .2% from the specification of the meter. I've never once heard or read such a statement so it is a little bit confusing.

author: wraper
author_rank: Unknown
timestamp: January 20, 2024, 03:49:05 am
content: I wrote 0.2% just as example, it's not about this particular case, the same as temperature having nothing to do with your measurement.