How To Use A Multi-meter To Measure Milli-ohm Resistors?

[@pault6533]

5V is enough headroom for 15 ohms DUT. I built this LM317t supplying 100 mA for milliohm measurements too. My source is a dissembled powerbank which provides 5V. I did not use an input diode, so no drop. I did an increasing resistance test and found the regulation dropout first occurs when the device under test is around 20 ohms when source is 5.000V. This can be de-rated to about 15 ohms to account for varying circuit components and test conditions. I see some people suggest increasing the input voltage. The additional voltage would create heat, leading to drift from the 100 mA value as components increase temperature. Once above 15 ohms, a good DMM with two low resistance wires will start to provide more accuracy than a home-made 4-wire setup.

[@diymade2175]

Thank you for the interested and like the way you did! The best way to understand the device's principle, master it, experience yourself and may be to do some improvement as well. I'm glad you did in so much details. Thanks again for your feedback! 👍👍👍

[@allinclusive5762]

lousy sound but good content

[@karlmartell9279]

He seem to live under a highway bridge and loves loud music.

[@diymade2175]

😎

[@DeeegerD]

You should lose the background NOISE.

[@ebasoalto]

Too many errors in this video. Resistor should be 15 ohms instead of 22 ohms in parallel with 100 ohm potentiometer instead of 200 ohms, to obtain a combined resistance of 12,5 ohms, for getting constant current of 100 mA, since reference voltage of LM317 is 1.25 volts. Calibration should be done using the 200mA scale in the meter instead of 10 Amp scale, to have more accurate calibration. The input voltage meter is unuseful. And the worst mistake, is when reading the millivolts drop in the tested resistor, the multimeter leads should be touching the resistor terminals and not the crocodile clips from the current source to avoid introducing the error of joint voltage drop. Use Kelvin 4-wire measuring scheme. And by the way, you can reduce background music volume.

[@diymade2175]

Thank you for the interested and agree with some of your points however I've to point out: It is nothing wrong about 22 ohms in parallel with 200 ohms potentiometer (its ohm value is adjustable), their combination value is between 0 ohm and 19.8 ohms which indeed, includes resistance of 12.5 ohms. The device was the best configuration by considering given the components involved. I recommended highly you'd make your own device like in the video showed and experience yourself. Thanks again for your effort!

[@ebasoalto]

I have done many constant current supplies, and those with LM317 has no secrets for me. Paralleling 22 ohms resistor with 200 ohms potentiometer will force to flow a little more than half of the 100mA total current thru the potentiometer and it will be in the lower part of his resistance, with a high power disipation and poor precision in the adjustment. The ideal is to have a potentiometer that can be adjusted about his midpoint.@@diymade2175

[@harrydelange1821]

Kelvin testclamps (4 wire) are rather cheap ~ 20 € or $ or and very usefull. And combine it with a ICL7107 based mV meter (there accurate) and you have a stand alone mOhm meter. And you will be able to set the decimal point at the right place for easy readings.

[@diymade2175]

@@harrydelange1821 Good on you Mate, that is the one of ways to get the result(s) you want. As an example: There are many multi meters on the market now days, you can choose the one which is the best suitable for yourself. Thanks for your interested! 👍

[@harrydelange1821]

@diymade2175 Thanks, personally I am charmed for a standalone solution but I realize that not everyone has the skills or interest to build a more complicated project and choose for building an extention using a multimeter (witch works fine of course).

[@chrisbrett1857]

I am completely puzzled by your choice of a 22 Ohm resistor when your target constant current from the LM317 is 100 mAmps. The nominal reference voltage of an LM317 is 1.25V ; so, to get 100 mAmps you should need a resistor of nominal value 12.5 Ohms. Even with your 200 Ohm callibrating potentiomer set at its maximum, in parallel with 22 Ohms this would only take the effective resistance down to just under 20 Ohms - which should result in a constant current of around 1.25V/20 Ohms = 0.0625 Amps, i.e. about 62 milliAmps. Even if your main resistor were 10% lower than its nominal value (i.e. actually only 20 Ohms), so giving (combined with the pot at its maximum) an effective resistance of 18 Ohms, and even if the reference voltage of the LM317 were actually 1.3V (the maximum possible, according to the datasheet), the current would still be nowehere near your target 100 mAmps. I can only assume your multimeter is wildly inaccurate when measuring in the tens of milliAmps range, and you chose 22 Ohms emprically to make the 'callibration' work. I notice when you show the calibration process, your multimeter appears to be set on the 10 Amp range (??) - if so it is hardly likely to measure milliAmps very accurately.

[@diymade2175]

Good on your effort and I agree with the most comments you made however I have to point out the one: "Even with your 200 Ohm calibrating potentiomer set at its maximum, in parallel with 22 Ohms this would only take the effective resistance down to just under 20 Ohms" ---- This was incorrect statement! In fact, their combination value of resistance IS BETWEEN 0 AND 19.8 OHMS (it is adjustable) which includes a resistance's value of " nominal value 12.5 Ohms" as you indicated. Thanks for your interested!

[@chrisbrett1857]

@@diymade2175 Apologies - my stutupid mistake - you are absolutely right !! Adjusting the pot to 29 Ohms will indeed give you your 100 millAmps. Puzzle resolved. Thanks for taking the time to correct my error.

[@adeyemibabatunde1814]

Great video. The shematic diagram is blurred and DMM of 5 digits wound be better. The supply voltage can be higher. Kindly post a better schematic diagram. Can you include an ammeter to confirm constant current value?

[@diymade2175]

Thank You for the info ! The device's working principle is base on a circuit for the constant current of LM 317. In the video at 3:03, the device was calibrated for 0.1 ampere (100 mA) so no need to have an ammeter connected. For combination reason, the best supply voltage range is 5V to 7V (I leave to you for the reasons). Thanks again.

[@nerddub]

The supply voltage is relevant here in terms of power dissipated by the voltage drop of higher supplies. Eg: you put 18v in and you need to do something with the wasted voltage, which gets lost as heat through the LM317

[@KB1UIF]

What is the point of the music and ocean sounds when it masks your voice? It started at a reasonable level but gets louder and louder until its louder than your voice !! Your video would be fine with just a voice over. Calibration should be done on the 200mA meter range and the measured voltage should me on the 200mV range. Otherwise its got good information and a simple design to implement.

[@diymade2175]

Good points. Thanks for interested!

[@M0GLU]

Why needs the digital panel meter? Connect the fix and stable 5Volt and that's it. Why measured 100mA current in 10Amp. level?

[@diymade2175]

Well, the digital panel meter is not a "must" component in the circuit for the principle operation however it is handy and convenience during testing operation if a DPM or Led installed in the circuit. Thanks.

[@jorgeescalantegonzalez1178]

Congratulations, it seems to be a good circuit. I have a doubt: what is the use for the display?

[@diymade2175]

Thank you for the interested! The display? You mean the mini voltage meter in the circuit: it is not a MUST component in the circuit, its function are showing the device has the power on & indicating the power level, just for the convenience in the practical experience. Thanks.

[@Manticore1960]

Why not use the 200mV range on meter? Using the lowest range possible will give more accurate meter reading.

[@diymade2175]

You could if you wish.

[@marcelomigueldasilva7444]

🎉

[@diymade2175]

[@naf4494]

Nice sir, how much maximum input source voltage we can use? Like 9v common battery

[@diymade2175]

You can use 9v but the best range is DC 5V to 7V. Thanks.

[@mmghv]

Why not just use the 3-wire digital voltage display to measure the output voltage drop so you don't need the multimeter? you would have 4 wires, 2 for the constant current and 2 for the voltage measurement using the digital display

[@diymade2175]

Good on you Mate! About digital voltage display: Well, that'll be the one of different ways to get the same result. The reason that Multi-meter used is, I believe, the most people in general are well acknowledged the Multi-meter than the digital voltage display so far, of course this might be changed in the future. Thanks.for your comment.

[@rkalle66]

This device is measuring in a 4-wire configuration like Kelvin clamps. Multimeter and constant current pcb are not grounded together but at the measure point eliminating wire resistance.

[@TeenaKalia-p4j]

why we have multiplied with 10 in the formula.......

[@diymade2175]

The device's working principle is based on the one of circuits from Voltage Regulator of LM 317 functions which is the constant output current. In the video, the device for output is calibrated for 100 mA (0.1 A), which is considered for the best configuration given the components involved for the device. As an example: You have 1 ohm resistor to be tested, according to ohm law - output voltage cross the resistor = 0.1 A x 1 ohm = 0.1 V which is also a multi meter's tested result. we use only figure of 0.1 for testing purpose and omit V. The 0.1 needs times 10 to get result of 1, this is "why we have multiplied with 10 in the formula". On the other hand, you might have asked: why not the device calibrated to 1 A instead of 0.1 A? Well, it wouldn't be practical to have it done simply & efficiently. Thanks for your interested!

[@HYTMUAE]

Sadly i couldn't find a 200-ohm potentiometer to do this project

[@diymade2175]

Thanks for your interested. In the most case, you are able to find a way or ways to solve the problems. One example is here: You can use a 1 K ohms potentiometer and a 250 ohms resistor in parallel, the total result is 200 ohms and you also can try the different ways to get the same result. Good luck!

[@HYTMUAE]

@@diymade2175 Thanks for the replay ... I'm trying because I like this project

[@rubmel98]

I completly apreciate this video and how has explained. But, I'm don't understand at all how it really works. Thanks a lot buddy!

[@diymade2175]

Okay, you might need to refresh what you had learnt otherwise no worries! As an example for the people, they know how to use a multi meter but the most of them don't understand how it works. Thank you very much for your interested!

[@rubmel98]

@@diymade2175 I've watched other similar videos, and backing to this, I finally understood. Really thanks. I don't see the time to build my own! It's really useful to locate electric shorts in electronic repairs

[@rubmel98]

@@diymade2175 Yes, I'm an example of that people, I know to use multi-meter, but don't know how it works en digital, but in analog, I can understand how it is

[@DeeegerD]

It uses Ohm's law. The 317 is set to a fixed current. The voltage is measured. Knowing those two the resistance can be calculated.