I'm using a component tester to measure a resistor which I have removed from a circuit. The tester gives a reading of around 0.46 - 0.50 Ω. I have 7 other identical resistors (in circuit) and they all give the same readings.

The colour bands look to be red/red/silver/gold which would be 0.22 Ω. Allowing for discolouration, I still can't figure out how 0.47 Ω yellow/violet/silver/gold in anyway relates to the resistor that I have.

I also checked with a multimeter on 200 Ω setting, probes together give a reading of 1.2 Ω, on the resistor 1.6 Ω, so I deducted the 1.2 which gives approx 0.4 Ω - so consistent(ish) with the component reader I guess.

What or where am I going wrong?

The meter is a "T7" - it is cheap and cheerful and widely available on most Chinese marketplaces.

enter image description here

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    \$\begingroup\$ Are you certain these are resistors, and not inductors? \$\endgroup\$ Commented May 15 at 19:12
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    \$\begingroup\$ For a 2-terminal measurement, 0.22 ohms is identical to 0.47 ohms. You need to do a 4-terminal measurement at that low level to get results you can believe. \$\endgroup\$
    – Neil_UK
    Commented May 15 at 19:53
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    \$\begingroup\$ What reading do you get when you just short the meter's leads together? \$\endgroup\$
    – td127
    Commented May 15 at 22:03
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    \$\begingroup\$ I by-passed the probe leads and plugged the component directly into my tester - now I'm getting a reading of 0.33ohms - so I guess there is 0.15 ohms of resistance being added. I'm not sure i can do a 4-terminal test with this unit? \$\endgroup\$
    – Alan A
    Commented May 15 at 22:03
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    \$\begingroup\$ If you can solder 5 in series and then measure you will reduce your measurement error after you divide the total by 5. I also would investigate the possibility they are inductors as the body colour is uncommon for a resistor. \$\endgroup\$
    – KalleMP
    Commented May 16 at 15:56

5 Answers 5


With your meter set to the 200Ω scale, you are at the lowest 1% of the range, so you cannot expect perfect linearity. Besides, for such low-value resistors, even things like mild corrosion on the leads will affect your reading. To accurately measure such small values, you need a meter with a comparable scale, and one that can do four-wire resistance measurements.

I would say that getting a reading of half an ohm on a (nominally) .22Ω resistor is already better accuracy than we can expect.


I share @evildemonic's suspicion in the comments: this is most likely an inductor you are measuring. Not a resistor.

See also this post: How to visually tell the difference between axial inductors and common resistors?

It's not always the case, but often these bright green packages such as the one in your picture are inductors. The component tester is making a mistake, and you are following up on that mistake by measuring resistance with your multimeter, instead of measuring inductance.

Red-Red-Silver-Gold would be an inductor of 0.22 µH with a tolerance of ±5%.

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    \$\begingroup\$ And if those bands are orange, it is a 0.33 µH. The seafoam green color instantly makes me suspicious it is an inductor. I have seen power resistors (for current sensing) this color too, though. I wonder what the designation on the PCB is? L = inductor, R = resistor. \$\endgroup\$ Commented May 16 at 15:34
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    \$\begingroup\$ Apart from the green colour, another telltale sign of a likely inductor is the hourglass shape. Resistors are typically quite cylindrical, often with clearly visible end caps, while inductors often have this smooth transition between the wider ends and the narrower middle. \$\endgroup\$
    – TooTea
    Commented May 17 at 10:47
  • \$\begingroup\$ @TooTea Now that is interesting! I did not know that! A quick search on Google images appears to confirm what you say, but do you maybe have some kind of source on the shape thingy? \$\endgroup\$
    – Opifex
    Commented May 17 at 14:45
  • \$\begingroup\$ Try using a magnet. If it's an inductor, the core will be attracted to the magnet \$\endgroup\$
    – StevenH
    Commented May 17 at 15:52
  • \$\begingroup\$ @Opifex I'm afraid I don't have any authoritative source for that, just personal experience (which is of course limited to my corner of the EU). So take it with a grain of salt. \$\endgroup\$
    – TooTea
    Commented May 17 at 16:04

The resistors are 5% tolerance, and the meter probably has 10x as much error close to the 0Ω point. These meters are not too great - their firmware has mistakes and wrong assumptions. The circuit they use is OK, they just didn't do the rest of their metrology right.

To measure such low resistances reasonably well, you'll need either a multimeter with a 20Ω range, or a home-made milliohmmeter circuit consisting of a current source and a DC amplifier. It doesn't need to be expensive or complex. Let's say we use 1% resistors:


simulate this circuit – Schematic created using CircuitLab

The readout error is a couple percent without any gain adjustments, just using 1% resistors from a reliable source (DigiKey, not AliExpress).

To null the meter, press the ZERO button, and adjust RV1 until the voltmeter reads 0V. Then release the ZERO button. The adjustment can remove up to ±5mV of input offset voltage of OA2. To double the adjustment range - should it be necessary - remove R5, and change R7 to 200kΩ.

The voltmeter can be any multimeter. The whole thing costs a couple US dollars total in parts. It's not a precision meter - more of an indicator - but it still beats the pants off those cheap testers.

The 5V source should be stable - say a bench power supply. The supply voltage directly affects the gain of the circuit - the scaling factor from ohms to volts. Adjusting the voltage up increases gain, adjusting it down decreases gain.

4-wire sensing is not used because this is just an indicator, and the sense current is fairly low - just 10mA.

  • \$\begingroup\$ Is my analysis of this circuit correct? I gather OA1/Q1 is just a current source, but closer to 13 mA than 10 mA. For a rough, back-of-the-hand calculation, I see RV1+R7 in parallel to R6 and so can be ignored since it's >>R6. Similar for R4||R5 and R6. The input impedance of stage 2 is also >100k so the voltage at V- is basically (R1+R2)/(R1+R2+R6)*(5-0V) or 3.69V. That means that the current through R3 would be about 13.12 mA. Is this a reasonable analysis? \$\endgroup\$
    – penguin359
    Commented May 22 at 0:23

Typical clip leads from budget/low-margin sellers are on the order of 0.1-0.3 ohms. Maybe even more. They may use CCA (copper clad aluminum) or even steel wires, or just use very much less metal than you'd expect in a wire of given size.

To wit, I have a DMM here with brand name leads plugged in, and I'm fortunate to have it read "000.0" ohms when shorted. So, ±0.1Ω give or take. With a clip lead inserted, I read 0.1Ω. With an alligator lead, 0.2-0.3Ω -- the latter are thicker, and have larger connectors, so you would expect them to handle more current, but you would be easily fooled if you didn't check!

The problem is compounded with unknown instrumentation: they may cheat where they display 0.0 when it's not actually (the internal reading might be + (or -!) 0.something), or when near other calibration points (like 100, 1000, etc.). (Dumb? Scary, even? You bet! But cheap weight scales have been caught doing this, rounding calibration weights towards the nearest exact reading. It wouldn't surprise me if an electronic meter might do the same.) The scales might not be linear, the calibration might vary with range/setting, etc. Cute, multi-function instruments like pictured, can be good to get a quick idea what something may be, but beware they can make incorrect determinations (e.g. resistor or inductor ambiguity; semiconductor identification ambiguity or error, especially if a component is faulty; etc.), and it should almost always be used as a hint, say for which instrument to reach for next and get a more accurate and representative measurement with.

That said: that it's measuring a very plausible figure, given the setup (0.22Ω resistor plus ∼0.1Ω leads), is at least very encouraging. Whether it really has 10mΩ resolution, or precision, I guess who knows, but 100mΩ at least seems plausible.

To really have confidence in it, you would have to calibrate at many more points, or cross-check with a better-calibrated meter (transfer calibration). Still more tests would be desirable, including a wide range of component types and values, to get a feel for how it decides what a component is or isn't (and what parameters affect that, e.g. test frequency, voltage or current, if selectable at all). There may, in fact, be an accurate and representative instrument in there; just knowing what's accurate about it -- and how to use it accurately -- is where work is required.

For a basic example, short the leads -- if it reads 0.0Ω, that's probably a bad sign (even good clip leads with connectors free of oxidation, should read some 10s of mΩ). If it increases smoothly as you add more resistance (including in similarly small increments), then it probably has the resolution, the accuracy -- if maybe not the precision. If the zero reading is reliable, then you can simply subtract it out of subsequent readings by eye.


As others have said, multimeter linearity is often poor at low resistances.

If you have a lab power supply with adjustable current limit, you can do this instead:


simulate this circuit – Schematic created using CircuitLab

The multimeter should be set to voltage mode and connected directly to the resistor leads. This way resistance of the connecting wires does not affect measurement accuracy.

Setting the power supply current to 1 A is convenient, as you can directly read millivolts from multimeter as milliohms of resistance. For larger resistances you'll want to use lower current setting to avoid heating the resistor too much.

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    \$\begingroup\$ I suppose this method may not be so good with fusible resistors, considering the not-so-usual resistor body colour of the component in the question. \$\endgroup\$ Commented May 16 at 18:55
  • \$\begingroup\$ For extra accuracy, use two multimeters to monitor voltage and current separately, as the ammeter on most lab PSUs are not designed for doing accurate measurements. \$\endgroup\$ Commented May 18 at 15:05

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