I noticed this behaviour on two different multimeters (different models and brands as well). At first I did not put a multimeter in the way to measure how much voltage changed for the different scales of the meter: I realized that by using my own tongue (hell, yeah). For both the multimeters I own, I could definitely feel that the tingle was getting stronger when the scale was smaller.

So: I tried to measure voltage applied to the probes of one multimeter at various resistance scale reading levels, by using the second multimeter one to read Volts. I am quite impressed by the results.

Here's what I read. On the left side is the "measured" multimeter scale setting, on the right the voltage I read:

  • 200Ω -> 2,96V
  • 2kΩ -> 2,95V
  • 20kΩ -> 2,93V
  • 200kΩ -> 2,69V
  • 2MΩ -> 1,48V (what a drop!)

If I switch the meters things are even more confusing to me:

  • 200Ω -> 2,71V
  • 2kΩ -> 2,69V
  • 20kΩ -> 0,35V (!!)
  • 200kΩ -> 0,32V
  • 2MΩ -> 0,18V

Can please anybody make clarity about why does this happens? I would expect that a higher voltage should be applied to measure bigger resistance. Just before hitting "Post" I chose to measure current as well - for different ohmmeter scales level. Guess what: those definitely dropped as well, but not with the same ratio as voltage. I am confused as heck. Thanks!

  • 2
    \$\begingroup\$ Please stop using your tongue to measure voltage or you will end up like this lizard: chat.stackexchange.com/transcript/message/11118485#11118485 \$\endgroup\$
    – jippie
    Sep 7, 2013 at 11:56
  • \$\begingroup\$ Attach a known resistor (in range) to the meter and see how the voltage changes with them. \$\endgroup\$
    – jippie
    Sep 7, 2013 at 12:03
  • 1
    \$\begingroup\$ @jippie thanks dude :) but being aware that my meters are powered by a 9V battery I was very aware that nothing bad could happen. \$\endgroup\$
    – Dakatine
    Sep 7, 2013 at 12:17
  • \$\begingroup\$ I think I solved the top example in my edited answer. \$\endgroup\$
    – jippie
    Sep 7, 2013 at 14:41
  • \$\begingroup\$ Plz don't use your body as multimeter, people actually died because of that (darwinawards.com/darwin/darwin1999-50.html), this was also only 9V... Alternatively you could use a LED, with or without resistor. Anyhow still better burning the LED than yourself... \$\endgroup\$
    – magu_
    Nov 28, 2014 at 7:23

2 Answers 2

  1. I think the voltage drop in your top example is caused by the voltmeter's input impedance (probably around 10M) that slowly gets into range of the ohm-meter.
  2. For range 20k and up it is again the voltmeter's input impedance issue. I think the 200Ω range is related to the diode measurement which requires a similar current source at a relatively high voltage. That leaves the 2kΩ range which is probably implemented in a cost effective way based on the current source for the 200Ω range.

Only with the circuit diagram the answer can be 100% sure.

Your multimeter will attempt to measure ohms by sending a known/set current through the attached resistor. This set current varies with the range your meter is in. However your multimeter has no ideal current source on board, but rather attempts to implement a current source from your battery voltage and a couple semiconductors, hence the open clamp voltage will never rise beyond the battery voltage.

Unsure why the voltage drops so much for the higher ranges, this will have to do with the way the current source is built. Notice that the 'high' voltage is not useful (forth column below) when you realize that the product of range times measurement current is much lower than the open clamp voltage (second column).

Also notice that the voltage measured in the lowest resistance range is identical to the voltage used for diode measurements for all three meters. For diode measurement you want a relatively high voltage to test the relatively high voltage drop across a diode. In that case you still use a constant current, but you are no longer interested in the resistance rather than the actual measured voltage. Useless to build two separate current sources for more or less the same current. On the other hand it is easier to build an accurate current source if you allow yourself a higher voltage drop across the current source and you don't need the voltage anyway (forth column).

Below are the results for my meters. For two out of three the input impedance of the voltmeter (10MΩ) was lower than the ohm-meter's range, so I skipped that value. The columns are as follows:

  1. range
  2. open clamp voltage
  3. measurement current
  4. maximum voltage required for measurement (range × current), notice how that voltage is reasonably constant!

DVM2000 (6V battery) \begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 3.25\text{V} &\Rightarrow& 785\text{µA}\\ 500Ω &\Rightarrow& 3.25\text{V} &\Rightarrow& 785\text{µA} &\Rightarrow& 500Ω × 785\text{µA} = 400\text{mV}\\ 5\text{kΩ} &\Rightarrow& 1.19\text{V} &\Rightarrow& 91.5\text{µA} &\Rightarrow& 5\text{kΩ} × 91.5\text{µA} = 460\text{mV}\\ 50\text{kΩ} &\Rightarrow& 1.18\text{V} ^{*)} &\Rightarrow& 11.5\text{µA} &\Rightarrow& 50\text{kΩ} × 11.5\text{µA} = 575\text{mV}\\ 500\text{kΩ} &\Rightarrow& 1.09\text{V} ^{*)} &\Rightarrow& 1.1\text{µA} &\Rightarrow& 500\text{kΩ} × 1.1\text{µA} = 550\text{mV}\\ 5\text{MΩ} &\Rightarrow& 614\text{mV} ^{*)} &\Rightarrow& 0.1\text{µA} \text{(last digit)}\\ 50\text{MΩ} &\Rightarrow& ? ^{*)} &\Rightarrow& ?\\ \end{array}

*) The open clamp voltage for ranges > 5kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 1.20V.

SBC811 (3V battery)

\begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 1.36\text{V} &\Rightarrow& 517\text{µA}\\ 200Ω &\Rightarrow& 1.36\text{V} &\Rightarrow& 517\text{µA} &\Rightarrow& 200Ω × 517\text{µA} = 103\text{mV}\\ 2\text{kΩ} &\Rightarrow& 645\text{mV} &\Rightarrow& 85.4\text{µA} &\Rightarrow& 2\text{kΩ} × 85.4\text{µA} = 171\text{mV}\\ 20\text{kΩ} &\Rightarrow& 645\text{mV} &\Rightarrow& 21.7\text{µA} &\Rightarrow& 20\text{kΩ} × 21.7\text{µA} = 434\text{mV}\\ 200\text{kΩ} &\Rightarrow& 637\text{mV} ^{*)} &\Rightarrow& 3.71\text{µA} &\Rightarrow& 200\text{kΩ} × 3.71\text{µA} = 742\text{mV}\\ 2\text{MΩ} &\Rightarrow& 563\text{mV} ^{*)}&\Rightarrow& 0.44\text{µA} &\Rightarrow& 2\text{MΩ} × 0.44\text{µA} = 880\text{mV}\\ 20\text{MΩ} &\Rightarrow& ? ^{*)} &\Rightarrow& 0.09\text{µA} \text{(last digit)}\\ \end{array}

*) The open clamp voltage for ranges > 2kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 645mV.

DT-830B (9V battery)

\begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 2.63\text{V} &\Rightarrow& 1123\text{µA} \\ 200Ω &\Rightarrow& 2.63\text{V} &\Rightarrow& 1123\text{µA} &\Rightarrow& 200Ω × 1123\text{µA} = 224\text{mV}\\ 2\text{kΩ} &\Rightarrow& 299\text{mV} &\Rightarrow& 70\text{µA} &\Rightarrow& 2\text{kΩ} × 70\text{µA} = 140\text{mV}\\ 20\text{kΩ} &\Rightarrow& 299\text{mV} &\Rightarrow& 23.0\text{µA} &\Rightarrow& 20\text{kΩ} × 23.0\text{µA} = 460\text{mV}\\ 200\text{kΩ} &\Rightarrow& 297\text{mV} ^{*)} &\Rightarrow& 2.95\text{µA} &\Rightarrow& 200\text{kΩ} × 2.95\text{µA} = 590\text{mV}\\ 2\text{MΩ} &\Rightarrow& 275\text{mV} ^{*)} &\Rightarrow& 0.35\text{µA} \text{(near scale low end)} &\Rightarrow& 2\text{MΩ} × 0.35\text{µA} = 700\text{mV}\\ \end{array}

*) The open clamp voltage for ranges > 20kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 300mV.

  • \$\begingroup\$ Thanks for your explaination, this is informative overall, but I am still clueless about why is the voltage dropping. Can you experience the same with yours? \$\endgroup\$
    – Dakatine
    Sep 7, 2013 at 12:20
  • \$\begingroup\$ I added some more detail and I think the diode measurement notice is interesting. \$\endgroup\$
    – jippie
    Sep 7, 2013 at 13:08
  • \$\begingroup\$ Thank you for testing with your meters, @jippie. I am getting closer to understand. Some thoughts: * Voltage is dropping for you as well - and there are some big "jumps" between some of the ranges while the drop is small across some other. Still, it's always going down or equal, never up. * Actually you latest column is "reasonably constant" only for your first meter. I can see big variations for the others, especially the second. * Most important: I can't understand that latest column. "maximum voltage required for measurement". Why 224mV is the max to measure 200 ohm, and 130mV for 2kohm? \$\endgroup\$
    – Dakatine
    Sep 7, 2013 at 13:13
  • \$\begingroup\$ Because the current used for the measurement is constant. \$\endgroup\$
    – jippie
    Sep 7, 2013 at 13:13
  • \$\begingroup\$ I think the best partial explanation for your 'problem' is in italics. \$\endgroup\$
    – jippie
    Sep 7, 2013 at 13:15

A nice "linear" way to measure resistance is to feed a known amount of current through the resistor and measure the voltage. Since the voltage will be proportional to the resistance, a meter whose reading is proportional to voltage will thus read a value proportional to the resistance.

Because resistors vary over many orders of magnitude, there's no single amount of current which will work optimally for measuring all resistances. A current of one microamp would cause a 1M resistor to drop a volt, but would cause a one ohm resistor to only drop a microvolt. A meter with a single current source that was limited to 2 volts and whose voltage reading on the finest range was only accurate to a microvolt would be unable to measure any resistances larger than 2 megs, and could only measure small resistances accurate to the nearest ohm. If instead of using a single 1uA current source a meter were to use a 0.1uA current source and a 100uA current source, then the smaller current source would be able to measure resistors up to 20 megs, while the larger one would allow a meter that could read voltages accurate to a microvolt to resolve resistances to the nearest 0.01 ohms.


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