There are a variety of ways of making the actual measurement.

One of the more traditional is a moving-coil micro-ammeter, consisting of a winding on an iron core that rotates between the poles of a permanent magnet, working against the force of a spring. The meter circuitry and selector connects various resistors to scale the input voltage to a tiny current. A crude, related instrument can be made by winding a few turns of wire across a plastic-body magnetic compass, with the introduced field forming a vector sum with the earth's field and resulting in a new pointer angle.

The next major development consisted of using a high impedance vacuum tube amplifier between the circuit under test and the meter movement, producing the Vacum Tube Volt Meter or VTVM. Later the tube was replaced with a field effect transistor.

The third major development would be replacing the meter movement with an analog to digital converter. This usually consists of a comparator which compares the unknown input against known reference voltages; either sequentially as the reference voltage is changed by a digital to analog converter under a search algorithm, or by charging a capacitor while measuring time, or for high speed applications bu using a number of of comparators and voltage sources acting in parallel to produce a faster answer (though that form is more likely to be found in a high speed instrument such as an oscilloscope than a typical digital multi meter).

- 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.
- 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:

- range
- open clamp voltage
- measurement current
- 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.

## Best Answer

Apply a known voltage over a series resistor. This resistor in combination with the internal resistance will form a voltage divider. Say you apply 5V over a 1M series resistor, and the DMM shows it as 2.5V, then the internal resistance is 1M.

editNow that I reread it, I guess it's not completely unambiguous. By "applying a voltage over a series resistor" I meant you connect the + to the resistor and the - to the ref. input of the DMM.