# Why does my multimeter show a wrong voltage over a large resistor? I am having a hard time answering one particular question about our experiment.. In our experiment, R1 and R2 were set to 1Meg each and later to 10k... I understand the need for R1 and R2 a bit. Without R1 and R2, the voltage sharing wouldn't exactly be 50-50 for both D1 and D2 because no two diodes are completely identical. D1 and D2 will both have the same leakage currents (without R1 and R2) since they are just in series. However, they probably will have non-identical IV curves, so this particular leakage current will result to V@D1 /= V@D2.

The question I am having a hard time is that, why is V@R1 + V@R2 /= 10v when R1 = R2 = 1Meg?... One the other hand, those two voltages add up (to 10v) when R1 = R2 = 10k... I included the 60 ohm source resistance in my diagram for completeness. However, as I can see, both D1 and D2 are reversed biased and thus, they offer a very large (reverse resistance) which should be much greater than the 60 ohms. Even with the parallel combination of 1Meg and D1 reverse resistance, it should still be much greater than the 60 ohms. I tried thinking of an answer in terms of the RD1reverse//R1 = Req1 and RD2reverse//R2 = Req2. Req1 + Req2 (series) should still be much more than 60ohms and I thought that the 10v should still show up at the node of D1 cathode. Yet in our experiment, V@R1 + V@R1 < 10v.

Can anyone point me out if I am thinking this in a wrong way? Some tips/first step hint would really be appreciated

Edit: question answered thanks to @CL. Assuming D1 and D2 are open during reverse bias for simplicity and noting that Rmultimeter = 10Meg, V@R2 (shown on multimeter) = 10v * (1Meg//10Meg)/((1Meg//10Meg)+1Meg+60) = 4.76v measured.

• What was the purpose of this experiment? Feb 20, 2017 at 14:31
• why is V@R1 + V@R2 /= 10v when R1 = R2 = 1Meg? You need to be more precise. When R1 = R2 = 1Meg and V1 is 10 V the voltages across R1 and R2 will be 5 V each, 10 V in total. Maybe you're measuring that in the wrong way ? Perhaps with a 10 MOhm input impedance meter ? Feb 20, 2017 at 14:33
• Our data is this: 1.) when R1 = R2 = 1Meg, V@R1 = 4.82v. V@R2 = 4.75v. 2.) when R1 = R2 = 1k, V@R1 = V@R2 = 5v.. @CL, sorry I don't know the exact value of the input impedance of the multimeter Feb 20, 2017 at 14:36
• digital multimeter Well that explains it then, look up what the impedance is of that digital multimeter (my bet: 10 Mohm) so when you measure across a 1 MOhm resistor, the total resistance drops a bit below 1 MOhm (around 900 K ohm or so). Feb 20, 2017 at 14:38
• I don't know the exact value of the input impedance of the multimeter Well, look it up then, type the model number in Google and look at the specs. That's what engineers do, they don't guess, they look it up. Feb 20, 2017 at 14:40

The input impedance of your multimeter changes the circuit: With 10k resistors, the difference would not matter, but the 1M resistors pass so little current that the additional current through the multimeter has a noticeable effect.

If you knew your multimeter's input impedance, you would be able to calculate the voltage that you would get without it.

• Took me a couple of mins, but I finally got what you said. This a fault on my own not taking into account the multimeter impedance... V@R2 = 10v*(1Meg//10Meg)/((1Meg//10Meg) + 1Meg + 60) = 4.76v -> assumed diodes were open when reverse biased though technically they are not really open, but just for simplicity of calculation... Things make sense now thanks Feb 20, 2017 at 15:08

In addition to the multimeter issue, your 1 $M\Omega$ resistor likely has 5% tolerance. To figure out the range of voltages you may see, assume the values of the resistors may vary by as much as 10% (2*5%)

To see whether this or the multimeter is the problem, measure the voltage drop across the bottom resistor, switch the resistors and repeat the measurement. If the measurement is the same, the problem is the multimeter impedance. If different, the issue is resistor tolerance.

Another possibility is if you're touching the probes while making the measurement, in which case you become a parallel resistor.

To get rid of the effect of the multimeter's input resistance, try making a measuring bridge. You put something like a 1k precision pot across the voltage source and measure the voltage between its wiper and your measuring point. Then you adjust the pot until the measured voltage is 0V. At a voltage of 0V, there will be no current through the multimeter influencing the measurement. Afterwards, you measure the voltage at the wiper as compared to 0V. As the resistance of your pot is much lower than that of your multimeter, the result will be reasonably exact.

why is V@R1 + V@R2 /= 10v when R1 = R2 = 1Meg?... ...Yet in our experiment, V@R1 + V@R1 < 10v.

Depending on how theoretical you want it to be. In theory,

V@R3 + V@R1 + V@R2 = 10v. So in theory, V@R1 + V@R2 < 10.

However, since the current in the circuit is so small (approximately 10v/(R1 + R3 + R2) = 5ua), the voltage drop over R1 = 5ua * 60R = 300uv << 10v.

So V@R1 + V@R2 = 10v, for practical purposes.

that doesn't hold when R1+R2 is sufficiently close to R3, or your meter is sufficiently precise, or your experiment is sufficiently picky.

For practical reasons, assume that a modern multimeter (of the powered, digital type) will in itself behave like a 10 or 20 Megaohm resistor; this will change the pictured voltage divider by 5 or 10%.

Analog ones that work without their own power supply for voltage measurements will usually have a lower input resistance that is also dependent on what measuring range is set.

Voltmeters that have a far higher input resistance exist, but these are more typically lab grade than portable field equipment since they are easily confused (showing nonzero values with the probes connected to nothing) or even damaged by static electricity.