simulate this circuit – Schematic created using CircuitLab

I have 3 Red LEDs and a 3 volt cr2032 coin cell. All the LEDs are in series with the cell. Obviously, none works as the combined forward voltage is more than supply voltage. But voltage across each LED is 0.12 volt, but it should be around 1 volt. The main problem is with sum of the voltages; the supply voltage is showing 2.88 volts on multimeter whereas adding the voltage across each led gives me only 0.36 volts. Where is the rest of the voltage? since Total voltage = V1 + V2 +V3. Is there any theory behind it? The same problem occurs with LEDs of different forward voltages with the same cell. For example, white in series with red.

I think that when these LEDs are in series there is a small current flowing through the circuit. And according to its V-I curve, for that small current, LED gets a particular voltage across it. Because when only 2 LEDs of same forward voltage (same colour) are in series they satisfy V=V1 +V2. Please help with this one and thanks already.

  • \$\begingroup\$ How did you measure it? Draw a circuit diagram. \$\endgroup\$ – winny Nov 18 '16 at 11:18
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    \$\begingroup\$ If each LED is off, acting as a resistor of about 100 Megohms, then there will be 1V across each. However when you measure that voltage with a 10 Megohm DMM, you will only see about 0.1V. Also, that voltage reduction puts more voltage across the others, bringing them closer to conduction, reducing their resistance. \$\endgroup\$ – Brian Drummond Nov 18 '16 at 11:23
  • \$\begingroup\$ The circuit diagram is there and i measured with a DMM in parallel to each led as shown. @Brian but with only 2 leds i get 1.45 V across each LED. But why this changes with 3 LEDs in series? \$\endgroup\$ – U.sir Nov 18 '16 at 11:39
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    \$\begingroup\$ @dim totally possible. The meter impedance is loading the remaining LEDs. Now if he used three meters to measure the three LED voltages at the same time and the readings differed, that would be impossible (give or take small reading errors). \$\endgroup\$ – Neil_UK Nov 18 '16 at 12:10
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    \$\begingroup\$ @BrianDrummond as i understand, when i have 3 LEDs the current flowing through them is obviously less than the 2 LEDs in series. And in 2 LEDs in series, the LEDs are almost near conduction as 1.6V is the fwd voltage drop and hence they have less resistance and lesser than my DMM and so i get the accurate reading. But with 3 LEDs, the LEDs are far away from conduction and hence higher resistance, can be higher than my DMM and hence the voltage that i see is because the efective resistance is now dropped and as @.winny said, i must get a DMM with high i/p impedance. Is this the reason? \$\endgroup\$ – U.sir Nov 18 '16 at 13:57

The term non-conducting is misleading. Even in reverse polarity a LED's internal resistance is not infinity but still has a (very) high resistance that can be measured or found in the datasheet as I-sub-R (reverse current). Replacing the LEDs and the DMM by their resistive values it is easy calculating their parallel (LED and DMM) and their serial resistances. It is a simple Voltage-divider circuit, but you should be aware that the voltage measured at each node is depending on the DMM connected parallel to it or not.

If you put DMMs parallel to each LED at the same time the sum of the voltages will add up to the input voltage. But still the measured voltages are different to those that exist without the DMMs.

E.g.: 100 MΩ parallel 10 MΩ = ~9 MΩ. 100 MΩ || 100 MΩ = 50 MΩ

If the DMMs internal resistance is not substantially higher than the internal resistance of the subject to be measured then the result is rather far away from reality.

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  • \$\begingroup\$ But the sum of the voltages across each led does not equals the battery voltage. why? \$\endgroup\$ – U.sir Nov 18 '16 at 12:20
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    \$\begingroup\$ Because every time you measure the voltage across another LED the Voltage indicated is influenced by the DMM parallel to it. In order to learn the real voltages across each LED you have to include the internal resistance of the DMM into your calculation. \$\endgroup\$ – Dekafive Nov 18 '16 at 12:30
  • \$\begingroup\$ so if i put 3 DMM simultaneously across every led, i should get the desired reading. and also why the same thing does not happen with only 2 LEDs in series? \$\endgroup\$ – U.sir Nov 18 '16 at 12:31
  • \$\begingroup\$ So, i got 3 DMMs and did it. Yes, i got 1 volt across each led. But couldn't understand why the same thing doesn't happens with 2 LEDs. Even with 1 DMM i get 1.47 V and it goes upto 1.5 V when 2 DMMs are used \$\endgroup\$ – U.sir Nov 18 '16 at 12:51
  • \$\begingroup\$ @U.sir Yes. But at that point you can remove the LEDs as well since you are only measuring the voltage distribution between the multimeters, not the LEDs. You need a multimeter with even higher imput impedance. \$\endgroup\$ – winny Nov 18 '16 at 12:53

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