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Please take a moment to checkout the simple circuit below:

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, this is a simple circuit consisting of three LEDs in series driven by one 6 volt DC power supply. The LEDs are white LEDs with a forward voltage drop or around 2.6 volts according to my multimeter.

Based on my measurements (using the 6 Volts power supply) there is no current flowing through the circuit at this point and the voltage drop for each of the LEDs is as follows:

  • LED1=1.642V
  • LED2=1.702V
  • LED3=1.607V

schematic

simulate this circuit

Ok, now that we have laid out the facts, I was hoping someone could help me out with the following questions.

First question: Why is there a voltage drop between the LEDs? I am asking this because it was my understanding that an LED (diode) that is not conducting current (forward voltage has not been reached) is like having an open connection. Basically, it was my understanding that the circuit shown above is equivalent to the following circuit:

schematic

simulate this circuit

And if that is the case, the voltage drop between all the open connection should be zero right? So why is there a voltage drop between the LEDs? What am I missing?

Second question: Why do the voltages between all the LEDs not add up to 6 volts? If I add the LED voltages the result comes up to 4.951 total dropped voltage. This is more than one whole volt missing here. Where did that volt go?

Thank you.

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    \$\begingroup\$ Welcome to wittgensteins ladder, your understanding of diodes is wrong. Or at least incomplete. Have a look at wikipedia and some good datasheets about LEDs, especially intresting to complete your understanding might be the amps vs. Vf curves. \$\endgroup\$ – PlasmaHH Feb 16 '15 at 16:54
  • \$\begingroup\$ How did you measure the Voltage drop with "no current flowing through the circuit" ? This is relevant because the voltage drop of a diode depends on the current flowing through it. Not linearly like Ohm's law, but non-linearly such that you can approximate it as on/off at a threshold voltage; but that is only an approximation. \$\endgroup\$ – akellyirl Feb 16 '15 at 17:08
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The LEDs do conduct some very low current even with only two volts or less across them - they may appear as resistors of a few megohms.

When you place your meter across an LED, it adds a 10 Mgeohm resistance in parallel with the LED, reducing the measured voltage - that accounts for the measured voltages not adding up to six volts.

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it was my understanding that an LED (diode) that is not conducting current (forward voltage has not been reached) is like having an open connection.

OK

And if that is the case, the voltage drop between all the open connection should be zero right?

No. An open circuit can have any voltage at all across it.

A (ideal) short circuit will have zero volts across it (with any amount of current).

An open circuit has zero current through it (with any voltage).

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  • \$\begingroup\$ Isn't an open circuit just a very high resistance? Air does conduct electricity, it's just very difficult to measure so we call it infinity and say it drops all the applied voltage. Is it a typo where you said: can (or can't) have any voltage? \$\endgroup\$ – Misunderstood Jul 3 '18 at 0:22
  • \$\begingroup\$ @Misunderstood, an open circuit is a resistance high enough that whatever tiny current goes through it is not enough to affect the rest of the circuit significantly. \$\endgroup\$ – The Photon Jul 3 '18 at 1:49
  • \$\begingroup\$ Okay. What I did not understand was this: An open circuit can have any voltage at all across it. ----- I would think it would have a very distinct voltage, like the supply voltage. Reads as if can should have been can't. Was not sure. Do not interrupt my inquiry as being disrespectful. You are one of the few I respect here. \$\endgroup\$ – Misunderstood Jul 3 '18 at 1:58
  • \$\begingroup\$ @Misunderstood, an open circuit has a very fixed current through it: 0 A, but does nothing to control the voltage across it. A short circuit has a fixed voltage across it (0 V) but does nothing to control the current through it. \$\endgroup\$ – The Photon Jul 3 '18 at 2:09
  • \$\begingroup\$ Originally I misinterpreted your "any voltage" to mean it was unpredictable and could be any voltage what so ever, like pick any value out of thin air. I was also thinking of the OP's circuit. If one of the LEDs were open circuit, the voltage across the open LED would be 6V rather than any voltage. But I now understand what you are trying to say. \$\endgroup\$ – Misunderstood Jul 3 '18 at 2:47
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To answer your first question, the voltage drop has to do with the barrier potential (for any P-N junction, not just LEDs) and the chemistry of that P-N junction. For instance, white LEDs typically have a higher forward voltage drop (Vf in datasheets) than say red or green LEDs. This is a fixed voltage due to the chemistry and only changes with temperature, albeit a small change which also depends on the chemistry of the LED.

The second question - a good one! By KVL I think you're assuming that the voltage rises equal the voltage drops. This is true but relies on an assumption that the voltage source has zero impedance (that it can provide infinite current to maintain the voltage). In practice this is not true. The impedance of the voltage source will keep trying to drop to a point (increasing the current) to maintain the ideal KVL balance and will probably exceed the maximum forward current of the LEDs.

For a constant brightness, you want to power the LEDs with a current source who's voltage is > the maximum Vf of each of the three LEDs in series.

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Diodes vary current allowed at any given voltage. Your leds, if we assume 2.6V at 20mA typically will have a curve where at 1.6 V, they open enough to conduct a few micro amps, causing a voltage drop as that's how diodes work, but don't conduct enough to be visible, the primary use of an Light Emitting diode. Even if they were reversed bias, there is still some leakage voltage, so your open circuit example still wouldn't be completely correct.

the others have explained the imperfect multimeter results. Keep in mind the battery or voltage source equivalent series resistance as well, it might not be 6V under load.

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