I was recently learning about how a diode works, and I understand how a diode, in it's natural state, develops a sort of "pushback" voltage that pushes against any voltage that is applied, effectively lowering the energy of any electrons that pass (forward voltage). This part makes pretty good sense to me.

However, I fail to understand how to measure this "pushback voltage." Using a multi meter, I tried the setting which I use to measure the voltage out of a battery (200 mV setting), and I noticed that I got different values when testing my small LED's: 3.4 for the Red Led, nothing for the Blue Led, and 9.8 for the green LED. I'm not sure what's going on here. I expected the 3.4 Voltage drop to be normal, but the other LED's don't seem correct. And how can a multi meter even measure the forward voltage of a diode? As in, wouldn't any electrons that had the potential energy to cross against the induced electric field have already made the jump from the n-junction to the p-junction? Any ideas?

  • \$\begingroup\$ I believe the 3.4mV and 9.8mV you are measuring are the open circuit voltages the PN junction is generating when it is illuminated. The threshold voltage is measured in volts, not millivolts. And it is not 'visible' if there is no forward current flowing. \$\endgroup\$ Jul 17, 2020 at 22:19
  • \$\begingroup\$ You can’t measure the junction potential at the leads, otherwise it would be a free source of energy, right? \$\endgroup\$ Jul 18, 2020 at 1:53
  • \$\begingroup\$ @SpehroPefhany I mean I would imagine it would only be for an extremely short period of time, because the free electrons floating around would all be gone... \$\endgroup\$
    – Dude156
    Jul 18, 2020 at 2:55

2 Answers 2


None of those values make sense. Red is usually around 1.8V, green is a little over 2 V.

You can't measure the forward voltage of an LED with just a voltmeter. You have to connect your LED to a current source and measure the voltage across the diode when current is flowing through it.

Many multimeters have a diode test function that does exactly what I've described - it pushes a low current through the diode and measures the voltage across it.

Whatever you were measuring, it wasn't the forward voltage of the LED. On the 200mV scale, your numbers would be millivolts, not volts as you wrote.

I expect you were getting a tiny little bit of current from the light hitting the LED. That could be enough for your meter to register a few millivolts.

  • \$\begingroup\$ Aha, I believe you are right indeed! I just hooked it up to a basic circuit (with a resistor of course) and observed a voltage drop of 1.92 for the red LED. However, I am still a bit confused. Why can we measure a voltage across a battery without hooking it up, but aren't able to measure the voltage across a diode when it's not hooked up? \$\endgroup\$
    – Dude156
    Jul 17, 2020 at 22:26
  • 2
    \$\begingroup\$ Because a diode isn't a voltage source, or a current source either. Except that if you shine enough light on a diode junction then it acts like a tiny solar cell. Solar cells are basically big, flat diodes optimized to catch light and deliver current. \$\endgroup\$
    – JRE
    Jul 17, 2020 at 22:36
  • 1
    \$\begingroup\$ You're not going to measure a voltage in a circuit which has no power source. \$\endgroup\$ Jul 17, 2020 at 22:39
  • \$\begingroup\$ But doesn't the diode have an intrinsic electric field between the p and n junction that pushes back? Shouldn't this cause a voltage to arise? Thanks for the help by the way, I'm just having great trouble with understanding this. \$\endgroup\$
    – Dude156
    Jul 17, 2020 at 22:40
  • \$\begingroup\$ An old fairly dim green LED had a forward voltage of about 2.2V. But a modern very bright green LED uses the same chemistry as a blue and white LED with a forward voltage of about 3.3V. \$\endgroup\$
    – Audioguru
    Jul 17, 2020 at 22:45

If anyone else comes across this question and still has confusion after reading the chosen answer, check out this answer: https://physics.stackexchange.com/questions/108314/why-isnt-there-a-potential-difference-across-a-disconnected-diode

  • \$\begingroup\$ I think this is a better explanation now that you've made clear what you were talking about. You were trying to measure the barrier potential. \$\endgroup\$
    – JRE
    Jul 18, 2020 at 7:58

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