# Difficulty measuring forward voltage for a LED

Discussion in 'LEDs and Optoelectronics' started by Lol999, Apr 11, 2019.

1. ### Lol999

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0
Feb 16, 2017
I have a couple of LED's, both 12 volt or so and one is in fact a 3w cluster.
I was toying with the idea of running one of them off a 12v battery but need to calculate the resistor value I would need, and for this I need the forward voltage.
I got my multimeter out, which I admit is a cheapie, and set it on "diode". On both led's the display never changed from 1. The single led lit up so I was doing something right.
Despite switching polarities nothing changed on the display so is it my multimeter or am I doing something wrong?

Last edited by a moderator: Apr 12, 2019
2. ### Ylli

339
94
Jun 19, 2018
Ohmmeters on the diode range usually would not have enough voltage to turn forward bias an LED - particularly as series string of LEDs. If the LED assemblies are truly rated at 12 volts, then they would likely already have a series resistor installed and no addition resistor would be needed.

If you want to measure the forward voltage of a discrete LED, then you need to supply it with nominal operating current and use the voltmeter.

3. ### Bluejets

4,264
907
Oct 5, 2014
As for resistors on 12v, forget it and run a small constant current module.(Ebay...couple of dollars)
Reason being as voltage changes even slightly, the power dissipation requirement of the resistor change quite a bit.

Harald Kapp likes this.
4. ### Harald KappModeratorModerator

10,582
2,360
Nov 17, 2011
If you prefer the resistor method instead, use a 12 V battery and a 100 Ω series resistor (for the 12 V rated LEDs) to measure the approximate voltage drop. Calculate the required resistor and repeat the measurement using the newly calculated resistor value. After 1 or 2 iterations you should reach an aceptable level of current/brightness.

5. ### Bluejets

4,264
907
Oct 5, 2014
If you watch this video it shows the variations I explained above.

6. ### dave9

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250
Mar 5, 2017
Quick and dirty way I would do it is if it's a white LED, figure about 3.2V/die, and since 12V LED, "probably" 3 dies in series (even if there are parallel series for a multiple of 3 dies in total), then calculate for 3W or whatever your power source and/or heatsink will support. That might be off by a significant amount so I'd have a multimeter ready to measure resultant current and Vf. In the few seconds it takes to measure this, it shouldn't be off enough to overheat and damage the LED, assuming you work fast and/or heatsink it properly.

However a battery is possibly an issue here, might be 12.6V fully charged if lead acid (so a difference of roughly 3V between that and the LED Vf) but then as it discharges down to about 11V where you might want to terminate the discharge, you'd have only a 1.4V difference between battery voltage and LED Vf. This very large % change in voltage and corresponding decrease in LED drive current using a series resistor, makes a current regulating driver highly preferred if you don't want the LED to dim to less than half as bright  