Determining Forward Voltage of LED's




About: Autistic person who's interests include utility cycling, recreational cycling, cycling safety, electronics, gardening, Arduino, and LEDs.

In this Instructable, I will show you how to measure the forward voltages of LED's if it isn't provided. I recently ordered 200 pieces of assorted 20 mA LED's but there was no information on the forward voltages. This may also work for higher power LED's too but you have to be really quick.

Step 1: What You Will Need

LED's to test
Forward current of the LED's
Voltage source higher than the LED's forward voltage
Multimeter (having two would be very useful if you are going to test many LED's)
Alligator clips or anything to hold on to the test leads
Solid wires (or leads from burned out components)
500 or 1000 ohm variable resistor

Step 2: Measuring the Forward Voltage

To measure the forward voltage, set the multimeters to their proper settings (ie. current and voltage). Always set the resistance to the highest value before testing it to avoid frying the LED. It may be easier to clamp the multimeter leads by inserting solid core wires to the breadboard. Lower the resistance until the current is up to 20 mA and record the voltage and current.

The forward current of the blue LED was 3.356 V at 19.5 mA. If you are powering it with 3.6 V, the resistor value to use is the next higher value for R=(3.6V-3.356V)/0.0195A)=12.5 ohm.

Measuring high power LED's with >350 mA forward current can be a bit tricky because when they heat up quickly, their forward voltage drops continuously. This means the current will be higher at a given voltage. To measure high power LED's, follow the same procedure and set the current. Quickly hold the value on both multimeters. It if's too late, you would have to let the LED's cool down to room temperature before taking the measurement again. For this reason, current limiting resistors are not used to power them. They must have a constant current source.

Don't know the forward current?
If the LED is part of an indicator light of a device, you can desolder one of the LED's lead and measure its current. This may not be necessary because you can measure its forward voltage across the leads while it is on.

Increase precision of current adjustment
I was unable to set it to 20 mA because the variable resistor had a large range of 50k ohm. You may use 500 ohm or 1k ohm. To allow coarse and fine adjustments, you may connect a higher and lower range variable resistor in series.

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    18 Discussions


    8 months ago

    Funny, the comments are a thousand times more helpful than the "tutorial". The tutorial needs a lot of work.


    4 years ago on Introduction

    You don't need to be quick testing high powered LEDs and should NOT try to be quick!

    The relevant forward voltage is the one it actually has in your implemented design, based on the actual temperature it will be in that design. As such, you want it on the same heatsink, same everything within reason, and to let it heat up to the maximum temperature it is expected to see.

    This is the only way the data can be valid in a real use, or of course if you did have the datasheet and it had a graph for that, well then none of this was necessary but if there is no graph you still need to do that kind of test to verify your design falls within the parameters you want.

    Perhaps it is time we started trying to have frank discussions with LED manufacturers about a more sane baseline temperature than 25C, since it is practically never that you would have a design putting full rated current through an LED and have it stay at 25C die temperature. IMO, 70C would be a better value to start the discussion with, especially for the purposes of lifespan rating and lumen derating.


    4 years ago on Introduction

    What does the "diode test feature" icon on the meters look like?

    Because, I have a meter and would like to use it instead.


    2 replies

    4 years ago on Introduction

    What does the "diode test feature" icon on the meters look like?

    Because, I have a meter and would like to use it instead.


    1 reply

    Looks like the symbol being pointed to on the DMM in this picture. It looks like the normal diode symbol.

    I notice the two meters you have there have a diode test feature, which displays the forward voltage on the screen automatically, is there a reason why this "manual" way is more effective or are you just demonstrating so that anybody with any meter can find the forward voltage of a diode?

    10 replies

    What does the "diode test feature" icon on the meters look like?

    Because, I have a meter and would like to use it instead.


    The diode test function does not work for LED's although they glowed dimly. The reading was 1.7 for the red LED. For the diode test, the auto ranging multimeter puts out 0.9 mA @1.6 V and the manual ranging meter puts out 1.2 mA @2.7 V.

    I see. I have found reasonable results using them. They can't deliver the current for high brightness, but it gives you a good idea of forward voltage. From there you can select a resistor for 10mA @ X volts, then judge the brightness and find a more suitable resistance. LEDs are pretty forgiving, for a short time at least.

    It id useful to note that the flat side of regular LEDs is the cathode (negative side).

    I think most DMMs output only a limited current on the diode test range, typically about 1mA, so the LED will be dim.

    Older or cheaper DMMs may have a diode test function limited to 2V (intended for ordinary diodes), so may not light white or blue LEDS which usually need at least 2.5V to start glowing.

    If using an analog test meter, they usually have a reversed polarity on the resistance ranges (to simplify the internal circuit), and you may need a resistor in series with the LED because the output current can be higher.

    ALWAYS have high power LEDs mounted securely on a heatsink before testing, unless the current is kept very low, such as 10 - 20mA.
    If using 12V or less for testing, a 1 k Ohm resistor in series will be adequate because it will keep the current low enough for complete safely.

    High power LEDs, such a the 3W or 10W "Star" used in many torches, are very easy to kill.  Do not supply anything like full power to the Star LED without it being in the torch, or very  firmly clamped to a lump of aluminium or copper, or it will quickly overheat and die.    (I speak from experience !)

    Yes you are right about the high power LEDs. I normally build a constant current circuit to drive them since the resistors used to current limit from a voltage source are gigantic and a waste of energy, especially when using batteries.

    I always hear that resisters as opposed to a constant current source are wasting energy, but I wonder if there really is a difference between the two.
    If I feed a led from a 9 volt battery and I choose my resister such that there is 20mA flowing through the LED, I use 180mW to feed that LED.
    Now if I have a constant current source that I configure for 20mA and I feed it with 9 Volts, I am also using that same 180mW.
    So unless I overlook something... the only reason to build a constant current source is because it is less picky in the voltage... it will still deliver that same current (unless the Vcc is too low) and it might be cheaper for large currents.
    If one takes that same 9 volts again but now for a 1 Amp RED LED with a forward Voltage of 2 Volts, you will need a resistor capable of handling (9-2)x1x1=7 Watt, while the resistor in an average constant current source only needs to be 0.7x1x1=0,7 Watt
    (resister value following out Vbe/I = 0.7/1=0.7 Ohm, P=I2xR= 1x1x0.7)

    That makes a lot of difference in price and size of the resistor but not in the power used by either circuit

    If I feed a led from a 9 volt battery and I choose my resister such
    that there is 20mA flowing through the LED, I use 180mW to feed that

    No, you would be using 20mA x the voltage of your LED, for example 20mA x 2V = 40 mW to feed that LED, the remaining 140mW of power would be used entirely by the resistor.

    In the other case, you would probably set your constant current source to something less than 9V, in this example to 2V and 20mA.

    So in one case you would be using 180mW and in the other 40 mW.

    However as mattthegamer463 said, if your CC source is using an equivalent internal resistor, the result will be the same !


    sorry, i didn’t make my self clear. I meant I needed in total 180mW to get that LED to work, even if the LED only 'uses' 40mW

    If your CC source is using a linear regulator to do the current regulation, then it is not really any different than using a resistor. Look up how linear regulators work and you'll see they're just a zener diode and a resistor. They waste the excess power as heat just the same, the difference is that CC sources are stable current whereas a resistor will vary with manufactured accuracy, temperature, and LED characteristics.

    A proper CC source using a switch mode supply or DC/DC converter is very efficient for running LEDs, since the input voltage is efficiently dropped to the range needed for the diode and the current is regulated so no resistor is required at all, really. If running a high power LED on batteries, a circuit like this pays for itself in a few hours most likely.

    Example of a great IC for LED driving, likely used in a lot of flashlight designs:

    Note how the LED has only a 0.3 ohm resistor in its current path. This chip claims up to 96% efficiency.

    Tnx for the info :-) No i was not referring to just a resistor and zenerdiode, but it doesnt really matter what circuit -unless there is a voltage converter, as there seems to be in the TPS63000- the loss is similar.
    The TPS63000 is certainly an interesting chip. Thanks for pointing it out to me, but i would classify it as a DC/DC converter Buck driver, not truly a 'constant current source', but who cares as long as it gets the job done :-)
    It seems the IC delivers 3.3 Volt at 1200 or 800mA

    A quick calculation then learns that with a yellow Led with a forward Voltage of 2V, feeding it with 1200mA there still would be a loss of 1.3x1.2 is 1.56 Watt that needs to be dissipated in the resistor. But at least that is still a fairly current and affordable size :-)
    Anyway....... I feel I am touching upon 'off topic' ass yr great article is on forward voltage.

    I definitely learned something, where I thought my MM would just test if a diode is working I now know it shows me the forward voltage (not saying every MM wil do that)