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I was wondering what the easiest way to figure out a LED's forward voltage is, using measurement tools. I know we can assume red LEDs to be around 1.8V - 2.2V, and that we have similar information for other LED colors, but I was wondering if there's a way to figure out without assuming that.

I purchased several LEDs that don't have a data-sheet with their specifications - so as an exercise I would like to write that information down. (I'm learning)

Most answers I see start by connecting the LED in series to a resistor, but I want to make sure the resistor is right before connecting it.

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    \$\begingroup\$ This is too vague, and it sounds like you are making everything way too complicated. What do you mean by "right?" If you have a particular current you want to put through the LED, then you can either use a power supply with regulated current output, or build a simple current sink circuit. Hook it up, measure Vf with a VOM. You can measure If, too, if you want. Just make sure the Ammeter voltage drop is not included in the Voltage measurement. After you know Vf at your desired If, you should be able to easily calculate an appropriate resistor value. \$\endgroup\$ – mkeith Jun 1 '15 at 21:17
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    \$\begingroup\$ It's also possible my understanding is wrong, so feel free to point out conceptual mistakes in my question. My doubt is: I'd like to figure out the Vf of a LED before connecting it to a resistor. The reason is that I'm a newbie and I'd like to avoid making any assumptions regarding that. Measure first -> calculate -> implement. \$\endgroup\$ – diegoreymendez Jun 1 '15 at 21:37
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    \$\begingroup\$ You should always be safe a small current, so you can start with a resistor that you know will be too large, then try smaller and smaller resistors (therefore more and more current). Stop when you are happy with how bright it is. \$\endgroup\$ – Austin Jun 1 '15 at 23:05
  • \$\begingroup\$ The Wikipedia article LED circuit has a good explanation about the various calculations and considerations needed for simple LED circuits. \$\endgroup\$ – Richard Chambers Nov 16 '17 at 4:56
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I agree with some of the others here... you're trying too hard.

As others have mentioned, the forward drop of an LED varies with its bias current, but for almost every application a hobbyist will get in to this isn't something you have to spend a great deal of time worrying about.

Almost every handheld multimeter has a diode setting. It will tell you the forward voltage of a diode at the meter's testing bias level (usually a few mA). This will put you in the right ballpark very quickly.

Determining LED Forward Drop (easy way)

  1. Set meter to diode setting (i.e. #14 in this picture).

this picture

  1. Connect the LED to the meter leads, verifying correct polarity
  2. Meter will indicate forward drop (usually 1V-3V for most LEDs.) Note that the LED may glow.

Now that you have the LED's forward voltage drop you can figure out how much voltage everything else in the "chain" will need to drop. For very simple circuits it may just be a limiting resistor. For more complex circuits it may be a bipolar or field-effect transistor, or maybe even something more esoteric. Either way: The voltage through a series circuit will be distributed through all the elements in the circuit. Let's assume a very simple circuit with a red LED, a resistor and the supply.

If the meter indicated 1.2V Vf for the LED, you know your resistor will have to drop 5V - 1.2V or 3.8V. Assuming you want about 10mA through the LED it's now a simple matter of applying Ohm's law. We know that in a series circuit the current through all elements must be identical, so 10mA through the resistor means 10mA through the LED. So:

R = V / I
R = 3.8V / 10mA
R = 380 ohms

If you connect your LED to your 5V supply with a 380 ohm resistor in series, you will find the LED glowing brightly as you intended. Now can your resistor handle the power dissipation? Let's see:

P = V * I
P = 3.8V * 10mA
P = 38mW

38mW is well within the dissipation spec for any 1/4 or 1/8W resistor. Generally speaking, you want to stay well under the power rating for a device unless you know what you're doing. It's important to realize that a resistor that is rated for 1/4W will not necessarily be cool to the touch when dissipating 1/4W!

What if you wanted to drive that same LED with a 24V supply? Ohm's law to the rescue again:

R = V / I
R = (24V - 1.2V) / 10mA
R = 22.8V / 10mA
R = 2280 ohms (let's use 2.4k since it's a standard E24 stock value):

And a power check (using an alternate power equation just to change things up):

P = V^2 / R
P = 22.8V * 22.8V / 2400 ohms
P = 217mW

Now you'll notice that by driving the applied voltage up we have driven the voltage across the resistor up, and that in turn causes the total power dissipated by the resistor to go up considerably. While 217mW is technically under the 250mW a quarter-Watt resistor can handle, it will get HOT. I'd suggest moving to a 1/2W resistor. (My rule of thumb for resistors is to keep their dissipation to under half their rating unless you're actively cooling them or have specific needs laid out in the specification).

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  • \$\begingroup\$ Sublime answer, except that C-E drop at saturation in a transistor doesn't need to be (and often ins't) 1.4V, otherwise everything perfectly explained and worked out. \$\endgroup\$ – Asmyldof Jun 1 '15 at 22:02
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    \$\begingroup\$ @diegoreymendez Yes, my meter has them both on the same setting as well. Usually there will be another button to select between modes on the same selection switch position. Consult your meter manual for the details for your meter. \$\endgroup\$ – akohlsmith Jun 1 '15 at 22:18
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    \$\begingroup\$ @diegoreymendez you're right, you do want to know what the Vf is in order to drive it correctly but what we're all saying here is that a very rough "rule of thumb" is appropriate for this kind of application. A lot of newbies get caught up in being very precise when there is no need for more than a bit of precision. When you must be precise and when you can be general comes with experience and usually with destroyed parts as well. :-) \$\endgroup\$ – akohlsmith Jun 1 '15 at 22:19
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    \$\begingroup\$ @diegoreymendez: Your meter apparently has a maximum reading of 1.99 volts in the diode test mode, so it can't indicate the Vf if it is higher than that. You'll have to use one of the methods mentioned in other answers for the higher-voltage LEDs. \$\endgroup\$ – Peter Bennett Jun 1 '15 at 23:28
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    \$\begingroup\$ @Ned64 without a datasheet you're reduced to experimenting. A light meter and variable resistor will give you a curve showing input current vs light output. It'll be a little more complex than this, but really there is no such thing as "perfect current" - all components will vary with PVT. Pick a value that works for the 80% case (which is really probably 99%) and run with it. \$\endgroup\$ – akohlsmith Jun 2 '15 at 19:45
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If you have a power supply with adjustable current limit (like this one), then it becomes very easy.

  1. Set the output voltage to around 5V and dial the current limit all the way down.
  2. Connect the diode directly to the power supply, with no resistor. Don't worry! You've already limited the current!
  3. Dial up the current until it reaches your target (say, 20mA).

The power supply is limiting the current through the LED to the dialed-in limit. The voltage display will show you what voltage is required to push that much current. That's your forward voltage!

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  • \$\begingroup\$ that doesn't seem fast or easy, but it will work. For me, fast and easy is using my meter's diode drop setting and measuring the LED. :-) \$\endgroup\$ – akohlsmith Jun 1 '15 at 21:32
  • \$\begingroup\$ @akohlsmith That's true! You should put that in an answer! I find that my meter only gives a rough estimate, since the Vf can be considerably different at 1-2mA than at 20mA... \$\endgroup\$ – bitsmack Jun 1 '15 at 21:34
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    \$\begingroup\$ There's no need to short the output, just crank the current limit down to zero, connect the LED across the supply, then adjust the current for the desired LED current and read the supply's output voltage. That'll be Vf. \$\endgroup\$ – EM Fields Jun 1 '15 at 23:10
  • \$\begingroup\$ @EMFields Hah, I've been doing that extra step for years! My thanks to you. I'll edit the answer... \$\endgroup\$ – bitsmack Jun 1 '15 at 23:15
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    \$\begingroup\$ @diegoreymendez Yes, the reason for the series resistor is to limit the current flowing through the LED. Your "standard" LED, called "5mm" or "T 1-3/4", can generally handle 20mA. Some surface-mount LEDs can only handle a few mA, wheareas a 3W LED can handle a full amp (at Vf = 3V!). The more current, the brighter the illumination. I usually find that 5mA is sufficient for a 5mm LED, even though it can take 20mA. It depends on your application. \$\endgroup\$ – bitsmack Jun 1 '15 at 23:24
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Most common LEDs can handle at least 20 mA, so if you select a resistor value that will pass 20 mA when connected directly across your power supply, a LED will not be damaged when connected in series with that resistor. Then just measure the voltage across the LED to get the LED's forward voltage. The LED voltage will vary slightly with current, but the current you eventually choose to use is not at all critical.

I generally assume that common red, yellow and green LEDs are about 2 volts, and I aim for about 10 mA current (although I recently had some extremely efficient green LEDs where I had to reduce the current to under 1 mA to get the desired brightness (dimness?)). No real need to get extremely scientific about it!

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    \$\begingroup\$ I, in fact, recently sampled a couple of reels from a Chinese fab and the deep red, emerald green, aqua and orange LED were so bright, in normal test-boards to indicate logic signals I use 0.05mA to avoid blindness in the case of a full byte 1's at once. \$\endgroup\$ – Asmyldof Jun 1 '15 at 21:32
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To expand on Peter Bennett's answer: take your LED, add a 1k resistor, and apply 12 volts (making sure to get the polarity right). Now measure the voltage across the LED. This will give you Vf at about 10 mA. If you want to know Vf at 20 mA, us a 500 ohm resistor. If you want to know Vf at 1 mA, use a 10k. None of these numbers is super precise, but knowing Vf precisely is not a generally useful idea. At the very least, Vf will vary with temperature, so obsessing about it won't get you anywhere.

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You misunderstand how an LED works in that Vf isn't the voltage you put across an LED to make it work, it's the voltage that appears (is dropped) across an LED when current is forced through it.

If you look at a proper data sheet you'll see Vf(min), Vf, and Vf(max) specified for a particular current, and what that means is that if you force the specified current through the LED, you can expect Vf to fall anywhere between Vf(min) and Vf(max), With Vf being the typical value.

So, the answer to your question is:

enter image description here

The power supply is any voltage-variable supply, R provides a ballast for the LED, decreasing its sensitivity to power supply variations.

That'll keep the LED from releasing its magic smoke if you inadvertently crank up the supply too far, and its value [R] isn't critical, within reason.

For example, if you use a 1000 ohm resistor and you're trying to push 20 mA through the LED, that 20 mA also has to go through R, so R will drop:

$$ \text{ E = IR} = 0.02A \times 1000 \Omega = \text {20 volts,} $$

and you'll need some headroom on top of that for the LED.

"A" is an ammeter used to measure the current through the LED, and "V" is a voltmeter used to measure the voltage across the LED.

In use, what you'd do would be to start the supply off at zero volts and then crank it up until the ammeter read 20 milliamps, then the voltage displayed on the voltmeter would be Vf for that particular diode at that particular current and ambient temperature.

Referring back to your question, the way to determine what value of series resistance is "right" for your LED is first to determine its Vf at the desired forward current (If) and then to use Ohm's law to determine the value of the resistance, like this:

$$ \text {R = } \frac{Vs - Vf}{If} $$

Assuming, then, that Vs (the supply voltage) is 12 volts, that Vf is 2 volts, and that If is 20 mA, we'll have

$$ \text {R = } \frac{12V - 2V}{0.02A} = \text{500 ohms} $$

Then, to determine the power the resistor will dissipate we can write:

$$ \text{Pd = (Vs - Vf)} \times \text{If}\ = \ \text{10V} \times \text{0.02A} = \text{0.2 watts} $$

510 ohms is the closest E24 (+/-5%) value that'll keep If on the conservative side of 20mA, and a 1/4 watt resistor should be fine.

Duck soup, eh? ;)

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    \$\begingroup\$ It does go both ways though. If you had a perfect voltage source with V = the LED's Vf at some current, and nothing else, the LED would pass that current. Of course, the bad news is if you're a little over (imperfect supply, drift due to temperature, etc.) your current might be significantly higher and blow out the LED, so it's not a good way to regulate things in practice. \$\endgroup\$ – hobbs Jun 2 '15 at 4:19
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Build a constant current source because common bench supplies wont go down that low.This could be an easy 1 or two transistor circuit . It is easy because it doesn't need to be accurate . A sensible current would be the current that you intend to drive the led with . Now your DVM will give you the forward volts measurement and you wont blow any LEDS

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