LED forward voltage/current - the proper way to determine without datasheet - Page 1
Forum section: electronics / beginners

author: okw
author_rank: Frequent Contributor
timestamp: September 20, 2023, 12:21:11 pm
content: How do I properly determine forward voltage and current of a LED (no datasheet available)?Question has been asked and asked, but I don't get a definite answer from any of the posts.I usually set my power supply to 5V (for a single LED), then start at 1-2mA and slowly increase current until I don't see any significant increase in brightness (of course making sure not to burn it). At that point, I note the voltage and current, and use that as basis for calculating the LED resistor. But what is "enough current"? Question is, where do I stop? LED becomes slightly brighter with increasing current, but at some point it will drastically reduce its lifetime, or burn out.I read somewhere that using the diode test on a multimeter gives the forward voltage, but it seems to not work properly. When I set this forward voltage and increase current from say 2 to 20mA, it barely glows. As an example I have 3.1V @ 8mA for a white LED. Diode test gave 2.57V, but at this voltage LED is almost dead.What is the proper way?

author: Psi
author_rank: Super Contributor
timestamp: September 20, 2023, 12:40:24 pm
content: You look at the LED shape/size and say "That looks like it can handle x mA"Then, if it fails, you shrug and say "oh well.." and grab another from the box of 100.Small ones are not worth enough to care about measuring, and large ones you can measure their heatsink temp and determine safe current from that.Perhaps if you had some vintage/rare LEDs it might make sense though.But to answer your actual question, you probably need a thermal camera with macro lens to watch the die temp.

author: Veteran68
author_rank: Frequent Contributor
timestamp: September 20, 2023, 01:16:33 pm
content: Quote from: okw on September 20, 2023, 12:21:11 pmThis depends on the DMM. Some won't light up white or blue LEDs at all. Some will light them but not be able to measure a forward voltage drop. Others will light all LEDs AND display forward voltage. Typically the cheapest meters have the poorest LED testing performance (though not always). If you watch Darren Walker's DMM reviews, he has a little LED test fixture he made that demonstrates both the illumination and forward voltage measurement capabilities of the meters he tests. Since he reviews a lot of cheap DMMs, you often see them struggling to measure forward voltage with the higher voltage LEDs, even when they can light them.The forward voltage is the voltage drop across the diode at the point it starts fully conducting (i.e. it's depletion region is fully collapsed). It has nothing to do with the max voltage a diode can sustain, nor how bright the LED can possibly get. The brightness of the LED is proportional to the current flowing through it, which continues to increase as voltage is increased beyond the forward voltage.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 20, 2023, 03:25:37 pm
content: The forward voltage depends on the test current. For LEDs this is usually some 10 mA, maybe 1 mA or 20 mA. With the normal diode curve shape the change with the current is not that large (some 60 mV morefor 10 x the current). So a test at 1 mA may be good enough even if later used at 10 mA or so. Also the temperature can be an issue (some -1 to -3 mV/K) and this can be an issues with higher current.With some DMM the test current is around 1 mA and the diode test votlage can thus be OK. However at the upper end with blue / white LEDs the current may drop for some, expecially cheap meters. So it depends on the meter.The max current for smaller THT LEDs is usually some 20 mA - if they are not bright enough at that current, chances are they are old and bad and would look for a new brighter one anyway.With modern ones one rarely needed the nominal 20 mA and 1 mA may be plenty for many application like signal lights.

author: BeBuLamar
author_rank: Super Contributor
timestamp: September 20, 2023, 04:01:05 pm
content: The forward bias voltage is the voltage that makes the LED barely glows. This voltage doesn't increase although if you measure across the LED you would see higher voltage when the current is higher but this is due to the voltage drop across the bulk resistance of the LED.

author: TimFox
author_rank: Super Contributor
timestamp: September 20, 2023, 05:05:16 pm
content: Simple test:1. Assume temporary value ofVF= 2 V.2. Wire appropriate resistor in series with LED and 9 V battery to give, say, 10 mA.\3. Measure voltage across LED.4. Re-calculate current given that voltage.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 20, 2023, 05:48:30 pm
content: Quote from: BeBuLamar on September 20, 2023, 04:01:05 pmThe LED voltage still goes up with the current, very much like the normal diode equation, not just the bulk resistance. At low currents, like the onset of seeing light the bulk resistance is usually not relevant. The first light can often be seen at rather low current (e.g. 10 µA) and thus some 180 mV below a more normal 10 mA operation point. Many modern LEDs work OK to rather low current and the eye can be pretty senstive if it is dark. The point were the LED starts to glow can also scatter a lot (in voltage and current) between samples amd it thus not a good reference point.

author: okw
author_rank: Frequent Contributor
timestamp: September 20, 2023, 05:58:49 pm
content: Quote from: Kleinstein on September 20, 2023, 03:25:37 pmSo this is perhaps what I'm looking for. Some ratio between the increase of current, resulting in slight increase of forward voltage.Let's say when current is increased say 2x times, but forward voltage only increases slightly. At this point, it's time to stop increasing the current. Any ratio would indicate the sweet spot.It would of course be referenced from the forward voltage measured with a DMM.I guess most LEDs have different characteristics, but some ballpark number?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 20, 2023, 07:20:01 pm
content: The exponential I/V curve is pretty similar between diodes / LED. So the 60 mV (und thus a little under 20 mV for doubling) per decade is ballpark figure that is good for most diodes, from 1N4148 over 1N5403 and LEDs and even many shottky or germainium diodes. For thermodynic reasons it is rarely less, sometimes a little more (like 2 x). The exonential curve can be valid over a large range (e.g. like 1 nA to 1 A for a 1N5403) so there is no "sweep" spot from the I/V curve.The forward voltage is normally needed for a typical opration point, like 1 mA, 10 mA or 20 mA.

author: Sredni
author_rank: Frequent Contributor
timestamp: September 20, 2023, 07:27:00 pm
content: There is no "point of transition" or knee in a diode characteristic. It is a (translated) exponential function, and as such is smooth and self-similar.If you plot the ideal exponential characteristic on a log scale you will find a straight line with no breakaway point whatsoever.The knee you see on a curve tracer is just an illusion,, dictated by the scale on the axes.

author: TimFox
author_rank: Super Contributor
timestamp: September 20, 2023, 07:54:51 pm
content: Quote from: okw on September 20, 2023, 05:58:49 pmI don't understand your question here.Very, very, roughly the light output (in photons/sec) is proportional to the LED current and the voltage across the LED itself is approximately constant.The voltage is a function of the LED's chemistry, and therefore its color (wavelength).Therefore, as you increase the current the light output and the power dissipated increase roughly proportionately.At some point, the power dissipated is too high and the device fails: that should be far above the amount of light you need for an indicator.Therefore, you should adjust the current until the light output suffices for your application.You should never drive an LED (or other solid-state diode in the forward direction) directly from a "stiff" voltage source, since the current will not be determined and will either be too low or too high (maybe destructive).Always drive an LED from either a constant-current source or a voltage in series with a reasonable resistance.

author: TimFox
author_rank: Super Contributor
timestamp: September 20, 2023, 07:56:35 pm
content: Quote from: Sredni on September 20, 2023, 07:27:00 pmSpecifically, it is because the curve tracer's vertical axis has alinearscale, not alogscale.Paper graphs of the I-V curve should use a log scale for the current and a linear scale for the voltage.

author: IanB
author_rank: Super Contributor
timestamp: September 20, 2023, 08:28:46 pm
content: Quote from: okw on September 20, 2023, 12:21:11 pmI think it is because your question is confusing and ambiguous. What is "proper", in your estimation?Firstly, you need to know what is a suitable current to operate the LED. You start this process by slowly increasing the current until you have the brightness you need. You continue the process, especially for larger LEDs, by monitoring the temperature of the die or the heat sink. You need this to be cool enough that the LED will not have a short life.When you have settled on a suitable current, then you find out what voltage that needs, and make sure your choice of LED driver can accommodate it.Generally speaking, if the LED is a miniature encapsulated kind, then you assume the current is 10-20 mA max and go with that.If the LED is a bigger, power LED on a star or heat sink, then it is best to know the manufacturer part number and go by the data sheet.

author: LinuxHata
author_rank: Frequent Contributor
timestamp: September 21, 2023, 06:02:39 am
content: LED light amount is never proportional to the power being delivered to it.This can be verified easily by checking datasheets - Bridgelux, Cree, Osram - all have datasheets with output light vs current curves.So say if led is rated at 1000 lumens at 1A, it won't deliver 500 lumens at 0.5A or 2000 lumens at 2A.It rises proportionally to certain power but then flattens. And also depends on the temperature.So this is why all professional, led based light fixtures used in movie-cinema-science industrial field use optical feedback to have exact amount of light at output.

author: EPAIII
author_rank: Super Contributor
timestamp: September 21, 2023, 10:21:21 am
content: The reason why you have not been able to get a definite answer to your question is if you don't know what the specifications for your LED are, then you are forced to make some assumptions. This is why manufacturers issue data sheets on their products. But LEDs, unlike many other components, are difficult to mark with their IDs. A resistor can have colored bands or numbers. A transistor can have a part number. An IC can have a part number. A capacitor can have a value and Voltage rating.But most LEDs do not have any markings and once they are separated from a bag or other container with the part number, they are just a mystery part. Operating current? Unknown! Light level output? Unknown! Forward Voltage? Exact value is unknown! If you have enough of them you can run your own tests to the point of destruction. But if you have only one or a very few, then you are forced to just GUESS. Unfortunately, LEDs are usually so small that it is very difficult to impossible to observe things like the die temperature as they are operated at increasing current levels. So total failure is usually the first indication that you will have when testing them.So let me guess. You have purchased a bargain bag of assorted LEDs and want to use them. I have done the same thing. And I just GUESS. Most LEDs will work with a current of around 15 or 20 mA. So that is where I would start. I set my lab power supply for it's minimum current (0 - 5 mA) and the Voltage to 5 to 10 Volts. Most LEDs will have a forward Voltage under 5 Volts so that should be high enough to allow them to draw the current as set by the supply's limit. I apply this limited current power and observe the LED. If it is bright, I assume that current is where it should be operated and stop. If it is dim I crank up the current limit until it appears to be bright enough. How bright is that? Well, you have seen other LEDs, haven't you. It is a GUESS!If you have several of the same LED, you can crank up the current, carefully watching the current meter, until the first one blows. Then turn the current control down and hook up a second one. Crank it up to 1/2 the current where the first one released the smoke and leave it on overnight. If it is still OK in the morning, that current is probably OK. If it also burned out, then try again at 1/4 the original current. Repeat as needed.Other than destructive testing, there is no way to determine the exact characteristics of an unknown LED.SBD! (Sorry Bout Dat!)And if you want to work with definite specifications then you need to buy known parts with available specification sheets. That's just the way things are. And brand, spanking, NEW LEDs with known part numbers are DIRT CHEAP! Well, usually, but don't ask.

author: EPAIII
author_rank: Super Contributor
timestamp: September 21, 2023, 10:31:37 am
content: Well, I am not an LED expert, but the Voltage/current or current/luminosity curves I have seen do not exhibit any such characteristic.Perhaps others can offer something here but I don't think it exists as a general principle.Quote from: okw on September 20, 2023, 05:58:49 pm

author: TimFox
author_rank: Super Contributor
timestamp: September 21, 2023, 04:25:52 pm
content: Quote from: LinuxHata on September 21, 2023, 06:02:39 amIn my reply above, I used the wording "very, very roughly" to describe the relationship between light output and diode current.I checked a typical datasheet for "indicating" LEDs in the current range relevant to the original post, and the light output increases faster than proportional to current at these current levels.https://www.vishay.com/docs/83006/tlhg440.pdffor Vishay parts in "T-1" (3 mm or 1/8" diameter) package.Specifically, when increasing the current from 2 mA to 10 mA (factor of 5:1), the light output increased by a factor of 10:1 (0.1 to 1 in "relative units", referred to the individual specification point at 10 mA).Over the entire range graphed in Fig 7 for a red LED, the light output increased by 280:1 (0.025 to 7 relative to 10 mA) when increasing the current by 30:1 (1 mA to 30 mA).When referring to higher-power devices (> 1 A, for example) used for room illumination, etc., the junction temperature increase will be an important factor.