# How test LED?

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I have a light consisting of one super-bright white LED which plugs into the USB port of my PC. It was flickering and now does not illuminate.

The supply leads to the LED are delivering 5v from the USB port but my meter wouldn't show the sort of intermittency which might cause flicker.

On dismantling the light, I see the leads of the LED are the same length and are unmarked.

(1) In general, how can I test such an LED using my multimeter and/or battery? (If it helps, my multimeter has a socket marked E.B.C.E which I think can somehow test transistors.)

(2) If I apply something like 5 volts or perhaps 8 volts to just the LED in reverse to its normal polarity then will I cause the LED damage?

• posted

Check it with a 1000 ohm in series and a 9v battery

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Alex Coleman wrote in news:Xns981678CFC5BA471F3M4@127.0.0.1:

Never connect it directly to a battery, even in reverse, Bad habit. An LED depends on control of current, not of voltage.

The formula for finding the right resistor for an LED is (Vs - Vf) / If.

Vs = Supply voltage in volts. Vf = LED's rated forward voltage in volts. If = LED's rated forward current in AMPS. Take care, LED's are rated in mA

Plug those values into the formula to get the series resistance in ohms.

If you need a controlled current and a variable input supply range, look at the datasheet for the LM317 to see how to use it as a current regulator. It's the simplest way, but take care to calculate power as well, or a large supply voltage will burn the regulator.

More info than you want, perhaps, but hopefully good info you can use.

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Most diode checkers built into multimeters don't provide enough forward bias voltage to check a LED.

I quick-check LEDs by using a 1k resistor in series with a 9 volt battery as mentioned by another poster here. I have never damaged a LED using this method even if I reverse bias a LED.

One way to quickly tell the polarity of the leads is by looking inside the LED. The lead attached to the chip carrier is the cathode or negative terminal. The chip carrier is the larger of the two pieces of metal inside the LED. The other terminal has a fine wire running over to the chip on the carrier. You may need a powerful magnifying glass to see the wire and the chip but the size of the cathode in relation to the anode is unmistakable.

I have only seen one or two exceptions for identifying the cathode this way with some special, high-wattage LEDs.

Dorian

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Simply use a 3 volt coin cell to test it. Same as your pc clock battery. They use them in the small keychain lights with a white led.

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The LED keyfobs I've opened used 2x 2016 coin cells for white LED and 1x

2032 for red.
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dumb ist.

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Cheap white LEDS are a bit failure prone, flickering before end of life not unknown.

As suggested battery and resistor, white blue and true green LEDs need at least 3.something V to light, so a single 1.5V cell wont work , and even 2 may not be quite enough to reach minimum voltage hence most LED torches 3 cell, 9V and 1K would be fine, drop it to 390R if you want bright.

Stick with the 1K until you work out polarity.

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Like already mentioned, never connect to a constant voltage without a series resistor. But a comment on connecting in reverse polarity:

In the basic theory, in normal use, where you limit the current by a resistor, there is no "the highest possible _supply voltage_", as long as you equip it with a resistor calculated correctly according to the voltage (as it gets dropped on the correctly calculated resistor). But notice that if the LED is accidentally connected in reverse polarity, there will be no current, which results in no drop in the resistor, so there will be the full supply voltage across the LED, regardlessly to the resistance of the resistor.

So then if the supply voltage is higher than the rated max reverse voltage, connecting in wrong polarity will destroy the led even if there is a correct-resistance series resistor.

So short answer: don't connect in reverse polarity.

• posted

That's not entirely correct. Many LEDs have a typical breakdown far in excess of the max rated reverse voltage, so it's more correct to say that it "might" or "could" damage the LED. Many 5V rated LEDs have a typical reverse breakdown in excess of 30V. I also think that it's not likely to damage the LED unless the power dissipation is too high, even if the LED conducts some reverse current.

I have yet to see an LED or LED array s.t. probing it with a regulated

5V source in series with 1K could cause any damage. Some are apparently sensitive to ESD, so take care with that as well.

Best regards, Spehro Pefhany

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"Simoc" wrote in news: snipped-for-privacy@b28g2000cwb.googlegroups.com:

No. Volts are EMF, they move the force that is represented by current. While the voltage appears to be high across the LED in reverse, it's the current that does the actual damage, if any be done, and we already know that there is no current.

Even ESD is only dangerous because it instantly and momentarily causes high current.

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The thing with those ESD-sensitive LEDs and GaN and InGaN ones in general is some extreme intolerance to conducting any current as a result of reverse breakdown. Reverse current conduction due to reverse breakdown apparently causes some destructive elecrolysis effect that forms some sort of short or partial short.

At least some other LEDs don't like reverse breakdown, due to current conduction and heating being uneven and localized. But at least it appears to me unlikely that damage will occur that way from forcing a milliamp or a fraction of a milliamp through in the wrong direction. Some LEDs also have issues of ingredients diffusing out of place as a function of voltage, temperature and time. But I surely don't expect a few seconds with 9, 12 or 15 volts with theb chip at 30 degrees C to be any problem in that area. That is mainly a problem with elevated temperatures combined with long time, although a few LEDs can degrade a little significantly from normal forward voltage within just hundreds of hours at higher allowed temperature. Thankfully that one progresses at a rate slowing greatly with progressing degree of damage - a (formerly) HP app note appears to tell me that if the LED prone to this degradation mechanism degraded 10% from that mechanism in the first 100 hours, it will degrade another 10% in the next 900 hours and another 10% in the following

9,000 hours. That app note advises reducing duty cycle rather than instantaneous current if you want average current less than 10 milliamps, and appearing to mainly have that being more important if the LEDs will be experiencing higher temperatures. This app note was specific to InGaAlP chemistry. One thing I also suspect is that these LEDs having wear-and-tear on them could have high resistance partial shorts or something reducing their ability to work well at lower currents.

- Don Klipstein ( snipped-for-privacy@misty.com)

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"ian field" wrote in news:VQ8Bg.2077\$ snipped-for-privacy@newsfe7-win.ntli.net:

... and don't be fooled. The led probably has a built in current source, otherwise you could not connect the batteries to the LED without a resistor!

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A CR2032 has around 50-100 ohms of internal resistance at some tens of mA, and that's what they are depending on.

Best regards, Spehro Pefhany

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Spehro Pefhany wrote in news: snipped-for-privacy@4ax.com:

I don't think those figures will establish a resonable current range ,tolerance or predictable battery life. I do know current limiters are used in those LED's.

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"Dave M." wrote in news:Xns981B92450D406DaveM@

216.196.97.131:

...however a current limiter would not work with in this application because of a barely existing compliance voltage, so you must be right.

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We're talking about cheap here, not predictable or reasonable.

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I happen to have disassembled keyfob lights and used different LEDs in them and interchanged their LEDs in other applications. Keyfob lights in my experience use LEDs that don't have any components or circuits added within the LED bodies. Keyfob lights in my experience usually lack resistors or added circuits. In my experience keyfob lights do depend on internal resistance of the battery. With fresh batteries the LED current sometimes (as in often) exceeds the maximum rating for the LED. My guess is that keyfob light manufacturers depend on at least one and maybe both of the following being true:

1) The battery heatsinks the LED

2) The amount of time throughout the product's life spent with the LED being overpowered does not degrade it to a noticeable extent, or does not with a high percentage of users.

- Don Klipstein ( snipped-for-privacy@misty.com)

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This is greatly assisted by the fact that most of these units are so badly made that they fall apart after a few weeks anyway...

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The keyfobs I bought came in 2 versions - one of which was surprisingly well made! Aside from the ones I've cannibalised for the LEDs and batteries, I do actually use one for its intended purpose and use it often enough to wear out the battery.

The better made version has a button that can be pressed or also works as a slide switch for continuous operation, the less well made version is simply

2 flexible polythene shells that are squeezed to light the LED - these are the first to get stripped when I want the LED or the battery, the LED on my heyring is still working after 3 or 4 new batteries.

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