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I have several cylindrical LED flashlights that have a large number of LEDs in them (the one I'm looking at has 9, which I think is typical.)

They take 3 AAA batteries, wired in series, to give around 4.5 volts (More, on fresh alkaline batteries, but voltage probably drops to 4.5 or less under load)

I've had a couple fail and taken them apart. I don't see any current limiting resistors or any components. The little PC board the LEDs are wired into is crowded so it's hard to tell what's going on, but I think they are wired in series.

I gather that if you wire your LEDs in series and can get the total voltage drop across your LEDs to equal the supply voltage you can omit a current limiting resistor? But I don't see how you get a drop of exactly 4.5 volts out of 9 LEDs. With LEDs with a 1.2 volt drop, 5 LEDS would equal 6 volts, so that would work.

Actually, I just looked at the PC board and the LEDs appear to be wired in parallel. Are there ultra-bright LEDs that you can drive directly off of 4.5 volts without a current limiting resistor? Or are there special purpose ultra bright white LEDs made for 4.5 volt supply that have internal current limiting resistors?

Follow-up question: Does anybody know if the 12 volt LED bulbs that are in landscape lights have a voltage regulator in them? I'm interested in using them as room lights for a photography setup, where I can quench the lights as the camera shutter opens. If they are simply LEDs and resistors, the'll quench well within the approx. 50 MS before the camera shutter opens. If there's a regulator in the system, they probably won't.

EDIT: I looked more closely at the PCB, and it's wired with all the LEDs in parallel. The flashlights that have failed have had a failure in the switch/wiring, not in the PCB assembly. The one I photographed got messed up by leaking alkaline batteries. I was able to clean it off and repurpose it as a subject light for my photography. It's "grotty" appearance is the residue of the battery electrolyte paste, and the wires are wires I soldered on in order to power it directly from a trio of AA batteries. (The original flashlight uses 3 AAA cells wired in series.

Here is the back of the PCB, showing the traces:

LCD flashlight PCB back

And here is the front, showing the LED lenses:

enter image description here

There is nothing else to the flashlight aside from the metal barrel, the battery holder, and a switch. I guess it's possible there is a current limiting resistor in the battery holder that I missed, but I doubt it. Plus I've driven the LED assembly for 15 minutes or more from 3 AA batteries, so if it was going to heat up and fry, I would think it it would have happened already. My guess is that these are ultra bright white LEDs that have a 4.5 forward voltage drop. Is there such a thing as LED packages with built-in current limiters?

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  • \$\begingroup\$ A photo of the opened up flashlight would help. \$\endgroup\$ – Vladimir Cravero Aug 8 '14 at 13:55
  • \$\begingroup\$ I'll work on a photo. The PCB is tiny and crowded, so it's going to be tough to get a clear picture. \$\endgroup\$ – Duncan C Aug 8 '14 at 14:20
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    \$\begingroup\$ I think it's thougher to guess what you have on your pcb without even seeing it... \$\endgroup\$ – Vladimir Cravero Aug 8 '14 at 14:38
  • \$\begingroup\$ @VladimirCravero, I just posted a couple of images. It's quite clear that the PCB has the LEDs wired in parallel, with no resistors. \$\endgroup\$ – Duncan C Aug 8 '14 at 15:31
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    \$\begingroup\$ Once I was shocked to discover several ohms in a battery pak's springs. \$\endgroup\$ – user56208 Oct 14 '14 at 19:05
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Background:
I have designed a number of LED lighting products which are manufactured in China.

I have several cylindrical LED flashlights that have a large number of LEDs in them
... Are there ultra-bright LEDs that you can drive directly off of 4.5 volts without a current limiting resistor? Or are there special purpose ultra bright white LEDs made for 4.5 volt supply that have internal current limiting resistors?

No and no, unfortunately.
Many LED lights are constructed as you describe, with multiple white LEDs wired in parallel and connected essentially directly across the battery.

They are junk.
They are not "designed".

They build them this way "because they can" and they work well enough to be able to sell them.
When supplied with 4.5V + the LEDs are driven well above their maximum design rating and their lifetimes are greatly shortened. The LEDs used are typically low lifetime low cost devices.

Follow-up question: Does anybody know if the 12 volt LED bulbs that are in landscape lights have a voltage regulator in them?

The 12 volt LED strips usually use 3 LED die in series plus a series resistor.
Turn on / turn off time is liable to be sub `1 microsecond if capacitors are not used downstream of the switch.

Current is set to be "about right" at 12 Volts so will vary substantially if used in an automotive context where several volts of variation occurs. Many strips use individual LEDs but some use 3 die per package LEDs with all 3 independent die wired in series. It is possible but not certain that strips with individual LEDs will run somewhat cooler due to a lower concentration of Energy per package.

Lifetime of these LEDs may be better as the series resistor means that they are somewhat more properly driven. I have seen very substantial variations in output of similarly appearing strips. The brightness bears no obvious relationship to LED specifications and a brighter strip may simply reflect a manufacturers 'marketing decision'. You can get a range of LEDs per metre but current drain and number of LEDs are not directly related.


White LEDs are typically have a voltage drop in the 3.0 - 3.5V range at rated current.
Current increase tends to be exponential with voltage and at 4.5V almost any LED would self destruct almost instantly. The "saving grace" (if it can be called that) is that the combination of small batteries and many LEDs means that the batteries are unable to produce more than 'vastly too much' current when the batteries are new. Any light constructed in this manner demonstrates a total lack of concern and/or knowledge by the manufacturer.

Adding even a single common series resistor makes a substantial improvement in voltage/current profile and a resistor per LED would greatly assist current balancing between LEDs.


Added May 2016

Harper commented:

OP is asking about LED bulbs, not strips. Those are commonly made as screw-in replacements for incandescents. Some have a resistor, but many have a switching buck converter which will accept a range of voltages from 12-30V or higher. The LED series voltage is quite close to 12V actual, so if voltage drops much below 12V the buck converter will go to 100% duty cycle and simply pass the voltage through, causing the LEDs to dim rapidly.

My answer addressed LED strips as I noted, which the OP did not ask about, as Harper noted :-).

Harper's comments above are correct where applicable. I have not seen a bulb with a buck converter internally, but no doubt that exist. White LEDs have Vf typically in the range 2.8V - 3.5V. 2.8V is unusual and usually only seen in reaqsonably modern LEDs or ones operated well under full power. At 12V nominal, 4 LEDs have 12/4 = 3V each available. Allowing a small voltage drop in connectors and wiring 4 LEDs with Vf of 2.8V to 2.9V would be able to be operated at full power. In real world situations with Vin able to be somewhat below to substantially above 12V, 4 LEDs in series will often work but 3 x LEDs in series is 'safer'. Bulbs may not match strips in configuration, but all 12V LED strips that I have seen use 3 LEDs in series.

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  • \$\begingroup\$ That the pictures are from such a flashlight where the batteries leaked is further evidence of junk. Likely pulling so much current from the batteries made them run warm, in excess of their specifications, resulting in their destruction. \$\endgroup\$ – Phil Frost Aug 8 '14 at 15:41
  • \$\begingroup\$ OP is asking about LED bulbs, not strips. Those are commonly made as screw-in replacements for incandescents. Some have a resistor, but many have a switching buck converter which will accept a range of voltages from 12-30V or higher. The LED series voltage is quite close to 12V actual, so if voltage drops much below 12V the buck converter will go to 100% duty cycle and simply pass the voltage through, causing the LEDs to dim rapidly. \$\endgroup\$ – Harper May 20 '16 at 0:06
  • \$\begingroup\$ I'm not sure why newcomers to SE EE with rep ~= 0 and no posts are editing posts and, along with very minor useful changes, are making punctuation changes which are based on stylistic preferences or are just plain wrong. \$\endgroup\$ – Russell McMahon May 20 '16 at 11:47
  • \$\begingroup\$ @Harper You are correct re him not asking about strips. My answer addressed his Q1 but not Q2. I have added material, quoting your input. \$\endgroup\$ – Russell McMahon May 20 '16 at 11:57
  • \$\begingroup\$ "They are not "designed"." - Au contraire, they are designed very well! This is the cheapest way to make them, so maximum profits remain for the manufacturer. As a huge bonus, they are likely to die within a year or two, and the chances of the consumer taking the effort of a warranty claim are virtually nil; they'll just end up buying another one. Eternal repeat customers! \$\endgroup\$ – marcelm Apr 12 at 18:25
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The 9 LED flashlight I have definitely has the LEDs wired in parallel.

White LEDs typically have a forward voltage of 3.2 volts, or so, so with a 4.5 volt supply, they can't be wired in series.

The flashlight apparently depend on the internal resistance of the batteries to limit current through the LEDs to a safe (or at least, non-fatal) value. There is no attempt to balance currrent between the LEDs.

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  • \$\begingroup\$ Wow, so they are over-driven from 3.2 to 4.5 volts? So the LEDs will have shortened life then, right? \$\endgroup\$ – Duncan C Aug 8 '14 at 15:35
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    \$\begingroup\$ @DuncanC when you account for the internal resistance of the battery, the terminal voltage will no longer be 4.5V. You could validate this with a simple experiment: try powering them with a 4.5V source with a low internal resistance, like a bench supply. My guess: the supply's current limit will kick in, or the LEDs will melt. \$\endgroup\$ – Phil Frost Aug 8 '14 at 15:39
  • \$\begingroup\$ Hmm. How do you estimate the internal resistance of a typical alkaline battery? And how bright do you thing the LEDs in these lights are? I'd like to get a bucket of LEDS and wire up my own LED light bank on a perf board, but I don't have a real-world idea how bright 1000 mcd is. I found these white LEDs at Jameco: jameco.com/webapp/wcs/stores/servlet/… They're rated at 3.8V and 5500 at 20mA maximum current. \$\endgroup\$ – Duncan C Aug 8 '14 at 16:08
  • \$\begingroup\$ @DuncanC, You can just place a resistor across the battery and measure the voltage. Measure the resistance of your resistor as well. If you know the voltage on the battery before placing a load on it then you can use these measurements and determine the source resistance. \$\endgroup\$ – sherrellbc Aug 8 '14 at 17:33
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It might be as simple as three sets of three series LEDs wired in parallel That would allow for 1.5V across each LED and would work fine assuming the batteries can supply the current. If the forward voltage is any lower than 1.5V then the LEDs are going to draw exponentially more current than they otherwise would and damage the device, so I'm sure there are some limiting resistors on the PCB. Then again, you mentioned having a few fail, so that may be the reason.

To be any more confident than a simple conjecture we'll need to see a diagram, or at the very least a photo as Vladimir has suggested.

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  • \$\begingroup\$ > If the forward voltage is any lower than 1.5... You mean the forward voltage drop on the LEDS? \$\endgroup\$ – Duncan C Aug 8 '14 at 14:20
  • \$\begingroup\$ I think shrellbc meant ..any higher.. \$\endgroup\$ – Vladimir Cravero Aug 8 '14 at 14:39
  • \$\begingroup\$ @DuncanC, I mean if the actual forward voltage of the LED (for that device) is lower than 1.5V (the voltage you are applying across them 4.5/3), then you will have a much larger current draw. If it the forward voltage was 1.5V then you would be fine, and an exorbitant amount of current draw would not occur. If the forward voltage is lower than 1.5V, then you are over-driving it to some degree. \$\endgroup\$ – sherrellbc Aug 8 '14 at 14:55
  • \$\begingroup\$ @sherrellbc, I posted a couple of images to my original question. After looking at the images its very clear that the PCB is wired with all the LEDs in parallel. Am I right that what happens is that with increasing temperature, the internal resistance goes down, the current flow goes up, and you have a run-away increase in current until the LEDs burn out? \$\endgroup\$ – Duncan C Aug 8 '14 at 15:33
  • \$\begingroup\$ @DuncanC, it is really hard to say, but yes that could be happening to some degree. Certain devices exhibit negative temperature coefficients thereby lending themselves to exactly what you describe. The likely culperate I would say, however, is over-voltage across the LEDs and subsequent over-current/over-heating. As you suggest in your first post, you can drive a 1.5V LED with 1.5V and no limiting resistor, but if you over-drive it by a small margin then it will over-heat and destroy itself due to current drawn (characteristic of the non-linearity of diodes). \$\endgroup\$ – sherrellbc Aug 8 '14 at 19:59

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