AA batteries v LEDs

[ss]

I have a sep of LEDs (xmas lights) Original batteries were Alkaline 1.5V. Are the LEDs likely to run from 1.2V nickel metal hydride rechargeables?

[tabbypurr]

Yes, but dimmer.

NT

[ss]

ok, thanks

[Brian Gaff]

I would suspect they will in most cases, but obviously one might expect them to be a bit dimmer. Brian

[PeterC]

I've at least a dozen torches and push-switch lights that are running on NiMH and they are pretty good. With luck, the lower voltage will prolong the LEDs' lives. One torch can bee seen shining on a pale object about 100 yds away - enough for most prposes. The Oral-B toothbrush goes for about 5 weeks on a pair of Eneloop Pros, but I change the cells every month, just as a routine.

[Cursitor Doom]

Obviously? Why obviously? Most small LEDs turn on at less than one volt forward voltage and it's the current through them rather than the voltage across them that is the predominant consideration WRT brightness. Consequently they may well run less brightly when powered by the non alkaline cells, but that's due to the difference in the cells' inherent characteristic internal *resistances* rather than their terminal voltages.

[Chris Green]

Hmm. I sort of know what you're saying, but:- It's the voltage across the LED that will determine the current going through it.

If there is 1.35 volts across the LED that fixes the current going through it, the LED doesn't know whether it's an alkaline cell or a NiMh one.

What I *think* you're saying is that the alkaline cell may well have a higher internal resistance so that at (say) a current drain of 100mA its output voltage is *less* that that of the NiMh cell at the same current.

[Cursitor Doom]

You're really not going to achieve anything useful here by focusing on the voltage drop across the device which is a function of the diode's characteristic *barrier height* as opposed to the applied voltage. An avalanche effect occurs - a sudden rapid increase in current - at an applied voltage well below the output of either type of cell (could be as little as 0.6V) so *current* is the dominant criterion we must consider.

[The Natural Philosopher]

No, they don't.

RED is 1.4v and green is more. Cant remember what blue is. (looks it up) Oh. 3-4V or more.

Because of the exponential nature of the curve, almost no current will flow until the turn on threshold is nearly reached.

[The Natural Philosopher]

Utter bollocks.

Almost no current occurs in even an IR LED below 1.2v, and for a blue, that could be up around 3V+

so *current* is the dominant criterion we must consider.

That bit is correct, but don't expect a blue LED to light up on a single

1.5V battery at all.

Red and amber maybe, but that's about it.

[Chris Green]

But it's *still* the voltage across the device that determines the current going through it. Yes, it's a non-linear device so there is a 'knee' in the current/voltage characteristic but this has no bearing on the brightness from Alkaline versus NiMh batteries. At a given voltage the brightness will be the same, if the Alkaline battery puts a higher voltage across the LED then it will be brighter. The fact (or not) that the NiMh battery *could* supply more current at a given voltage makes no difference, the LED will only take the current it takes at a given voltage.

[Cursitor Doom]

Yours is a hopeless approach that takes no account of the parameter spread one encounters with different devices, even those from within the same batch. This is a common problem we come across all the time in electronic design. If you have a multimeter with a diode test function, select a bunch of seemingly identical diodes and test them. You will find there is a spread of results from maybe 0.65V to say 0.73. Hooking these up to the same voltage will produce differing brightnesses and quite possibly destroy those devices with a lower barrier height. OTOH, rigging up series resistance in accordance with the device's datasheet to limit the *current* to the same value will guarantee any variation in brightness between devices to be so small as to be imperceptible and no devices will be damaged or destroyed. This is the correct approach to use. Try googling "monte carlo analysis" and you'll soon get the picture.

[Chris Green]

A hopeless approach to what?

[Cursitor Doom]

A hopeless approach to understanding and predicting the behaviour of these devices at a fundamental level. With much of electronics it wouldn't matter, but with diodes and bipolar junction transistors you really have to consider them from the current perspective, not voltage.

[The Natural Philosopher]

I am beiginning to wonder if tesco have put something in the Xmas sherry.

The number of post claiming outrageously erronous horseshit to be perfectly true and valid has increased dramatically over the last few days.

[Chris Green]

That isn't what was being discussed. What originally made me comment was the idea that somehow (because of its greater current capability) a NiMh battery would push more current through the LED. I was just pointing out that for a particular LED the current through the LED will be the same for a given voltage across the LED, it makes no difference what current the voltage source is capable of.

I agree that junction transistors and LEDs are 'current' devices, but that wasn't particularly relevant to my original comment.

[tabbypurr]

yup

of course it has!

That's not really how LEDs work

I've done that.

36x Silicon diodes at 0.86mA: 0.540v - 1.350v 3x Germanium diodes: 0.326v 0.329v 0.336v.

Connecting silicon diodes to anything does not produce any brightness.

LEDs are not the same thing as rectifing diodes. LED Vf is about 1.6v - 4v. The LED christmas lights I've used on NiMH instead of alkaline presumably have inbuilt resistance, and all the colours worked fine on both battery types.

NT

[Chris Green]

??

[Dave Plowman (News)]

Surely a decent torch would have electronics which drove the LED at constant current? Easy enough to do these days. So would give a consistent brightness until the battery was exhausted?

[Cursitor Doom]

Yes, in theory. But that regulation doesn't come for free. And it will be linear regulation not switching so even less efficient. There are losses involved so the batteries wouldn't last as long as with the "Chinese Method" of basically just hooking them all up in parallel to a voltage source and hoping for the best.