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Another LED Question (Was: AC/DC adapter for LEDs)

All:

Based on the feedback I received from you guys (thanks for that!) and after referencing a few web pages, I created the following circuit (mind you I've got two of these paralleled across the cap for 10 LEDs total):

470ohm 5 LEDs in series (Specs show 3.6V drop @ 20mA) (+)---------+----/\/\/----|>|-|>|-|>|-|>|-|>|--+ | 1/2W 18V Drop | | | Transformer 1000U | | Vin = 10.5V 50V ----- Filter Cap reads | w/ no load ----- 27-28V w/ no load | | | | | | | (-) -----------+-------------------------------+

Now, assuming I understood everything correctly up to this point, I should be getting approx. 21mA current in the LEDs (28V-18V = 10V / 470ohm ~=

21mA), right? So, why is it then that when I check the current across the LEDs with the multimeter, I read around 42mA (around 22mA across one LED)?

Also, when I light these LEDs up with a 3.6volt power source (three nicad batteries), they are much brighter than when used in the circuit shown above. Does anyone know why this would be?

Am I missing a crucial piece of the puzzle here?

Thanks, Kev> All:

> I am trying to light up 10 white LEDs using an AC/DC transformer. > Originally, this transformer was used to charge batteries. The input is > 120volt/60Hz/8watts (US) and the output is supposed to be 15volt > DC/100mA. I would like to wire 4 of the LEDs (3.6 volt drop across each > one, 20mA) in series with a 47ohm resistor (just to be safe with the > LEDs). This should give me slightly less light output at 15 volts due > to the reduced current. > > Now for the question: > > Based on readings from my multimeter, the output sits right around 10.5 > volts. However, in experimenting with the adapter, I have been able to > put 5 LEDs in series with the 47 ohm resistor with no visible loss of > light (versus not having the resistor there - Quick note: I was able to > see a drop in light output upon putting a 220ohm resistor in series with > the LEDs). Why am I able to drive 5 LEDs when the volt meter is reading > only 10.5 volts DC? What's wrong with this? 5*3.6volts=18volts + 47ohm > * .02A = 18.94 volts? This is way more than 10volts! Is this due the > to fact that this adapter was used for charging batteries? Should I > maybe put a 12volt DC voltage regulator in the circuit? > > I'm pretty confused... > > Thanks for any help! > > Regards, > Kevin
Reply to
Kevin Dressel
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You say you checked the current ACCROSS one led and it is 22ma and ACROSS the lot it is 42ma. That's about expected results but very wrong practice. Yo must measure current in SERIES with the load not ACCROSS it. When you measured across one led you more or less removed that led from the circuit and measured the current thru the rest. When you measured across all the leds you more or less removed them from the circuit and measured the current thru just the resistor and 42ma multiplied by 470 Ohms gives about 19.7 volts which is about right for the supply volts you quote. If you had measured the current across the lot including the resistor the current would have only been limited by the powersupply or the meter selfdestructing whichever was the weaker.

-- Wot's Your Real Problem? John G.

after

I've

drop

Reply to
John G

Thanks for the reply. That explains the differences between the two readings. I rechecked it and got about 19mA through the entire circuit. I figure that based on this, I could brighten things up a bit by reducing the resistor to around 430 ohms, which should take the current up to around 20mA.

Regards, Kev> You say you checked the current ACCROSS one led and it is 22ma and ACROSS

Reply to
Kevin Dressel

In article , snipped-for-privacy@charter.net mentioned...

You cannot measure across the LEDs with the DMM on the current range, because it's almost a dead short. You must measure the current by inserting the DMM between the resistor and LEDs, in other words in series.

If you put the LED across the 3.6V, you should *still* put a current limiting resistor in series with it. If you don't, then the battery is probably pushing much more than the 30 mA maximum thru the LED. Watch out for the overly hot LED.

[snip] 
--
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Reply to
Watson A.Name - 'Watt Sun'

I figured that was the reason. I did put a 10 ohm resistor in series with the

3.6v source and the LED, but it still appeared brighter (of course, that is based on my eyes, which are far from perfect).

For the series of 5, I figure I can get by with a 430ohm resistor (maybe a 390ohm if necessary for brightness, though that will bring the current to around

21-22mA). Does anyone have a quantitative feel for how the higher current will affect LED life? Based on the "perfect" scenario, they should last around 50,000-100,000 hours (these are probably standard numbers). I won't need them on full time, so if I could get a quarter of that, it would more than suffice. Heat dissipation should not be a problem.

later, Kev> In article , snipped-for-privacy@charter.net

[snip]

Reply to
Kevin Dressel

In article , snipped-for-privacy@charter.net mentioned...

based

390ohm

on

Heat

For the white LEDs, I use a 33 ohm resistor for 4.5V = three AA cells. That gives about 30 mA, and with fresh batteries, it's probably closer to 40 mA. The lifetime may be shorter by 50%, but I figure that it'll take a lifetime to put that many hours on them, being they get used a few hours a year. So it really doesn't matter. And the tradeoff of increased light output is well worth it.

I've read that some flashlight makers run their LEDs at 2 or 3x the max, or something like 50 to 90 mA.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS?   Check HERE First:###
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My email address is whitelisted.  *All* email sent to it 
goes directly to the trash unless you add NOSPAM in the 
Subject: line with other stuff.  alondra101  hotmail.com
Don't be ripped off by the big book dealers.  Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com  You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
Reply to
Watson A.Name - 'Watt Sun'

I suppose I can see the need with the flashlights, being that the alternative is more LEDs (more costly upfront / bulkier).

I agree that based > In article , snipped-for-privacy@charter.net

the

based

390ohm

will

them on

Heat

Reply to
Kevin Dressel

In article , snipped-for-privacy@charter.net mentioned...

is more

it'll be

and

From my own personal experience, by the time the LEDs burn out, I will have lost the light and had to replace it with another. Last week I lost one of my regular mini Maglites, so I'm only out ten bucks. Glad it wasn't the one with the LED conversion, 'cause that cost another $30. :-/

the

is based

390ohm

will

them on

suffice. Heat

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS?   Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted.  *All* email sent to it 
goes directly to the trash unless you add NOSPAM in the 
Subject: line with other stuff.  alondra101  hotmail.com
Don't be ripped off by the big book dealers.  Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com  You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
Reply to
Watson A.Name - 'Watt Sun'

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