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

K

Kevin Dressel

Jan 1, 1970
0
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,
Kevin
 
J

John G

Jan 1, 1970
0
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.
 
K

Kevin Dressel

Jan 1, 1970
0
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,
Kevin
 
W

Watson A.Name - 'Watt Sun'

Jan 1, 1970
0
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?

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.
Thanks,
Kevin

[snip]

--
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###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 <at> 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
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
 
K

Kevin Dressel

Jan 1, 1970
0
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,
Kevin

Watson A.Name - 'Watt Sun' said:
mentioned...
[snip]

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.
Thanks,
Kevin

[snip]

--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###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 <at> 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
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
 
W

Watson A.Name - 'Watt Sun'

Jan 1, 1970
0
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.

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.

later,
Kevin

Watson A.Name - 'Watt Sun' said:
mentioned...
[snip]

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.
Thanks,
Kevin

[snip]


--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###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 <at> 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
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
 
K

Kevin Dressel

Jan 1, 1970
0
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 on usage, if I can get 25,000 hours of use out of them, it'll be
more than adequate. Very likely that by the time the LEDs burn out, some new and
improved light will be out anyway.

Watson A.Name - 'Watt Sun' said:
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.

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.
later,
Kevin

Watson A.Name - 'Watt Sun' said:
mentioned...
[snip]

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.

Thanks,
Kevin

[snip]

--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###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 <at> 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
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
 
W

Watson A.Name - 'Watt Sun'

Jan 1, 1970
0
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 on usage, if I can get 25,000 hours of use out of them, it'll be
more than adequate. Very likely that by the time the LEDs burn out, some new and
improved light will be out anyway.

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. :-/

Watson A.Name - 'Watt Sun' said:
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.

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.
later,
Kevin

:

[snip]

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.

Thanks,
Kevin

[snip]

--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###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 <at> 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
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
 
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