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99 LEDs....

Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

Thanks, any help is greatly appreciated!

Alex
 
D

Don Lancaster

Jan 1, 1970
0
Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

Thanks, any help is greatly appreciated!

Alex

Are the LED's carrefully factory matched?
If not, you do not have a prayer.


--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: [email protected]

Please visit my GURU's LAIR web site at http://www.tinaja.com
 
Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

http://www.dotlight.de/datasheets/L-53HD.pdf

You've had bad advice, as you will be albe to see from the voltage
versus current curves in the data sheet.

The worst case forward votlage at room temperature at 20mA is 2,5V.
2.25V is typical, so a 12V power pack probably can't drive a string of
more than 5V diodes.

You could probably set up a current mirror to put a well-defined 10mA
through a string of four LEDs running from a 12V supply, but this is a
long way short of 99, which you'd have to manange with 25 parallel
strings, controlled by 25 current mirrors.
 
Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

Thanks, any help is greatly appreciated!

Alex

Here's what I would do, not knowing exactly what your requirements are.
These are red, so about 1.7V forward voltage drop. Connect 6 or 7 in
series, along with a resistor 20-80 ohms. I'm just making wild
guesses; you'd have to measure the current draw to get the appropriate
value. Yeah, it could be calculated but it's easier to experiment.

You'll end up with about 15 of these strings. Just connect them all to
your battery. One string will have a different amount of LEDs, just
use a different value resistor.

It's not precise, you could use about 15 current regulators, but it's
quick and dirty and all you probably need.
 
B

Ban

Jan 1, 1970
0
Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

Thanks, any help is greatly appreciated!

Alex


You can make a current regulator from a 2N2222 transistor with a single
resistor to the base. For 12V
you need between 100k and 200k for 20mA, but you have to choose it for each
transistor individually. This regulator only takes 0.4V forward voltage, so
you can wire 6 LEDs in a row. Usually when they get warm, the forward
voltage also lowers and they will pull much more current, which might
shorten lifetime. This is prevented with the transistor, which driven in
this way will limit the increase to 10% or so.
 
G

GregS

Jan 1, 1970
0
Hello!

I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

Thanks, any help is greatly appreciated!

Alex

Crank up the voltage to at least 18. You can probably get one at 24 volds DC also.

greg
 
F

Fred Bloggs

Jan 1, 1970
0
I'm wiring 99 LEDs together and want to power it by battery....

I have a very basic knowledge of circuitry and was wondering if anyone
could help me get started?

The LEDs that I have are:

Kingbright description: L-53HD
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak : 700nm
Power dissipation PT: 120mW
Light output min.@ 10mA: 2mcd
Light output typ.@ 10mA: 5mcd

and I have been told that I could power them without a resistor from a
12v power pack.

That's true but the informant failed to mention that means a Royer
Converter. Now some nut is going to come on here and rage about Peter
Baxandall, but it is the Royer that you really want.
 
T

Tam/WB2TT

Jan 1, 1970
0
Here's what I would do, not knowing exactly what your requirements are.
These are red, so about 1.7V forward voltage drop. Connect 6 or 7 in
series, along with a resistor 20-80 ohms. I'm just making wild
guesses; you'd have to measure the current draw to get the appropriate
value. Yeah, it could be calculated but it's easier to experiment.

You'll end up with about 15 of these strings. Just connect them all to
your battery. One string will have a different amount of LEDs, just
use a different value resistor.

It's not precise, you could use about 15 current regulators, but it's
quick and dirty and all you probably need.
Assuming some kind of hobby project, this makes the most sense. Although 6
in a row might be pushing it. L-53HD does not come up at the Kingsbright web
site; so, no way to tell for sure what the forward drop is. Also, it makes a
difference based on what kind of "12 Volt" supply there is. They tend to
come 12.0, 12.6, and 13.8 V. A car battery with the engine running will be
at least 13.8. Depending on how the car is wired, the cigarette lighter
socket could go as high as 15V.

If you want to do something fancier, build a DC/DC converter with an output
of about 180(?)V, and set the current limit to 20 ma. All LEDs in series.
Tam
 
P

Paul Mathews

Jan 1, 1970
0
Don said:
Are the LED's carrefully factory matched?
If not, you do not have a prayer.

This used to be true. However, the resistive component of many high
efficiency LEDs is large enough, and their matching is often good
enough 'right out of the bag', that parallel combinations work out just
fine, particularly if you don't really need brightness matching. It's
as if they're self-ballasted to a degree. If the battery pack has some
significant amount of source impedance, you can sometimes get away
without any resistors. Try it.
Paul Mathews
 
G

GregS

Jan 1, 1970
0
This used to be true. However, the resistive component of many high
efficiency LEDs is large enough, and their matching is often good
enough 'right out of the bag', that parallel combinations work out just
fine, particularly if you don't really need brightness matching. It's
as if they're self-ballasted to a degree. If the battery pack has some
significant amount of source impedance, you can sometimes get away
without any resistors. Try it.
Paul Mathews

The 1 watt and 5 watt Phillips Luxeons are very well matched out of the bag.
A group of American Bright I used were very well matched.

greg
 
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