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'[EE]: powering LEDs from 24V DC'
2001\09\27@102955 by ckchan

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Greetings,

what is the most efficient way to drop the voltage from 24V DC to safe
enough to power 5 leds in parallel ? thanks.


regards,
ckchan

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2001\09\27@104221 by cision Electronic Solutions

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Power them in series.

-----Original Message-----
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Subject: [EE]: powering LEDs from 24V DC


Greetings,

what is the most efficient way to drop the voltage from 24V DC to safe
enough to power 5 leds in parallel ? thanks.


regards,
ckchan

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2001\09\27@104635 by cision Electronic Solutions

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Power them in series AND use a current diode tied to 24V.  You can make one
from a DMOS FET and a resistor.

{Original Message removed}

2001\09\27@104845 by t F. Touchton

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The most efficient way to drop power is to average it, not dissipate it.

A low loss PWM driver followed by a low pass filter (no resistance) will
extract the DC average (Class D type system).  Analogous to a buck
converter.

Any R's in the system waste energy... so if you can feedback on the current
in the LED's to control the average voltage this should yield the minimal
loss.

Scott F. Touchton
1550 Engineering Manager
JDS Uniphase



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Greetings,

what is the most efficient way to drop the voltage from 24V DC to safe
enough to power 5 leds in parallel ? thanks.


regards,
ckchan

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2001\09\27@105048 by David VanHorn

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>
>what is the most efficient way to drop the voltage from 24V DC to safe
>enough to power 5 leds in parallel ? thanks.

Put the leds in series.

After that, you may not need any more than just a resistor.

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2001\09\27@105458 by t F. Touchton

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Well.... there are several different ways to define efficient in this
sentence... I guess I'm still oriented in running a circuit forever from a
tiny lithium battery!!!!!

For minimal parts... in series with a single dropping resistor.

Scott F. Touchton
1550 Engineering Manager
JDS Uniphase



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The most efficient way to drop power is to average it, not dissipate it.

A low loss PWM driver followed by a low pass filter (no resistance) will
extract the DC average (Class D type system).  Analogous to a buck
converter.

Any R's in the system waste energy... so if you can feedback on the current
in the LED's to control the average voltage this should yield the minimal
loss.

Scott F. Touchton
1550 Engineering Manager
JDS Uniphase



                   ckchan
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Greetings,

what is the most efficient way to drop the voltage from 24V DC to safe
enough to power 5 leds in parallel ? thanks.


regards,
ckchan

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2001\09\27@113901 by Allen Mahurin

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Agreed.  I hear too many peolpe that want to put LEDs
in parallel.  While it works, they don't have the same
current through each, and if nothing else, it will
vary the brightness level (I've proven this to myself,
before I took any electronics classes that is).   ;)

If they're "standard" LEDs, putting all 5 in series
will only drop about 10V.  So just use one current
limiting resistor (in the range of 700 - 500 ohms for
20 - 30 mA current) should suffice.  As others have
said (I'm just reiterating), this would probably be
the best way to go.

Good luck,

ATM
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2001\09\27@124353 by Olin Lathrop

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> what is the most efficient way to drop the voltage from 24V DC to safe
> enough to power 5 leds in parallel ? thanks.

Probably a buck converter.  Do you really want to connect the LEDs in
parallel?  Connecting diodes in parallel is usually not a good idea.  Series
would guarantee each LED is receiving the same current.  This would also
have all the LEDs together require less current at higher voltage.  A simple
resistor from 24V would be much more efficient this way, and possibly good
enough depending on your requirements.


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2001\09\27@125212 by Michael Vinson

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Scott F. Touchton wrote, in part:
>[How to drive 5 LEDs from 24Vdc]
>For minimal parts... in series with a single dropping resistor.

LEDs work well on, what, about 20mA? If so, and if 5 of them each
have a 0.6V forward drop, then the resistor will be about 1 k ohm,
and it will dissipate around 420 mW of heat. So the resistor will
still get hot (though nothing like wiring them in parallel: if you
do it with one resistor, it'll dissipate like 2.4 W (yes, 2,400 mW)
of heat! Ouch.).

I like your PWM approach, which is energy/heat-efficient even if it
isn't parts-efficient.

Michael

Thank you for reading my little posting.


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2001\09\27@135239 by Allen Mahurin

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One thing to note is:

1)  Silicon diodes typically drop 0.6 - 0.7V.
2)  A standard (red, green, or amber T-1 or
   T-1 3/4) LED will drop around 1.8 - 2.1V.
3)  White and blue LEDs are along the lines of
   4 - 5V.

ATM

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2001\09\27@140521 by Michael Vinson

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Allen Mahurin wrote:
>[In response to my note in which I incorrectly guessed the  forward drop of
>an LED to be 0.6V:]
>
>One thing to note is:
>
>1)  Silicon diodes typically drop 0.6 - 0.7V.
>2)  A standard (red, green, or amber T-1 or
>     T-1 3/4) LED will drop around 1.8 - 2.1V.
>3)  White and blue LEDs are along the lines of
>     4 - 5V.

Oops. In that case, taking 2.0V for the forward drop, the series
case would require a 700-ohm resistor which would dissipate
420mW of heat. In the parallel case, one common resistor for
all 5 LEDs would dissipate 2200mW.

Michael

Thank you for reading my little posting.


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2001\09\27@162220 by Fok, Ben

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It depends on your current requirement.  Let's assume for a moment that each
of your LED has a forward bias voltage of 2V. Then 5 of them in series will
require 10V to operate.  On the other hand, if each LED require 20mA to
operate, then you need a 24V power supply with at least 100mA supply current
capability, if you are going to connect all your LED in parallel.

But either way you need a ballast resister in series to limit the current.
In the series case, assuming 20mA operation and 2V forward bias voltage, you
need (24-10)/0.02 = 700 ohm resistor.  This means the resistor itself will
dissipate (24-10)*0.02 = 0.28W.  In the parallel case, you need (24-2)/0.1 =
220ohm.  This means the resistor itself will dissipate (24-2)*0.1 = 2.2W.
So if you don't want a big 2.2W resistor hanging around you should go with
the series configuration.

However, if you want the simpliest connection scheme, parallel connection is
the best.

Ben

{Original Message removed}

2001\09\27@180556 by Olin Lathrop

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> LEDs work well on, what, about 20mA? If so, and if 5 of them each
> have a 0.6V forward drop,

Most "normal" LEDs have about 2V forward drop.  Some, like the blue ones,
have more.


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2001\09\27@190530 by Olin Lathrop

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> >1)  Silicon diodes typically drop 0.6 - 0.7V.
> >2)  A standard (red, green, or amber T-1 or
> >     T-1 3/4) LED will drop around 1.8 - 2.1V.
> >3)  White and blue LEDs are along the lines of
> >     4 - 5V.
>
> Oops. In that case, taking 2.0V for the forward drop, the series
> case would require a 700-ohm resistor which would dissipate
> 420mW of heat. In the parallel case, one common resistor for
> all 5 LEDs would dissipate 2200mW.

There's still something wrong with you series case.  5 LEDs at 2v each drop
10V.  This leaves 14V for the resistor.  14V * 20mA = 280mW.  I can't see
where you got the 420mW from.


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2001\09\28@084820 by Olin Lathrop

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> However, if you want the simpliest connection scheme, parallel connection
is
> the best.

Huh!?  Care to justify this statement?  Either way, it's 5 LED's and one
resistor.


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2001\09\28@114545 by Michael Vinson

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Olin Lathrop wrote, in part:
>There's still something wrong with you series case.  5 LEDs at 2v each drop
>10V.  This leaves 14V for the resistor.  14V * 20mA = 280mW.  I can't see
>where you got the 420mW from.

You are right of course, Olin. I just typed in the wrong number
from my scribbled notes on the back of an envelope. (The 420mW
was the answer from the (wrong) 0.6V forward drop).

Michael

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'[EE]: powering LEDs from 24V DC'
2001\11\25@102533 by Spehro Pefhany
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At 10:31 PM 9/27/01 +0800, you wrote:
>Greetings,
>
>what is the most efficient way to drop the voltage from 24V DC to safe
>enough to power 5 leds in parallel ? thanks.

The most efficient way is to use a switching regulator. Some are quite
inexpensive. But, why not put them in series? Then a dropping resistor
wouldn't be so bad.

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2001\11\26@060824 by Vasile Surducan

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Or, if you don't like serial ideea ( which is the best one )
and need 5 LEDs connected at two kilometers one each other,
use some FETs like constant current source.
Regards, Vasile

On Sun, 25 Nov 2001, Spehro Pefhany wrote:

{Quote hidden}

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