1 watt luxeon of 12 volts dc

[mark krawczuk]

hi , i am using 2 1 watt luxeon red leds for brake lights , can i run
these of 12 volts via a resistor, or do i need a driver circuit ?
what value resister would i have to use , (if it can be done this way ./.)

thanks, mark k
--


---------------------------------------------------------------------
"Are you still wasting your time with spam?...
There is a solution!"

Protected by GIANT Company's Spam Inspector
The most powerful anti-spam software available.
http://mail.spaminspector.com

 

[Ian Stirling]

mark krawczuk <krawczuk@dodo.com.au> wrote:

hi , i am using 2 1 watt luxeon red leds for brake lights , can i run
these of 12 volts via a resistor, or do i need a driver circuit ?
what value resister would i have to use , (if it can be done this way ./.)

What colour of luxeon?

You need to find the forward voltage of them.
If it's for example 3.5 volts, and you are putting them in series (a
good idea) then the total voltage across them is some 7V.
Take this number away from 12V, and you get the voltage that
you need on the resistor.
This is some 5V.

If the current is .3A, then you need 5V/.3A = 15 ohms.
And the power needed for that resistor is .3A*5V = 1.5W.

 

[Robert Monsen]

mark krawczuk wrote:

hi , i am using 2 1 watt luxeon red leds for brake lights , can i run
these of 12 volts via a resistor, or do i need a driver circuit ?
what value resister would i have to use , (if it can be done this way ./.)

thanks, mark k
--

According to the datasheet at

http://www.lumileds.com/pdfs/protected/DS23.PDF

luxeon reds have a forward voltage of between 2.25 and 3 volts,
depending on the current you want to put through them.

If you want 300mA, then they will be at about 2.8V each, according to
the current to forward voltage graph. So, put them in series, that'll
give you 5.6V. Then, you need a resistor to drop the voltage
sufficiently. That is 300mA over 6.4V, which works out to about 22 ohms.
However, the power dissipated in the resistor will then be

P = 6.4^2 / 22 = almost 2 W. You'll need a big resistor.

You can use a network of resistors if you can't get 2W ones; if you use
10 220 ohm resistors in parallel, you can use standard quarter watt
resistors. The amount of heat generated will be the same, but it'll be
spread over the whole set. If you mount them so there is a bit of space
between them, it should be OK.

Another option would be to use a switching power supply, like the 'roman
black' step-down supply, shown here: http://www.romanblack.com/smps.htm.
Its simple to build, and you can easily make it put out 5.6V (which will
cover your two LEDs) by increasing the zener to a 6.2V one. The nice
thing about this is that its much more efficient; this may not be a
concern for you, though.

Regards,
Bob Monsen

 

[Ban]

Don Klipstein wrote:

hi , i am using 2 1 watt luxeon red leds for brake lights , can
i run these of 12 volts via a resistor, or do i need a driver
circuit ?

http://www.lumileds.com/pdfs/protected/DS23.PDF

luxeon reds have a forward voltage of between 2.25 and 3 volts,
depending on the current you want to put through them.

snip resistor method

What to do: Use a switching current regulator (or any current
regulator), or use a voltage source significantly more than the
voltage of the LED (or series string thereof) and use a "dropping
resistor". Maybe set the "roman black" power supply to 6.9 volts,
and drop the 1.4 volt difference with a 4.7 ohm resistor.

The necessity of limiting LED current has been discussed here before
enough times. I mention a bit more of this (and calculating dropping
resistors) in a web file of mine, http://www.misty.com/~don/ledd.html

Much better would be a fixed current output switching regulator such as the
LT1618. For a bike it is important not to waste any energy in form of heat.
You can even have the options of current adjustment and automatic on/off
switch. Only a few external components needed, no power resistor. See design
notes in last EDN.

--
ciao Ban
Bordighera, Italy

 

[Ban]

Ban wrote:

Much better would be a fixed current output switching regulator such
as the LT1618. For a bike it is important not to waste any energy in
form of heat. You can even have the options of current adjustment and
automatic on/off switch. Only a few external components needed, no
power resistor. See design notes in last EDN.

I copied the schematic which is a bit difficult to explain, because a boost
converter is used as a buck. Also that original is faulty, as the SW is not
connected, I guessed it goes to the other end of the coil. Efficiency is
claimed as being 70% with a 3.6V/700mA Lumiled. Hope I got it right.
+9...16V
o-+--------------------------------+
| |
| ___ 700mA ___ |
+----+--|___|--+--->|--UUU-+-->|-+
| | 0.07R | LED 47uH | PMEG2010
| | | | schottky
| | .-------------. |
| | | ISN | |
| | | | |
| +--|ISP LT1618 | |
| | | | |
| +--|Vin SW|----+
| | |
| | |
| o---|SHDN\ FB|-----------+
| | | |
| | | |
| |Iadj GND Vc | |
| '-------------' |
| | | | ___ |
--- | | +-|___|---+ |
--- | | | 10k | |
|4u7 | | --- --- |
| | | --- --- |
| | | |220p |2n2 |
+---------+---+----+---------+----+
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
view\font\fixed

--
ciao Ban
Bordighera, Italy

 

[Don Klipstein]

In article <ZM23d.6164$wV.428@attbi_s54>, Robert Monsen wrote:

mark krawczuk wrote:
hi , i am using 2 1 watt luxeon red leds for brake lights , can i run
these of 12 volts via a resistor, or do i need a driver circuit ?
what value resister would i have to use , (if it can be done this way ./.)

According to the datasheet at

http://www.lumileds.com/pdfs/protected/DS23.PDF

luxeon reds have a forward voltage of between 2.25 and 3 volts,
depending on the current you want to put through them.

If you want 300mA, then they will be at about 2.8V each, according to
the current to forward voltage graph. So, put them in series, that'll
give you 5.6V. Then, you need a resistor to drop the voltage
sufficiently. That is 300mA over 6.4V, which works out to about 22 ohms.
However, the power dissipated in the resistor will then be

P = 6.4^2 / 22 = almost 2 W. You'll need a big resistor.

You can use a network of resistors if you can't get 2W ones; if you use
10 220 ohm resistors in parallel, you can use standard quarter watt
resistors. The amount of heat generated will be the same, but it'll be
spread over the whole set. If you mount them so there is a bit of space
between them, it should be OK.

Another option would be to use a switching power supply, like the 'roman
black' step-down supply, shown here: http://www.romanblack.com/smps.htm.
Its simple to build, and you can easily make it put out 5.6V (which will
cover your two LEDs) by increasing the zener to a 6.2V one. The nice
thing about this is that its much more efficient; this may not be a
concern for you, though.

Do not apply fixed/regulated voltage to an LED - the current will be
unpredictable. The LED has a tolerance in its voltage drop (assuming the
amount of current specified in the datasheet for characterizing voltage
drop). The actual voltage drop *at specified current* has some
significant tolerance, and varies with temperature. Since the voltage
drop does not vary much with current, a tolerance in voltage drop
characteristics can mean current being far from what you want if you apply
a fixed voltage.
If you apply a fixed voltage, the amount of current usually varies
greatly with temperature. Such current variation with temperature can
reinforce the temperature change, leading to the current changing even
more. Sometimes this even results in "thermal runaway", which is the
situation where increasing temperature and increasing current reinforce
themselves - sometimes in a spectacular "runaway spiral", sometimes to a
lesser extent but to an extent great enough to seriously overheat the LED.
The LED may fail in minutes or may merely have a seriously shortened life.

What to do: Use a switching current regulator (or any current
regulator), or use a voltage source significantly more than the voltage of
the LED (or series string thereof) and use a "dropping resistor".
Maybe set the "roman black" power supply to 6.9 volts, and drop the 1.4
volt difference with a 4.7 ohm resistor.

The necessity of limiting LED current has been discussed here before
enough times. I mention a bit more of this (and calculating dropping
resistors) in a web file of mine, http://www.misty.com/~don/ledd.html

- Don Klipstein (don@misty.com, http://www.misty.com/~don/ledx.html