Running a 1 watt LED of 6 volt lantern battery

[Dave.H]

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 at www.dse.com.au.

 

[Chuck]

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 at www.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck


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[Dave.H]

On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

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Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

 

[Dave.H]

On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:



On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

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Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

 

[Dave.H]

On Jan 24, 12:38 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:



On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

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----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

I forgot to mention I was talking about the resistor for the 9 volt
supply in my previous post.

 

[default]

On Wed, 23 Jan 2008 05:38:09 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:



On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----http://www.newsfeeds.comThe#1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

A higher power resistor will work. For 6 volts it calculates to 10.9
ohms / 1.33 watts and for 9 volts 19.4 ohms / 2.38 watts

Where are you getting 16?

6 volts minus 2.2 volts across the led = 3.8 volts across the
resistor, divided by .35 amps = 10.857 ohms

9 volts minus 2.2 volts = 6.8 volts across the resistor, divided by
..35 amps = 19.4 ohms

The battery voltage may drop a little when you pull current - a lot if
your intention is to use a 9V transistor radio battery.

A resistor will work but they say to use a constant current source in
the ad. And the ad itself is suspect - you notice all the leds are
rated at one watt and only the blue and white are actually one watt ?
Actually 1.225 watts versus .77 for the red.

You may burn it out because they don't show the data. The led itself
will be dissipating .77 watts and depending on the physical size, that
heat has to be dissipated without raising the temperature of the die
above ~100 degrees C.

You need to look at the data sheet to see if they will do what you
want.

It is normal to derate leds to compensate for ambient temperature -

I use 4 one watt leds on my motorcycle. Each one is mounted to a 1"
diameter aluminum slug to spread the heat, all four are mounted to 20
square inches of 1/8" aluminum. I run them at 10% power for dim and
100% for bright and calculated the ambient for around 37 degrees C.

Mine are called one watt leds made by Cree and dissipate 1.4 watts
(650 milliamps) and cost $8 each in the US.

What Dick Smith is showing appears to be a bare small surface mount
chip - it will take some work to make it dissipate a watt if that's
the case. Unless you just turn it on briefly . . .
--

 

[Chuck]

On Wed, 23 Jan 2008 05:41:59 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 24, 12:38 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:



On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

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----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

I forgot to mention I was talking about the resistor for the 9 volt
supply in my previous post.

Yes. Again, I would view these resistance values as minimums. Unless
you really neet to extract maximum output from the LED, you might go
as high as 37 ohms with the 9 volt battery. Ideally, you would start
with a high resistance and see what brightness results. You could then
reduce the resistance by say 10% at a time and observe the
differences. Yeah, it's much easier said than done. But you should
make sure that the voltage and current to the LED do not exceed the
maximums.

Chuck

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[Dave.H]

On Jan 24, 1:09 am, default <defa...@defaulter.net> wrote:

On Wed, 23 Jan 2008 05:38:09 -0800 (PST), "Dave.H"



the19...@googlemail.com> wrote:
On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----http://www.newsfeeds.comThe#1Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

A higher power resistor will work. For 6 volts it calculates to 10.9
ohms / 1.33 watts and for 9 volts 19.4 ohms / 2.38 watts

Where are you getting 16?

6 volts minus 2.2 volts across the led = 3.8 volts across the
resistor, divided by .35 amps = 10.857 ohms

9 volts minus 2.2 volts = 6.8 volts across the resistor, divided by
.35 amps = 19.4 ohms

The battery voltage may drop a little when you pull current - a lot if
your intention is to use a 9V transistor radio battery.

A resistor will work but they say to use a constant current source in
the ad. And the ad itself is suspect - you notice all the leds are
rated at one watt and only the blue and white are actually one watt ?
Actually 1.225 watts versus .77 for the red.

You may burn it out because they don't show the data. The led itself
will be dissipating .77 watts and depending on the physical size, that
heat has to be dissipated without raising the temperature of the die
above ~100 degrees C.

You need to look at the data sheet to see if they will do what you
want.

It is normal to derate leds to compensate for ambient temperature -

I use 4 one watt leds on my motorcycle. Each one is mounted to a 1"
diameter aluminum slug to spread the heat, all four are mounted to 20
square inches of 1/8" aluminum. I run them at 10% power for dim and
100% for bright and calculated the ambient for around 37 degrees C.

Mine are called one watt leds made by Cree and dissipate 1.4 watts
(650 milliamps) and cost $8 each in the US.

What Dick Smith is showing appears to be a bare small surface mount
chip - it will take some work to make it dissipate a watt if that's
the case. Unless you just turn it on briefly . . .
--

I'm looking at some Cree LEDs at Jaycar Electronics
(*www.jaycar.com.au) Cost a bit more but probably better quality then
the generic DSE ones.

 

[default]

On Wed, 23 Jan 2008 06:26:40 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 24, 1:09 am, default <defa...@defaulter.net> wrote:
On Wed, 23 Jan 2008 05:38:09 -0800 (PST), "Dave.H"



the19...@googlemail.com> wrote:
On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----http://www.newsfeeds.comThe#1Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

A higher power resistor will work. For 6 volts it calculates to 10.9
ohms / 1.33 watts and for 9 volts 19.4 ohms / 2.38 watts

Where are you getting 16?

6 volts minus 2.2 volts across the led = 3.8 volts across the
resistor, divided by .35 amps = 10.857 ohms

9 volts minus 2.2 volts = 6.8 volts across the resistor, divided by
.35 amps = 19.4 ohms

The battery voltage may drop a little when you pull current - a lot if
your intention is to use a 9V transistor radio battery.

A resistor will work but they say to use a constant current source in
the ad. And the ad itself is suspect - you notice all the leds are
rated at one watt and only the blue and white are actually one watt ?
Actually 1.225 watts versus .77 for the red.

You may burn it out because they don't show the data. The led itself
will be dissipating .77 watts and depending on the physical size, that
heat has to be dissipated without raising the temperature of the die
above ~100 degrees C.

You need to look at the data sheet to see if they will do what you
want.

It is normal to derate leds to compensate for ambient temperature -

I use 4 one watt leds on my motorcycle. Each one is mounted to a 1"
diameter aluminum slug to spread the heat, all four are mounted to 20
square inches of 1/8" aluminum. I run them at 10% power for dim and
100% for bright and calculated the ambient for around 37 degrees C.

Mine are called one watt leds made by Cree and dissipate 1.4 watts
(650 milliamps) and cost $8 each in the US.

What Dick Smith is showing appears to be a bare small surface mount
chip - it will take some work to make it dissipate a watt if that's
the case. Unless you just turn it on briefly . . .
--

I'm looking at some Cree LEDs at Jaycar Electronics
(*www.jaycar.com.au) Cost a bit more but probably better quality then
the generic DSE ones.

Yeah, those are mounted to aluminum heat spreaders. I only saw white
high power leds are you going for white or red? Z4251 Is red

http://www.dse.com.au/cgi-bin/dse.storefront/4797468104a33ad8273fc0a87f9c071c/Product/View/Z4251

That might explain the difference in resistors we calculated
--

 

[Ecnerwal]

In article
<904f5e87-56b8-4aa1-a26b-4df351c06430@q39g2000hsf.googlegroups.com>,
"Dave.H" <the1930s@googlemail.com> wrote:

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the

Do you have any interest in max light out, or max battery life?

This is basics, so perhaps just slapping in a resistor is as far as you
want to go - but you could run 2 in series from the 6 volt battery, and
3 from the 9 volt battery, at no increase in power drawn - just making
more into light, and less into heat from resistors.

Alternatively, you could get somewhat more complicated and build a small
switch-mode driver circuit to get 2-3 times the battery life, with the
same light out, by converting more voltage at less current into less
voltage at more current, again reducing the power lost as heat from
resistors.

If you need more power dissipation, you can use resistors in parallel
and/or series to increase power without having to go track down a
high-power resistor. i.e., if you need 20 ohms at 2 watts, you can
parallel 2 1-watt 40 ohm resistors (39 in practice), or run 4 5 ohm (5.1
in practice) 1/2-watt resistors in series.

--
Cats, coffee, chocolate...vices to live by

 

[Dave.H]

On Jan 24, 1:53 am, Ecnerwal <LawrenceSM...@SOuthernVERmont.NyET>
wrote:

In article
904f5e87-56b8-4aa1-a26b-4df351c06...@q39g2000hsf.googlegroups.com>,

"Dave.H" <the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the

Do you have any interest in max light out, or max battery life?

****max light output

If you need more power dissipation, you can use resistors in parallel
and/or series to increase power without having to go track down a
high-power resistor. i.e., if you need 20 ohms at 2 watts, you can
parallel 2 1-watt 40 ohm resistors (39 in practice), or run 4 5 ohm (5.1
in practice) 1/2-watt resistors in series.

**** I was sitting here trying to remember where I read about that, I
think it was a radio restoration website.

 

[Dave.H]

On Jan 24, 1:49 am, default <defa...@defaulter.net> wrote:

On Wed, 23 Jan 2008 06:26:40 -0800 (PST), "Dave.H"



the19...@googlemail.com> wrote:
On Jan 24, 1:09 am, default <defa...@defaulter.net> wrote:
On Wed, 23 Jan 2008 05:38:09 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----http://www.newsfeeds.comThe#1NewsgroupService in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

A higher power resistor will work. For 6 volts it calculates to 10.9
ohms / 1.33 watts and for 9 volts 19.4 ohms / 2.38 watts

Where are you getting 16?

6 volts minus 2.2 volts across the led = 3.8 volts across the
resistor, divided by .35 amps = 10.857 ohms

9 volts minus 2.2 volts = 6.8 volts across the resistor, divided by
.35 amps = 19.4 ohms

The battery voltage may drop a little when you pull current - a lot if
your intention is to use a 9V transistor radio battery.

A resistor will work but they say to use a constant current source in
the ad. And the ad itself is suspect - you notice all the leds are
rated at one watt and only the blue and white are actually one watt ?
Actually 1.225 watts versus .77 for the red.

You may burn it out because they don't show the data. The led itself
will be dissipating .77 watts and depending on the physical size, that
heat has to be dissipated without raising the temperature of the die
above ~100 degrees C.

You need to look at the data sheet to see if they will do what you
want.

It is normal to derate leds to compensate for ambient temperature -

I use 4 one watt leds on my motorcycle. Each one is mounted to a 1"
diameter aluminum slug to spread the heat, all four are mounted to 20
square inches of 1/8" aluminum. I run them at 10% power for dim and
100% for bright and calculated the ambient for around 37 degrees C.

Mine are called one watt leds made by Cree and dissipate 1.4 watts
(650 milliamps) and cost $8 each in the US.

What Dick Smith is showing appears to be a bare small surface mount
chip - it will take some work to make it dissipate a watt if that's
the case. Unless you just turn it on briefly . . .
--

I'm looking at some Cree LEDs at Jaycar Electronics
(*www.jaycar.com.au) Cost a bit more but probably better quality then
the generic DSE ones.

Yeah, those are mounted to aluminum heat spreaders. I only saw white
high power leds are you going for white or red? Z4251 Is red

http://www.dse.com.au/cgi-bin/dse.storefront/4797468104a33ad8273fc0a8...

That might explain the difference in resistors we calculated
--

Red. They also have blue versions but they're about 1/3 as bright as
the red/white.

 

[John Fields]

On Wed, 23 Jan 2008 05:41:59 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 24, 12:38 am, "Dave.H" <the19...@googlemail.com> wrote:
On Jan 24, 12:28 am, "Dave.H" <the19...@googlemail.com> wrote:



On Jan 23, 11:58 pm, Chuck <nos...@nospam.at.all> wrote:

On Wed, 23 Jan 2008 02:40:57 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so
it will get hot. A 2 watt resistor would be appropriate.

The battery voltage will drop slightly under load but you might prefer
to operate the LED at a lower current, say 300 mA or even 250 mA to
give the LED longer life. It is not cheap!

A similar calculation can be done for the 9 volt supply.

Chuck

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Thanks, the only resistor I can get that is over 1 watt is a 5 watt
wirewound, I can't see why that wouldn't work.

if I've calculated it correct I would need a 16 ohm resistor. Is this
right? I don't want to risk damaging the LED!

I forgot to mention I was talking about the resistor for the 9 volt
supply in my previous post.

---
Regardless of the supply voltage, the formula for calculating the
value of the resistor is:

Vs - Vled
Rs = -----------
Iled

Where Rs is the resistance of the resistor in ohms,
Vs is the supply voltage, in volts,
Vled is the _minimum_ specified forward voltage of the LED, and
ILED is the desired current through the LED.

Also,

Pd(rs) = (Vs - Vled) * Iled

where Pd(rs) is the power which will be dissipated in the series
current-limiting resistor.

Teach a man to fish...

JF

 

[Ben Jackson]

On 2008-01-23, Dave.H <the1930s@googlemail.com> wrote:

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this?

You got a lot of answers to your specific question, but a better question
might have been: How should I power this LED from a lantern battery?

If you're converting to LED to save power, it doesn't make a lot of sense
to burn more power in the current limiting resistor than the LED.

--
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/

 

[David L. Jones]

On Jan 23, 9:40 pm, "Dave.H" <the19...@googlemail.com> wrote:

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC.  What value resistor would I need for this?  I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

Don't use a resistor, that's going to suck, brightness will drop with
the rapid fall in battery voltage.
Use a proper DC-DC 350mA constant current converter designed for these
LED's.
A 6V converter is not easy to find, but 3V converters designed to run
off two D cells are readily available, like this one:
http://shop.ata.org.au/cart.php?target=product&product_id=16286&category_id=320
If you used one of those D cell latern battery holders like this:
http://www.jaycar.com.au/productView.asp?ID=PH9224
then you can use the cheaper D cells and have room to mount a module.

For 9V use get another module like this one desing for the higher
input voltage:
http://cgi.ebay.com.au/1W-LED-Driver-for-Luxeon-White-Green-Blue_W0QQitemZ230213812311QQihZ013QQcategoryZ66954QQrdZ1QQssPageNameZWD2VQQcmdZViewItem?_trksid=p1638.m122

Dave.

 

[David L. Jones]

On Jan 24, 2:00 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 24, 1:53 am, Ecnerwal <LawrenceSM...@SOuthernVERmont.NyET
wrote:> In article
904f5e87-56b8-4aa1-a26b-4df351c06...@q39g2000hsf.googlegroups.com>,

"Dave.H" <the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the

Do you have any interest in max light out, or max battery life?

****max light output

In that case you may be very disappointed indeed.
These cheap asian copy 1W LED's aren't nearly as bright as the genuine
Luxeon ones, and even the luxeon ones are old hat now compared to the
Cree brand. The Cree are much more efficient than Luxeon, and Luxeon
in turn are much more efficient than the cheap copies.
All "1W" LEDs are not the same, they can have *vastly different* light
outputs.

Also, unless you drive the LED with it's maximum current, light output
will be very significantly reduced. That's why you should be using a
constant current driver, so the current remains the same as the
battery voltage drops.

Dave.

 

[Dave.H]

On Jan 24, 10:41 am, "David L. Jones" <altz...@gmail.com> wrote:

On Jan 24, 2:00 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 24, 1:53 am, Ecnerwal <LawrenceSM...@SOuthernVERmont.NyET
wrote:> In article
904f5e87-56b8-4aa1-a26b-4df351c06...@q39g2000hsf.googlegroups.com>,

"Dave.H" <the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the

Do you have any interest in max light out, or max battery life?

****max light output

In that case you may be very disappointed indeed.
These cheap asian copy 1W LED's aren't nearly as bright as the genuine
Luxeon ones, and even the luxeon ones are old hat now compared to the
Cree brand. The Cree are much more efficient than Luxeon, and Luxeon
in turn are much more efficient than the cheap copies.
All "1W" LEDs are not the same, they can have *vastly different* light
outputs.

Also, unless you drive the LED with it's maximum current, light output
will be very significantly reduced. That's why you should be using a
constant current driver, so the current remains the same as the
battery voltage drops.

Dave.

What is a constant current driver, and How would I go about building
one? Is this similar to regulated circuits or the same thing?

 

[ehsjr]

Dave.H wrote:

I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 at www.dse.com.au.

I didn't see a constant current solution posted, probably
because it is costs more than using just a series resistor.
But constant current was mentioned, so here's a constant
current circuit that uses 1 watt resistors and an LM317
chip, and works either 6 or 9 volts input. It provides
about 347 mA to the LED.

-----
+ ----Vin|LM317|Vout---+-----+
----- | |
Adj [3R9] [47R]
| | |
+----------+-----+ I = 347 mA
|
[LED]
|
Gnd --------------------------+


The chip needs to be installed on a heat sink. I would omit the 47
ohm resistor to reduce the current to about 320 mA. That's kinder
to the LED.

Dick Smith has the heatsink for $1.25, the LM317 for $2.10 and
the resistors for 10 cents each.

Ed

 

[David L. Jones]

On Jan 24, 11:13 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 24, 10:41 am, "David L. Jones" <altz...@gmail.com> wrote:



On Jan 24, 2:00 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 24, 1:53 am, Ecnerwal <LawrenceSM...@SOuthernVERmont.NyET
wrote:> In article
904f5e87-56b8-4aa1-a26b-4df351c06...@q39g2000hsf.googlegroups.com>,

"Dave.H" <the19...@googlemail.com> wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the

Do you have any interest in max light out, or max battery life?

****max light output

In that case you may be very disappointed indeed.
These cheap asian copy 1W LED's aren't nearly as bright as the genuine
Luxeon ones, and even the luxeon ones are old hat now compared to the
Cree brand. The Cree are much more efficient than Luxeon, and Luxeon
in turn are much more efficient than the cheap copies.
All "1W" LEDs are not the same, they can have *vastly different* light
outputs.

Also, unless you drive the LED with it's maximum current, light output
will be very significantly reduced. That's why you should be using a
constant current driver, so the current remains the same as the
battery voltage drops.

Dave.

What is a constant current driver, and How would I go about building
one? Is this similar to regulated circuits or the same thing?

Use Google, plenty of circuits available, here is one that uses a
purpose designed chip:
http://www.linear.com/pc/productDetail.jsp?navId=LTC3490

http://en.wikipedia.org/wiki/Current_source

A constant current regulator puts out a constant current regardless of
the input (battery) voltage.
Yes, it is a regulator, but in this case it's a current regulator
instead of a voltage regulator.

No need to build one, plenty of ones you can buy off the shelf.

Dave.

 

[David L. Jones]

On Jan 24, 12:05 pm, ehsjr <eh...@bellatlantic.net> wrote:

Dave.H wrote:
I'm looking to convert a flashlight I have to LED, the flashlight uses
a 6 volt "lantern battery but the LED's max voltage rating is 3.1
volts DC. What value resistor would I need for this? I also want to
run it off a 9 volt supply, again I need to know the value of the
resistor.

The LED is part # Z4251 atwww.dse.com.au.

I didn't see a constant current solution posted, probably
because it is costs more than using just a series resistor.
But constant current was mentioned, so here's a constant
current circuit that uses 1 watt resistors and an LM317
chip, and works either 6 or 9 volts input. It provides
about 347 mA to the LED.

-----
+ ----Vin|LM317|Vout---+-----+
----- | |
Adj [3R9] [47R]
| | |
+----------+-----+ I = 347 mA
|
[LED]
|
Gnd --------------------------+

The chip needs to be installed on a heat sink. I would omit the 47
ohm resistor to reduce the current to about 320 mA. That's kinder
to the LED.

That circuit is not entirely suitable for use with a 6V battery, as
the constant current will not be maintained once the battery voltage
drops low enough.

Dave.