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LED Lighting PSU

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Bigus

Jan 1, 1970
0
Hi

I want to experiment with powering LEDs from a 220V mains supply. I note you
can get GU10 style LED bulbs these days but was wondering what circuitry
they have in them. Has anyone seen/got a circuit diagram for such a thing?

Presumably, if using standard LEDs then the circuit could consist of a
bridge rectifier, capacitor & then a suitable resistor?

I want to also play with Luxeon LEDs aswell. Trouble is a single Luxeon 3W
white LED connected to the mains and driven at 800mA would mean a power
dissipation of 175W, so that would be transformer territory wouldn't it?

Regards
 
T

T

Jan 1, 1970
0
Hi

I want to experiment with powering LEDs from a 220V mains supply. I note you
can get GU10 style LED bulbs these days but was wondering what circuitry
they have in them. Has anyone seen/got a circuit diagram for such a thing?

Presumably, if using standard LEDs then the circuit could consist of a
bridge rectifier, capacitor & then a suitable resistor?

I want to also play with Luxeon LEDs aswell. Trouble is a single Luxeon 3W
white LED connected to the mains and driven at 800mA would mean a power
dissipation of 175W, so that would be transformer territory wouldn't it?

Regards

You could use a step down transformer before you hit the rectifier that
way you can get the voltage down towards the 5V or a little over using a
voltage regulator.

The thing with LED's is current. You need to supply enough for the
junction to work.
 
B

Bigus

Jan 1, 1970
0
Electronworks.co.uk said:
"Bigus" <[email protected]> wrote in message
Not sure what your electronics knowledge is like, but I would use Power
Integrations. Here is such an applicaiton circuit:

http://www.powerint.com/sites/default/files/PDFFiles/di74.pdf

However, this will give you a non isolated circuit. If you want something
simpler, use a transformer to bring the voltage down to something sensible
(5Vac) ... and isolated... then use the rectifier capacitor and resistor
you mention.

If the forward voltage across your LED is 3V and you have 7V coming out of
your transformer (5V rectified = 5 x 1.41), then your resistor is I =
(7-3)/R

Hi. Thanks for the link to that circuit. My knowledge isn't brill - I can
create a circuit from a diagram but things get hazy when it comes to
modifying a circuit to vary things, like the number and types of LEDs.

That circuit looks like it has a current limit of 100mA, so it sounds like
20 LEDs would be the limit there (some LEDs are 20mA so might squeeze a few
more in a I suppose). I guess that's the kind of circuit that'd be used in a
GU10 LED array as it probably can be made pretty small.

For currents like the Luxeon III though, with current demands of 800mA for
just one 3W LED (and I was ganting to try 3 at a time in that department!),
do you think a transformer would be the only way?

Regards
SPencer
 
B

Bigus

Jan 1, 1970
0
You can do 800mA easily with an offline switched mode power supply like
those sold by Poower Integrations. However, if your knowledge of
electronics is only basic, stick with a transformer. You can get these in
varieties of voltages and currents and they are easier to design.

Have a look at:
http://www.electronworks.co.uk/techarticles/linearpsu/linearpsu.htm

Thanks, that's a good (easy to follow!) article. I've bookmarked your site
too :)

One other question. I just bought a switchable mains adapter that I was
going to use for initial testing of the LEDs because it said it can handle
up to 1.2 Amps. Looking at the label though it says the following:

Input: 230V / 0.28A MAX
Output: 3/4.5/5V = 1.2A, 6/9V = 1A, 12V = 0.9A

This is my patchy electronics knowledge showing but I don't understand why
the input says 0.28A. I thought there was just one current in an electronic
device, that which the load creates - i.e: in the case of the Luxeon 3W LED
around 800mA (it can actually take up to 1A). Where does the 0.28A come into
it and will it be OK for driving the LED?

Thanks
Bigus
 
B

Bigus

Jan 1, 1970
0
John Fields said:
So, the LED will be using (dissipating):

P = IE = 0.8A * 3.5V = 2.8 watts.

In order to supply that, the power supply must take 2.8 watts from the
230V mains and transfer it to the LED at 3.5V.

With the mains at 230V and the power needed by the LED at 2.8 watts, we
can rearrange to solve for the mains current:

P 2.8W
I = --- = ------ = 0.012 amperes
E 230V


It'll me more because of the efficiency of the supply being less than
100%, but you get the idea, yes?

Yes! Awesome, thank you. I love it when someone explains something v.clearly
the penny finally drops :)

When I first looked at the supply current of the Luxeon and saw it could
take a max current of 1A, I was thinking "blimey, so I could only run 13 of
them on one mains plug!!!" (given a 13A fuse). In actual fact one could run
more like 1000 of them :)
Now, on to your mains adapter.
[..]
Here's the power it can deliver into a load at the different settings:

VOLTAGE CURRENT POWER
VOLTS AMPS WATTS
----------+--------+--------
12 0.9 10.8
9 1 9
6 1 6
5 1.2 6
4.5 1.2 5.4
3 1.2 3.6

Now, since the input voltage is 230V and the maximum current it will
draw from the mains is .28A, it'll be taking 64 watts (Volt-Amperes,
actually, but that's for another post) from the mains when the 12V is
fully loaded.

So you can see that the load current doesn't have to be the same as the
current being taken from the mains, and now you know why, Yes?

One thing about your supply; notice that for 10.8 watts out it's taking
64 watts from the mains, which means it's running with an efficiency of
about:

Pout * 100 10.8 * 100
E(%)= ------------ = ------------ = 16.9%
Pin 64

I see. Theoretically then, the current on the 230V input side when supplying
12V at 0.9A, should be about 47mA. So, you think the fact it says 0.28A max
on the input side is a clue to the inefficiency of the adaptor rather than
some kind of surge rating?

If I did build a power supply myself, with a fixed supply and using the
transformer/rectifier/resistor approach, would that have an inherent level
of inefficiency as well?

Regards
Bigus
 
G

greg

Jan 1, 1970
0
Bigus said:
If I did build a power supply myself, with a fixed supply and using the
transformer/rectifier/resistor approach, would that have an inherent level
of inefficiency as well?

There will be some inefficiency, particularly if you use
a very small transformer, since the efficiency of a
transformer gets better the bigger you make it.

I'd be surprised if it were *that* inefficient, though.
17% sounds rather apallingly bad!

In any case, you'll get better efficiency if you choose
a transformer with a somewhat higher power rating than
you intend to use. (Not *too* much higher, though, since
the no-load loss goes up with the transformer size as
well.)
 
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