LED Lighting PSU

Not sure what your electronics knowledge is like, but I would use Power Integrations. Here is such an applicaiton circuit:

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

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.

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

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

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:

If you need help

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

--- Ahhhhh... It's clear now. :-)

In this instance we're talking about the current going into the load, which is different from the current going into the supply.

Here: (View in Courier)

What's happening here is that the power supply is transferring _power_ from its input to its output by converting the voltage at the input to the voltage the load needs to work properly, and supplying the current the load needs as well.

For example, if you're working with 240V mains and the LED needs to have the supply generate 3.5V when there's 800mA through the LED, then the LED needs:

P = IE

where P is the power in watts, I is the current in amperes, and E is the voltage in volts.

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

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

Now, on to your mains adapter.

What you have there is something that probably looks like this:

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

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

Ouch!

JF

Hi Bigus

If you take a current from the output, that current has to come from somewhere. With a transformer, they are 'constant power' devices, so power supplied to your load is equal to the current the mains input has to supply.

So if you multiply the output current by the output voltage you will get the POWER taken by the load, since P = I x V. If the device is a constant power device, this is the power that the input has to supply.

For a 12V output, your transformer can supply a maximum current of 0.9A. Thus your maximum load is 10.8W.

With a normal transformer, with 230V input, this would imply an input current of:

I cannot work out why your input current is specified at 0.28A. Maybe this is a surge rating, but with a 230V input, your input current should be 47mA. OK it will be a little bit larger as your transformer will not be 100% efficient, so you have to put in more current...

In summary, just look at the output figures and do not worry about the 230V side (especially DO NOT try to measure the 230V current).

Good to see you are using a ready made transformer - keeps it much safer

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 :-)

I see. Theoretically then, the current on the 230V input side when supplying

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

Thanks again for the explanation, as per my reply to John, it's all clear now! I'm curious though - why shouldn't one try to measure the current on the 230V side (as it happens my multimeter is only specified to measure DC current anyway). As it happens I have one of those current/voltage.power/kWh measuring devices that plugs into your mains socket and the device plugs into it, so that would probably reveal the input current.

Regards Spencer

Safety -- both yours and your multimeter's.

For you, the consequences of accidentally contacting the mains could be very bad, so the less opportunity you give yourself of doing that, the better.

As for your multimeter, even if it apparently has a high enough voltage or current scale for the job, the mains can carry voltage spikes *much* higher than the nominal voltage. Unless your meter's manual says explicitly that it is designed for direct measurement of the mains, you risk destroying it. (And keep in mind that the mains is capable of supplying a *lot* of energy for the purpose of destroying things!)

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.

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.)