LED Lighting PSU

B

Bigus

Guest
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
 
"Bigus" <a_person@anyolddomain.fake> wrote in message
news:4so5m.13965$Il.7959@newsfe16.iad...
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
Click to expand...
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


--
Bill Naylor
www.electronworks.co.uk
Electronic Kits for Education and Fun
 
In article <4so5m.13965$Il.7959@newsfe16.iad>,
a_person@anyolddomain.fake says...
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
Click to expand...
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.
 
"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:3PidnUa6HtD1e8vXnZ2dnUVZ8s2dnZ2d@bt.com...
"Bigus" <a_person@anyolddomain.fake> 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
Click to expand...
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
 
"Bigus" <a_person@anyolddomain.fake> wrote in message
news:LPL5m.27491$Wx7.18455@newsfe04.iad...
"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:3PidnUa6HtD1e8vXnZ2dnUVZ8s2dnZ2d@bt.com...
"Bigus" <a_person@anyolddomain.fake> 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
Click to expand...
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

If you need help

--
Bill Naylor
www.electronworks.co.uk
Electronic Kits for Education and Fun
 
"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:vvadnbc6JLS3FMXXnZ2dnUVZ8gidnZ2d@bt.com...

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
Click to expand...
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
 
On Sat, 11 Jul 2009 15:42:25 +0100, "Bigus" <a_person@anyolddomain.fake>
wrote:

"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:vvadnbc6JLS3FMXXnZ2dnUVZ8gidnZ2d@bt.com...

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?
Click to expand...
---
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)


+---------+
IN>----| |----->OUT>---+
| POWER | |
| SUPPLY | [LOAD]
| | |
IN>----| |----->OUT>---+
+---------+

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
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?

Now, on to your mains adapter.

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


+-----+ S1 +------+
MAINS>--------| 12|---O<---O--|~ +|----+---->+OUT
| | | | |
| 9|---O | | |
| 6|---O | | |+
| | | | [BFC]
| 5|---O | | |
| 4.5|---O | | |
| 3|---O | | |
| | | | |
MAINS>--------| |-----------|~ -|----+---->GND
+-----+ +------+
XFMR FWB


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


Ouch!

JF
 
"Bigus" <a_person@anyolddomain.fake> wrote in message
news:qx16m.20340$%02.9616@newsfe15.iad...
"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:vvadnbc6JLS3FMXXnZ2dnUVZ8gidnZ2d@bt.com...

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
Click to expand...
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:

10.8/230 = 0.047A (47mA).

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

--
Bill Naylor
www.electronworks.co.uk
Electronic Kits for Education and Fun
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:vgch55l2q852ppsgl6bfivcpm65c4rs62g@4ax.com...
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?
Click to expand...
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
Click to expand...
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
 
"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:2P-dnejlH9yMfsXXnZ2dnUVZ8uSdnZ2d@bt.com...
"Bigus" <a_person@anyolddomain.fake> wrote in message
news:qx16m.20340$%02.9616@newsfe15.iad...

"Electronworks.co.uk" <newsgroups@electronworks.co.uk> wrote in message
news:vvadnbc6JLS3FMXXnZ2dnUVZ8gidnZ2d@bt.com...

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


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:

10.8/230 = 0.047A (47mA).

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).
Click to expand...
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
 
Bigus wrote:

I'm curious though - why shouldn't one try to measure the current on
the 230V side
Click to expand...
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!)

--
Greg
 
Bigus wrote:

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?
Click to expand...
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.)

--
Greg
 
On Wed, 15 Jul 2009 08:49:59 +0100, "Bigus" <a_person@anyolddomain.fake>
wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:vgch55l2q852ppsgl6bfivcpm65c4rs62g@4ax.com...
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 :)
Click to expand...
---
Yup!

2990W / 2.8W ~ 1068
---

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?
Click to expand...
---
Yes, but it's unusually low, so I suspect that 280mA also includes the
reactive current caused by the adapter's power factor being non-unity.
---

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?
Click to expand...
---
Yes.

There will be I˛R losses in the transformer windings, the rectifiers,
and the resistor, and there will be eddy current losses in the
transformer core as well.

JF
 
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