LED 240V AC

[George Bray]

I've just read a previous thread on this subject, on this group, dated
Dec 2003.

I should be grateful if anyone could advise on a simple circuit and
components values for 240VAC wiring, using 70 (approx) LEDs in series,
a bit like Christmas tree lights.

Please assume the LEDs are 3.5v 20ma and my UK mains supply is
nominally 240VAC.

I'm interested in using a capacitor, as previosuly suggested, to
rectify the mains. Or is there a better solution, would you say?

In any event, will I need at least one resistor, or perhaps 70
resistors, to protect each LED?

Regards
George

 

[Stefan Heinzmann]

I've just read a previous thread on this subject, on this group, dated
Dec 2003.

I should be grateful if anyone could advise on a simple circuit and
components values for 240VAC wiring, using 70 (approx) LEDs in series,
a bit like Christmas tree lights.

Please assume the LEDs are 3.5v 20ma and my UK mains supply is
nominally 240VAC.

I'm interested in using a capacitor, as previosuly suggested, to
rectify the mains. Or is there a better solution, would you say?

In any event, will I need at least one resistor, or perhaps 70
resistors, to protect each LED?

If we had a FAQ document this topic would certainly qualify for an entry.

A capacitor does not rectify mains. It limits the AC current in your
application. Diodes (and LEDs are diodes) *do* rectify the current.

You need to protect each LED from overcurrent in the forward direction
and overvoltage in the reverse direction.

A convenient way to protect a LED from overvoltage in the reverse
direction is to use LEDs in antiparallel pairs. In such a pair the LEDs
light up alternatively in each half cycle of the mains voltage.

In your case you'll end up with 35 pairs of LEDs in a string. If each
LED has a 3.5V forward voltage (you probably use blue or white LEDs),
you end up with 122.5V voltage drop for the whole string. If using a
current limiting series resistor it would need to drop about 120V at
40mA (twice the current because each LED only lights up half of the
time). This results in 3kOhm at 5Watt. This is actually greatly
simplified - the LEDs are in reality not a linear load.

To achieve the same effect with a series capacitor it would have to have
3kOhm impedance at 50Hz. You could try 1µF. But again that's a
simplification. You need a non-polarized capacitor with sufficient
voltage and current ratings, of course. It is wise to use an additional
100 Ohm series resistor to limit transient turn-on currents.

Finally be reminded that you're dealing with potentially lethal
voltages. Safety is not optional.

Siemens (now Osram) had an Application Note on this once, maybe you can
still find it on the net (Appnote 6).

 

[John Smith]

George Bray wrote:

To achieve the same effect with a series capacitor it would have to have
3kOhm impedance at 50Hz. You could try 1µF. But again that's a
simplification. You need a non-polarized capacitor with sufficient
voltage and current ratings, of course. It is wise to use an additional
100 Ohm series resistor to limit transient turn-on currents.

A word of caution: High frequency spikes can get through the capacitor
easily and cause very high LED currents. I suggest the use of a series R-C.

Just my 2 cents worth.

 

[Fred Bloggs]

I've just read a previous thread on this subject, on this group, dated
Dec 2003.

I should be grateful if anyone could advise on a simple circuit and
components values for 240VAC wiring, using 70 (approx) LEDs in series,
a bit like Christmas tree lights.

Please assume the LEDs are 3.5v 20ma and my UK mains supply is
nominally 240VAC.

I'm interested in using a capacitor, as previosuly suggested, to
rectify the mains. Or is there a better solution, would you say?

In any event, will I need at least one resistor, or perhaps 70
resistors, to protect each LED?

Regards
George

We have been through this before and it is not good to run the LEDs off
the AC, the reason being that the mains frequency is too low which
causes noticeable flicker as well as requires you to derate the LED
output from the equivalent DC due to heating effects. The best option
for the LEDs is to rectify with a bridge diode, filter with capacitor to
produce approximately 340VDC, then current limit the output into the
LEDs. You have 70 x 3.5V=245V so that the current limiting resistor
becomes (340-245)/20ma=4.7K ohm dissipating 4.7K x (20ma)^2=1.88Watts-
make this 2x 2.4K at 5W flameproof in series with each end of the LED
string. The capacitor should be 50/(2pi*50*17K) or 10.6uF, make this a
22uF at WVDC 450V or more. Use a bridge rectifier rated for 600PIV and
1.5A IF,avg, and place a 15 ohm bulk ceramic/ ceramic composition 2W
resistor in series with the AC input to the bridge. These additions come
in at maybe 5% the cost of your LED string.

Please view in a fixed-width font such as Courier.


OHMITE OY
15 OHM 2W 1.5A 600PIV *
+------------+ 2.4K 5W

--/\/\-----| ~ + |----+----------/\/\----+

| | | |
| -|>|- | | 22U LED x70
LINE | | === 450WVDC STRING blue
| BRIDGE | | |

-----------| ~ - |----+----------/\/\----+

+------------+ *
2.4K 5W
* Metal Oxide Flameproof

 

[George Bray]

Fred and others

Thank you for your considerable help re. my question.

My objective is to help light a room with these 70 (approx.) white
LEDs so I wouldn't want to lose half the illumination if they're only
on for half the time due to AC affects, running almost directly off
the mains. Is that the case?

Presumably, the bridge rectifier would provide a 'solid' DC voltage,
so I'd get full brightness out of the LEDs, and no flicker either?

As an alternative, might it be better to use a small 12v regulated
power supply, of the type used all around the house for various
appliances.

After checking that the PSU does actually produce 12v (normally they
do not):
Stage 1: Wire three 3.5v LEDs in series with a resistor (82 ohms for
12v) and connect this "array" to the 12VDC output. The LEDs should
illuminate.

Stage 2: Build 22 further arrays like the first one, and wire them all
in parallel; all 23 arrays supplied from the same 12VDC source.

I calculate that with 20ma LEDs, the overall current consumption
should be quite small (about 5 watts) even with 69 LEDs.

Will a small, cheap PSU suffice? Not too cheap, perhaps, so it is
relatively stable. Is the 12v approach safer and less liable to blow
the LEDs in the event of mains voltage spikes?

Regards
George

 

[Stefan Heinzmann]

Fred and others

Thank you for your considerable help re. my question.

My objective is to help light a room with these 70 (approx.) white
LEDs so I wouldn't want to lose half the illumination if they're only
on for half the time due to AC affects, running almost directly off
the mains. Is that the case?

The idea was that you run the LEDs at a higher peak current, so that the
average current and hence average brightness would be the same again.
The flickering could still be a problem, though.

Presumably, the bridge rectifier would provide a 'solid' DC voltage,
so I'd get full brightness out of the LEDs, and no flicker either?

The bridge rectifier would steer both half-waves through the LED,
doubling the flicker frequency (100Hz normally isn't noticeable
anymore). For a "solid" DC voltage you need to add the capacitor, as
Fred's schematic has shown.

As an alternative, might it be better to use a small 12v regulated
power supply, of the type used all around the house for various
appliances.

After checking that the PSU does actually produce 12v (normally they
do not):
Stage 1: Wire three 3.5v LEDs in series with a resistor (82 ohms for
12v) and connect this "array" to the 12VDC output. The LEDs should
illuminate.

Stage 2: Build 22 further arrays like the first one, and wire them all
in parallel; all 23 arrays supplied from the same 12VDC source.

I calculate that with 20ma LEDs, the overall current consumption
should be quite small (about 5 watts) even with 69 LEDs.

Will a small, cheap PSU suffice? Not too cheap, perhaps, so it is
relatively stable. Is the 12v approach safer and less liable to blow
the LEDs in the event of mains voltage spikes?

It is most certainly a lot safer, since you get mains isolation. If you
don't mind the additional cost for the PSU, then by all means go for it.
It also gives you considerable latitude regarding your LED arrangements.
For example, you could also go for a 24V supply or even 48V supply,
depending on what you can easily get. More volts mean fewer LED chains,
and hence fewer resistors. Watch the resistors' power rating, though.

At 12V you need about 0.5A with your setup. That is available as a
fairly cheap standard wall wart.

 

[John Woodgate]

I read in sci.electronics.design that George Bray

I'm interested in using a capacitor, as previosuly suggested, to rectify
the mains. Or is there a better solution, would you say?

Yes. Think again. If you believe a capacitor is a rectifier, you are
quite likely to kill yourself or someone else with this project.

 

[John Woodgate]

I read in sci.electronics.design that Fred Bloggs <[email protected]>

Please view in a fixed-width font such as Courier.


OHMITE OY
15 OHM 2W 1.5A 600PIV *
+------------+ 2.4K 5W
| | | |
| -|>|- | | 22U LED x70
LINE | | === 450WVDC STRING blue
| BRIDGE | | |
+------------+ *
2.4K 5W
* Metal Oxide Flameproof

While technically excellent, this is a suicide kit. The whole LED string
is at a lethal voltage to earth.

DON'T DO IT!

 

[John Woodgate]

I read in sci.electronics.design that George Bray

As an alternative, might it be better to use a small 12v regulated power
supply, of the type used all around the house for various appliances.

Yes, this is VERY MUCH BETTER AND SAFER APPROACH. But please read on.

After checking that the PSU does actually produce 12v (normally they do
not): Stage 1: Wire three 3.5v LEDs in series with a resistor (82 ohms
for 12v) and connect this "array" to the 12VDC output. The LEDs should
illuminate.
Yes.

Stage 2: Build 22 further arrays like the first one, and wire them all
in parallel; all 23 arrays supplied from the same 12VDC source.

You now have 23 x 20 mA = 460 mA current drain. Make sure that your
power supply is rated at at least 500 mA, preferably a bit more.

I calculate that with 20ma LEDs, the overall current consumption should
be quite small (about 5 watts) even with 69 LEDs.

Will a small, cheap PSU suffice? Not too cheap, perhaps, so it is
relatively stable. Is the 12v approach safer and less liable to blow the
LEDs in the event of mains voltage spikes?

Yes, and it won't kill you or your family, as the direct-on-line
suggestions could easily do.

 

[Fred Bloggs]

I read in sci.electronics.design that Fred Bloggs <[email protected]>
wrote (in <[email protected]>) about 'LED 240V AC', on Wed, 4
Feb 2004:



While technically excellent, this is a suicide kit. The whole LED string
is at a lethal voltage to earth.

DON'T DO IT!

That is not nonsense- it is quite safe. By your reasoning even extension
cords would be banned. It is a "power limited" transmission to the LED
string which I assume would use line-rated insulated wiring, and is MUCH
safer than an extension cord as ordinarily used. The resistors are
deliberately grossly underrated to avoid hot surfaces and in the event
of a catastrophic fault will open the circuit very quickly. Add a
fast-blo 50mA, or even pair, of fuses or PTC's in series with each 2.4K
if you're so worried about some trivial amount of current delivery. As
for the line transients- a bipolar TVS at 240V line rated in shunt with
the bridge ac-inputs will take care of that.

 

[Spehro Pefhany]

I read in sci.electronics.design that Fred Bloggs <[email protected]>

While technically excellent, this is a suicide kit. The whole LED string
is at a lethal voltage to earth.

DON'T DO IT!

Yeah, that was my first thought too. If you have to ask... I'd
recommend using one of those cheap halogen electronic transformer
units (12VAC isolated output).

A similar circuit to the above could be used; it will still require a
physically large capacitor because this kind of PS has a
100Hz-modulated-output of maybe 20 or 30kHz. The LEDs will have to be
in a number of series/parallel strings, but the series resistors could
then be lower power dissipation (eg. 1/4W).

Best regards,
Spehro Pefhany

 

[John Woodgate]

I read in sci.electronics.design that Fred Bloggs <[email protected]>

That is not nonsense- it is quite safe.

The OP is a beginner and we now know he wants to spread the LEDs around
a room. Your solution would be quite safe in your hands or mine, because
we know here not to put our hands, but we need to be extra-cautious when
giving advice.

Anyway, he's going for a low-voltage solution, which is far better.

 

[Fred Bloggs]

Spehro Pefhany

Yeah, that was my first thought too. If you have to ask... I'd
recommend using one of those cheap halogen electronic transformer
units (12VAC isolated output).

A similar circuit to the above could be used; it will still require a
physically large capacitor because this kind of PS has a
100Hz-modulated-output of maybe 20 or 30kHz. The LEDs will have to be
in a number of series/parallel strings, but the series resistors could
then be lower power dissipation (eg. 1/4W).

Best regards,
Spehro Pefhany

That is just more hysterical bull. There is no distinction to be made
between isolation achieved magnetically or through wiring insulation.
You might get a clue that your transformer isolation is ultimately
dependent on insulation as commonly constructed, also.

 

[Spehro Pefhany]

That is just more hysterical bull. There is no distinction to be made
between isolation achieved magnetically or through wiring insulation.
You might get a clue that your transformer isolation is ultimately
dependent on insulation as commonly constructed, also.

There is a BIG difference between insulation installed properly in a
well-run factory, by the hundreds of thousands, and verified by hi-pot
testing after assembly, and what some fellow who has to ask about
Ohm's law might come up with in his basement. And you know it- you
just want to argue, and will probably take the other side next time
this inevitably comes up.

Your circuit's fine, it's the implementation of it that could kill the
user.

Best regards,
Spehro Pefhany

 

[no_email_addy@no_email_addy.com]

That capacitors gona have some clout if your fingers get near it, though it
won't matter much if all you got between you and the mains is a 15R resistor, or
rather 0R if the - bridge input is connected to live.

Besides, wheres the fuses ?

Bare in mind, Caps often go short circuit.

 

[Fred Bloggs]

That capacitors gona have some clout if your fingers get near it, though it
won't matter much if all you got between you and the mains is a 15R resistor, or
rather 0R if the - bridge input is connected to live.

Besides, wheres the fuses ?

Bare in mind, Caps often go short circuit.

That whole package goes into an enclosure that plugs into the AC
receptacle, and the output is a wire pair going to the LEDs. What are
you thinking? When the capacitor blows or the bridge shorts, the input
resistor opens.

 

[Fred Bloggs]

Spehro Pefhany

There is a BIG difference between insulation installed properly in a
well-run factory, by the hundreds of thousands, and verified by hi-pot
testing after assembly, and what some fellow who has to ask about
Ohm's law might come up with in his basement. And you know it- you
just want to argue, and will probably take the other side next time
this inevitably comes up.

Your circuit's fine, it's the implementation of it that could kill the
user.

Best regards,
Spehro Pefhany

Point taken- I was not considering ineptitude.

 

[George Bray]

I read in sci.electronics.design that George Bray


Yes. Think again. If you believe a capacitor is a rectifier, you are
quite likely to kill yourself or someone else with this project.

I should have said I was interested in using a capacitor AS PART OF
THE CIRCUITRY to rectify the mains, as proposed in the Siemens appnote
6. I believe a diode would provide rectification. But you're right - I
know relatively little about electronics.

Why is the proposed ring of LEDs any more dangerous than Christmas
tree lights?

I certainly wouldn't touch any of the circuitry whilst it's connected
to the mains. The LEDs wiring and associated cicuitry would all be
well out of reach of anybody. But I'm already moving in favour of a
12v transformer. They're more reliable and longer lasting than an
electronic halogen-type PSU aren't they?

Regards
George

 

[John Woodgate]

I read in sci.electronics.design that George Bray

Why is the proposed ring of LEDs any more dangerous than Christmas tree
lights?

They have to pass safety tests. I don't suppose you plan to send your
set-up to a safety test house.

I certainly wouldn't touch any of the circuitry whilst it's connected to
the mains.

No, but other people almost certainly will, when you get it working.

The LEDs wiring and associated cicuitry would all be well out
of reach of anybody.
Sure?

But I'm already moving in favour of a 12v
transformer. They're more reliable and longer lasting than an electronic
halogen-type PSU aren't they?

Well, there are two sorts of halogen PSU - a transformer and an
electronic device. You don't need one anyway; they have much higher
current ratings than the 500 mA that you need for the good series-
parallel arrangement that you described earlier.

 

[George Bray]

I read in sci.electronics.design that George Bray


They have to pass safety tests. I don't suppose you plan to send your
set-up to a safety test house.

You're right. I wouldn't send my circuit to a safety test house but
that doesn't make it any more dangerous. Neither does a test itself
make Christmas tree lights any safer, if they are not safe already.
Can nobody tell me why Christmas tree lights are so safe, as designed?

No, but other people almost certainly will, when you get it working.

How can you say that? Any more than suggesting that visitors to my
house might unscrew the back panel of my TV, to deliberately
electrocute themselves whilst my back is turned.

Sure?

Yes, I am sure.

Well, there are two sorts of halogen PSU - a transformer and an
electronic device. You don't need one anyway; they have much higher
current ratings than the 500 mA that you need for the good series-
parallel arrangement that you described earlier.

Thank you.

Regards
George