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'Using LED in 220VAC circuit [OT]'
2000\03\14@085756 by

Hello everyone..

What is the easies way of powering LEDs straight off a 220 V line?? The
obvious options to me are:
1. Led in series with 5 W resistor (current limiting) and 1N4007 or similar
diode (prevent reverse breakdown)
2. Use a relay switched from 220 V to power the LED from a lower voltage.

The first of these methods requires a lot of heat dissipation from the
resistor, and the second is not very elegant either.. Is there a better
solution?

Just to pre-empt a question: I am using LEDs and not neon or something
better suited to 220 VAC because most of my circuit and LEDS run on 5 V DC
..

thanks
Roland

{Quote hidden}

I believe that you can use a series cap, suitably rated, and use its
reactance to limit the current, again using diode across led to limit
reverse voltage.
andy farrar

At 02:48 PM 3/14/00 +0200, you wrote:
>Hello everyone..
>
>What is the easies way of powering LEDs straight off a 220 V line?? The
>obvious options to me are:
>1. Led in series with 5 W resistor (current limiting) and 1N4007 or similar
>diode (prevent reverse breakdown)
>2. Use a relay switched from 220 V to power the LED from a lower voltage.
>
>The first of these methods requires a lot of heat dissipation from the
>resistor, and the second is not very elegant either.. Is there a better
>solution?

1) You could use a higher efficiency LED and cut down on the resistor
power dissipation. For example, at 1mA average, the power would only
be 1/4W (use a 1/2W resistor). I suggest an inverse-parallel diode
across the LED to carry leakage current from the 1N400X, for example
a 1N4148.

2) You could use a capacitive dropper

LED
220 A ------C1 --- R1------ BR1 ~    + ----|>|--

|                                       |
R2                                      |
|                                       |
220 B --------------------- BR2 ~    - ---------

Eg. R1 = 200R  1/2W flameproof MOF (it dissipates < 0.1W)
R2 = 1M    rated for line voltage (dissipates < 0.1W)
C1 = 0.22uF/630V safety approved for line voltage
BR1 = W02M bridge rectifier (voltage rating not important)
The LED gets around 15mA with the above circuit.

Note that you have to be careful to prevent contact between
people and any part of the circuit in 2), for safety reasons.
DEATH or injury can occur if contact does take place.
The plastic on LED's may not be enough to insulate the
circuit safely. If in doubt, DO NOT use this circuit.
Similar cautions apply to the line connected portions of
all of these circuits.

3) You could use an optocoupler to detect the AC line and switch
low voltage.

4) A small transformer will work as well, and is the
most reliable and safe solution.

>Just to pre-empt a question: I am using LEDs and not neon or something
>better suited to 220 VAC because most of my circuit and LEDS run on 5 V DC

5) You can use a neon. ;-)

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
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speffinterlog.com
Fax:(905) 271-9838                      (small micro system devt hw/sw + mfg)
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

<x-flowed>At 02:48 PM 3/14/00 +0200, Roland Andrag wrote:
>Hello everyone..
>
>What is the easies way of powering LEDs straight off a 220 V line?? The
>obvious options to me are:
>1. Led in series with 5 W resistor (current limiting) and 1N4007 or similar
>diode (prevent reverse breakdown)
>2. Use a relay switched from 220 V to power the LED from a lower voltage.
>
>The first of these methods requires a lot of heat dissipation from the
>resistor, and the second is not very elegant either.. Is there a better
>solution?
>
>Just to pre-empt a question: I am using LEDs and not neon or something
>better suited to 220 VAC because most of my circuit and LEDS run on 5 V DC

What is powering the rest of your circuit?  Can all parts of your circuit
be considered to be isolated from any external contact?

If your circuit is suitable for a direct line power circuit (*NO*
isolation), you can use a capacitive power supply with a twist.  My
standard capacitive PSU consists of a resistor and capacitor in series from
the line to a clamp zener diode followed by another series diode to the
reservoir capacitor.  The added twist is a LED and anti-parallel connected
diode in series with the dropping components with a transistor shunting the
LED.  Assuming the zener is 5.6V, you will get about 5 Vdc from the
supply.  If you cannot tolerate ripple, set the zener voltage to about 6.8V
and use a LDO to drop down to 5V.  I normally don't bother with this but if
you are using the onboard a/d with the supply rail as the reference, you
might have to.

The resistors feeding the base of the transistor are sized to guarantee
that the transistor is OFF when the pic pin is HI and saturated when the
pic pin is LO.

The usual warnings apply with this circuit - any contact with humans or any
grounded object is dangerous and can be lethal.  If any portion of your
circuit can be touched accidently by the outside world, *DO NOT* use this
type of power supply!

Horrid ascii art follows (use mono spaced font for viewing)

4k7           10k
-/\/\-+--------/\/\/------- to pic pin
|      |  pnp
|   e --- c
+ ---/   \--+
-----/\/\/----||---+ ---->|----+---->|-------+-----+dc to reg or to circuit
+ ----|<----+             |
led    |             |
|            ---
\---\          ---
zener/ \           |
|             |
-------------------------------+-------------+---- circuit common

dwayne

Dwayne Reid   <dwaynerplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax

Celebrating 16 years of Engineering Innovation (1984 - 2000)

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Do NOT send unsolicited commercial email to this email address.
This message neither grants consent to receive unsolicited
commercial email nor is intended to solicit commercial email.

</x-flowed>
> I believe that you can use a series cap, suitably rated, and use its
> reactance to limit the current, again using diode across led to limit
> reverse voltage.

Except if there is any series harmonic content on the line.....

If the customer hooks the item to a square wave mains inverter...... Poof
!!!...

........................ Zim

I just wanted to thank everyone who too the trouble to reply to my question!
Looks like I will have to do a bit of maths to figure out how the cap in
series with rectifier works.. I will be very glad to get rid of 10 W
resistors though!

Thanks again
Roland

In message <001101bf926f\$27c65e00\$df36fea9@roland>, Roland Andrag
<randragICON.CO.ZA> writes
>I just wanted to thank everyone who too the trouble to reply to my question!
>Looks like I will have to do a bit of maths to figure out how the cap in
>series with rectifier works.. I will be very glad to get rid of 10 W
>resistors though!

I would be interested to see how your maths works in practice, many
years ago a particular range of Ferguson B/W TV's used a capacitor to
feed the heater chain (rather than the usual huge ballast resistor).
Being an inquisitive type I worked out the reactance of the capacitor at
the 50Hz mains frequency - it was nowhere near the required value to set
the current at the required 300mA for the valve (tube for our USA
readers!) heaters. I've always presumed there was more to it than simply
the reactance of the capacitor, but have never really understood why?.
--

Nigel.

/--------------------------------------------------------------\
| Nigel Goodwin   | Internet : nigelglpilsley.demon.co.uk     |
| Lower Pilsley   | Web Page : http://www.lpilsley.demon.co.uk |
| Chesterfield    | Official site for Shin Ki and New Spirit   |
| England         |                 Ju Jitsu                   |
\--------------------------------------------------------------/

Nigel Goodwin <nigelgLPILSLEY.DEMON.CO.UK> said:
> I would be interested to see how your maths works in practice, many
> years ago a particular range of Ferguson B/W TV's used a capacitor to
> feed the heater chain (rather than the usual huge ballast resistor).
> Being an inquisitive type I worked out the reactance of the capacitor at
> the 50Hz mains frequency - it was nowhere near the required value to set
> the current at the required 300mA for the valve (tube for our USA
> readers!) heaters. I've always presumed there was more to it than simply
> the reactance of the capacitor, but have never really understood why?.

The reactance of the capacitor is 90 degrees out of phase with the reactance of the
heater chain so they don't simply add together when placed in series.

The total reactance of the Heaters in series with the capacitor will be

square_root of ( Heater_resistance squared plus capacitor_reactance squared )

Hence to reduce the current the reactance of the capacitor has to be higher than
you expected when you thought they simply added.

Brian Gregory.
briangcix.co.uk

1uF passes approx 70mA

> I just wanted to thank everyone who too the trouble to reply to my
> question ! Looks like I will have to do a bit of maths to figure out how
> the cap in series with rectifier works.. I will be very glad to get rid of
> 10 W resistors though!

> In message <001101bf926f\$27c65e00\$df36fea9@roland>, Roland Andrag
> <randragICON.CO.ZA> writes

> >I just wanted to thank everyone who too the trouble to reply to my
question!
> >Looks like I will have to do a bit of maths to figure out how the cap
in
> >series with rectifier works.. I will be very glad to get rid of 10 W
> >resistors though!

> I would be interested to see how your maths works in practice, many
> years ago a particular range of Ferguson B/W TV's used a capacitor to
> feed the heater chain (rather than the usual huge ballast resistor).
> Being an inquisitive type I worked out the reactance of the capacitor
at
> the 50Hz mains frequency - it was nowhere near the required value to
set
> the current at the required 300mA for the valve (tube for our USA
> readers!) heaters. I've always presumed there was more to it than
simply
> the reactance of the capacitor, but have never really understood why?.
> --

I'm digging back a long way in the memory vault here so I could be
wrong, but ISTR that many old TVs had a line output valve that used
600mA heaters. In this case the other valves were arranged in a
series-parallel config to make the whole heater chain 600mA.
Dunno if your Ferguson was designed like this though.

Also the capacitor is almost pure reactance so there's a j component you
need to consider when calculating the drop.

</hazy memory>

.

In message <027901bf9341\$882bea80\$6e1e86d4@oemcomputer>, andy howard
<musicaUKONLINE.CO.UK> writes
>I'm digging back a long way in the memory vault here so I could be
>wrong, but ISTR that many old TVs had a line output valve that used
>600mA heaters. In this case the other valves were arranged in a
>series-parallel config to make the whole heater chain 600mA.
>Dunno if your Ferguson was designed like this though.

No, the heater chain was 300mA (and was in almost all European TV's -
I've never seen a 600mA one), they used 'P' series valves like 'PL81',
where the 'P' means 300mA heater, the 'L' means Pentode, the '8' means
B9A base, and the '1' means it's not a '2' :-). Some of the larger
valves had fairly high voltage heaters, in order to produce sufficient
wattage with only 300mA - the PY500 (colour boost diode) had a 42 volt
heater.

>Also the capacitor is almost pure reactance so there's a j component you
>need to consider when calculating the drop.
>
></hazy memory>

But how do you calculate it?.

<equally hazy memory>
--

Nigel.

/--------------------------------------------------------------\
| Nigel Goodwin   | Internet : nigelglpilsley.demon.co.uk     |
| Lower Pilsley   | Web Page : http://www.lpilsley.demon.co.uk |
| Chesterfield    | Official site for Shin Ki and New Spirit   |
| England         |                 Ju Jitsu                   |
\--------------------------------------------------------------/

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