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'[EE:] Obtaining 5 Volts from live wire'
2004\05\26@050232 by roines reenig

picon face
Hi,

I was just looking at this board that is intended to allow for remote control of a mains load. Meaning, let's say you've got a lamp. It's got just one wire coming in and one wire going out. That's the live-wire in, and live-wire out and in-between, you (the lamp in this case) are the load. This board goes in between the live-wire in and the lamp. This board appears to have some sort of processor, looks like a samsung 8 bit processor, some eeprom and some digital IO. I didn't get a chance to examine the board any further before it was taken away from me.

Now, it's powering itself from the mains live-wire in and I presume grounding itself using the live-wire out, therefore increasing the leakage current. I assume that it's not leaking so much that a flourescent lamp would start to flicker. Any ideas how this power supply mechanism works? The CPU's very likely to be a 5Volt part, so they must be able to generate the 5 volt DC voltage and current from the mains. I realize that there are high voltage switching power regulators like the VB409 that could generate 80mA and regulated 5Volt from mains. Is that what they are likely to be using? Or do they have some special stuff since they don't have access to ground and neutral wires.

Thanks,
Roines


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2004\05\26@053006 by Shawn Wilton

flavicon
face
They're probably just using a step down transformer, and a rectifier
circuit.


Shawn Wilton
Junior in CpE
MicroBiologist

Phone: (503) 881-2707
Email: spam_OUTshawnTakeThisOuTspamblack9.net

http://black9.net


roines reenig wrote:
{Quote hidden}

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2004\05\26@053628 by techy fellow

picon face
I am no expert in EE but, I ever came across a method of converting AC to DC using just capacitors ! I think it is called as, full-wave rectifer. The capacitor I came across is quite large (about the size of an eraser) and looks like it is made of ceramic/ plastic. Perhaps, you can visit this web address to see is this what you want;

http://teachers.web.cern.ch/teachers/archiv/HST2002/smallexp/Vogt/diode.htm


roines reenig <.....roinesreenigKILLspamspam@spam@YAHOO.COM> wrote:
Hi,

I was just looking at this board that is intended to allow for remote control of a mains load. Meaning, let's say you've got a lamp. It's got just one wire coming in and one wire going out. That's the live-wire in, and live-wire out and in-between, you (the lamp in this case) are the load. This board goes in between the live-wire in and the lamp. This board appears to have some sort of processor, looks like a samsung 8 bit processor, some eeprom and some digital IO. I didn't get a chance to examine the board any further before it was taken away from me.

Now, it's powering itself from the mains live-wire in and I presume grounding itself using the live-wire out, therefore increasing the leakage current. I assume that it's not leaking so much that a flourescent lamp would start to flicker. Any ideas how this power supply mechanism works? The CPU's very likely to be a 5Volt part, so they must be able to generate the 5 volt DC voltage and current from the mains. I realize that there are high voltage switching power regulators like the VB409 that could generate 80mA and regulated 5Volt from mains. Is that what they are likely to be using? Or do they have some special stuff since they don't have access to ground and neutral wires.

Thanks,
Roines


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2004\05\26@055535 by David Duffy

flavicon
face
Shawn Wilton wrote:

> They're probably just using a step down transformer, and a rectifier
> circuit.
>

No, he said there's no neutral return. It sits in series with the mains
supply
and the load. Maybe a zener to ensure a min voltage for the micro. That
way you're keeping a few volts to generate the stable micro supply from.
The touch lamp dimmers that came out years ago did something like this.
An internal diagram of one of those chips would show how it's done.
http://www.lsicsi.com/pdfs/LS7631_LS7632.pdf shows one of them.
David...

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2004\05\26@055914 by Russell McMahon
face
flavicon
face
> I was just looking at this board that is intended to allow for remote
control of a mains load. Meaning, let's say you've got a lamp. It's got just
one wire coming in and one wire going out. That's the live-wire in, and
live-wire out and in-between, you (the lamp in this case) are the load. This
board goes in between the live-wire in and the lamp. This board appears to
have some sort of processor, looks like a samsung 8 bit processor, some
eeprom and some digital IO. I didn't get a chance to examine the board any
further before it was taken away from me.
>
> Now, it's powering itself from the mains live-wire in and I presume
grounding itself using the live-wire out, therefore increasing the leakage
current. I assume that it's not leaking so much that a flourescent lamp
would start to flicker. Any ideas how this power supply mechanism works? The
CPU's very likely to be a 5Volt part, so they must be able to generate the 5
volt DC voltage and current from the mains. I realize that there are high
voltage switching power regulators like the VB409 that could generate 80mA
and regulated 5Volt from mains. Is that what they are likely to be using? Or
do they have some special stuff since they don't have access to ground and
neutral wires.

From your description they are EITHER

1.    Dropping a small voltage across the device and powering the
electronics from that
OR
2.   Turning the controlled circuit off occasionally, and using the
resultant full mains voltage as  a supply and then turning back on
OR
3    Using part of each cycle near zero with switch turned off so rising
voltage is applied to their power supply, then turning on switch to short
this power source out for rest of cycle.
OR
4.    Some combination of all of these :-)

When the circuit is off it has to withstand full mains power across itself,
so something like a combination of 2 & 3 sounds useful.

Not too hard to engineer once you decide which way to go.
Helps to have an IC to put all the pieces in :-)


       RM

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2004\05\26@060745 by Shawn Wilton

flavicon
face
Ok, reread.  I would still say that you could use a transformer by
simply inlining it.  But I'm not sure what he means by remote control of
the main.  Does that mean this device could potentially cut off the
power flowing out of it?  Or is it simply inlined and using the power
coming in?  I suppose if it's meant to act as a switch, then a
transformer wouldn't work.  But if it were simply pulling its power from
the line, then it would definitely be doable with a 2 in 2 out step down
transformer.

Roines, I'm curious, why did "they" yank the board from you?  Can you go
ask "them" how it works?  I would be curious to know what "they" did end
up using.  What sort of product is this?

This is a great question by the way.


Shawn Wilton
Junior in CpE
MicroBiologist

Phone: (503) 881-2707
Email: shawnspamKILLspamblack9.net

http://black9.net


David Duffy wrote:
{Quote hidden}

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2004\05\26@061543 by Jinx

face picon face
>
http://teachers.web.cern.ch/teachers/archiv/HST2002/smallexp/Vogt/diode.htm

Is it just me or is there something "up" with that page ?

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2004\05\26@100830 by llile

flavicon
face
There is a common circuit used to drop mains voltage to reasonable levels,
that works generally like this:

A capacitor is connected to the AC mains on one end and to a full wave
bridge rectifier on the other.  It can be called a dropping capacitor. The
output of the full wave bridge can be connected to a zener diode or some
other suitable regulator.  The circuit will not work correctly with a
single diode, only a full wave rectifier will do.

There are a couple of additional subtleties:
A high value (~ 100K) resistor across the dropping capacitor to discharge
it quickly when the mains are disconnected.
A small series resistor (~ 10 to 22 ohm) is placed on the input.  This is
usually a flameproof metal oxide power resistor about 1/2 watt.  It
generates nowhere near this amount of power, it is used as a fuse, short
circuit current limit, and surge limit device.  It is required to pass
certain UL tests.

The dropping capacitor needs to be able to handle AC, and it usually a
metalized polyester 250V (for 120V service) about 1 uF or so.

This is basically a cheap trick, only suitable for use in cheap mass
produced stuff.  It has many disadvantages, including no mains isolation,
the capacitor conducts high frequency transients directly into the power
supply (instead of attenuating them like a transformer would do) there are
questions about the life of the capacitor when used like this (abused?). I
do not use this circuit often, except in a million or so steam irons built
in the last 7 years <sarcasm off>

-- Lawrence Lile





techy fellow <.....techyfKILLspamspam.....YAHOO.COM>
Sent by: pic microcontroller discussion list <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU>
05/26/2004 04:36 AM
Please respond to pic microcontroller discussion list


       To:     PICLISTspamspam_OUTMITVMA.MIT.EDU
       cc:
       Subject:        [EE:] Obtaining 5 Volts from live wire


I am no expert in EE but, I ever came across a method of converting AC to
DC using just capacitors ! I think it is called as, full-wave rectifer.
The capacitor I came across is quite large (about the size of an eraser)
and looks like it is made of ceramic/ plastic. Perhaps, you can visit this
web address to see is this what you want;

http://teachers.web.cern.ch/teachers/archiv/HST2002/smallexp/Vogt/diode.htm


roines reenig <@spam@roinesreenigKILLspamspamYAHOO.COM> wrote:
Hi,

I was just looking at this board that is intended to allow for remote
control of a mains load. Meaning, let's say you've got a lamp. It's got
just one wire coming in and one wire going out. That's the live-wire in,
and live-wire out and in-between, you (the lamp in this case) are the
load. This board goes in between the live-wire in and the lamp. This board
appears to have some sort of processor, looks like a samsung 8 bit
processor, some eeprom and some digital IO. I didn't get a chance to
examine the board any further before it was taken away from me.

Now, it's powering itself from the mains live-wire in and I presume
grounding itself using the live-wire out, therefore increasing the leakage
current. I assume that it's not leaking so much that a flourescent lamp
would start to flicker. Any ideas how this power supply mechanism works?
The CPU's very likely to be a 5Volt part, so they must be able to generate
the 5 volt DC voltage and current from the mains. I realize that there are
high voltage switching power regulators like the VB409 that could generate
80mA and regulated 5Volt from mains. Is that what they are likely to be
using? Or do they have some special stuff since they don't have access to
ground and neutral wires.

Thanks,
Roines


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2004\05\26@115832 by Martin McCormick

flavicon
face
llile@SALTONUSA.COM writes:
>There is a common circuit used to drop mains voltage to reasonable levels,
>that works generally like this:
>
>A capacitor is connected to the AC mains on one end and to a full wave
>bridge rectifier on the other.  It can be called a dropping capacitor. The
>output of the full wave bridge can be connected to a zener diode or some
>other suitable regulator.  The circuit will not work correctly with a
>single diode, only a full wave rectifier will do.

       This is a clever circuit all right but it kind of scares me.
It's proper operation as opposed to smoke-venting mode depends upon
the reactive resistance of a 1uf capacitor at 120 volts at 60 or 50
HZ.  It is possible for something like a brush motor or switching
power supply to dump energy back in to the line at all sorts of
frequencies.  Not only that, but the wave form of all that trash could
be anything.

       Do these circuits get fried very often by sustained radio
frequency-like hash on mains wiring?  The circuit is basically a
high-pass filter being operated at the extreme edge of its roll-off.

       When one gets in to radio frequencies, that 1uf capacitor will
begin to pass most of the signal appearing at the input.

       If there is any energy there, it's going to heat stuff up
until something pops.  It is probably that flame-proof resistor, but I
am surprised more of these types of voltage dropping circuits don't
vaporize more often.

       Oh well- Who am I to argue with a million steam irons?:-)

Martin McCormick WB5AGZ  Stillwater, OK
OSU Information Technology Division Network Operations Group

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2004\05\26@132628 by llile

flavicon
face
I had the same reaction when I first saw it.  Several long diatribes on
the PIClist have railed against the idea for these very reasons.  However,
I am not the only one to use this trick, most any auto-shut-off steam iron
has a similar circuit.  The reason is, there is no place to dump heat from
a dropping resistor inside a steam iron chassis (it is already hot) and
there is no money for a transformer, nor room for it.   You will also find
this in the electronics of many other cheap appliances.  I am not aware of
any problems due to the power supply in any of these units.

-- Lawrence Lile





Martin McCormick <KILLspammartinKILLspamspamDC.CIS.OKSTATE.EDU>
Sent by: pic microcontroller discussion list <RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU>
05/26/2004 10:58 AM
Please respond to pic microcontroller discussion list


       To:     spamBeGonePICLISTspamBeGonespamMITVMA.MIT.EDU
       cc:
       Subject:        Re: [EE:] Obtaining 5 Volts from live wire


TakeThisOuTllileEraseMEspamspam_OUTSALTONUSA.COM writes:
>There is a common circuit used to drop mains voltage to reasonable
levels,
>that works generally like this:
>
>A capacitor is connected to the AC mains on one end and to a full wave
>bridge rectifier on the other.  It can be called a dropping capacitor.
The
>output of the full wave bridge can be connected to a zener diode or some
>other suitable regulator.  The circuit will not work correctly with a
>single diode, only a full wave rectifier will do.

       This is a clever circuit all right but it kind of scares me.
It's proper operation as opposed to smoke-venting mode depends upon
the reactive resistance of a 1uf capacitor at 120 volts at 60 or 50
HZ.  It is possible for something like a brush motor or switching
power supply to dump energy back in to the line at all sorts of
frequencies.  Not only that, but the wave form of all that trash could
be anything.

       Do these circuits get fried very often by sustained radio
frequency-like hash on mains wiring?  The circuit is basically a
high-pass filter being operated at the extreme edge of its roll-off.

       When one gets in to radio frequencies, that 1uf capacitor will
begin to pass most of the signal appearing at the input.

       If there is any energy there, it's going to heat stuff up
until something pops.  It is probably that flame-proof resistor, but I
am surprised more of these types of voltage dropping circuits don't
vaporize more often.

       Oh well- Who am I to argue with a million steam irons?:-)

Martin McCormick WB5AGZ  Stillwater, OK
OSU Information Technology Division Network Operations Group

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2004\05\26@133914 by Harold Hallikainen

face picon face
Isn't there also a current limit resistor in there somewhere so if they
happen to plug it in at the AC line peak, they don't get "infinite
current"? Also, it seems there should be a bleeder resistor across the
capacitor so someone doesn't get shocked by touching the AC plug once it's
unplugged.

Harold


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2004\05\26@140317 by Eisermann, Phil [Ridg/CO]

flavicon
face
pic microcontroller discussion list wrote:
> Isn't there also a current limit resistor in there somewhere so if
> they happen to plug it in at the AC line peak, they don't get
> "infinite current"? Also, it seems there should be a bleeder resistor
> across the capacitor so someone doesn't get shocked by touching the
> AC plug once it's unplugged.
>

correct on both account. Both of those should be there. IIRC, UL has
a requirement that the voltage discharge below some threshold value
in some amount of time for exactly that reason. I don't remember the
voltage, but the time was on the order of 0.5 second.

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2004\05\26@150321 by Martin McCormick

flavicon
face
Harold Hallikainen writes:
>Isn't there also a current limit resistor in there somewhere so if they
>happen to plug it in at the AC line peak, they don't get "infinite
>current"? Also, it seems there should be a bleeder resistor across the
>capacitor so someone doesn't get shocked by touching the AC plug once it's
>unplugged.

       Right on both counts.  The original message  contained:

> There are a couple of additional subtleties:
> A high value (~ 100K) resistor across the dropping capacitor to discharge
> it quickly when the mains are disconnected.
> A small series resistor (~ 10 to 22 ohm) is placed on the input.  This is
> usually a flameproof metal oxide power resistor about 1/2 watt.  It
> generates nowhere near this amount of power, it is used as a fuse, short
> circuit current limit, and surge limit device.  It is required to pass
> certain UL tests.

       While it still makes me queasy, I can see that it must be a
valid solution due to the fact that it has been successfully deployed
countless times.  It is safe as long as there is _NO_ chance of
unexpected contact between any part of the circuit and anything else
conductive that could accidentally contact Earth or a person who is
also contacting Earth.

       One could even have external connections to such a circuit if
they are optically or inductively isolated.  I think some X10 modules
are built this way.

Martin McCormick

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2004\05\26@163501 by Richard.Prosser

flavicon
face
You can protect against spikes & r.f. etc by placeing an additional cap to
neutral on the output side of the feed cap. This then forms a voltage
divider and all high frequency noise etc. is divided accordingly.  Since it
is operating on the low voltage side of the feed cap, it doesn't need to be
anything special (assuming the feed cap is appropriately rated).

Richard P




RemoveMEllilespam_OUTspamKILLspamSALTONUSA.COM writes:
>There is a common circuit used to drop mains voltage to reasonable levels,
>that works generally like this:
>
>A capacitor is connected to the AC mains on one end and to a full wave
>bridge rectifier on the other.  It can be called a dropping capacitor. The
>output of the full wave bridge can be connected to a zener diode or some
>other suitable regulator.  The circuit will not work correctly with a
>single diode, only a full wave rectifier will do.

       This is a clever circuit all right but it kind of scares me.
It's proper operation as opposed to smoke-venting mode depends upon
the reactive resistance of a 1uf capacitor at 120 volts at 60 or 50
HZ.  It is possible for something like a brush motor or switching
power supply to dump energy back in to the line at all sorts of
frequencies.  Not only that, but the wave form of all that trash could
be anything.

       Do these circuits get fried very often by sustained radio
frequency-like hash on mains wiring?  The circuit is basically a
high-pass filter being operated at the extreme edge of its roll-off.

       When one gets in to radio frequencies, that 1uf capacitor will
begin to pass most of the signal appearing at the input.

       If there is any energy there, it's going to heat stuff up
until something pops.  It is probably that flame-proof resistor, but I
am surprised more of these types of voltage dropping circuits don't
vaporize more often.

       Oh well- Who am I to argue with a million steam irons?:-)

Martin McCormick WB5AGZ  Stillwater, OK
OSU Information Technology Division Network Operations Group

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2004\05\26@174308 by hael Rigby-Jones

picon face
{Quote hidden}

Properly designed, these circuit shouldn't rely purely on the reactance of
the cap for the very reasons you mention.  Usualy part of the drop will be
shared with a resistor to ensure some impedance at high frequencies.


Mike

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2004\05\26@201957 by William Chops Westfield

face picon face
On Wednesday, May 26, 2004, at 10:26 US/Pacific, llileSTOPspamspamspam_OUTSALTONUSA.COM
wrote:

> It's proper operation as opposed to smoke-venting mode depends upon
> the reactive resistance of a 1uf capacitor at 120 volts at 60 or 50 > HZ.

Don't forget though, that "extra safe" capacitors are made in this
value range, and appear in the rf filter sections of nearly every
switching power supply in existance...  (of course, you oughta use one
of those,
and not a generic cap from some junkbox...)

BillW

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2004\05\26@204725 by William Chops Westfield

face picon face
I've seen a number of circuits (in shavers and wall warts, I think)
that use a cap and/or resistor in series with the primary of a (quite
small but otherwise conventional) transformer.  Presumably it works the
same way with some added filtering/voltage conversion from the
transformer?

(so, anybody know where I can GET those sorts of transformers?)

BillW

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2004\05\26@214005 by roines reenig

picon face
Hi,

I think we have drifted a little bit from what I attempted to describe initially. I drew up some diagrams to explain the original question better.

http://www.geocities.com/roinesreenig/picture1.JPG

If you look at picture 1, above, it's just a picture of how a normal house light switch is wired.

http://www.geocities.com/roinesreenig/picture2.JPG

Picture 2, above is what this board that we're all trying to guess at is. I've replaced what used to be a simple switch with this control device. On this board, the control device was a samsung 8 bit CPU with either RF or an IR interface for remote control. So basically, you click on a remote control, and this control device dimms up/down and/or turns on/off your load. The load could be anything, typically, it'd be a set of lamps or fluorescent tubes. That part I think is typical, what's atypical is how it's powering itself from just the one-wire mains live wire.

http://www.geocities.com/roinesreenig/picture3.JPG

Picture 3, above is my interpretation of what Shawn Wilton described, ie: step down and a bridge rectifier to power a micro that controls the load. David Duffy pointed out the fact that things are more complicated and mentioned the LS7631 (discussed as picture 4). Now, Shawn's circuit could work, I think, except there's that question mark above the load. The question-mark-box, QMB would need to somehow turn fully on when we want the load to be on and turn somewhat off but still allow for dropping a sufficient current to power the CPU. Then as Russell McMahon pointed out, it's possible that this QMB, under the control of the CPU, or maybe without, could be switching on and off with respect to the input mains at a sufficient frequency. Thus, the QMB would prevent the load from appearing to be turned on, (for example, a flourescent light should not be flickering) but it would be able to charge up a capacitor enough to provide a reasonably clean 5 Volt DC to the CPU. That's where I
thought they might be using something like a VB409. Link to that datasheet below. But the problem is that the VB409 still needs a return neutral or at least a sufficient potential difference across itself, so it's not clear if that's feasible without a return neutral or earth.

http://www.st.com/stonline/books/pdf/docs/6941.pdf

Issue 2 with what Shawn described is the mechanical design. The board/package I saw was somewhat "thin". Coudn't have been more than an inch and a half in height. Maybe about 2 inches on both sides in terms of area. So if it was using transformers in there, which is possible, they were definitely small. A mains to around 5V step down would be pretty big woudn't it? I don't know how many turns it would have but a ratio of (120/5=24) 1:24 or 1:48 if 240V supply.

Basically, the key piece of information is that you don't have access to a neutral return and/or earth but you still need to generate 5 Volts of potential difference in order to power a controller/cpu. I think Russell's answer listed the right needs except he left figuring out the details as a challenge to the reader with the "not too hard to engineer" :-). I certainly don't know how to do it. Anybody else?

http://www.geocities.com/roinesreenig/picture4.JPG

Picture 4, is extracted from the LS7631 datasheet that David Duffy mentioned.

http://www.lsicsi.com/pdfs/LS7631_LS7632.pdf

I drew it up, hoping that I could understand how it worked. But I don't. I can see that the Zener would allow a potential to develop when the mains was moving up to and down from the 5.6V mark, this would charge up C5 (flowing through D1 and charging C2 as well) which could then power the control device during the negative part of the cycle. But it's not clear to how C1, R1, C2 and D1 participate. Further, this circuit would have to use the LS7631 since it's LS7631 that would modulate the power to control the load. Maybe the LS7631 in conjuction with a PIC controlling it's external input would be the answer? I think that this board isn't a typical design and does appear to have something unique about it's power supply mechanism.

Thanks,

Roines

ps: nope Shawn, I don't think I can get access to the board again. A sales guy brought it in to show it and I started taking it apart at which point it was taken back. I saw it work so I know it somehow works. It had a bicolor LED with which it reflected the state of it's load (in the demo, a regular light bulb), so if the load was on, it was green, load was off, it was amber.


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2004\05\26@232002 by Shawn Wilton

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Roines, you can get transformers that will step sufficiently in either
direction in smd packaging that is no more than a 1cm^2.

Here's a site of such a company producing these transformers:
http://www.tmpco.com.tw
or
http://www.tmpco.com.tw/kmportal-deluxe/front/bin/ptlist.phtml?Category=100038



Shawn Wilton
Junior in CpE
MicroBiologist

Phone: (503) 881-2707
Email: spamBeGoneshawnSTOPspamspamEraseMEblack9.net

http://black9.net


roines reenig wrote:
{Quote hidden}

ere I
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2004\05\27@012639 by William Chops Westfield

face picon face
On Wednesday, May 26, 2004, at 18:39 US/Pacific, roines reenig wrote:

> I think we have drifted a little bit from what I attempted to describe
> initially.

I was wondering about that.  You were mostly interested in how to get
power at (for instance) a lightswitch, when you only have access to the
"hot" side of the connection, and the load, right?

Most things that do this (ie X10 dimmer switches, or regular dimmers,
for that matter) rely on the load being essentially a slightly
resistive path to the return (a 100W lightbulb has a resistance of only
130 ohms or so, so it's pretty easy to get 50mA worth of 5V out of
there somehow.
That's why they "only" work on resistive loads.  Sneaky, eh?

BillW

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2004\05\27@035334 by roines reenig

picon face
I agree with your observation about the need for a resistive load. I'm having difficulty getting my head around how the control circuit (some type of CPU) controls the load. so in series:

live_in --- control --- live_out --- load --- neutral

i think that control probably has to have a triac that it is phasing on and off in order to dimm/turn on/off the load. at the same time it's got to use that same line to power itself. i can't get my head around that part. i think i'll do some searching for how dimmers power themselves from the live line. hopefully one of them has a CPU in them so then it'll all be clear to me.

thanks,
roines

William Chops Westfield <KILLspamwestfwspamBeGonespamMAC.COM> wrote:
On Wednesday, May 26, 2004, at 18:39 US/Pacific, roines reenig wrote:

> I think we have drifted a little bit from what I attempted to describe
> initially.

I was wondering about that. You were mostly interested in how to get
power at (for instance) a lightswitch, when you only have access to the
"hot" side of the connection, and the load, right?

Most things that do this (ie X10 dimmer switches, or regular dimmers,
for that matter) rely on the load being essentially a slightly
resistive path to the return (a 100W lightbulb has a resistance of only
130 ohms or so, so it's pretty easy to get 50mA worth of 5V out of
there somehow.
That's why they "only" work on resistive loads. Sneaky, eh?

BillW

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2004\05\27@100722 by Charles Craft

picon face
There's a couple web sites where people reverse engineered the X10 schematics.

And then lo and behold, they're posted on the Microchip site now! (when did this happen)
ww1.microchip.com/downloads/en/AppNotes/00236a.pdf
(pg. 19 - transformerless power supply)


Here's the original Microchip whitepaper/tech brief on transformerless supplies:
ww1.microchip.com/downloads/en/AppNotes/91008b.pdf
Transformerless Power Supply - TB008

A search on "transformerless" at http://www.microchip.com brought up 5 documents:

AN236 - X-10 Home Automation Using the PIC16F877A  Release Date: 10/30/02
TB008 - TechBrief Tranformerless Power Supply  Release Date: 8/26/97
DD2016 - Freezer Protector  Release Date: 4/22/98
DD1009 - PIC12C508-Based Timer  Release Date: 4/22/98
DD2009 - Programmable Lights  Release Date: 4/22/98


And just for kicks, the Agilent Powerline modem IC data sheet has one on the last page:
cp.literature.agilent.com/litweb/pdf/5989-0402EN.pdf
Agilent HCPL-800J  PLC Powerline DAA IC



{Original Message removed}

2004\05\27@103216 by Randy Ott

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On Wed, 26 May 2004 02:03:12 -0700, roines reenig wrote
{Quote hidden}

Roines,

If you're talking about a light dimmer, it steals power through the load
while the load is switched off.  This is stored in a capacitor.  If you look
at the voltage fed to the load with an oscilloscope you will find that the
load is never switched on 100% even at full brightness.

Randy


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2004\05\27@105046 by Matthew Brush

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There is also some transformerless power supplies on Discover Circuits
website:
http://www.discovercircuits.com

Great website too!

MJ Brush

----- Original Message -----
From: "Charles Craft" <EraseMEchuckseaspamEraseMEMINDSPRING.COM>
To: <@spam@PICLIST@spam@spamspam_OUTMITVMA.MIT.EDU>
Sent: Thursday, May 27, 2004 10:06 AM
Subject: Re: [EE:] Obtaining 5 Volts from live wire


> There's a couple web sites where people reverse engineered the X10
schematics.
>
> And then lo and behold, they're posted on the Microchip site now! (when
did this happen)
> ww1.microchip.com/downloads/en/AppNotes/00236a.pdf
> (pg. 19 - transformerless power supply)
>
>
> Here's the original Microchip whitepaper/tech brief on transformerless
supplies:
{Quote hidden}

the last page:
> cp.literature.agilent.com/litweb/pdf/5989-0402EN.pdf
> Agilent HCPL-800J  PLC Powerline DAA IC
>
>
>
> {Original Message removed}

2004\05\27@111329 by Dave Tweed

face
flavicon
face
Charles Craft <spamBeGonechuckseaspamKILLspamMINDSPRING.COM> wrote:
> A search on "transformerless" at http://www.microchip.com brought up 5 documents:
> AN236 - X-10 Home Automation Using the PIC16F877A  Release Date: 10/30/02
> TB008 - TechBrief Tranformerless Power Supply  Release Date: 8/26/97
> DD2016 - Freezer Protector  Release Date: 4/22/98

Three of these documents recommend connecting the circuit to the grounding
(safety) conductor of a 3-wire plug, and then putting a fuse or other
component between the grounding conductor and the grounded (neutral)
conductor.

THIS IS A MAJOR SAFETY RISK AND IS BLATENTLY ILLEGAL IN THE US AND
PROBABLY ELSEWHERE AS WELL!

The grounding conductor must never intentionally carry load currents.
This type of transformerless power supply must be completely isolated
from the safety ground!

-- Dave Tweed

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2004\05\27@122222 by Byron A Jeff

face picon face
On Thu, May 27, 2004 at 11:13:29AM -0400, Dave Tweed wrote:
> Charles Craft <.....chuckseaspam_OUTspamMINDSPRING.COM> wrote:
> > A search on "transformerless" at http://www.microchip.com brought up 5 documents:
> > AN236 - X-10 Home Automation Using the PIC16F877A  Release Date: 10/30/02
> > TB008 - TechBrief Tranformerless Power Supply  Release Date: 8/26/97
> > DD2016 - Freezer Protector  Release Date: 4/22/98
>
> Three of these documents recommend connecting the circuit to the grounding
> (safety) conductor of a 3-wire plug, and then putting a fuse or other
> component between the grounding conductor and the grounded (neutral)
> conductor.
>
> THIS IS A MAJOR SAFETY RISK AND IS BLATENTLY ILLEGAL IN THE US AND
> PROBABLY ELSEWHERE AS WELL!
>
> The grounding conductor must never intentionally carry load currents.
> This type of transformerless power supply must be completely isolated
> from the safety ground!

Agreed. Also it'll trop GCFI systems which specifically detect current in
that leg. It's really bad engineering.

So let's get back to the issue. I have a simple engineering question: Why
isn't the neutral available? Other than an academic exercise, I can't conceieve
of a situation where you only have the hot wire and no physical access at all
to the neutral. Even in a 3way or 4way switch situation, both hot and neutral
are available right?

You'll have to leak current through the load like this.

Hot -> Control -> X1 -> Switch -> X2 -> Load -> Neutral

Where X1 and X2 are connected so that the control and switch can be powered
through the load even when the switch is disengaged leaking a minimal amount
of current through the load. You can even use a resistor or cap and a zener
to limit the voltage and current going around the switch so that the load
doesn't get enough power to activate.

But again other than the academic exercise, why bother. Connect X1 directly to
the neutral and be done with the task.

Or am I missing something significant?

BAJ

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2004\05\27@125335 by Bob Ammerman

picon face
The neutral is not available a good percentage of the time at a standard
single pole switch in a home. In many cases a two-wire (plus safety ground)
cable brings hot to the switch and switched-hot back to the load.

> You'll have to leak current through the load like this.
>
> Hot -> Control -> X1 -> Switch -> X2 -> Load -> Neutral
>
> Where X1 and X2 are connected so that the control and switch can be
powered
> through the load even when the switch is disengaged leaking a minimal
amount
> of current through the load. You can even use a resistor or cap and a
zener
> to limit the voltage and current going around the switch so that the load
> doesn't get enough power to activate.
>
> But again other than the academic exercise, why bother. Connect X1
directly to
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2004\05\27@134403 by Byron A Jeff

face picon face
On Thu, May 27, 2004 at 12:52:41PM -0400, Bob Ammerman wrote:
> The neutral is not available a good percentage of the time at a standard
> single pole switch in a home. In many cases a two-wire (plus safety ground)
> cable brings hot to the switch and switched-hot back to the load.

I don't profess to be an electrician and I don't have a lot of experience.  I
guess it shows. Every box I have in my 30+ YO home has hot, neutral and
safety ground (SG). I'm just trying to envision a wiring diagram where only
hot and switched hot (SH) are available in a box. I guess I can see it: hot,
neutral, and SG come into the light fixture. A separate wire with hot and
SH goes to the switch box. So there's three wires, but no neutral. BTW
how is such a line (especially the white wire) supposed to be marked?

Actually I do have such a problem in my den fan fixture that would be helped
by such a situation. It's a 3way switch wired normally with a neutral going
from the switch to the fixture. It would be better if there was a straight
hot to the fixture so that the fan can be run all the time without being
affected by the switch. In other words there is only a SH at the fixure.  So
if it were layed out where a only a hot/SH went to the switch and there was
always a hot wire in the fixture, I could connect the fan directly to the hot
wire and the fan light to the SH that goes to the switch. I'm a firm believer
that switches should only control lights, not fans. My current thought is
replacing the current 3 wire conductor between the fan box and the switch
with a 4 conductor with both the SH and hot wires heading out to the fan
fixture. But as you (or was it me? ;-) pointed out, hot going directly to the
fan box would be helpful.

So it makes sense, though I don't have it in my own place.

BTW am I right about having to leak current through the load? And what the
heck do you do if the load isn't resistive?

BAJ

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2004\05\27@145426 by Scott Fraser

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At 10:43 AM 5/27/04, you wrote:
>On Thu, May 27, 2004 at 12:52:41PM -0400, Bob Ammerman wrote:
> > The neutral is not available a good percentage of the time at a standard
> > single pole switch in a home. In many cases a two-wire (plus safety ground)
> > cable brings hot to the switch and switched-hot back to the load.
>
>I don't profess to be an electrician and I don't have a lot of experience.  I
>guess it shows. Every box I have in my 30+ YO home has hot, neutral and
>safety ground (SG). I'm just trying to envision a wiring diagram where only
>hot and switched hot (SH) are available in a box. I guess I can see it: hot,
>neutral, and SG come into the light fixture. A separate wire with hot and
>SH goes to the switch box. So there's three wires, but no neutral. BTW
>how is such a line (especially the white wire) supposed to be marked?

That is the case in many switched lighting outlets.
Hot and neutral in the ceiling box, hot going to the switch box and switch
leg returning to the lamp.

Intermatic makes digital lamp time switches that you can program with a
half dozen on/off times.  It gets it's power by putting part of the circuit
in series with the incandescent lamp while off.  While the lamp is on is
another story.  I happen to have a burnt out unit in my hands (don't know
why I never threw it away).

It has 15 resistors, six caps of varying sizes, 9 diodes, five transistors,
one triac and a toroidal inductor to support the display.
If I get some time, I can see about throwing together a rough schematic of
at least the power section so you can see the "magic"

Then again, one could always pick up one of these at your local home
improvement place and modify to meet your needs.


>BTW am I right about having to leak current through the load? And what the
>heck do you do if the load isn't resistive?

The intermatic switch has a stamp on the back "Incandescent Loads Only"

Scott

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2004\05\27@151842 by 4HAZ

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----- From: "Scott Fraser" <kn6f@D
-snip-
> >how is such a line (especially the white wire) supposed to be marked?

The National Electrical Code  says that the White conductor of a switch-leg
should never be switched, this means the white wire is the HOT side and the
black is the switched-hot, leaving the bare for safety ground.

> That is the case in many switched lighting outlets.
> Hot and neutral in the ceiling box, hot going to the switch box and switch
> leg returning to the lamp.

The trick is simple enough, the triac will have either full voltage across
it when the circuit is off, or a small voltage across it when on, stack in a
little more voltage drop in the on-state and you can have 5v to your circuit
and the remainder to the load (say 115 - 120v) think of a string of 6a
diodes with one 6a diode reverse parallel as a very large zener.

KF4HAZ - Lonnie

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2004\05\27@152607 by Spehro Pefhany

picon face
At 01:43 PM 5/27/2004 -0400, you wrote:
>On Thu, May 27, 2004 at 12:52:41PM -0400, Bob Ammerman wrote:
> > The neutral is not available a good percentage of the time at a standard
> > single pole switch in a home. In many cases a two-wire (plus safety ground)
> > cable brings hot to the switch and switched-hot back to the load.
>
>I don't profess to be an electrician and I don't have a lot of experience.  I
>guess it shows. Every box I have in my 30+ YO home has hot, neutral and
>safety ground (SG). I'm just trying to envision a wiring diagram where only
>hot and switched hot (SH) are available in a box. I guess I can see it: hot,
>neutral, and SG come into the light fixture. A separate wire with hot and
>SH goes to the switch box. So there's three wires, but no neutral. BTW
>how is such a line (especially the white wire) supposed to be marked?

I believe you are supposed to wrap black electrical tape around the white
wire at both ends in this case.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
TakeThisOuTspeff.....spamTakeThisOuTinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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2004\05\27@160621 by Martin McCormick

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face
       Some of those electronic light switches that are for
incandescent bulbs only use a SCR instead of a triac so the bulb runs
on half-wave pulsating DC.  If they do that in the circuit, there is
the other half-cycle of AC that is un-touched and will be at about the
same voltage no matter if the light is on or off.  The bulb will burn
at a slightly lower brightness because the over all energy level is
.707 times the RMS level, but it will basically work normally.

       You might even see some flicker at 60 or even a little more at
50 HZ.  Hook that up to a fluorescent tube ballast and you may see the
flicker of flames unless the SCR burns out instantly and shorts at
which time your fluorescent lamp will be perpetually on.

       In other words, "no inductive loads" really means just that.

       While this is slightly off of the topic, I remember my days as
an audio visual technician in the eighties in which we had a brand of
overhead projector that, for some strange reason, used a bulb in
series with a diode.  Occasionally, one of these would come in after
the customer would have blown two or three $20 bulbs in quick
succession and caught on to the fact that something else was wrong.
We would check the diode and find it shorted such that the 85-volt
bulb was now getting the full 120 volts full-wave.  I never understood
the engineering logic behind that design rather than just putting a
120-volt bulb in the socket like everybody else did.

       The other half-cycle, BTW, was just wasted except to drive the
cooling fan.  There was no electronics other than that stupid diode.
I bet the bulb manufacturers came up with that design to sell more
bulbs.:-)

Martin McCormick WB5AGZ  Stillwater, OK
OSU Information Technology Division Network Operations Group

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2004\05\27@185512 by David Duffy

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Martin McCormick wrote:

>        Some of those electronic light switches that are for
>incandescent bulbs only use a SCR instead of a triac so the bulb runs
>on half-wave pulsating DC.  If they do that in the circuit, there is
>the other half-cycle of AC that is un-touched and will be at about the
>same voltage no matter if the light is on or off.  The bulb will burn
>at a slightly lower brightness because the over all energy level is
>.707 times the RMS level, but it will basically work normally.
>
>

"Slightly lower brightness" ?? Try it for yourself one day.
Running a lamp with half wave is more than slightly reduced!
David...

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2004\05\27@185927 by Howard Winter

face
flavicon
picon face
Byron,

On Thu, 27 May 2004 13:43:33 -0400, Byron A Jeff wrote:

>...<
> I don't profess to be an electrician and I don't have a lot of experience.  I
> guess it shows. Every box I have in my 30+ YO home has hot, neutral and
> safety ground (SG). I'm just trying to envision a wiring diagram where only
> hot and switched hot (SH) are available in a box. I guess I can see it: hot,
> neutral, and SG come into the light fixture. A separate wire with hot and
> SH goes to the switch box. So there's three wires, but no neutral.

This is absolutely standard nowadays in the UK - it's known as "loop-in" wiring.  Light fittings ("ceiling
roses") have in and out connections for each of Line, Neutral, Earth, then a twin-and-Earth cable goes to the
light unit, and another twin-and-Earth goes to the switch.  The power cable goes from one rose to another,
"looping in" and out of each one, usually supplying all of the lights on one floor.  The pair going to the
switch has Line on one wire and obviously returns Switched Line on the other.  Older houses (like mine!) have
small junction boxes mounted on the celing joists doing the loop-in thing, with the ceiling roses having just
the two connections.

> BTW how is such a line (especially the white wire) supposed to be marked?

Our installed cables have red for Line and black for Neutral.  The Switched Line (black) wire has a red sleeve
(or tape!) placed on the ends to denote that it's not one you want to touch (or connect to the Neutral
anywhere!  :-)

The upshot is that our switches don't usually have a Neutral available, so X10 switches have to rely on there
being a resistive load beyond them, and get their power by leaking through it.  This means that there are X10
"light" and X10 "appliance" switches - the former are for resistive-load-only and don't need Neutral, the
latter are any-load-you-like but must have a Neutral connection.  Consequently "low energy bulbs", that are
actually flourescent lamps with electronics controlling them, can't be used with X10 "light" controllers.

Cheers,

Howard Winter
St.Albans, England

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2004\05\27@220326 by M. Adam Davis

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The basic idea is that when the unit is off you have 120v potential -
you can pass many milliamps without brightening the lamp.

When the load is on there is still a voltage potential across the
triac.  If that potential is not enough (usually these devices need
/very/ little current, probably less than a mA) then you can add a very
low resistance wire to add a bit more voltage with lower current lamps.

Either way, since the circuit draws only a few milliamps, you can simply
use a zener and resister across the dropping element (the triac in this
case).  When off the zener forces the resister to drop hundreds of
volts, but at only a milliamp that's still less than a 1/2 watt
resister.  When on the resister may only be dropping a few volts.

In really cheap circuits the you don't use a bridge, the zener simply
passes all the current of one half the wave through the resister.  The
capacitor keeps the circuit running.

The touch element, when there is a wire for people to touch in such a
switch, is capacitively coupled to the circuit.  Incidently, the X-10
radio receivers have an antenna that is capacitively coupled - the
antenna ends in the plastic of the case, and on the other side of the
plastic wall is a wire, so the case itself is the capacitive element.
Some people have modified them for greater range by using a capacitor
soldered between the antenna and the wire.  Hope the cap is mains rated,
and fail 'safe' (open).

-Adam

roines reenig wrote:

{Quote hidden}

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2004\05\28@003728 by Russell McMahon

face
flavicon
face
> So let's get back to the issue. I have a simple engineering question: Why
> isn't the neutral available?

In this specific case I have no idea. But in general it allows you to locate
and intercept a load carrying wire on either side of a load without
reference to the return path.
Where supply and return are always run as a pair it SHOULD be academic.
Where they may be split, it can be useful to not need the other wire.

An example where this is meaningful in another type of application is
controlling a high level stop lamp in a car. If you can cut one wire without
reference to ground, you can place the circuit anywhere in the circuit,
without having to contend with what the manufacturers have done for ground
return.


       Russell McMahon

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2004\05\28@111210 by Randy Ott

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face
On Fri, 28 May 2004 08:52:32 +1000, David Duffy wrote
> Martin McCormick wrote:
>
> >        Some of those electronic light switches that are for
> >incandescent bulbs only use a SCR instead of a triac so the bulb runs
> >on half-wave pulsating DC.  If they do that in the circuit, there is
> >the other half-cycle of AC that is un-touched and will be at about the
> >same voltage no matter if the light is on or off.  The bulb will burn
> >at a slightly lower brightness because the over all energy level is
> >.707 times the RMS level, but it will basically work normally.
> >
> >
>
> "Slightly lower brightness" ?? Try it for yourself one day.
> Running a lamp with half wave is more than slightly reduced!
> David...
>
That's because the lamp sees 1/2 the voltage not .707 as was suggested.

Randy

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2004\05\28@112623 by hael Rigby-Jones

picon face
{Quote hidden}

No, it's running at half *power* (ignoring the reduction in filament
resistance from running cooler).  RMS voltage is indeed reduced to SQR(2)/2
= 0.707

Regards

Mike




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2004\05\28@184515 by David Duffy

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Michael Rigby-Jones wrote:

{Quote hidden}

The filament resistance won't be any different when running with half wave.
It doesn't change much over most of the range. The largest change is in the
first few % of the voltage range. Half wave or 50% phase control setting
will have the lamp typically at much less then half brightness. The curve of
a typical incandescent lamp is very non linear. More like an S curve. Not
as much change in light output at either end than at the middle. (vs
Voltage)
David...

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2004\05\29@135619 by Peter L. Peres

picon face
> I'm a firm believer that switches should only control lights, not fans.

That is very true. Most light switches will not take the inductive
kickback from even a small (30W) fan for too long.

I have seen a scheme where a small load is powered from the live wire in
two ways: When the load is off, a small leakage through the load is used
to obtain the power. When the load is on, the voltage drop on the switch
is used. The scheme I saw that used this, powered a single blinking led.
This could be expanded easily using a dc/dc upconverter from 1-1.5V to 3
or 5V. The voltage drop in operation was obtained from two bridge
rectifiers 'misused' as four (eight ?) antiparallel diodes. With the
switch off, the power came from a capacitive voltage reducer (Xc etc).
Neat ? Ideally the switcher would work from 0.5V, to use a single diode
drop.

Peter

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2004\05\29@135621 by Peter L. Peres

picon face
> Some of those electronic light switches that are for incandescent bulbs
> only use a SCR instead of a triac so the bulb runs on half-wave
> pulsating DC.  If they do that in the circuit, there is the other
> half-cycle of AC that is un-touched and will be at about the same
> voltage no matter if the light is on or off.  The bulb will burn at a
> slightly lower brightness because the over all energy level is .707
> times the RMS level, but it will basically work normally.

I have never seen such a single halfperiod switch, but I know that many
electronic switches will only turn on the load well into each half period
(say by the time the voltage is >=5V absolute). This is for technical
reasons (triac gate drive powered by anode voltage). So simply rectifying
the voltage across the switch may get you going with something, even if
the switch is on.

The other circuit I wrote about was used in series and parallel with a
normal mechanical switch, with zero (almost) drop across it.

Peter

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