> 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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> roines reenig 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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> 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...
>
>> roines reenig 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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>>
>>
>>
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> 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 <RemoveMEmartinTakeThisOuTDC.CIS.OKSTATE.EDU>
> Sent by: pic microcontroller discussion list <PICLISTEraseME.....MITVMA.MIT.EDU>
> 05/26/2004 10:58 AM
> Please respond to pic microcontroller discussion list
>
>
>         To:     EraseMEPICLISTMITVMA.MIT.EDU
>         cc:
>         Subject:        Re: [EE:] Obtaining 5 Volts from live wire
>
>
> RemoveMEllileEraseMEEraseMESALTONUSA.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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>-----Original Message-----
>From: Martin McCormick [RemoveMEmartinTakeThisOuTspamDC.CIS.OKSTATE.EDU]
>Sent: 26 May 2004 16:58
>To: EraseMEPICLISTspamspamBeGoneMITVMA.MIT.EDU
>Subject: Re: [EE:] Obtaining 5 Volts from live wire
>
>
>RemoveMEllileKILLspamSALTONUSA.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.

> 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 wh

>  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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> 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
>

> 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 neutral and be done with the task.
>
> Or am I missing something significant?
>
> BAJ
>
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> http://www.piclist.com hint: The list server can filter out subtopics
> (like ads or off topics) for you. See http://www.piclist.com/#topics
>

>
> > 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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> >
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>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 <TakeThisOuTwestfwKILLspamspamMAC.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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>
>
>

>-----Original Message-----
>From: Randy Ott [.....randyRemoveMETHEOTTS.NET]
>Sent: 28 May 2004 16:12
>To: RemoveMEPICLISTspamBeGoneMITVMA.MIT.EDU
>Subject: Re: [EE:] Obtaining 5 Volts from live wire
>
>
>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

>>-----Original Message-----
>>From: Randy Ott [TakeThisOuTrandyspamTHEOTTS.NET]
>>Sent: 28 May 2004 16:12
>>To: PICLISTEraseMEMITVMA.MIT.EDU
>>Subject: Re: [EE:] Obtaining 5 Volts from live wire
>>
>>
>>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
>>
>>
>
>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
>
>