Low voltage power from a UK live only light switch

[JohnW2]

The two wire dimmer does not need continuous power to operate a wireless receiver so it can "steal" a volt or two for a power supply source.

Yes, that's essentially what I was talking about.

Some wireless ON-OFF remote control switches use a remnance relay which only requires a short pulse of power to latch it ON or OFF and thus this can operate for many months on a button cell battery. The wireless receiver is designed to use the minimum possible current. Many turn on the receiver for a few milliseconds about ten times a second. If there is no wireless signal the receiver turns off. Thus the receiver is only on for about 1/10 of the time which extends battery life by almost 10 times.

Sure, that's like mobile phone technology - and I think they've managed quite a bit better than 10:1, haven't they. However, my understanding is that the OP is going to have a 'mini computer' in there, which I imagine has to be at least partially powered-up for most/all of the time - I certainly can't see it coming on and off many times per second!

Kind Regards, John

 

[SimonH2]

To get power when "on", you can delay turn on until the voltage across the switch is a few volts - and charge a capacitor. That gives you a quick charge to a cap every half second (ie 100 times a second). If you look at the area under the voltage curve, and work out the power a resistive load will use during this period, then it'll make naff all difference to maximum brightness - for an incandescent light.
But, some LED or CFL lights might detect this delay as being from a dimmer and not go to full brightness.

When "off", you can simply charge a cap from the voltage across the switch.

So circuitry wise, bridge rectifier across the switch, allowing it to charge a capacitor - needs to be suitably rated as it'll charge to around 370V DC.
Now comes the really tricky bit to design - a buck converter to get a low voltage supply from a cap that may be charged to only (say) 10V or up to 370V. That's going to be a really tricky bit to design, and I suspect a hysteresis regulator may be appropriate - but it's well outside my field of experience.

But, as pointed out, the current you take in the off state is likely to cause problems with CFL and LED lights. Do a search and you'll find loads of thread in there about flashing lights
This can be "fixed" by putting a resistor across the light to take the current away from the light - but this is another component to fit, and will create heat.

But remember that you will have anything up to 370V DC in your circuit. This is a serious voltage that must be treated with respect.

 

[JohnW2]

To get power when "on", you can delay turn on until the voltage across the switch is a few volts - and charge a capacitor. That gives you a quick charge to a cap every half second (ie 100 times a second).

[I take it that you mean "...every half cycle"].
Is the 'switch' you are talking about a dimmer? If it were a real (on/off) switch there obviously would/should never be any significant voltage across it when "on" (switch closed).

Kind Regards, John

 

[SimonH2]

Yes, half cycle - never type when tired
By "switch", I mean whatever (solid state) device he is proposing to switch the load with.

 

[ban-all-sheds]

TBH, I would have thought that the world would be, if not awash, then at least splishy-splashy with designs of, and fora discussing, Arduiono-based lighting automation systems.

 

[bernardgreen]

that's like mobile phone technology - and I think they've managed quite a bit better than 10:1, haven't they.

Some systems run at 1 /100 but at the expense that the transmitter has to be transmitting each command for long enough to ensure the receiver(s) have switched on during the command. Mostly this is achieved by sending a pre-amble which will keep the receiver(s) active until the command(s) have been sent.

However, my understanding is that the OP is going to have a 'mini computer' in there, which I imagine has to be at least partially powered-up for most/all of the time - I certainly can't see it coming on and off many times per second!

There are many micro processors that are normally asleep until woken up by a stimulus ( from a wireless receiver ) , perform what ever function is needed and then go back to sleep.

This can be "fixed" by putting a resistor across the light to take the current away from the light - but this is another component to fit, and will create heat.

Better to use a capacitor and low ohm resistance in series across the lamp. This will produce hardly any heat.

 

[ban-all-sheds]

There are many micro processors that are normally asleep until woken up by a stimulus ( from a wireless receiver ) , perform what ever function is needed and then go back to sleep.

They still need power, even in sleep states.

 

[JohnW2]

By "switch", I mean whatever (solid state) device he is proposing to switch the load with.

In that case, certainly if the switching is on/off (rather than dimming), the voltage across the switch ought to be very low (almost zero with a mechanical switch, but also very low with a solid state one) in the 'on' state, so I don't really understand the 'few volts' you expect to find across the switch (in it's 'on' state) to use for brief pulsed charging of capacitor/battery. What am I missing?

Kind Regards, John

 

[JohnW2]

However, my understanding is that the OP is going to have a 'mini computer' in there, which I imagine has to be at least partially powered-up for most/all of the time - I certainly can't see it coming on and off many times per second!

There are many micro processors that are normally asleep until woken up by a stimulus ( from a wireless receiver ) , perform what ever function is needed and then go back to sleep

Sure, but, as I said/implied, waking up takes a finite time (seconds in the case of a laptop/PC) so I just can't see it being able to come on/off umpteen times per second (or even umpteen times per minute), could it? (and it clearly would need some non-volatile storage) ... and, as BAS has said, it would need at least some power even when asleep, otherwise it would not be able to wake up (and/or know when to wake up).

Kind Regards, John

 

[ban-all-sheds]

and, as BAS has said, it would need at least some power even when asleep, otherwise it would not be able to wake up (and/or know when to wake up).

Nor be able to do anything, having lost the contents of all its registers, caches, program counter, instruction decoders etc.

 

[JohnW2]

and, as BAS has said, it would need at least some power even when asleep, otherwise it would not be able to wake up (and/or know when to wake up).

Nor be able to do anything, having lost the contents of all its registers, caches, program counter, instruction decoders etc.

Well, as I said, it could transfer all of that into some sort of non-volatile storage before going to sleep and then restore it all on waking up - but that would presumably take 'significant' time and would therefore presumably preclude very rapid asleep/awake cycling.

Kind Regards, John

 

[SimonH2]

In that case, certainly if the switching is on/off (rather than dimming), the voltage across the switch ought to be very low (almost zero with a mechanical switch, but also very low with a solid state one) in the 'on' state, so I don't really understand the 'few volts' you expect to find across the switch (in it's 'on' state) to use for brief pulsed charging of capacitor/battery. What am I missing?

You can't do it with a mechanical switch, but with an electronic switch you delay the turn on for each half-cycle until there is enough voltage across the switch to charge your capacitor - basically it's a dimmer with a fixed "near enough full on" setting.
For a resistive load, the effect of this delay (unless it's quite a long one) is negligible in terms of reduction in lamp brightness.

Now, unless i've cocked things up badly, if you wanted (say) around 10V on your cap, and used low forward volt drop diodes, then you could hold off until the mains was up to (say) 12V. That's about 3.25% of mains peak voltage, and occurs less that 2˚ into the cycle. I can't be bothered right now, but I think you'll find that if you integrate sin(v)^2 from 2˚ to 180˚, and convert to an RMS value - you'll find the difference is very little from doing the same thing over the full half cycle.

There are many micro processors that are normally asleep until woken up by a stimulus ( from a wireless receiver ) , perform what ever function is needed and then go back to sleep.

They still need power, even in sleep states.

But that power is generally "incredibly small" for devices designed for that type of application. In fact, certain CMOS devices take virtually unmeasurable power (it's only the leakage current through the gate insulation) in a static state. There are certainly processors which take a matter of nW in sleep modes.

As an example, picked at more or less random ...
http://ww1.microchip.com/downloads/en/DeviceDoc/41291G.pdf, starting on page 249
RAM Data Retention Voltage 1.5V Device in Sleep mode
Power-down Base Current Typ 0.05µA Max 1.2µA

So it's possible to power down the, and keep it's memory intact, with typically 50nA @ 1.5V min. Normally you'd keep the supply voltage above it's minimum operating voltage (only 2V for low clock speeds), that way when a stimulus arrives, the device can "spring to life", do something, then power down again.
If you chose (say) 3V, that's still only (typically) 150nW, max 3.6µW in sleep mode.

That's how so much battery operated equipment works these days - small devices which shut down the oscillator when idle and go into a DC mode where CMOS takes naff all power. It's a basic issue with MOS devices that the bulk of the power for a gate is typically taken in charging and discharging the capacitance of the gate (the gate and channel effectively form a capacitor, with the metal oxide insulation layer as the dielectric) - the more often you do that (ie the faster you run a circuit), the more power it dissipates.


That's why one of the earlier suggestions was to charge some form of storage (whether rechargeable cell or supercap) when in the off state to run the device when on. If the power budget is sufficiently small then that might work. If the light was left on, then the device would need to turn it off every so often to recharge - but with the fast charge ability of a cap, this might only mean running the load off the charging circuit for half a cycle every so often.
To elaborate, when the cap is running down, on one half cycle you leave the switch off and charge the cap using as much current as the load will pass until the cap is charged. Then turn on the switch. This won't affect the load as it'll still see a supply current, just passing through the charging circuit (with say 10-20V drop) rather than the switch element.

Still an interesting design challenge - to make a supply circuit that will "fast charge" a cap with only a few volts across it (to avoid flashing the light each charge cycle), but while also coping with 370V peak across it when the light is off.
Actually I can see one way of doing it. Turn a transistor on at each zero crossing or whenever the rectified mains is below the desired cap charging voltage, and turn it off whenever the cap is up to the desired voltage - perhaps with a bit of hysteresis. I suppose it's a variation on a hysteresis regulator, but using the load impedance as the current limiter rather than an inductor.
It would also mean only drawing current through the load when the voltage is low, which might possibly be enough to prevent CFLs or LEDs glowing or flashing.

Certainly an interesting design challenge.

 

[bernardgreen]

waking up takes a finite time (seconds in the case of a laptop/PC)

micro-processors can wake up in a few micro-seconds. The RAM contents can be kept intact while asleep. Some do this without use any power at all. In others a small amount of RAM is kept powered with a couple of micro ( or pico ) amps of power. Provided the programmer writes code that stores vital data in this RAM before the SLEEP command then nothing is lost.

A PC has to read the hard drive to load program code into RAM before the machine can access the code to start running the code. A micro-processor accesses program code from ROM and need not load it into RAM.

 

[SimonH2]

Well, as I said, it could transfer all of that into some sort of non-volatile storage before going to sleep and then restore it all on waking up - but that would presumably take 'significant' time and would therefore presumably preclude very rapid asleep/awake cycling.

You posted that while I was writing the previous post. With CMOS processors designed for the job, you don't lose memory or registers in sleep mode.
So you can go to sleep, use a really tiny amount of power to keep the state, and then wake up exactly where you left off.

 

[JohnW2]

You can't do it with a mechanical switch, but with an electronic switch you delay the turn on for each half-cycle until there is enough voltage across the switch to charge your capacitor - basically it's a dimmer with a fixed "near enough full on" setting.

Fair enough - but, as a general observation, we seem to be involved in (albeit very interesting) discussions about all sorts of clever/sophisticated solutions - all for the sake of not running a neutral to the switch position.

If the OP is trying to develop a commercial '2-wire' product, that might be worthwhile. However, if it's just a 'DIY project' I might question the approach!

Kind Regards, John