Resistor hot in bathroom extract

[EFLImpudence]

As bernard has said, it's down to the powering of the 'electronics' which produce the run-on delay. If the supply for the electronics was derived from the S/L., rather than the permanent L, the fan would probably come on when the light was switched on, but the fan would then go off immediately the light was switched off, because the electronics sustaining the run-on would also have lost its supply.

To get around that would not be impossible, but would introduce more complexity, since a way would have to be devised to maintaining 'alternative' supply to the electronics after the light had been switched off - either with a battery (or very large capacitor) or by 'switching' to a supply derived from the perm L during that period.

Maybe I haven't thought this through well enough, but what I don't really understand is why they do not (as with cheap LED 'lamps') use a capacitive 'dropper', rather than a resistor, in which case the heat (and most of the 'energy wastage') would go away.

I don't know much about the electronics.

However, it is my belief that a timer fan is always switched on by the timer connection (even when set to no time delay). That is why the Switched live has to be connected to both PL and SL(LT) when only L and N are available. The fan then stops when the supply is removed from both - because there is no supply..

So, with three wires: PL, SL & N, could not a timer be triggered by the SL being switched on which then connected the PL to whatever needed it until the timer switches off - when it would be disconnected until the next time?

Could that not be achieved by the electronic equivalent of a timer and a relay?
Well, yes, I know it could but why isn't it? How much can it cost?

Apart from being very cheap, if that is the reason, why would something be designed to be drawing current continuously if used possibly only once or twice a day?

 

[JohnW2]

However, it is my belief that a timer fan is always switched on by the timer connection (even when set to no time delay).

That is true. The only thing switching the (P/L) supply to the fan is the triac - which, in turn, is only controlled by the electronics.

That is why the Switched live has to be connected to both PL and SL(LT) when only L and N are available. The fan then stops when the supply is removed from both - because there is no supply..

Indeed so.

So, with three wires: PL, SL & N, could not a timer be triggered by the SL being switched on which then connected the PL to whatever needed it until the timer switches off - when it would be disconnected until the next time?

Yes, as I said, it would be possible. As I said, it would require that the timer electronics were initially supplied with power derived from the S/L but, after the light was turned off (no longer any voltage on S/L) would then have to 'switch' to the electronics being powered by a second 'power supply' (fed from the P/L) for the duration of the timed run-on period. Relatively simple, but would probably require 'duplication' of all the 'power supply' components (all of those in bernard's 'red box' in his post #62] - certainly some increased cost but, maybe more to the point, perhaps more components that could easily be fitted into the available space.

I haven't thought to deeply, but one could probably avoid that 'duplication' of components' by doing the switching (of source for the power supply) with a relay - but that, again, might be a problem getting a relay to fit in the available space.

Apart from being very cheap, if that is the reason, why would something be designed to be drawing current continuously if used possibly only once or twice a day?

As I said, what I haven't worked out whether there is a sensible reason why they don't use a capacitive dropper. The "current being drawn continuously" would still be roughly the same but, because of its phase, the energy consumption would be minimal (i.e. very low PF), without the need for additional components for your alternative approach.

Kind Regards, John

 

[bernardgreen]

As I said, what I haven't worked out whether there is a sensible reason why they don't use a capacitive dropper.

A capacitor would have zero impedance and thus when the rising positive sine wave was greater than the Zener diode voltage the Zener diode would begin conduct. As the voltage increased there would be nothing to limit the current passing through the Zener hence a larger more expensive Zener diode would be needed.

 

[JohnW2]

A capacitor would have zero impedance ...

Eh? The whole point of using a 'capacitive dropper' is that it has a relatively high reactance/impedance.

At 50Hz, a capacitor of about 0.14 μF would have about the same impedance as the 22kΩ - 24kΩ resistor used in these modules.

and thus when the rising positive sine wave was greater than the Zener diode voltage the Zener diode would begin conduct. As the voltage increased there would be nothing to limit the current passing through the Zener hence a larger more expensive Zener diode would be needed.

With a 50Hz supply, a 0.14 μF capacitor with 215V RMS across it (230V-15V) would limit the current (through zener and/or the 'electronics') to about 9.5ma RMS (about 13mA peak) maximum, in exactly the same as would a 22kΩ - 24kΩ resistor in the same position. The difference is that the resistor would dissipate a very significant amount of power (producing heat), whereas a 'perfect' capacitor would dissipate zero power.

As I said, this is exactly what happens in a cheap LED 'lamp'.

What, if anything, am I missing?

Kind Regards, John

 

[DetlefSchmitz]

At 50Hz, a capacitor of about 0.14 μF would have about the same impedance as the 22kΩ - 24kΩ resistor used in these modules

What you suggest doesn't actually work (try simulating it to see why). However something like this might:

The 5k is a rough guess at the average load.

 

[JohnW2]

What you suggest doesn't actually work (try simulating it to see why). However something like this might: .... The 5k is a rough guess at the average load.

Maybe I'm getting old (I did have a birthday 10 days ago!)

I admit that I did 'simplify things in what I wrote (particularly in essentially 'ignoring' the rectification), but the concept of what I wrote still seems essentially right to me!

In what way does your "something like this" differ from what I was talking about, other than apparently having a 15 μF capacitor (am I reading that correctly?) ? With such a 'large' capacitor (and a ~5kΩ load) I would think that an awful lot of current would be going through the zener.

What am I still missing?

Kind Regards, John

 

[DetlefSchmitz]

Well, you need to play with the components. But the red line shows the current drawn (so also through the zener) scale at right hand side.

Your version (if I guessed it right) gave about 0V at the output.

 

[JohnW2]

Well, you need to play with the components. But the red line shows the current drawn (so also through the zener) scale at right hand side.

I realised that. I'm still lost and therefore continue to think/play

Your version (if I guessed it right) gave about 0V at the output.

What did you 'guess'? 'My version' was seemingly identical to yours, except for C1 = 0.14 μF rather than 15 μF (i.e. a bit over 100 times smaller than yours)..

Kind Regards, John

 

[DetlefSchmitz]

John, I only spent about 5 minutes trying 4 or 5 versions, and I accept that I might have made a mistake. What you see is the first arrangement that worked. I'll have a go tomorrow and see if I've maligned you. If you weren't aware this is LTspice, is free, and a good way to check things like this fast. I've had enough for today, so sorry for the wait.

 

[JohnW2]

John, I only spent about 5 minutes trying 4 or 5 versions, and I accept that I might have made a mistake. What you see is the first arrangement that worked. I'll have a go tomorrow and see if I've maligned you.

Fair enough. regardless of who is right, I certainly don't feel 'maligned', and it may well be me who has, at least in some senses, been wrong, so I look forward to what your further exploration reveals!

If you weren't aware this is LTspice, is free, and a good way to check things like this fast.

Yes, I have LTspice, on the recommendation of someone here (probably you!) a good few moons ago. However, I have to confess that I never managed to climb very far up the (seemingly fairly steep) learning curve, and therefore have used it very little. Maybe you have stimulated me to have another go

I've had enough for today, so sorry for the wait.

No problem - I know the feeling ! In the meantime, doing some 'thinking aloud' (very rough ballparks, again 'ignoring' the rectification) aloud ....

It seems to me that probably the only way in which there could be a flaw in my thinking is if I am misunderstanding and hence underplaying the effect the rectification has, since if one just sticks to thinking about ('full wave') AC, none of it is much more than Ohm's Law ....

Your assumed 5 kΩ, load (at 15V) represents 3 mA, which sounds fairly credible for the timer module.

If there were no current going anywhere else (like through the zener, when one has rectification), that 3 mA travelling through your 15 μF 'dropping capacitor' (reactance/impedance about 212 Ω at 50Hz) would result in a voltage drop of a mere 0.6V ! On the other hand, with my suggested 0.14 μF dropping capacitor (reactance/impedance about 22 kΩ at 50Hz), the voltage drop would be about 66V.

Those drops are still too low. However, if, not unreasonably, one assumed that, an additional, say, 6 mA of the current going through the dropper was going somewhere other than through the load (e.g. like through a zener, when there is rectification), hence a total current through dropper of 9 mA, then your 15 μF capacitor would still only drop about 1.9 V, whilst my suggested 0.,14 μF one would drop about 198V - very much closer to 'what the doctor ordered'

My assumption was that none of the above is appreciably conceptually changed by introducing the rectifier. At first sight it might seem that the (average or RMS) currents would be roughly halved, but I think that may well be to some extent offset by the fact that the large 'smoothing capacitor' across the load maintains current through the load (which has to come from somewhere) which is pretty constant throughout both half cycles of the supply. However, as I've said, it is here (the impact of the rectification) that their might possibly be a flaw in my reasoning.

However, I have to say that my thinking above (about capacitor values) is consistent with my practical experiences. As I've often mentioned, in my house I make fairly extensive use of capacitors for 'switched dimming of (usually 5W - 12W) cheap LED 'lamps'. My experience is that adding anything like 15 μF has no noticeable effect on LED brightness. Depending on the degree of dimming I want (and the LED wattage), I generally use capacitors in the range 0.015 μF to 0.22 μF or thereabouts.

Kind Regards, John

 

[bernardgreen]

The capacitor can be charged once and only once during the first positive cycle of the supply, it cannot lose that charge as the diode prevents any current from leaving the capacitor.

 

[DetlefSchmitz]

John, we were both wrong, for the reason that Bernard explained, and you later realised. If I continue the plot for a longer time, the supplied 15V just dies away. If I get time I'll try a few other circuit arrangements.

 

[JohnW2]

View attachment 312536
The capacitor can be charged once and only once during the first positive cycle of the supply, it cannot lose that charge as the diode prevents any current from leaving the capacitor.

Thanks. It's hard to argue with that

However, that immediately begs the question as to how/why cheap LED 'lamps' (and my dimming of them with a series capacitor) actually work. The answer, I presume, is that (as I had overlooked), they utilise full-wave, rather than half-wave, rectification, thereby providing the 'discharge path' for the capacitor which, as you say, is absent with half-wave rectification. This real-world situation therefore becomes this, which I would think does. will work, doesn't it? :

Returning to what started all this, if that does work, then there ought to be plenty of physical space to replace the rectifier diode with a small encapsulated bridge rectifier and to replace the infamous resistor with a 'small' capacitor, shouldn't there?

Kind Regards, John

 

[JohnW2]

John, we were both wrong, for the reason that Bernard explained, and you later realised. If I continue the plot for a longer time, the supplied 15V just dies away.

Indeed. It seems that I had overlooked the relevance of the difference between half-wave and full-wave rectification.

If I get time I'll try a few other circuit arrangements.

Thanks. You will have seen what I've just written to bernard. It would be interesting to know what LTspice has to say about the situation with a bridge rectifier - but I would still think that the value of the 'dropping' capacitor probably ought to be in the ballpark I have suggested.

Kind Regards, John

 

[bernardgreen]

It gets more complicated.

Given that two capacitors in series ( with no other components ) will have the same amount of charge ( current x time ) then the smaller capacitor will have a higher voltage ( V = charge / capacity ) than the larger capacitor

But with 1 Watt LED lamps with a full wave rectifier between the capacitors the circuit becomes a charge pump. The charge that passes through the 0.47 uF capacitor passes into the 10 uF capacitor. The rectifier bridge prevents that charge leaving the 10 uF when the AC reverses. Instead when the AC reverses the 10 uF capacitor gets another 0.47 uF worth of charge thus increasing the voltage on the 10 uF capacitor. Eventually the voltage on the 10 uF capacitor becomes high enough for the LED elements to light and also act as a pseudo Zener diode.

These LED lamps are rated at 230 V but two in series ( 115 V each ) appear be as bright as they are on 230 V.