My bathroom extract fan's wax actuator has failed. This is used to slowly open the shutters, when the fan is triggered. This one..
Anyone know please?
It happens that Harry Bloomfield formulated :
I've just had another idea - a 400v diode will reduce the consumption to around 50%, without the extra heat dissipation of a resistor.
Just trying to extend the working life of the actuator.
Good idea, but 400V doesn't give much margin on 240V RMS - go for a
1N4007 1kV diode. This used to be a trick for keeping a non-thermostatic soldering iron on standby, you put a diode in a torpedo switch.(Although the motor does say .02A which would make it ~5W @ 240V.)
Cheers
Clive Arthur submitted this idea :
Good point, I have plenty of 4004's around here.
Yes, no you mention it, I have used that trick myself for irons.
Yes, I spotted the .02A but I wasn't clear that it meant amps and it doesn't indicate .02A at which voltage.
As it was, it was quite quick to respond to open, it will open a bit slower now with the diode in circuit - once the replacement actuator arrives.
Harry Bloomfield explained on 05/06/2019 :
I meant 4007's
I doubt if what you want to do is going to make much difference. These things are designed to not get hot, if they do then they will be faulty. All a dropper will do is make a lot of heat wherever you site it, what is the point of that. I don't know how these devices work but it might well be that the actual mechanism has become stiff and hence caused the failure. Back in the old days Greenhouses had a purely mechanical device that opened and closed the vents. Brian
Brian Gaff pretended :
Quite the contrary, they are designed to get hot. The heat melts a high expansion wax, which pushes out a plunger. The work in a similar way to the wax thermostat used in car engine cooling systems - heat in the coolant, opens the valve. In this case a tiny heating element heat the wax, causing a plunger to extend, this then opens the fan cowl up.
It is designed to work with 120/240v and has heat destroyed it, my idea is to fit a replacement but halve the 240v here, to average 120v, using a 1N4007 diode. It will no doubt open up little slower, but no problem.
All Brian's points are incorrect except that they do often get stiff due to limescale or corrosion. A diode will give 240v x 0.707 rms volts, which is more than 120.
NT
Correct, the unit with a diode at 240V will consume half the power at
240V without.I suspect a diode is the cheapest and most reliable method but this is an ideal candidate for using a capacitive dropper.
It says on it 240V, 0.02A
Has it overheated because the connections to the spade terminals were not perfect?
Was it actually overheating or just discoloured due to it being used in a damp environment and dust in the atmosphere has just stuck to the casing over the years that it has been in use.
I would just replace like with like without modifications.
Will it not get to the same temperature but just take longer? Putting
2.5W or 5W into a device that cannot dissipate that power to heatsink or air flow will cause the temperature to rise above that required to push out the plunger to its maximum travel.If the shutters take longer to open will the actuator be in a static or a lower airflow for longer?
With that sort of actuator, even once it has opened, the power stays on, adding heat energy. It will also be losing heat to the environment. The temperature will stabilise when energy inflow equals energy outflow.
If you halve the inflow, you halve the outflow at equilibrium and as the rate of energy loss is proportional to temperature difference (body to air), that dictates that equilibrium will be established at a lower actuator body temperature - so it *WILL* run cooler.
The actuator will still be fully extended by the time it reaches its normal fully open temperature. It will continue to rise to its equilibrium temperature, but the vent is already fully open by then.
SteveW
no
obviously it does dissipate that
that's how they work
alan_m wrote on 06/06/2019 :
The wire is soldered onto the terminals with no sign of overheating there. The fan casing had become discoloured and brittle, the end of the actuator opposite the piston/plunger had actually burst away. It has quite a strong spring built into the actuator, to return it once it has cooled.
on 05/06/2019, snipped-for-privacy@gmail.com supposed :
No, it will be 50% of the AC waveform, less a bit for the fraction of a volt lost across the junction.
0.707 x the peak AC is the RMS voltage.If you know the RMS, then multiplying by 1.41 will give you the peak voltage. If you charged a cap via a bridge without any loading, that would achieve 240v x 1.41 = 338.4v peak, less junction losses.
do you ever get anything right?
Quite, so half the power? Same voltage but half the duration?
If you halved the voltage, what would happen to the power?
And 0.707 of the rms voltage gives you half the power.
<snip>
Of a sinewave, not a half-wave rectified sine wave.
Sinewave: 339V / sqrt(2) = 240V RMS
Half-wave rectified sinewave: 339V / 2 = 169.5V RMS
SteveW
Steve Walker expressed precisely :
169.5v x 0.707= 119.83v RMS..
Yes, quite often actually and I am right this time too.
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