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I need to replace an electronic timer lightswitch, and both the old one, and this example of a replacement, state that they have a minimum load of 40w.

Why would the switch have a minimum load, and what would happen if I used it with a lower wattage bulb?

This is a for a lantern style outdoor light (although the switch is indoor) and I'd ideally like to put a small energy saving led bulb (5-10 watts) in it.

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The timer is designed to replace a regular switch as shown on the left of the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

Since there is often no neutral connection in the switch wall box it presents a problem for the timer designer. It won't work with a live feed only.

The trick is to replace the switch with a triac which behaves as a fast acting electronic switch and can switch on the current to the lamp at any point in the half-cycle.

Triac waveform.

Triac modified sinewave. Delaying the turn-on point results in reduced voltage on the lamp.

If the timer circuit delays the turn-on point of the triac the voltage to the lamp is reduced. That leaves the remainder available to power the switch. The minimum 40 W lamp ensures that there is enough of a load to pass enough current to keep the timer running. The designers hope you won't notice that the lamp is ever so slightly dimmer.

When the lamp is off there is almost full voltage across the triac so the timer again has power.

These circuits are a little tricky to design as they have to work with full voltage and a fraction of supply voltage.

Note that if the bulb is removed that there is no return path for the timer so it will power down - although there may be a delay if it has some energy stored in a capacitor.

Battery

The battery is probably to keep the clock running during power cuts. As explained in the answer it can always steal enough power when the supply is there. The trigger delay is exaggerated in my sketch and probably 10 V or so would be enough to charge up a capacitor in the switch. The low voltage portions of the sine wave contribute very little to the power of the lamp. On a 240 V supply \$V_peak_ = \sqrt 2 \times 240 = 340 V\$. Since power is proportional to the square of the voltage, the power to the lamp is \$\frac{340^2}{10^2} = 1150 \$ times stronger at peak than at 10 V.

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  • \$\begingroup\$ For a basic RC-timer control circuit (timing the off-delay on every half-cycle), this is easily adequate, but I would think that even a small bit of logic would be too demanding to work there, in terms of getting consistent power in all cases. A battery may suffice, and maybe a clever charging scheme, but I don't see it powered directly like you've drawn it. \$\endgroup\$ – AaronD Jan 5 '16 at 23:17
  • \$\begingroup\$ I read the description a little closer on the example product (precious little to go on), and it appears that they do expect it to be wired as you drew it, which tells me that they either expect the user to change the 24-hour backup battery periodically or that it does harvest some energy from its position in the AC circuit. If it's the latter, I would not be surprised if it messed with the power level to create a more favorable condition to harvest that energy from. Incandescent bulbs won't care and may not even be noticed, but others might, like CFL's and LED's. \$\endgroup\$ – AaronD Jan 5 '16 at 23:24
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    \$\begingroup\$ The battery is probably to keep the clock running during power cuts. As explained in the answer it can always steal enough power when the supply is there. The trigger delay is exaggerated in my sketch and probably 10 V or so would be enough to charge up a capacitor in the switch. The low voltage portions of the sine wave contribute very little to the power of the lamp. On a 240 V supply \$V_peak_ = \sqrt 2 \times 240 = 340 V\$. Since power is proportional to the square of the voltage, the power to the lamp is \$\frac{340^2}{10^2} = 1150 \$ times stronger at peak than at 10 V. \$\endgroup\$ – Transistor Jan 5 '16 at 23:55
  • \$\begingroup\$ That's all true and relevant...for incandescent loads. The OP has an LED bulb, which draws much less current and may or may not have a DC power supply in it. Being a cheap, mass-produced thing, I wouldn't expect that DC supply to have anything fancy in it that isn't legally required, which means that what it can and can't handle by design or by accident are completely unknown. That changes things dramatically for both reasons, hence my comment that incandescents won't care but CFL's and LED's might. \$\endgroup\$ – AaronD Jan 6 '16 at 0:03
  • \$\begingroup\$ Sorry @AaronD. I got mixed up and thought you were the OP. I put all my effort into explaining how the switch is intended to work. You are quite right that an SMPS LED or CFL may not pass any current until powered up properly. Similarly they might misbehave on what's effectively a fixed dimmer switch. \$\endgroup\$ – Transistor Jan 6 '16 at 0:16
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It probably uses a triac as the switching element, which is not a simple switch. For example, it may have difficulty with reactive and electronic loads like motors and CFL's/computers.

The other quirk about triacs is that they latch on until the current stops. There is a threshold for that latching current, which when multiplied by the expected voltage, produces a number for power. I alluded to this here, but didn't actually say it: https://electronics.stackexchange.com/a/148084/53375

If the controller were to hold the triac on like I described in that other answer, then there would be no minimum load, but if it only "bumps" it on like a traditional manual dimmer, and relies on the latching behavior to do the rest, then there is.

If they're actually serious about the minimum load, and you give it something less, it may flicker with the "bumps" that try to turn it on, but will never be fully on. Anything besides an incandescent bulb, pure heater (no fan), or dirt-cheap LED rope light (the kind that has a bunch of LED's in series with a single resistor and clearly flickers at the line frequency when you wave it) will likely hate you for that.

I would recommend a mechanical timer for that circuit. It'll use a synchronous motor to drive the mechanism, so it's as accurate as the power line, which itself is tightly controlled simply to keep all the generators in sync across the country. And the important part is that it'll use a mechanical switch that really is simply on or off.

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