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I'm designing a light switch that will be controlled by a uC. The problem is that the uC will be connected to a coin battery cell (3,3v) and he can't waste energy in the switch circuit. So what would be a design that would have a low power consumption in the activation switch?

The light will be connected to the grid and it will draw a maximum of 10A.

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  • \$\begingroup\$ Something MOSFET-based that draws most of its power from the circuit. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 15 '16 at 2:55
  • \$\begingroup\$ Yes, i was thinking a N MOSFET and P MOSFET in anti-parallel but that won't work. \$\endgroup\$ – Lucas Arruda Nov 15 '16 at 2:58
  • \$\begingroup\$ You mean like a transmission gate? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 15 '16 at 3:01
  • \$\begingroup\$ Yes, like a transmission gate. \$\endgroup\$ – Lucas Arruda Nov 15 '16 at 3:06
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    \$\begingroup\$ If you are driving a light on the grid, why not draw power from the grid instead of a coin cell with 3k series impedance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 15 '16 at 3:12
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There are such things as Latching, Sequence and Impulse Relays. Most if not all such devices are designed to only require power when changing states.

There is also a 2 coil latching relay. I believe you momentarily energize one coil to turn it on and the other to turn it off.

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  • \$\begingroup\$ They require a large amount of power to switch. From the pdf there's a characteristic Minimum Switching Requirement: 100mA. That peak would be too high for the battery. But then again, is worth considering it if a use a more powerful battery. As for the 2 coil latching relay i can't find a current consumption. \$\endgroup\$ – Lucas Arruda Nov 15 '16 at 4:30
  • \$\begingroup\$ 2 different SSRs in a start/stop arrangement to energise a latching 110/240v relay for the light? \$\endgroup\$ – Joshyp00 Nov 15 '16 at 5:35
  • \$\begingroup\$ @LucasArruda if it is only for a short time, 100mA is not a problem, provided you have enough capacitance in parallel with the coin cell. \$\endgroup\$ – dim Nov 15 '16 at 9:43
  • \$\begingroup\$ Lucas, I think @Joshyp00 has a good point. You would control some sort of solid state relay (or other device like a triac) and switch mains power to flip the magnetic (latching) relays. I suspect the current necessary to flip a triac is much smaller than that needed to flip a magnetic relay. \$\endgroup\$ – st2000 Nov 15 '16 at 14:16
  • \$\begingroup\$ @st2000 well the latching would still consume a peak current from the coin cell. I think putting a parallel capacitor like dim suggest would do the trick. Thanks. And thank you Joshyp00 as well. \$\endgroup\$ – Lucas Arruda Nov 15 '16 at 15:16
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Finding a relay that has both contacts rated to 10A and a low coil wattage may be difficult/costly but I'm sure they are out there. The output signals from your uC would only be momentary and your battery life would depend on the number of times you planned on switching the circuit.

Could you add a powersupply/transformer that fed the uC from your mains supply?

P.S. This was my first time using circuit lab and had to rush to keep it in the free demo time!

enter image description here

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  • \$\begingroup\$ You might want to point out to Lucas that none of the relays in your answer are latching relays. Also, Lucas, if you use @Joshyp00's answer, you need to be careful about making sure you are using a continuous duty relay (one that is continuously left on). I would prefer to use a "Lamp Relay" that has 2 coils. One for activating (closing) the relay and another to deactivate (opening) the relay. Then wire one of the SSRs to each coil. That way you don't need to worry about getting a continuous active relay. Or burning the extra power. \$\endgroup\$ – st2000 Nov 16 '16 at 3:55
  • \$\begingroup\$ @Joshyp00 No I can't add a power supply, I only have one phase wire that it will go to directly to the light and anything that I put in the switch circuit it will be in series with the light. Except for the uC which will be drawing his power from the battery. In the schematic the coil of the lamp relay is connected directly to the neutral wire so is not viable. \$\endgroup\$ – Lucas Arruda Nov 16 '16 at 16:12
  • \$\begingroup\$ @st2000 I got it, but the coil of the relay would have to be in series with the lamp, I don't have access to the circuit in the lamp. All I have is the phase which is going directly to the lamp. So I can't control the coil from power coming from the grid. \$\endgroup\$ – Lucas Arruda Nov 16 '16 at 16:20
  • \$\begingroup\$ I think you have backed your self into a situation that can not be easily solved. You can try leaching power off the mains hot and load lines. But this technique depends heavily on the type of load. If an incandescent bulb it should be easy. If a high efficiency bulb this may or may not work. I would, at this point, seriously consider changing the scope of the project. As with its current constraints it sounds like it will only work for a limited number of operations (really small power source) or for only very specific loads (probably incandescent (which are nearly obsolete) ) bulbs. \$\endgroup\$ – st2000 Nov 17 '16 at 0:50
  • \$\begingroup\$ @st2000 what about using a low power latch relay, which only requires power to change state? Since the current to switch the relay is too high i would use capacitors so the current wouldn't drain the battery. The coin cell has a lot of power but it can't sustain high currents. \$\endgroup\$ – Lucas Arruda Nov 17 '16 at 3:01

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