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I'm investigating the lowest power approach to have a single momentary push-button switch function as an on/off switch for a high-wattage load (13.8V, max 5A)—i.e., a single push gives power to the load, another single push cuts power to the load.

I'd rather not use a switching relay unless I absolutely have to because of the additional current draw and heat that it will add to the setup when on.

I would also prefer to use an approach that draws as little current as possible when the button is not pushed and the load is not powered.

The duty cycle will be the load being powered up to an hour (contiguous) per day, some days several hours total / at a time. Current draw will generally be low (<0.5A) with occasional (not necessarily predictably timed) high-use peaks (max 5A) sustained for up to a few minutes at a time.

What solutions exist? I know some of the silicon-based approaches will require finding specific parts with appropriate ratings for the load.

I'm exploring this for a particular application, and can provide more details if it would be helpful.

EDIT 2024-04-22: addressed some of the clarifications requested in the comments.

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    \$\begingroup\$ How low is low enough? A latching relay consumes zero power unless it's changing state. \$\endgroup\$
    – vir
    Commented Apr 17 at 19:41
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    \$\begingroup\$ Does one push button turn the device on, and the device is turned off in another manner (perhaps another button)? Or does one push turn the button on, and a second push (perhaps a long push) turn the device off? Or do you want some other protocol? Can you use a latching push-button switch, or must it be a momentary push-button switch? \$\endgroup\$
    – Math Keeps Me Busy
    Commented Apr 17 at 21:02
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    \$\begingroup\$ @JonathanW. I personally like the latching relay approach, given what I've read from you and how I interpret it. But that is, of course, a mixture of mindsets and resonances. All of which may mean nothing much of use. I'll post up a simple circuit here. It consumes power while just sitting around. The latching relay does not. So it's better. But just in case a non-relay semiconductor answer might be of some use here. I have a bunch of these. But I'll select a very low power version. \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 13:28
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    \$\begingroup\$ @periblepsis It is possible to construct a solid state latching relay where the only quiescent current is leakage through mosfets. It can be implemented with many mosfets, but a lot of those could be replaced with cmos ICs. I am about to head out, and I have some work to do this afternoon, but if no-one posts such a circuit, before I get to it, I will post one. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Apr 25 at 14:52
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    \$\begingroup\$ @MathKeepsMeBusy Cool. More is better. ;) \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 14:55

3 Answers 3

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This is a semiconductor version of a latching relay. It should power-up off. (But I may want to think about this a bit more.)

schematic

simulate this circuit – Schematic created using CircuitLab

I also have no idea if your load would fit this arrangement. (In the case of a relay, it's almost certain you are fine. But I never know with circuits like this when there is too little data about the load arrangements.)

That's it. Maybe it helps. Maybe not.

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  • \$\begingroup\$ Do you know how much power this consumes when idle? Seems like it'd be significant enough that I wouldn't want to use it for this application. \$\endgroup\$
    – Jonathan W.
    Commented Apr 26 at 2:02
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    \$\begingroup\$ @JonathanW. Since I didn't know if you cared at all about it, I set R2 to 220k. At that level, it's like 60 uA. But you can multiply it by 10 and drop the current to 6 uA and it won't care. I am assuming the button is pressed for a few milliseconds for validation purposes in Spice. R5 and R6 could also be updated a bit, too. I just didn't want to put any extra thought into it. Since you weren't all that open handed about your application, I couldn't think for myself. So all I could be is some idiot. \$\endgroup\$
    – periblepsis
    Commented Apr 26 at 2:08
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    \$\begingroup\$ @JonathanW. Little enough when off that you won't be able to measure it with a bargain bin 3.5-digit multimeter. It consumes less power than most lead acid batteries waste on self-discharge. Seems like it'd be significant enough How much current do you think flows from 13.8V through a 1Mohm resistor? \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Apr 26 at 13:48
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Here is a circuit from Mosaic Industries that, if implemented with CMOS logic gates, appears to have only leakage current when it is quiescent, i.e. not switching. (I am of course not counting the current to the load).

enter image description here

The single pole switch is the momentary push-button. The double pole switch determines the behavior on power-up. It can be replaced by fixed wiring if it is known in advance that one always wants the circuit to power up with the load off, or with the load on.

The quiescent current of CMOS logic gates is only the MOSFET leakage current provided the loads are drawing no current. In this circuit, all the loads of the logic gates have capacitors that block DC current.

Word of warning. I have not simulated or tested this particular circuit. (I hope to get around to it). I have in the past been disappointed with circuits that I have previously found on this site, which needed tweaking to work properly. So this circuit may need tweaking. However, at the moment, I have other things on my plate.

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This latch draws nothing except leakage (nA) when load is off. enter image description here

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  • \$\begingroup\$ You have suggested parts for the Ns and the P that will be able to handle the load? Will standard ¼W resistors work? \$\endgroup\$
    – Jonathan W.
    Commented Apr 19 at 12:49
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    \$\begingroup\$ From left to right: 2n7000, BS250, IRF44Z. The 0.25W resistors are ok. Btw, this circuit isn’t controlled by one push button. You need one for On and one for Off. \$\endgroup\$
    – Michal Podmanický
    Commented Apr 19 at 13:29
  • \$\begingroup\$ Thanks! But yes, a single momentary push-button is a constraint I have to work with here. \$\endgroup\$
    – Jonathan W.
    Commented Apr 22 at 13:37

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