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I'm sewing a purse and I would like to put some LEDs inside of the bag that only draw on the battery and turn on when the bag is open.

The "switch" that I'm using are the metal snaps on the bag.

I have figured out how to make this work, but it slowly draws power all of the time. Is there a way to do that without that happening?

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    \$\begingroup\$ Please don't make us guess your circuit. Provide a schematic and list the part numbers you are using. \$\endgroup\$
    – Bort
    Commented Oct 24, 2018 at 22:44
  • \$\begingroup\$ What does your current circuit look like? You could use a low-leakage NFET in the low side with a very high value pull-up on that gate so that opening the purse un-grounds the gate and turns the LED on. \$\endgroup\$
    – John D
    Commented Oct 24, 2018 at 22:44
  • \$\begingroup\$ If its drawing power maybe when the purse is closed you can't see the LED turn on when the "switch" turns it on. What you need is a switch that closes (makes contact) when the purse opens. Got a picture of the purse? \$\endgroup\$ Commented Oct 24, 2018 at 22:50
  • \$\begingroup\$ No, it turns off when closed, I can see it through the fabric. I'm using a MOSFET. I just wanted to verify that having it work like a normal switch and totally cut the power is impossible. \$\endgroup\$
    – futurebird
    Commented Oct 24, 2018 at 22:53
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    \$\begingroup\$ "I'm using a mosfet". Okay. Some of the solutions below use a "mosfet". But that isn't very helpful without a schematic now is it? Perhaps you're already using the same circuit they have provided. But who knows, because you haven't provided a schematic. \$\endgroup\$
    – Bort
    Commented Oct 24, 2018 at 22:56

4 Answers 4

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The other answers provide electrical solutions. A simpler solution is mechanical. A reed switch and magnet, so that the connection breaks when the magnet is near is one solution. Another is a simple normally closed switch, that will be open when pressed. You would have to sew it on in a way that the closed purse pushes the switch. Micro switches (with or without a lever) are ideal for this. Both of these are 100% battery drain free (aside from self drain of course.)

enter image description here enter image description here

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    \$\begingroup\$ A reed switch is a great idea! \$\endgroup\$
    – futurebird
    Commented Oct 24, 2018 at 23:04
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    \$\begingroup\$ You'll want a normally-closed reed switch. They do exist, but they're a bit hard to find. It's probably worth specifying, because reed switches are almost always NO. \$\endgroup\$
    – Hearth
    Commented Oct 24, 2018 at 23:39
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    \$\begingroup\$ an SPDT reed switch could also be used. \$\endgroup\$ Commented Oct 25, 2018 at 0:33
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Obviously no.

The only way to distinguish a short circuit (closed switch) from an open circuit (open switch) is to pass a current through it.

The good news is that this current can be arbitrarily small — e.g., much smaller than the self-discharge rate of the battery. Use a MOSFET and a very large (many megohms) pullup resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

I've added diodes D1 and D2 to protect the MOSFET gate from ESD events, such as static discharge when someone touches the clasp. Select the value of R2 (and the battery voltage) to be suitable for your LEDs.

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  • \$\begingroup\$ Unfortunately this circuit may not work at all. The leakage current for the 1N1418 could be up to 10nA ….much greater than the current flowing through R1 ….so R1 effectively does nothing. In addition even if it were to work (diode leakage low enough) the RC time constant for R1 and FET gate capacitance is about 8 seconds, so it's likely that the LED light would take at least 6 seconds to turn on after the purse was opened. \$\endgroup\$ Commented Oct 25, 2018 at 5:43
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Here's a possible circuit. The pull-up could be even larger value, and the diode is for ESD protection, which you may need if the pull-up gets very large.

schematic

simulate this circuit – Schematic created using CircuitLab

The ESD diode, if needed, should have a breakdown voltage that's larger than the battery voltage and less than the gate max VGS rating. The diode leakage current should small enough to not affect the turn-on of the FET, given the pull-up resistor value. Without knowing more about the battery and LED it's hard to make recommendations for specific components.

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  • \$\begingroup\$ This is kind of what I have. But R2 is only 1M, and I don't have whatever the white arrow at the bottom is... (grounding) I think I need to do some more research in to the types of mosfets I could use. Seems like the trick is picking the right one. \$\endgroup\$
    – futurebird
    Commented Oct 24, 2018 at 22:56
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    \$\begingroup\$ @futurebird: the ground symbol is just there to make the designer (and possibly the simulation software) happy. The ground reference is an arbitrary thing. To add it to your circuit, just point at the negative terminal of the battery and say "that is ground". \$\endgroup\$
    – TimWescott
    Commented Oct 24, 2018 at 23:03
  • \$\begingroup\$ @futurebird: Define "slowly draws power" please. Have you measured the current draw with the thing in the off state? You shouldn't be pulling much more than the current draw of your 1M-ohm resistor. \$\endgroup\$
    – TimWescott
    Commented Oct 24, 2018 at 23:06
  • \$\begingroup\$ Just like the other answer you seem not to understand what the low current characteristics of a Zener diode look like. Below about 10V Zeners tend to have considerable slope to their on reverse current. So you would expect that you may not get the FET to turn on at all due to the low current from R2. You could increase the Zener voltage and get a much sharper conduction slope ….above 10V you may get down to 100nA, but even here, you may not be able to turn on the FET. \$\endgroup\$ Commented Oct 25, 2018 at 5:56
  • \$\begingroup\$ @JackCreasey I understand the leakage characteristics, I specifically said to check that wrt the pull-up resistor. Also, I specified an ESD diode, not a Zener. Some small ESD diodes have very low leakage. I also mentioned that it may or may not be needed. \$\endgroup\$
    – John D
    Commented Oct 25, 2018 at 14:01
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If the 'catch' of the purse were magnetic, it could be designed so that disengaging the catch raises the leakage magnetic field at one snap, and that could close a magnetic-activated reed switch.

This requires that the catch magnetic flux be steered in two ways, one of which keeps the purse closed and delivers low magnetic field strength to the axis of the reed switch, and the other of which has the aperture open but delivers higher field strength (and causes the reed switch to close).

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