maintain a switch position when arduin is off/sleep mode

I got one of these adafruit LCD Controllable Shutter Glasses:

If you put 5V to it, it goes dark, and stays dark even after removing the power. When you ground the two contacts or apply 0V, or add a resistor, then it becomes clear again.

I want to use this as "zero power indicator" for a mailbox. Basically showing a sign behind the valve when there is mail, and being black when there isnt. I can toggle the shutter using arduino digital pins, or I can use analog pin "pwm" to control the opacity. However, when I turn the arduino off, or even put it in sleep mode, the voltage is drained, and the screen becomes clear.

So now I wonder, how can I maintain the shutter state even with the arduino is off or in sleep mode? I basically need to disconnect the wire somehow. Any idea's?

HI,

Might it be as simple as putting a low leakage diode in series with the 5V to the LCD?

Keep in mind:

  • The display would only get ~ 4.5 volts
  • The leakage through the diode will drain the charge of the LCD so it might not stay dark when un powered as long as a mechanical switch.

I guess you could use a small relay.

Have you left the LCD "Valve" dark and un-powered for an extended period of time? I would think the charge would slowly dissipate but have no idea how long it would take (i.e. minutes, days etc).

Good luck.

I have not tested over a long time (trying now, and you seem to be right), but refreshing it every minute or so if fine, the arduino will need to wake up to make a measurement anyway.

4.5V is plenty too, it only needs ~4V to go fully dark.

But if I put a diode in series, how will I blank the screen?

Ive considered a relay too, but doesnt a relay need to be powered to maintain its on position?

The diode would allow current into the LCD but when the controller goes off it will not allow current out.

Relay would be easy, the connection to the LCD is only made when the relay is powered. So you power on the relay, set the LCD to the condition you want the unpower the relay (and arduino)

You’re right, that way a relay ought to work. But you where also right about the shutter charge self draining in a few minutes, and Im worried switching the relay every minute will cost more battery power than keeping the arduino on. Could I do the same with a transistor?

Yes you could probably accomplice it with a transistor or maybe FET but I think you should reconsider the diode. A quick google came up with

BAS416 Low-leakage diode

The leakage is stated as 3p amps (p = pico = 10^-12) which at 5V is the equivalent of 1 trillion ohms. Your board and wiring are likely to have more leakage (from contamination, no clean solder flux , moisture etc)

And I doubt you will find a transistor or Mosfet with less leakage, or even as low a leakage.

Another suggestion is a low power logic gate (suggestion by another forum member). Look at parts like the SN74AUP1G79 . I didn't read into the specs very far but the front page stated 0.9µA static current.

If you used this type of device you would not have to re-fresh the LCD every minute (it would be on all the time, or off all the time). So take a look at your power consumption and see what you think works best.