I am using a GeekPi 8 digit, 7-Segment display in my project controlled by a Nano.
I wanted to turn off the display, without turning off the Nano, when the game is OFF.
So, I placed the +5V across a N.O. contact on my "START" relay.
When I turn the game off. The display dims somewhat, but the digits remain illuminated.
Even if I disconnect the Power to the display the numbers remain.
I guess I'm getting a signal from the Nano through the CS, DIN and CLK pins.
Is this normal?
DO I have to turn off the Nano in order to turn off the display?
I wanted to shut off the readout only (without the software) when the arcade is off. I assumed removing the 5V to the display, though the "START" relay would easily accomplish this. The nano is still running in the background though.
If I must use the "lc.shutdown" command to do this through a pinState for start, then OK.
For my information, is this display defective, or is it operating normally when the 5V is off?
The wiring diagram is not anything fancy. CLK = Pin 13, CS = Pin 10, DIN = Pin11.
Through a pinState?
No, you just tell the display to turn off the lights with code.
The driver chip will still retain what is on the display.
Nothing is defective. "Phantom powering" is when the circuit still gets power from the signal lines through the input protection diodes, which is a bad thing. You could also stop this by changing all the signal lines of the display to INPUT, but that's not as easy as blanking the display.
The wiring diagram is a lot more than that. It should show where the power is coming from, how have you got the relay coil wired up along with the contacts. Words don't cut it in electronics we need schematics. Just a pen and paper schematic is fine.
I assumed removing the 5V to the display, though the "START" relay would easily accomplish this.
So you know why this is a mistake now with the other advice you have received.
The nano is still running in the background though.
So use the relay contacts as a switch on an Nano input, and when you detect it is closed / open ( depending on how you have wired it ) tell the chip to blank off the display.
Yes, make the pin mode an input and enable the internal pull up resistors. Then connect the relay contacts common to grounds and the normally closed to the pin. Then read the pin and if it returns LOW then the relay is off, if it returns HIGH the relay is on.
Get your relay circuit working by all means, you are learning lots of important stuff. But after that, you absolutely should think carefully about whether to remove it. When the display is on, the relay's coil is probably drawing as much if not more current than the display itself! If you use the shutdown method mentioned above, ok, that does not draw zero current when the display is off. But with the relay method, the current you save when the display is off could be less than the extra current drawn when the display is on, even though that is only for short periods.
If using the relay, its coil draws, say, 70mA when the display is on.
If you use no relay, the display, when in shutdown, draws 150uA.
So comparing the 2 options, having the display on for 1 minute using the relay uses as much energy as the display being shut down for over 8 hours if not using the relay.
Knowing that should help you to decide which would be best for you project.
Of course, we have not discussed using a MOSFET to switch off the power to the display...