Relay doing opposite to digital pin

Question

I have a Arduino UNO with a Ethernet shield which hosts a basic HTML page with a button on it which controls a Relay module.

I'm trying to connect to a computer's power button so I can turn it on remotely.

I have found out that the relay basically does the opposite to the digital pin's output

Normally open Relay
HIGH = Open
LOW = Closed

I connected it through a NOT Gate made using a NPN transistor and it works perfectly how I wanted it.

Now to the question... Is there a way to write the code to make the digital pin = HIGH straight away and fast enough to not allow the relay to close the circuit and accidentally turn the computer on/off when the Arduino restarts, so I don't have to use a NOT Gate.

Answer 1

You could use a less invasive way to turn on your computer: wake on LAN. In this case you only need to generate the magic packet.

It won't work in forcibly turning off the PC, but it's more generic and doesn't require HW hacking.

Answer 2

To ensure that relay doesn't switch between NO-NC when controller is away relay will need some memory.

Imagine having a 1 bit memory driving relay. If arduino dies/restart, its own GPIO may change, but relay's memory will remember last state and stay there, till controller instructs otherwise, granted relay circuit and its memory element don't lose power.

This also means there will be at least 2 lines required from uC to control this relay now.

I'd prefer to use some kind of serial shift register, e.g. CD4094, CD4042, CD4099 etc because that allows me to control more than one relays from same 2 lines on uC.

You may find some other part more suitable for your application.

Answer 3

The relay module you have is this one from SainSmart:

• http://www.sainsmart.com/arduino/arduino-components/relays/sainsmart-2-channel-5v-relay-module-for-arduino-raspberry-pi.html

The schematic is like this:

As you can see you connect VCC to +5V and IN2 to your IO pin. Setting the IO to LOW will then turn on the LED and the optocoupler together.

But that's not the only way you can connect it up.

Connect VCC to the IO pin and IN2 to GND instead. Then you set your output HIGH to turn it on.

The arrangement shown in the schematic is commonly done because some MCUs are able to sink more current than they can source thorugh an IO pin. That is not the case with an Arduino which has symmetric IO pins that can sink and source the same amount of current.

Answer 4

I've solved the problem, Thanks to everyone who helped.
I don't need a NOT gate anymore, here's what I added to make it work:

void setup() {
  startup();

}

void startup() {
digitalWrite(relay1, HIGH);
digitalWrite(relay2, HIGH);
}

void loop() {

}

(I removed the rest of the code to keep it simple and easy to understand)