Relay Shield requires +5V to turn relay OFF

Other Hardware Motors, Mechanics, Power and CNC
1

The Arduino relay shields are built such that +5V to IN1 will de-energize the coil of the relay, whereas grounding IN1 will energize the coil (providing you have +5V at JD-Vcc).

The implication is that the normal state of the relay is energized, and it is only de-energized when you send a +5V signal to the shield.

This is backwards from what I want; I want the relay to energize ONLY when I send a signal voltage.

here is a schematic of the relay shield:

If I wire the shield as follows:
-remove the jumper between Vcc and JD-Vcc and send +5V power supply to JD-Vcc
-ground IN0, (and, of course, GND on shield)
-send signal (+5v HIGH, 0v LOW) from Arduino Uno to Vcc
then it works as I would like it to.

My question is, can anyone see a problem with this?

2

Simply tie the Arduino output to 5v and make your output low when you want to turn on the relay.

Much easier to change logic level than re wiring the shield.

Sending an out put of "high" or "low" is still sending a signal.

Weedpharma

3

It is not "sending a signal" that bothers me, it is powering the relay coil when you are not doing anything that bothers me. The way you describe is the standard method which then requires backwards logic in the sketch.

In other words:
digitalWrite(pin#, HIGH) powers the relay OFF
digitalWrite(pin#, LOW) powers the relay ON.

If you have a relay load wired as C-N/O then you must send a HIGH signal constantly to the relay coil to maintain the load as disabled, until the moment you want to enable the load and then you must send a LOW signal. This just seems wrong to me and I don't understand why the shield was designed this way.

In my method the relay coil is de-energized until you send a HIGH signal to it at Vcc.

4

If there is no connection to the input of the relay board, the relay is not operated. Taking the input low will turn on the relay. Having the output pin at 5v is the same as no connection and the relay will be off.

It is very common to have active low switching.

Weedpharma

5

Hi,
What do you mean by "not doing anything"?

If you leave IN0 open circuit, in other words not connected to anything, the relay will stay deactivated, because there is not current flow through the input of the opto coupler.
I agree with the logic of 0 activates and 1 deactivates being weird, but you can program the controller to cater for this.

Tom...... :slight_smile:

6

thanks for your input guys, and I see your point.

So, if want to upload a new sketch that doesn't use the relay, yet I don't want to remove the wiring of the relay, then in the new sketch, I must not initialize the pin that the relay is wired to, b/c as soon as that pin is initialized to OUTPUT, the relay will be ON.