Thank you for this, I’m on holiday at the moment, therefore I cannot try with a multimeter what comes out of the NC/NO when alarm is on/off…
Just for my understanding, if I circuit C2 and NO2 (via nodeMCU), if the alarm is off, nothing will circulate; when I will arm/turn on the alarm, the relay will close and the NO2 will “touch” the C2, allowing current (electricity) to flow. (Is this right)?
In this scenario, the voltage cannot be more than 3.3 V (to avoid to fry the nodeMCU) right? So, I am sorry, but I do not understand why “this relay” should not output anything… Sorry my ignorant question… By circuiting (via nodeMCU) C2 and NO2 is not as put a circuit in parallel?
My electronic knowledge is a bit Rusty… 20 years ago at school…
Touching two ends a of copper wire together makes a circuit. Just like a switch.
Current can flow through the circuit when it is made (same as a switch being on).
But ‘can’ is not the same as ‘will’. Current will not flow through the circuit - unless there is some external power source, like a battery.
The relay contacts are just a switch, under the control of the alarm logic instead of a finger. They are literally two bits of brass that are either touching or not-touching.
When the switch is open, no voltage comes out and no current flows.
When the switch is closed, no voltage comes out and no current flows.
You can test this with your multimeter. Using ohms range, you can see a very large resistance (open circuit) or a very low resistance (closed circuit).
The way you’ll be using this relay contact, your nodeMCU will become the “power source” for the switch detection.
The pull-up resistor connects one contact wire to the nodeMCU power supply. So of course that is a suitable voltage for the nodeMCU. The resistor limits the current that can flow to a tiny amount, because we are only making a signal here, not powering a lamp or heater. This contact wire is also connected to the nodeMCU input pin, it will be our signal wire.
The other contact wire is connected to the nodeMCU ground. That’s zero volts, no matter what happens.
When the contact is open, the resistor ‘pulls up’ the signal voltage to the nodeMCU supply voltage. No current flows. The nodeMCU can detect this ‘high’ voltage at its input pin.
When the contact is closed, the resistor is short-circuited to ground. A very small amount of current flows, but the signal voltage is pulled down to ground, zero. The nodeMCU can detect this ‘low’ voltage at its input pin.
Ivan, actually I have your same model of alarm system and putting in place the same solution of this topic (using ESP32).
I had connected gpio and the NO/C, activated the ESP32 (nodeMCU) pullup resistor but it do not work.
On the ESP32 side it is ok (I checked changing input from NO to NC and the chips detects correctly ON and OFF).
The problem in on the Logisty system: if I switch on or off the alarm system, if I send internal commands to the output relay 1 or 2, the status of the relay does not change, so on the nodeMCU the on or off do not change either…they are fixed…
