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Can I use a relay module to switch on & off a door alarm (1.5 V × 3)? Basically, can a relay switch work in this low DC power?

Also, I'm trying to make an alarm system which emits high noise when a particular set of sensors are triggered, so what I have done is that I have salvaged an old door alarm which will emit the sound and in order to do so I have to switch it on or off as per the sensor.

So, I am thinking of using relays; or if you can suggest any better option you are most welcome.

EDIT:

  1. I'm using Arduino for the sensors & programming related to triggering the alarm & the alarm that I have salvaged is this one.

  2. The Arduino and the Alarm are independently powered.

  3. Since the alarm is independently powered by 1.5v * 3 batteries, can I use Relay just to switch on & off it via the Arduino? As all the videos & articles about relay are dealing with high voltage mains power, switching bulbs, fans etc can Relay handle low power/volt?

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    \$\begingroup\$ Sure. A relay is just a mechanical switch. But instead of something you move with your finger it has something moved by an electromagnet. \$\endgroup\$ – Majenko Jan 25 '19 at 14:03
  • \$\begingroup\$ I don't see what this has to do with Arduino though - it's a generic Electronics question. \$\endgroup\$ – Majenko Jan 25 '19 at 14:04
  • \$\begingroup\$ Generally yes. At extremely low voltages you may have trouble relating to what's called the "wetting voltage", but I think 4.5V should be enough for most relays. \$\endgroup\$ – Hearth Jan 25 '19 at 15:26
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Some good info here, but let's keep this simple, yeah!? First off, yes, the relays can switch small DC voltages and small currents. You can purchase any number of "Arduino Relay Modules" that already have circuitry built into the module so you don't have to worry about flyback diodes and whatnot. Here are some links:

All of these relay modules will work fantastic with just a 5V output from any pin on an Arduino. They all can switch any voltage from 0V up to ~120/240VAC, so you can switch the 4.5VDC you wish with ease. Another method to switch your alarm on without using a clunky relay is to use a NPN transistor as a switch. I won't explain how as there are many tutorials on it, but here's a great one if you're interested in that route:

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A relay would be fine, pick one that has low coil current if driving the coil from an Arduino, like this one.

Connect the coil+ to 5V, and the coil- to an Arduino pin, output a low to energize the coil. Put a diode like 1N4148 across the coil connections - anode to the Arduino pin, cathode to +5. That will protect the Arduino pin when the pin goes from low to high and the coil creates some current as it turns off, which will then dissipate harmlessly back into the 5V line.

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  • \$\begingroup\$ You also need to pick one where the contacts are rated for dc current at whatever low current value you need. Some relay contacts will "stick" when used with dc current, these are only suitable for ac. \$\endgroup\$ – Elliot Alderson Jan 25 '19 at 15:03
  • \$\begingroup\$ A signal relay should be good to support that. \$\endgroup\$ – CrossRoads Jan 25 '19 at 15:26
  • \$\begingroup\$ Also, be aware that low voltage relays typically need more current than a higher-voltage relay, and if your control circuits are being powered by 3 batteries, they will not last as long as you might like. \$\endgroup\$ – WhatRoughBeast Jan 25 '19 at 22:20
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Yes, pretty much any relay will work. It's better if it has not already been used to switch relatively high power.

Your question mentions a relay module, which is different from a bare relay. You should refer to the instructions that come with the relay module if you use a relay mounted on a PCB with optoisolator and/or drive transistor, LED etc.

To switch a bare relay directly with an Arduino you can use a circuit such as this (assumes +V is 5V):

schematic

simulate this circuit – Schematic created using CircuitLab

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  1. Working of the relay at the mentioned voltage depends on the type of relay module you are using. Just try & see if the relay makes a "click" sound when connected to your power supply.
  2. If your alarm is powered by mains voltage or is at a different voltage potential then you are bound to use a relay (a triac is also suitable) in order to turn it on/off.
  3. Do not use a relay if you want to turn the alarm on/off rapidly. Since relay is a mechanical switch and may wear out easily when used in this type of scenario.

After the edit in your question:

A relay is just a mechanical switch. It does not matter if you are using it with mains voltage or in your case with (1.5*3=4.5v).

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  • \$\begingroup\$ No, you don't need a relay to control mains voltage. A triac can do this nicely. \$\endgroup\$ – Elliot Alderson Jan 25 '19 at 15:01
  • \$\begingroup\$ I am not forcing the use of relay. \$\endgroup\$ – Electric_90 Jan 25 '19 at 15:09
  • \$\begingroup\$ You said "if your alarm is powered by mains voltage...then you are bound to use a relay". My comment was that no, you are not bound to use a relay. \$\endgroup\$ – Elliot Alderson Jan 25 '19 at 15:12
  • \$\begingroup\$ By saying so I didn't mean to say that relay is the only option here \$\endgroup\$ – Electric_90 Jan 25 '19 at 15:14
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    \$\begingroup\$ Is English not your first language? "you are bound to" is an expression that indicates inevitability, like there's no other option. \$\endgroup\$ – Hearth Jan 25 '19 at 15:28
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A simple transistor like BC637 can work for your application. enter image description here

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    \$\begingroup\$ Welcome to EE.SE but you need to expand your answer quite a lot before it would be of any use to anyone. There's a CircuitLab button on the editor toolbar that will allow you to add a schematic. \$\endgroup\$ – Transistor Jan 25 '19 at 16:33
  • \$\begingroup\$ You cannot use a transistor as drawn without some series base resistor to limit the current. Also if you place the collector and emitter right across the 1.5V supply the transistor will short out the supply when it turns on most likely causing it to let out its magic smoke. \$\endgroup\$ – Michael Karas Jan 25 '19 at 17:53
  • \$\begingroup\$ My mistake...., a load(i.e., alarm is missed between). \$\endgroup\$ – Praveen B Jan 25 '19 at 17:59

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