SparkFun Forums

[Loobieburt]

Hi all

I have a radio activated switch that is powering a siren.

The switch is powered by a 9v pp3 battery.

I wish to use a relay or similar between the switch and siren.

The switch works between 9v and 4.5 v dc the siren also works the same. But how do I select a relay that will work between those voltages? I am looking at one on farnell which works 6.75-9v but below that down to 4.5 it won't switch the relay? The siren draws 250ma and will only be on when the remote switch activates it for no longer than 15 seconds.

The reason is that the siren draws I believe too much current meaning once you have activated the switch you can not dis activate it so putting two batteries in each will then have it's own supply.

Any advice would be much appreciated

[Loobieburt]

However, just looking at RS website http://docs-europe.electrocomponents.co ... dbfca2.pdf this show a 5v relay
Must operate voltage 80% max. of rated voltage
Must release voltage 10% min. of rated voltage
Max. voltage 200% of rated voltage at 23°C

Does this mean 80% of 5 v would be 4v minimum and max voltage working would be 15v?

Any help is much appreciated.

[waltr]

Yes, that is how I read the specs.
In between Must operate and Must release Voltages in the "Hold" Voltage.

A circuit I use for relay coils is a resistor and cap in parallel which is in series with the relay coil power.
What this does is apply the full Voltage when turned on (the cap acts like a short so the Full voltage is applied to the coil). The cap will then charge and once it does the resistor now limits the current (drops the Voltage) to the coil.

The 5V relay in your link has a minimum Voltage to "Hold" is 10% of 5V or 0.5V. So a a 150 Ohm series resistor at 4.5V will put 2.37V across the coil which is way above the minimum "holding" Voltage.
80% of 5V is 4.0V so the relay will NOT "pull in" with just this series resistor. To ensure the does "pull in" at your minimum Voltage add a 1uF (or larger) cap across the series resistor. Now the Full 4.5V will be applied to the relay coil upon turn on.

You may need to play with the R and C values to get this working reliably but this will work.

[Loobieburt]

Thank you that makes a sense. In the past I have put a 1n4001 across the coil of the relay to stop the coil becoming latched is the resistor and capacitor the same effect?

Many thanks again

[waltr]

No, the diode across the relay coil is still required but for a completely different reason. The diode is to 'clamp' the high reverse Voltage produced from the coil when the magnetic field in the coil collapses.
On the 5V relay coil it is not unusual to see a negative hundred Volts or more when the power to the coil is removed. So always put a diode across a relay coil.
Do you have or can borrow an O'scope?
Try to do so and look at the coil Voltage when it is switched off.

[Loobieburt]

Thank you. I will order and have a play. Next question is I am currently using the output to drive a sounder without a relay using one pp3 9v battery. It seems there is not enough current to drive the circuit and the battery efficiently. My main concern is when pressing the transmitter it turns the output on but the output is drawing 200ma and is a loud warbling sound. But Ivan not turn the rf off again with the transmitter. If I use my power supply it works perfectly well. I will try the relay tomorrow and take the power from the same battery but any thoughts? Would a capacitor across the output work? I have measures the output which when alarming measures the same as the input but I think the sounder is drawing all the current from the battery?