Relay coil power: where to get a lower voltage switch/relay power source

Thread Starter

groos

Joined Oct 6, 2013
2
I never see in diagrams of relay wiring where the smaller power that runs through a switch or sensor is obtained. I'd like to use the same power source for running the large load and the control circuit. Is is possible to tap off and reduce the voltage and current from a "high power" (120VAC to 12VDC, 1-2A ) transformer with resistor(s) to the point here it will not damage a reed switch or low voltage sensor?

Clever about a very few things, obtuse about most....

Help very much appreciated, TIA
 

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LesJones

Joined Jan 8, 2017
3,133
The text of your question is not very clear but from schematic it is clear that the load (the pump motor.) works on the same voltage as the relay coil. You will need a relay with a 12 volt AC coil. DO NOT put a diode across the relay coil. (A diode would only be used on a DC relay) Also you don't need the resistor. You do need to check that the current taken by the relay coil is less than the current rating of the reed switch. If you can't get a relay that meets this requirement you would need to use a DC relay driven by a transistor. you would also need a bridge rectifier and smoothing capacitor.

Les.
 

Alec_t

Joined Sep 17, 2013
12,003
Welcome to AAC.
One thing missing from your circuit is a bridge rectifier to convert the transformer output to DC. Note there is a difference between a transformer (shown in the schematic) and an AC-DC converter (as per the third line of your post).
If you choose a sensitive relay with a 12V rated coil its coil current is unlikely to exceed the current rating of your float-switch (but you would need to check the ratings of both relay and switch to be sure), so no resistor would then be necessary.

Edit: Les beat me to it :)
 

Thread Starter

groos

Joined Oct 6, 2013
2
Welcome to AAC.
One thing missing from your circuit is a bridge rectifier to convert the transformer output to DC. Note there is a difference between a transformer (shown in the schematic) and an AC-DC converter (as per the third line of your post).
If you choose a sensitive relay with a 12V rated coil its coil current is unlikely to exceed the current rating of your float-switch (but you would need to check the ratings of both relay and switch to be sure), so no resistor would then be necessary.

Edit: Les beat me to it :)
Thanks. The transformer will be an AC120v to DC 12v. Relay less of a problem than the switch, switch is rated at 400ma, which is about the same as the pump draw, so the relay is to protect the switch. Cannot use higher rated switch unfortunately
 

Dodgydave

Joined Jun 22, 2012
9,938
Okay, so if you're using a DC relay coil, put a Bridge Rectifier across the coil and feed it with the ac from the transformer, the resistor is only if you want to use a lower coil voltage, like 9 or 6v.
 
The standard way would be to buy an enclosure with a sub-panel in the bottom. That panel is usually 1/4" aluminum. The box is usually hinged and is flange mounted. In there, you can mount a DIN rail power supply such as this http://www.trcelectronics.com/din-rail-power-supplies.shtml one.

You can use DIN rail terminals and relay housings to basically build with an electrical erector set. You use cable glands/conduit as needed to make the connections.

It's wise to build the controls with definite places (terminals) to put the inputs and outputs and to keep the 12 VDC and 120 VAC separated. In the box wire should be rated for 120 VAC unless there is physical separation.

Wire duct, provides a way to route wires. The DIN rail, duct etc is secured with tapped hoes in the sub-panel.
Only a few tapped holes are usually required.

What you need depends on the environment as well.
 

LesJones

Joined Jan 8, 2017
3,133
IF the item in your schematic that is drawn as a transformer is not a transformer but a 12 volt DC power supply then all you need to do is to remove the resistor and check that the current rating of the relay coild is less than the 400 mA rating of your reed switch. Leave the diode connected across the relay coil.

Les.
 
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