Choosing Relay for 5V DC & 12V Motor

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When selecting a relay for a 5V DC circuit and a 12V motor, ensure the coil voltage matches the circuit voltage (5V) and the switching voltage corresponds to the motor voltage (12V). The minimum current required to activate the relay can be calculated using the formula Icoil = Vcoil / Rcoil, and it's important to consider any additional resistance that may be included in the relay design. To protect the normally open (NO) switch, an RC series connection is recommended to prevent damage from the motor's current. When connecting an LED to indicate relay status, it should be wired in parallel to maintain the full 5V supply, and a reverse diode must be placed across the relay coil to absorb voltage spikes during deactivation. Properly implementing these components ensures the reliability and longevity of the circuit.
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I have a 5v dc circuit and a 12v dc motor. If I had to use a relay, when looking at the specifications, what do I look for ?, will the coil voltage be 5v and the switching voltage be 12v ?

Will these relays work ?
http://www2.mouser.com/Electromechanical/Relays/_/N-5g31?P=1z0x3tdZ1yyfbg3

Also, how do I know the minimum current required to activate the switch ?
 
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TheRedDevil18 said:
will the coil voltage be 5v and the switching voltage be 12v ?
Yes.
TheRedDevil18 said:
how do I know the minimum current required to activate the switch ?
By activation the current will be
Icoil = Vcoil / Rcoil
However you may insert a save resistance in the circuit, decreasing the coil current when the relay has been activated. Often this save resistance has already been included in a relay with a dc-coil.

Also you should protect the NO-switch ( the righthand in the attached ) by means of a RC series connection. Otherwise you may "burn" the switch, breaking the dc-motor current several times. The values for the RC depends on the motor current and how fast the NO-switch will open.
 

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Ok, I want to connect an LED to show when the relay is on, but because the LED drops 1.7V I would have to connect the relay in parallel so as to receive the full 5V supply voltage

Here's a schematic
IMG_20160322_153744.jpg


Would this connection be fine ?
 
Also, usually when you have a relay coil you insert a reversed diode across the coil to absorb the voltage spike when the coil shuts off.
 
wirenut said:
Also, usually when you have a relay coil you insert a reversed diode across the coil to absorb the voltage spike when the coil shuts off.
Not only could you damage the transistor when left out, in this case you could damage the LED also.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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