Is a Shunt Resistor Necessary for a 12VDC, 40W Electromagnet Setup?

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SUMMARY

The discussion centers on the necessity of a shunt resistor in a 12VDC, 40W electromagnet setup powered by a 5A supply. Participants recommend using a flyback diode instead of a shunt resistor to protect against arcing when the relay opens. The diode should be rated to handle the full coil current, approximately 3.3A, and connected in reverse bias across the electromagnet coil. This configuration prevents high voltage spikes that could damage the power supply or relay.

PREREQUISITES
  • Understanding of electromagnet specifications (12V, 40W)
  • Knowledge of relay operation and SPDT relay configuration
  • Familiarity with flyback diodes and their function in inductive loads
  • Basic electrical principles, including Ohm's Law and current calculations
NEXT STEPS
  • Research the specifications for suitable flyback diodes on Digikey or similar platforms
  • Learn about the impact of inductive kickback and how to mitigate it
  • Explore the characteristics and applications of SPDT relays in circuit design
  • Study the principles of current limiting and protection in electrical circuits
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Electronics hobbyists, electrical engineers, and anyone involved in designing or troubleshooting inductive load circuits, particularly those using electromagnets and relays.

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I have a 12vdc, 5Amp power supply and need to connect to a 12vdc, 40 watt electromagnet. I put a SPDT relay switch between the power supply and the electromagnet to turn it on and off. Do I need to put in a shunt resistor or something else to protect the magnet and the power supply? How do I figure out what value my resistor should be and where do I put it? Should I put it across the electromagnet or across the output of the power supply before the relay switch? Thanks
 
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kfpanda said:
I have a 12vdc, 5Amp power supply and need to connect to a 12vdc, 40 watt electromagnet. I put a SPDT relay switch between the power supply and the electromagnet to turn it on and off. Do I need to put in a shunt resistor or something else to protect the magnet and the power supply? How do I figure out what value my resistor should be and where do I put it? Should I put it across the electromagnet or across the output of the power supply before the relay switch? Thanks

Why are you thinking that you need a shunt resistor?

And maybe instead of the resistor, you should put a big catch diode across the electromagnet coil? If you don't, when you open the relay, you could get some arcing across the relay, and possibly hurt your power supply...
 
So, a diode is better to protect arcing? what value do I need? I am a beginner and trying to learn to do this right. Thanks.
 
kfpanda said:
So, a diode is better to protect arcing? what value do I need? I am a beginner and trying to learn to do this right. Thanks.

The diode will go across the coil in the opposite direction of the normal magnetizing current (point the cathode toward the +12V side). The diode will need to be able to support the full coil current briefly as the relay is opened. When the relay opens, the coil voltage will "fly back" in the reverse direction, and the diode will catch it at about -0.7V, keeping it from flying back to a high enough voltage to cause an arc. The coil current will be supported by the diode as it decays to zero. The time it takes to decay is approximately L/R, where R is the resistance of the coil.

So given the 40W power and 12V input voltage, what is the current of your coil when it is on? Pick a diode that can support that current as a peak current, and you should be okay. You can use the Digikey website, for example, to help you find candidate power diodes.
 
The spec of the electromagnet is 12v, 40watt. I am assuming 3.3 amp. it will use. Will I need a zener diode? Thanks.
 
kfpanda said:
The spec of the electromagnet is 12v, 40watt. I am assuming 3.3 amp. it will use. Will I need a zener diode? Thanks.

No, not a Zener diode, just a vanilla diode.

http://en.wikipedia.org/wiki/Flyback_diode

.
 
Did you pick up on the fact that you need to connect the diode so that it is reverse biased (i.e.'backwards') so it doesn't conduct when operating normally? It will, however, conduct when the relay breaks the circuit and a back emf is generated by the falling current in the coil. Any 'rectifier' / power diode will do (as opposed to a small signal 'detector' diode).
 
I think you might be worried that the power supply can deliver 5 amps but the relay only uses 3.3 amps.

This is not a problem, as the relay will only take what it needs from the power supply and the extra current doesn't have to go anywhere else.
 

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