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Made a simple circuit with a reed switch, wall power source and LED strip (for a bathroom occupied sign). Switch-melting Circuit Diagram

Worked as I’d hoped at first. But, after a while, THE REED SWITCH MELTS CLOSED--leaving the light stuck on.

Im guessing I need a resistor? Can you help me figure out what type and where it should go? I tried reading about this but I was quickly overwhelmed.

Or is the problem is more fundamental? (e.g., I read that I may want to use a relay to separate the LED/power source from the switch. Is that overkill?)

Thanks for holding my hand. I know it’s a very, very simple question--but ya'll are saving my life. :)

Bryant Smith
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    Your problem is right there in your diagram: "REED SWITCH: Max contact ratings: 30VDC, **0.2A**, 5W" and "LED STRIP: ... Working current/meter: **1.2A** ...". What you want is almost certainly a relay (and not a resistor, unless you intentionally want to make your LEDs very dim). – brhans Jan 30 '18 at 19:37
  • a 72W Occupied LED lite wow. – Tony Stewart EE75 Jan 30 '18 at 19:43
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    Am I reading it right, that your LEDs use 1.2 A/m, you have 1.2m, so you really need 1.44 A for the circuit? You definitely need a bigger relay. – pscheidler Jan 30 '18 at 19:55
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    And how well does getting 72W out of a 60W power supply work? –  Jan 30 '18 at 20:02
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    ... I use less wattage than that to light up the entirety of my deck in the evening. Who all is supposed to see this 72 watt "occupied" light?! Aircraft? – Bryan Boettcher Jan 30 '18 at 20:07
  • Please see this [answer](https://electronics.stackexchange.com/a/336479/38335) to a related question. In your case, put the LED strip in place of the inductor in that answer. I think it should be illuminating! – bitsmack Jan 30 '18 at 20:35
  • Oh, and you won't need to include the flyback diode... – bitsmack Jan 30 '18 at 20:37
  • @BryanBoettcher, doesn't it say 1.2 m of a strip that's 72 W per 5 m? That would make it about 18 W. – TonyM Jan 30 '18 at 20:45
  • I'd give it 20 minutes if I were you! –  Jan 30 '18 at 21:11
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    So where is the OP? Not a word from the OP since the initial question. Oh, well. – jonk Jan 30 '18 at 21:19
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    OP here. Hey guys! Thank you SO much for all the answers and help! And Im sorry for the confusion caused by my lack of understanding lol. I appreciate your patience. And sorry for the delayed response. I had no IDEA this community would be so quick to answer. I was expecting to wait days! Admittedly I'm using components left over from another office project (power supply and LED strip) so if it seems like I made some weird choices its because I was scavenging. I will try both the answer below and the answer linked in comments and reply with what worked. Thanks again all! – Bryant Smith Jan 30 '18 at 21:39

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First, your question shows that your power supply is rated for 60W. It kind of looks like your LED strip is rated at "72W per 5 meters"...Are you using 5 meters?? If so, I think we can all agree that 72 > 60. That will burn-out your power supply. That also means you were actually using 6A, not 1.2A.

For the rest of my answer, I will assume that you're only using 1.2 meters, which explains why your image shows "1.2A" (because 72W/12V/5 1meterStrips = 1.2A). Note that 1.2 meters would actually mean 1.2A*1.2meters = 1.44A.

Anyway, that reed switch is only rated for 0.2A, but you have 1.44A going through it!

To use that reed switch in your setup, you'll need to use it as a low-current sensor of sorts to activate a relay which will control the high current.

The circuit below shows an example:

reed switch driving relay

A typical cheap relay can handle say, 10 A. That leaves leaves plenty of margin for the LEDs (1.44A).

To activate the relay, the magnetic coil must be activated. Let's assume you have a 12V relay with a 200Ω coil...
12V/200Ω = 60mA of current in the coil. That also leaves plenty of margin for the reed switch (max of 200mA).

Note that relays come in many different flavors, and a relay with a resistance of 12V/200mA = 60Ω or less would overload that reed switch, and you're back to where you started. So just be sure you use one that works for your circuit.

Observe the flyback diode in the diagram. That is there to protect the circuit from energy spikes that occur when a relay is turned off quickly. It's not a simple phenomenon to study for beginners, but just realize that you should have the diode in the circuit because the relay's coil is an inductor.

Bort
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    OP states he is using 1.2m – Tyler Jan 30 '18 at 20:49
  • @Tyler - OP doesn't show a clear understanding of the numbers, and thus I would not be surprised if the numbers from that image were incorrectly copied from somewhere. – Bort Jan 30 '18 at 20:54
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    True. OP does not have a clear understanding of the numbers. OP writes software and is bewildered by real things in the real world. THANK YOU SO MUCH FOR THIS ANSWER. I'm gonna get to work building it now and then accept it. You are the BEST. – Bryant Smith Jan 30 '18 at 21:35
  • Your reed switch melted because you drove 1.2A (or more) while it was rated at 0.2A. Relays have in their datasheet the power in watts necessary to activate the coil. At 12V and 0.2A, you have 12/5=2.4W available. Enough to power most relays, but still be careful. Or use the formula written by Bort. – Fredled Jan 30 '18 at 23:48