How to control closet LED's

@ronalex4203

You could probably just use a MOSFET with whatever switch you use to turn the LEDs on/off.

@will Ok, how do I work in the magnetic reed switch?

@ronalex4203 

I think I'm missing something, the MOSFET was intended to replace the reed switch.

When you (or your other half) decides that there must be light in the pantry, how will you tell the 12V source to power the LEDs ? If you're using a reed switch, does that mean that you're wanting to automagically turn the LEDs on when the door opens (i.e. magnet on door triggers reed switch ?)

If that's the case, then connect the reed switch to the gate of the MOSFET so that it's triggered when the reed switch conducts when the door opens.

@ronalex4203 

I'm curious, would combining 3 reed switches in parallel drop the current flowing in each one (1/3 A) to less than the 0.5A limit work ? If you had a sufficiently powerful magnet to trigger all three simultaneously, that would probably be a better solution (since it requires almost 0 power and no extra compomnents).

@will Yes, open door, lights turn on.

How in bloody blue blazes do I find a mosfet with the correct specs if I even knew what they were. I know how to do that for relays, even the newfangled solid state relays, but I have never wired up a mosfet and until very recently have not seen or touched one. From what you and others have told me, it sounds like a great solution, but I simply have no frame of reference to get started. Also I think I read that the trigger(gate) HAS to be taken to ground to turn off the mosfet. Also, it sounds like the mosfet only wants to see 5V as a trigger. I just got some tiny 12V to 5V buck boost modules so that can be the trigger, but how do I get it to ground since I think I read that if I simply remove the 5V the mosfet keeps conducting.

@will I don't think so for at least 3 reasons. Mounting room, a magnet that big would rip off the switches, and arcing could still happen since there is no way to guarantee that all 3 switch at the same time.

Posted by: @ronalex4203

@will Yes, open door, lights turn on.

OK, then you'll use the reed switch to operate the MOSFET which will power the LEDs

How in bloody blue blazes do I find a mosfet with the correct specs if I even knew what they were.

There are a few key numbers that you need to find out, but they're mostly upper limits and almost any MOSFET you can name will exceed your requirements.

You need to look at the maximum voltage and the maximum current that will be passed through the MOSFET. Those will depend on your application.

You'll also need to know the minimum voltage required to drive the MOSFET through the gate. This is called the "Gate Threshold Voltage" or known by the acronym VGS(TH). Those called "logic" level require a voltage of (less than) 5 volts to fully turn on. You have 12 Volts available, so you're not constrained to using a logic level one, but you'll need to choose one that needs less than 12 volts.

The last is "Drain to Source On Resistance" or rDS(ON) which is the effective resistance through the MOSFET while operating with the gate fully turned on.

For example, the BUZ11 is an N-Channel MOSFET (which means that it operates when the gate is pulled ABOVE ground). These are for illustration only, I am not recommending them for this application ...

This MOSFET can handle throughput of max 50V and max current 30A. VGS(TH) is typically 3V and max 4V. rDS(ON) = .04 ohms; which means that the resistance through the MOSFET is effectively 0.04 ohms when the MOSFET is fully turned on (i.e. gate voltage is 4V)

So, in your case, 50V >> 12V required; 30A >> 1A required; you have 12V available to drive the gate (but you'll need to drop the voltage over a resistor or something) and you can expect the MOSFET to burn up  power = current*current*resistance = 1A*1A*.04 = .04 watts = 40 mW (approximately). This means you wouldn't even need a heat sink for use in this case

From what you and others have told me, it sounds like a great solution, but I simply have no frame of reference to get started. Also I think I read that the trigger(gate) HAS to be taken to ground to turn off the mosfet. Also, it sounds like the mosfet only wants to see 5V as a trigger. I just got some tiny 12V to 5V buck boost modules so that can be the trigger, but how do I get it to ground since I think I read that if I simply remove the 5V the mosfet keeps conducting.

Bill has dove a video on them that will help get you used to the idea. You'll find them extremely handy and versatile. I'd suggest you buy some logic and 12V samples from Amazon and have a play with them.

Here's a link to Bill's introductory article about them ...

https://dronebotworkshop.com/transistors-mosfets/

@will Thanks for the info Will. I will review Bill's video, but I really have a hard time absorbing new stuff like this where I have NO frame of reference. My electronics training was all tube based.

I have 12V, I have a tiny 12V to 5V Buck Boost chip, capable of maybe 600ma but maybe less as the same 600ma is spec'd for the 3.3 version as well that I have and maybe should use if the mosfet will turn on at 3.3v. What causes the mosfet to turn off?

The 12ft of leds only uses about 11W so at 12V it's about 1A. Almost any mosfet will do I understand, but I have yet to encounter a device so miraculous that any version of it is good in so many use cases. My old brain simply refuses to buy into that argument. I think I have a few of them and will breadboard what I have if I can rig up a 12V supply. I do have a 90W 19V @4.74A laptop brick, so that should stand in for the LED strip supply but I don't have a load that I can think of.

I do not know what to do with the dangling -5V or 3.3V lead coming off the converter.

@ronalex4203 

You should put a 10K resistor between the gate and source. That will force the MOSFET to shut down cleanly.

I think you should connect the reed switch between the 12V + and the input + of the buck regulator. You don't want that sucker to be drawing power until the LEDs are needed.

I expect that the regulator - can just tie back onto the MOSFET source.

You have +12V and -12V drawn on your diagram, but you're not really using 24V are you ?

I think I'd try using a 10K pot between the +12 and gate to see if you can just use it to find a suitable resistor to drop the voltage between 12V+ and the gate without needing the regulator. Worst case is just use a 5V regulator chip and a couple of caps. Those buck/boost converters are expensive and seem like overkill here 🙂

Think of MOSFETs as an efficient electronic version of a coil relay to start with. You'll find a million uses for them.

@will Excellent idea re the reed switch.

Posted by: @will

I expect that the regulator - can just tie back onto the MOSFET source.

That makes no sense to me, did you mis-speak?

Not 24V. What would you label it as, I know what I mean.

So are you saying all I need is to determine what resistor between Gate and +12V will allow it to turn on? I like the idea of not needing the converter.

Expensive? $13.59 for 5.

Posted by: @ronalex4203

@will Excellent idea re the reed switch.

Posted by: @will

I expect that the regulator - can just tie back onto the MOSFET source.

That makes no sense to me, did you mis-speak?

No, just that it's the same ground as the MOSFET's source pin.

Not 24V. What would you label it as, I know what I mean.

I'd say 12V and 0V or GND. A lot of sound circuits take + and - voltages as well as GND and some of their comments can be quite ... err ... "tart" if you use the wrong  (e.g. if you say minus when you meant ground 🙂

So are you saying all I need is to determine what resistor between Gate and +12V will allow it to turn on? I like the idea of not needing the converter.

Me too, but I'm not sure how (or even if) it would work. A resistor on the gate isn't always shown, but the gate acts as a capacitor and there can be a large inrush of current when it's fired. Many designers add the resistor just to limit the inrush current when the MOSFET is turned on. This isn't desirable if you're trying extremely fast switching with the MOSFET (as in PWM) since it will interfere with a pure square wave pattern.

Expensive? $13.59 for 5.

L7805 regulators 10.99 for 10 🙂

@will I am watching Bill's video and more importantly looking at the pdf. I just noticed that this thing is wired into the -ve side of the circuit, is that why it's called an N? Now it makes more sense. If I can find a suitable load I will breadboard it ans see what happens. I will not be using an arduino, I don't think it's required, just a voltage on the sig pin (using the module)

Best price I found on Amazon for the L7805 is $1.50 ea

Posted by: @ronalex4203

@will I am watching Bill's video and more importantly looking at the pdf. I just noticed that this thing is wired into the -ve side of the circuit, is that why it's called an N? Now it makes more sense.

Yes, N-channel means that it operates when the gate is positive for it to operate. P-channel MOSFETs operate when the gate is negative.

If I can find a suitable load I will breadboard it ans see what happens. I will not be using an arduino, I don't think it's required, just a voltage on the sig pin (using the module)

Correct, you don't need an Arduino.

Best price I found on Amazon for the L7805 is $1.50 ea

see (with $1.00 off coupon)

"https://www.amazon.ca/Voltage-Regulator-Wholesale-10pcs-LM7805/dp/B07YPZLF2W/ref=sr_1_8?crid=2C03EQMB9GEAD&keywords=5v+regulator+l7805&qid=1643846692&sprefix=5v+regulator+l7805%2Caps%2C115&sr=8-8"

@will I am slowly getting this. Looking at Bills circuit, he is controlling 12V like I will be with the IRF520 module (but I may build my own). He is using an Arduino to trigger however, that is 5V. I only have a 12V supply. I know if it is more than 5V it will turn on the mosfet, but will it hurt anything. If I knew the amperage on the G pin, I could add a resistor to drop 7 volts (I would experiment but only have a 10k pot). I have looked at a few spec sheets for the IRF520N and do not see what I need. It may well be there, in fact I am fairly sure it's there but I am not recognizing the nomenclature. I am looking for IG probably in ma which means a dropping R of 1400 ohms. I am excited at the prospect of building this and starting to understand this to me newfangled cool device called the mosfet. I am going out today to buy a 6V lantern to use as a dummy load. Any value I determine for 6V is readily converted to 12V use. My 12V power supply would be handy but not ready yet. Am I on the right track?