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I'm trying to use a unipolar hall effect sensor as a power switch to turn off and on a 12V strip of LEDs, powered by a 12V 6.6Ah battery. I just tested the current draw of the LED strip with a multimeter and got a value of 442mA (much less than the 5A the seller on ebay claimed... maybe they weren't taking into account the resistors that were already integrated in the strip?).

I've seen some Hall effect sensors that can handle up to 500mA, but some of them are a bit expensive and I already purchased some of these. They're rated for 4.5-24V, with a continuous Output current of 25mA. It's supply current is 9mA (not sure if that's important).

If I use a resistor to limit the current that the Hall effect sensor gets, that would also make the LEDs appear less bright too, right... or am I wrong? Using ohm's law, it looks like a 480 ohm resistor is what the hall effect sensor needs (12V/0.025A = 480 ohm resistor).

Would it be possible to use an N-Channel MOSFET between the 12V 6.6Ah battery, hall effect sensor and LED strip, or would I need a microcontroller (arduino/attiny) and a 5V regulator to do that?

I'd rather avoid using a microcontroller, and keep this as cheap as possible unless there's no other way to avoid majorly affecting the brightness of the LEDs.

Any advice would be appreciated, Thanks!

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2 Answers 2

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I just got a reed switch and it simplified things. just connect one end to the LED strips positive lead and the other end to the batteries positive terminal.

Here's an example of one

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Hitting a reed switch that's only rated for 500mA with 442mA frequently will pit the contacts on your switch, eventually causing it to fail. There's a number of cheap MOSFETs that will switch the low side. I picked up a handful of IRLZ44 parts for 29 cents each at a local surplus store ($2 each from DigiKey). Even cheaper is the AO Semicon AO3400 or AO3400A at 2 to 4 cents each from lcsc, although the SOT-23 package isn't good for homebrew circuits. Other possible MOSFETs are IRF540, FQP30N06L, IRLML2502, etc.

In the circuit below, connect your reed switch between 12V and the PWM_3.3V pin and it'll buffer the current from your reed switch, allowing you to drive up to > 20A worth of 12V LEDs. The circuit was intended for an ESP8266 running at 3.3V, but it'll work with 12V control levels and you'll get lower Rds(on) from the MOSFET (less dissipation). If you want to run 24V strips you'll need to replace the 10K resistor with a 8 to 10V zener and add a 1.5K resistor in series with your switch, as most of these MOSFETs don't like more than ~12V on the gate.

LED driver

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  • \$\begingroup\$ Welcome to EE.SE. Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$
    – Transistor
    Mar 22, 2020 at 10:58

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