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I have built a simple automatic switch that synchronizes IKEA LED strip with the main light in the room, eliminating the need to switch multiple switches in the evening.

Currently it works like this:

schematic

simulate this circuit – Schematic created using CircuitLab

It's a bit simplified but that's the gist of it. Now the relay is a bit noisy, it's a high power relay soldered within an Arduino module that I bought along with the driving circuit.

I have a smaller relay that I could use but if I do, I need a mosfet to switch that relay. And the mosfet I have is rated for up to 60V and 50A continuous current. That's way more than the LEDs use.

So my question is, is it safe to replace the relay and the relay driver with a 2SK2049 mosfet. And if it is, how to connect it properly?

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It should be fine to switch those LEDs with a MOSFET. As you say, the VDS and current rating for that part should be sufficient, and the VGS_on is <2.5V, so you should be fine to drive the gate directly with a 3.3V IO. The only thing you'll have to adjust is that you need to put the MOSFET on the low-side of the LEDs. So, the LED anode connects directly to 24V supply, the drain of the MOSFET should connect to the LED cathode, and the source of the FET connects to ground.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks for the answer. Why does the mosfet need to be on the anode? I am asking because I already have mangled the original wire so that only the VCC wire is interuptible. Is there a way connect the mosfet differently? \$\endgroup\$ – Tomáš Zato Apr 24 at 15:23
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    \$\begingroup\$ Mosfet on the Cathode. 2SK2049 is N-channel MOSFET, they like to 'sink' current to Gnd. P-channel MOSFET would be used to 'source' current from a supply line. Looking at the figure "Typical Drain-Source-On-State-Resistance vs. ID", the part appears to have an Rds of .025 ohm with Vgs of 3.5V and up to 20A of current flow. How much current does your LED string need? digchip.com/datasheets/parts/datasheet/169/2SK2049-pdf.php Is this part available anywhere? It has the appearance of being obsolete. The big distributors do not appear to carry it. \$\endgroup\$ – CrossRoads Apr 24 at 15:38
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    \$\begingroup\$ You need an N-channel FET on the low-side because it is the gate-to-source voltage (VGS) that determines if the FET is "ON". If you put it on the high side (i.e. source connected to anode) then the source voltage will rise to very nearly 24V when the LEDs are on. In order to keep the transistor on, your gate voltage will need to be above this (not very practical without a special driver). You can use a PFET for a high-side switch, BUT you will still need a gate driver to bring the voltage to nearly 24V to turn it off. \$\endgroup\$ – Jeff McBride Apr 24 at 15:56
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    \$\begingroup\$ @TomášZato google "high-side switching" and "low-side switching" - in short: what turns a mosfet on? The potential-difference between gate and source, i.e. the voltage, therefore V_GS. The voltage at the gate is 3.3 relative to ground. Now imagine, what will the potential at the source be? You will see, that when the mosfet is on the "low-side" the potential is essantially ground, so across gate and source are 3.3V. But what will the potential be on the "high-side"? Hint: What happens, when the Fet is on? Can it stay on? \$\endgroup\$ – Anonymous Anonymous Apr 24 at 16:09
  • \$\begingroup\$ Thank you all for explaining, I really appreciate it. The mosfet I have is harvested from scrap parts, so that's why it's probably obsolete. I guess I'll just either buy a new P-channel mosfet or use a relay along with the n-channel one that I have. \$\endgroup\$ – Tomáš Zato Apr 24 at 16:20

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