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This is a quick and simple question but I was wondering if this was the right way to wire a NodeMCU to an RGB LED strip. I am using IRLZ44N MOSFET's and 470ohm 1/2 watt resistors.

I currently have it wired like this with the resistors connected to ground: enter image description here I have also seen it wired like this with the resistors connected directly to the controller: enter image description here

I assume they both do the same thing but I might be wrong. Which way is better?

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  • \$\begingroup\$ What is the current flow for the RGB channels? You should also learn to use the schematic tool, breadboards are not circuit diagrams. \$\endgroup\$ – Jack Creasey Jul 22 '18 at 2:42
  • \$\begingroup\$ @JackCreasey, given that it's working the LED channels must expect a switch to ground \$\endgroup\$ – Jasen Jul 22 '18 at 3:10
  • \$\begingroup\$ @Jasen Yup, I can see it is a low switch, what is your point? Unfortunately the FET selected is likely not to be able to deliver more than about 3A while remaining saturated. If it does not achieve saturation the dissipation will make it hot ….how hot will depend on the VD to VS, but the op has not qualified the requirements. \$\endgroup\$ – Jack Creasey Jul 22 '18 at 4:32
  • \$\begingroup\$ sorry, mis-read your comment, that breadboard will melt with 2A \$\endgroup\$ – Jasen Jul 22 '18 at 4:55
  • \$\begingroup\$ I’m not great at this stuff. How do I find the current flow of the RGB channels? Please clarify what I should do. I am using a 3m 12v 5050 analog RGB strip with the same power supply powering the NodeMCU(voltage converter). \$\endgroup\$ – Username Jul 22 '18 at 6:18
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The two do different things.

The first one has pull-down resistors. This is used to make sure the mosfet gate is pulled down to ground while the Node is disconnected or powered off. It's to prevent floating gates. 470 ohm is fairly low/strong for a pull-down. 4.7k is more typical. If you see the leds are slow to change you may want to reduce them.

The second is series, current limiting resistors. This is to protect the Node from the mosfet if it passes current through the gate somehow. It's a bit unnecessary in this use, but doesn't hurt. The scenario where this is useful is unlikely to happen, and a replacement node is cheap. As Mosfets are voltage driven, it doesn't affect anything afaik.

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  • \$\begingroup\$ So I should replace my 470ohm resistors in the first diagram for 4.7k and everything should be fine? \$\endgroup\$ – Username Jul 22 '18 at 2:23
  • \$\begingroup\$ In my opinion, yes. Others may suggest different values. \$\endgroup\$ – Passerby Jul 22 '18 at 2:26

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