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I'm an electrical engineering student trying to build an 8x8x8 RGB LED cube as a summer project. Currently I'm designing the portion of the driver circuit that controls the anode layers of the cube. I have 8 individual MOSFET Transistors that I'm using as switches and I'm trying to select resistors. So far I've calculated that the transistor will need to be able to supply 3.84A to the LED (20mA per diode * 3 diodes per RGB LED * 64 LEDs per layer). I've selected a 60W power supply (5V, 12A) and am now trying to calculate the values needed for a drain resistor. If I follow the example in my textbook and the few I can find online I need to subtract the maximum voltage across the LED (3.4V for my LED's) from the supply voltage, and then divide this voltage by the maximum drain current ((5V - 3.4V) /3.84A = 0.417Ohm). Given the low value here can I ignore the need for a current limiting resistor on my drain pin? Or am I missing something a bit bigger? Any help is appreciated!

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    \$\begingroup\$ How do you want to control 512 RGB LEDs (that is 1536 LEDs) by 8 MOSFETs? Could you add some scheme here? \$\endgroup\$ Commented Aug 10, 2016 at 19:06
  • \$\begingroup\$ LEDs roughly follow a diode IV curve. As an EE student you should know that a diode curve is quite steep. So a low value resistor will not set the current accurately, it will depend heavily on the LEDs. For many LEDs in series you would need to drop a couple of volts to get a stable (over temperature etc) current. You should NEVER leave out the current limiting resistor ! \$\endgroup\$ Commented Aug 10, 2016 at 19:17

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You can't do it with 1 common resistor for a whole layer. You would very rarely have all LEDs illuminated in a single layer, so you would be overdriving the LEDs that were illuminated. You would need to put a resistor per LED on the cathode side.

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  • \$\begingroup\$ that makes a lot more sense to me, thanks a bunch! \$\endgroup\$ Commented Aug 10, 2016 at 22:09
  • \$\begingroup\$ No doubt you have realised, but you will not need a resistor for every LED in the cube, you will just need one for every column, since only one layer will be turned on at a time. \$\endgroup\$
    – HandyHowie
    Commented Aug 11, 2016 at 7:15

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