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This is another LED pixel project that I'm looking to power in a slighty different way. It's 110no. 5050 SMD pixels on a 300mm x 300mm board.

The long and short of it is that I'll use USB-C Power Delivery to supply 3A at 20V. The logic side of the circuit is all 3.3V, while the LED pixels are 5V. In total I have an absolute maximum of 6.6A worth of pixels.

To keep the PCB low profile, and to allow for better/easier heat disspiation, my idea was to use 3 or 4 integrated buck modules (e.g. TPSM84209) to power groups of pixels.

Is my understanding correct in that:

  • I can tie all of the grounds together (of the µC, USB-C and buck modules)? In that way, they all share the same reference and it means logic to the pixels will work?

  • The particular module I've linked (TPSM84209) will output 2A at 5V? My reading of the graphs (for Vin = 24V) suggests that at 2A it's about 80% efficient. Does that mean that if I draw 10W on a buck module, the module will draw 12.5W from the 20V supply, or 1.6A?

What additional drawbacks might there be with using multiple buck converters (aside from increase in cost and complexity)?

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  • \$\begingroup\$ You may need a way to synchronize the frequency output, or do something to isolate the high frequency parts of the circuits. \$\endgroup\$
    – Ron Beyer
    Dec 27, 2020 at 1:51
  • \$\begingroup\$ a multi-phase buck converter might be better - that's the way it's done in PCs \$\endgroup\$ Dec 27, 2020 at 2:23

1 Answer 1

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I can tie all of the grounds together (of the µC, USB-C and buck modules)? In that way, they all share the same reference and it means logic to the pixels will work

Yes

Does that mean that if I draw 10W on a buck module, the module will draw 12.5W from the 20V supply, or 1.6A?

Yes, for the 12.5W. But

12.5W/20V = 0.625A

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