Basic Electronics Question

  • Awaiting the arrival of my Batch 4 Duet WiFi - I've got the 0.8.5 in a CoreXY and an planning on using the WiFi for a Ultimaker clone - I'm super excited!

    I've got a basic electronics question. I was kicking around the idea of using a better PSU than the 12V LED supplies available on Amazon. I have a few 870W server supplies laying around, one that's got 70A on the 12V rail. It runs near silently and the form factor will work well with the new printer.

    I'm going to use a LattePanda in the printer (for those unfamiliar it's an Atom SoC with an ATMega Arduino compatible chip on board and GPIO broken out) to do some LED control, slicing and other various tasks.

    I want to be able to control the PSU with a relay on the Arduino side of the LattePanda. The PSU has a 12V always on standby rail and I can connect a few pins via relay to enable the main 12V rail. I will power the LattePanda (Arduino) with 5V stepped down from the 12V standby rail. My question is: the 12V standby rail has a max current rating of 1.5A, I require at least 5V 2A to power the LattePanda (Arduino), my 12V to 5V converter has a max current rating of 12A. How does the current calculation change with the change in voltage? Can I draw more current at 5V than I could at 12V (1.5A)?

    This is the 12V to 5V buck converter I purchased:

    Any insight is greatly appreciated. Thanks to the Duet developers for an awesome product / project, I've had so much fun with these things in the last year!

  • After posting this question I was immediately teleported back to junior year physics class…

    12V / 1.5A = 8 Ohms.

    5V / 8 Ohms = 0.625 Amps.

    Is this correct?

    I could wire the PSU to always be on and pull my 5V from the main 12V rail, but I thought it would be cooler to be able to trigger the PSU on / off.

    Well, to further confuse me, I found this little guy on Adafruit's site.
    The chart at the bottom seems to indicate that at 3A output the input current is only 1.43A, seemingly the opposite of my foggy high school calculations.

  • administrators

    Hi Matt

    You want to use power, rather than resistance:


    Without any conversion losses you need 2A at 5V:


    To get 10W from the 12V rail you need

    10/12 = I = 0.833A at 12V

    What that does not take into account is the power loss of your step down converter it is listed on amazon as "up to 95%" which is not much help, you need the data sheet if one exists to see what the efficiency is with your current requirements.

    The minimum efficiency that fits within your requirements can be calculated as follows:

    1.5A*12V = 18W

    10W/18W = 55% efficient as a minimum. Hopefully the converter will be better than 55% efficient in your case so your setup should work.

    The LattePanda looks like an interesting single board PC. Are you going to have a screen connected to it or access it over the nextwork?

  • Thank you Tony, I knew something didn't feel right about my numbers! Let's hope I can squeak more than 55% efficiency out of that board..

    The LattePanda is definitely interesting - I have not yet played with one, the Kickstarter reviews are so-so, so we'll see how it turns out. I do plan on printing a case for a 7 inch LCD/touchscreen. If the dimensions work out correctly I will mount that and the PanelDue drawer style under the printer and be able to slide them out when needed.. more often than not I'm sure with will be accessing the Windows side of the Panda via RDP if it's powerful enough to run simplify.

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