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I am sorry if this is too naive question in this site. I am from non-electrical-engineering major but I am trying to build an embedded system which will be in a real-time imaging system box along with camera. It will be possibly running 24 h/day

Currently, I have a PC/104 motherboard ADLGS45PC with Intel CPU on it, and the system will have 2-3 SSD or SATA HD with USB port to camera. Naturally, I will need to have a power supply unit for the system. But deciding the PSU seems like a complex puzzle to me.

Question 1. What would be right power supply for this system? Choices are

  1. Desktop PSU

  2. PC/104 Power supply unit DC/DC

  3. PC/104 Power supply unit AC/DC

Question 2. What is right way to decide total amount Wattage for the system?

Following is the list of power consumption for each component in the system I got for PSU unit.

  • Motherboard: 13-20W from motherboard manual

  • 64GB WD sillicondrive A100's power consumption is 5V 0.9W (Active) -> 2W

  • 512 GB Crucial SSD 6Gb/s CT512M4SSD2 ($400): 0.15W (Active) -> 2W

  • 320 GB WD Scorpio Black HD ($62): 1.75W (Active) -> 4W

  • USB 2.0: 5V x 500mA = 2.5W => 6 x = 15W

  • Network port: 5W

Totaling above items is about 60W + 20W (for some extension). Is this right way to calculate wattage for PSU?

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2 Answers 2

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You need to break down ALL of the supplies into Voltage, Amperage and Wattage. - why don't you tabulate that in the question?

Once you have that then you can calculate the total wattage, and compare it against the various options you have for required Voltages vs. the current at each voltage.

You should expect to over size each supply by 15% (although others may have a different opinion on that)

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    \$\begingroup\$ I'd plan on more power/current margin than 15% if this is a developmental project. You'd be surprised how often you need to add "just one more thing". Try for 30% or more if you can find a PSU that is appropriate. \$\endgroup\$
    – HikeOnPast
    Commented Oct 16, 2012 at 17:18
  • \$\begingroup\$ Thanks for your answer and @DeanB's comment. I can make the table with more info I need to search but could you explain more about the step how I can utilize those information to calculate total wattage? About oversizing, that's why I added 20W above. Thanks again. \$\endgroup\$ Commented Oct 16, 2012 at 17:28
  • \$\begingroup\$ Regarding question 1 above, do you have any suggestion on type of PSU I can go about? Thanks \$\endgroup\$ Commented Oct 16, 2012 at 17:31
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    \$\begingroup\$ The powers add, but just having a wattage rating is not enough if it has (for example) one voltage that doesn't have enough current supply for you. Each supply must have margin and then the total is the sum of all the powers. \$\endgroup\$ Commented Oct 16, 2012 at 17:35
  • \$\begingroup\$ Thanks for clarification. So the right PSU should provide not only enough wattage but also enough current and voltage. \$\endgroup\$ Commented Oct 16, 2012 at 17:51
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Question 1: Go with a PC104 PSU. It will be easier for you to package, and substantially reduces the cabling necessary to power the stack. They are more expensive than desktop PSUs, however, so my answer is biased toward the "clean" over the "cheap".

As for AC input or DC input, that depends on how you intend to use the system, whether there are regulatory compliance issues, etc. An AC input PSU will generally only require upstream current limiting (i.e. fusing), while a DC input PSU will require a separate AC/DC power supply (single output, generally 24VDC). If you already need a separate DC power supply for your project, this may be a preferential way to go. If the PC104 stack is the only thing that you need to power, you may be better served with an AC input PSU.

You don't list your camera in your power budget, so I'm assuming that this will be powered separately from the PC104 stack. If it can use a 12V or 24V supply, your design may be cleaner by going with a DC input PSU for the stack and powering both the camera and stack off a common AC/DC front end.

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  • \$\begingroup\$ Thanks a lot for your kind explanation. It's already helped me a lot. \$\endgroup\$ Commented Oct 16, 2012 at 17:34

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