Posted: 9/3/2009 7:58:21 AM EDT
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I have some radio equipment I use along with my ham radios that require 48V DC power. I want to build a battery bank + solar to power everything in the event of an emergency. While I would love to build a native 48V DC battery bank, I can't afford to do that right now, 12V will have to do. SO - I have found a 12VDC to 48VDC converter - no issues there (better than 12VDC to 120VAC to 48VDC!) ––- but in terms of calculating usage I am lost.... Is there a way I can calculate how much draw (watts/amps) it will have on the 12V side? (Looking at this: http://www.l-com.com/item.aspx?id=20899) It says output current is 0.8amps (at 48V would be 38.4Watts) –– so would the draw on the 12V system be 3.2amps? (38.4watts/12V = 3.2?) Thanks in advanced for your help! |
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The current drawn from the 12V supply would be roughly = (48V/12V) * I * 1.2 where I is the current drawn by your equipment and 1.2 is a fudge factor to account for the inverter not being 100% efficient at the conversion.
I don't know what 48V equipment you're trying to power but .8A from the inverter you've selected doesn't seem like much. Make sure your 48V equipment doesn't need more current than that or find another inverter. |
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Quoted: The current drawn from the 12V supply would be roughly = (48V/12V) * I * 1.2 where I is the current drawn by your equipment and 1.2 is a fudge factor to account for the inverter not being 100% efficient at the conversion. I don't know what 48V equipment you're trying to power but .8A from the inverter you've selected doesn't seem like much. Make sure your 48V equipment doesn't need more current than that or find another inverter. I don't see an amp rating on the PSU for the unit I have, but the spec sheet says "48VDC, 35W max) ––- so based on that (35W/48V == 0.73amps) I would say I am within the safe area.... hopefully. |