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I purchased a board to run an old laptop LCD screen. The description on ebay says I need a 12 V 3 A input source. I only have 12 V 2.5 A and an old laptop powersupply at 14 V and 4.74 A. I thought about using a DC DC (LM2596) converter which is rated at max. 3 A to get the laptop power supply down to 12 V.

This feels like a hacked solution instead of purchasing a 3A power supply. But as I wanted to save money with this project and already have the converter I figured this maybe a solution as well.

Are there any safety issues with this solution I should consider? Any reasons one should never ever do that?

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    \$\begingroup\$ The description probably is generic, check your actual display power consumption. In any case, always keep 10-20% headroom in the power supply. \$\endgroup\$ – FarO Feb 7 '18 at 9:40
  • \$\begingroup\$ 1) do not assume that the LM2596, commonly found in Chinese boards, is actually delivering the nominal current 2) with a 75% efficiency, it may not be better than a linear regulator with a big heatsink 3) for 2 euro difference, buy the power supply rated 5A. \$\endgroup\$ – FarO Feb 7 '18 at 9:47
  • \$\begingroup\$ I probably will go for a > 3A power supply as I mentioned in the other comments that a suitable DCDC isn't that much cheaper. \$\endgroup\$ – idkfa Feb 7 '18 at 10:06
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You should not use the 12 V, 2.5 A supply as it cannot be expected to deliver 12 V at 3A reliably. Sure it might work for a while but there is a change it will slowly deteriorate (damage) when you constantly overload it like that.

Indeed the LM2596 is rated for a 3 A maximum output current but loading it with 3 A also means that there is no margin. In engineering, that's always a bad thing. And here it is easily solved, get a 5 A DCDC converter.

However converting 14 V to 12 V at 3 A is on the edge of what most DCDC converters can do, the voltage difference is quite small.

I would examine the board which needs this 12 V, 3A and see if it has any on-board regulators. If it does I would use a couple of diodes in series to drop the 2 Volts. You will need a lot of diodes though or a few "beefy" ones as they would need to drop 2 V at 3 A which is 6 Watt.

Also 12 V 3A is 36 W, which is a lot of power. The board might actually use a lot less power but you would have to measure that to confirm.

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  • \$\begingroup\$ So I should use my 3A converter, measure the actual current draw and see if it's less, then I could get away with my solution? 5A converters aren't that much cheaper than a cheap 3A power supply, so if that doesn't work I'd probably buy a properly rated supply. There's really no trade off in wasting the board and possible the screen over saving 3 euro if I have to buy new parts anyway. \$\endgroup\$ – idkfa Feb 7 '18 at 9:32
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    \$\begingroup\$ I am sometimes amused by these kind of questions, where someone wants to save a few bucks on an one-off. Do youreally think the time it has taken you to ask, and others to answer, this question, isn't already costing you (and us) more than those wimpy 3€? \$\endgroup\$ – anrieff Feb 7 '18 at 9:46
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    \$\begingroup\$ @anrieff I sincerely hope people do not feel their time wasted answering "stupid" questions. Ofc I could have figured I need a 3A supply, I don't have one so I go and get one. But after all I'm new to electronics, I have some parts in stock and I was just curious if I'm not able to find a solution with what I have. It's not about the 3 € (new DCDC vs new psupply) but more about the learning why one should or shouldn't do it that way. Personally for the idea to check for on board regulators and drop the voltage with diodes alone I think it was worth asking the question for me. \$\endgroup\$ – idkfa Feb 7 '18 at 10:04
  • \$\begingroup\$ Asking questions to learn things is all fine and well. But then your real motives aren't to "save money", but to learn, which are different things. It is that money-saving part in your question that I find amusing, not the question per se. \$\endgroup\$ – anrieff Feb 7 '18 at 10:21

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