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Voltage Regulator 5V, 2A

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Jan 28, 2011
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Quezon City, Philippines, Philippines
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I'm at my wit's end trying to find a good solution for a simple problem... I need a regulator that accepts about 12 - 24V input and outputs 5V, and at least 2A of current.

I know I can get a linear regulator, but I have size constraints. The heatsink I'd need just won't fit. I'm thinking I could just use a switching regulator IC (step-down). I'm familiar with IC's that need an inductor, a cap, and a diode, but my friend said that there are some certain voltage step-down IC's that do not need any external components. Can anyone give me suggestions?... THANKS!!!

Wow, thanks for the replies!!!

I think I'm going to go with the LM2596. I can't buy premade boards because the regulator has to be part of a bigger board.

Now a follow up question. I don't think I'll be going beyond 2A. So, if the LM2596-5.0 has 80% efficiency, and I'll be drawing 2A max, will it always dissipate 2.5W??? I calculated that like this:

output of 5V * 2A = 10W
efficiency 0.8 so, 10W/0.8 = 12.5W of input needed for 10W.
12.5W - 10W = 2.5W

My question is, do i still need a heatsink for that?...

Yes! ...but only a modest one.

The datasheet indicates the thermal resistances for a number of package/mounting configurations:

Looking at the worst case scenario - the TO220 package standing vertically without a heatsink - the semiconductor junction will rise 50 degrees above ambient for every watt of power dissipated. 2.5W = 125 degrees above ambient. Although the device is rated for operation to 150 degrees, this is too hot - elevated temperatures adversely affect reliability/safety margins and take a toll on surrounding components and casings too. Option (10) presented is getting better [although still too warm for my liking], and option 11 a vast improvement. It's not hard to do better than 20 degrees/watt with even a tiny piece of aluminium (without necessarily using a "formal" heatsink such as: - perhaps there's a nearby lid/case/chassis you could exploit?
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