Voltage Regulator 5V, 2A

[daneloctober]

Hello!

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!!!

 

[alexan_e]

LM2596 is a nice option but instead of making your own board you can buy one form ebay for less than $2
https://www.ebay.com/sch/i.html?_trksid=p2047675.m570.l1313&_nkw=LM2596&_sacat=0&_from=R40

- - - Updated - - -

The chip can provide up to 3A but will need better cooling so for tat case you will need to use a version in which you can attach a heatsink

 

[thylacine1975]

These little (reconditioned) modules from Ebay are great: https://www.ebay.com/sch/i.html?_sacat=0&_from=R40&_nkw=kis-3r33s&_sop=15 (and cheap
There's an abundance of info on the net about changing their output voltage - I've used them for 1-3 Amps at voltages from 3 - 12 V.

 

[daneloctober]

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?...

 

[thylacine1975]

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: https://www.farnell.com/datasheets/17590.pdf) - perhaps there's a nearby lid/case/chassis you could exploit?