Vreg heat issue and PCB power consumption questions

[Tetsu0]

Hi everyone!
I have a board that takes 24VDC and steps it down to 5VDC with a MCP1827 Voltage regulator. The board uses a few IC's and lights 8LED's for status indications.
In testing i noticed that the VREG put off a LOT of heat- to the point where the chip just shut down and discolored.
I put a heat sink on a new regulator and it's still gets pretty hot but it works. I'm pulling 1/4 Amp on the 24DC side of things...

I'm wondering if there's a better way to step 24DV down to 5V, and if there are any recommendations for low power/high brightness LED's that can ease the current draw of my board.

Thanks

 

[(*steve*)]

Check the specs. An MCP1827 is rated for a maximum 6V input voltage.

Are you sure it's a MCP1827?

 

[Tetsu0]

Ah you're right, i'm sorry its a MC7800 in the TO-220 package

I was using the MCP for a Vref regulator. Wrong datasheet

 

[BobK]

19V at 250mA is almost 5W of wasted power. A switch mode regulator would be far better.

Bob

 

[Tetsu0]

I have no experience with switch mode regulators. Do they need heat sinks? I'm working with a limited foot print on this board.
A switch mode would be able to take the 19v drop?

 

[GonzoEngineer]

What is the maximum power requirement for the 5V supply?

 

[Tetsu0]

Unfortunately i'm not 100% sure what the power requirement is but the regulator is supplying 13 IC's and 8 LED's
I tried doing a power requirement calculation before but i don't think i went about it properly, and I don't remember the result

 

[(*steve*)]

How much board real estate can this regulator use?

GonzoEngineer has a very valid point, and you're going to have to do that calculation -- properly -- before you can be certain how to go ahead.

But let's assume that 250mA is close to the maximum for the sake of argument.

You should be able to achieve 80% efficiency in your regulator without any trouble. So for 5V 250mA (1.25W) output, you would expect less than 300mW of heat generated by a switchmode regulator. You won't need a heatsink. (Much of that dissipation will be in the inductor and the diode, not the regulator itself in any case)

One option (for testing at least) is a pre-built module. You can get them all over eBay for a couple of bucks and occupy less than 2 square inches (and less then 1 cubic inch of volume). Typically they claim you can use them for up to 15W of output without the need for a heatsink, so 1.25W is well within this.

 

[CocaCola]

One option (for testing at least) is a pre-built module. You can get them all over eBay for a couple of bucks and occupy less than 2 square inches (and less then 1 cubic inch of volume). Typically they claim you can use them for up to 15W of output without the need for a heatsink, so 1.25W is well within this.

You can also get a drop in switching package in the TO-220 form factor... A little costly for what it is but dead simple to refit into an existing design...

http://www.dimensionengineering.com/products/de-sw050

 

[(*steve*)]

I'd forgotten about those!

If you already have a board designed and you don't want to get another produced (and especially if this is a one off project) then CocaCola's suggestion would be the way I would go.

 

[Tetsu0]

I'm still quite new to designing circuitry like this; i really appreciate all the help.
I have found this on DigiKey, would it be suitable?

http://www.digikey.com/product-detail/en/BP5277-50/BP5277-50-ND/2776226

Even if i'm drawing 1/4Amp, this should work, and i already have a very large heatsink.

 

[GonzoEngineer]

I'm still quite new to designing circuitry like this; i really appreciate all the help.
I have found this on DigiKey, would it be suitable?

http://www.digikey.com/product-detail/en/BP5277-50/BP5277-50-ND/2776226

Even if i'm drawing 1/4Amp, this should work, and i already have a very large heatsink.

That should be more than enough power. A bit pricey, but solves all your problems.

 

[GonzoEngineer]

You can also get a drop in switching package in the TO-220 form factor... A little costly for what it is but dead simple to refit into an existing design...

http://www.dimensionengineering.com/products/de-sw050

Looks like they wrapped a label on it and doubled the price!...assholes!

( I meant them, not you CocaCola!)

 

[Tetsu0]

That should be more than enough power. A bit pricey, but solves all your problems.

Great!
I ran through the power calculations, and it seems the total power usage is 431mW.
At 5 volts, the board will draw 86.2mA off the regulator. The IC's are all low power, most of the power usage stems from the LED's and the current-limiting resistors. I have some 10K pullup resistors that i didn't factor in, but 86mA seems like such a low current that it shouldn't really make a large difference right?

I'm still confused as to why the other Voltage regulator was getting so hot though.

 

[BobK]

I think this was Steve's test: How long can you leave your finger on it?

Bob

 

[Tetsu0]

Not long at all. My IR thermometer says 210Deg F - Unacceptable in my opinion.

 

[CocaCola]

I'm still confused as to why the other Voltage regulator was getting so hot though.

Goes back to that energy can't be created or destroyed principle, when using a linear regulator it dissipates any excess power as heat, so you start with 24V and reduce to 5V and now have to dissipate 19V@whatever current as heat...

 

[Tetsu0]

Right right.
I did some more research and i came across

http://www.digikey.com/product-search/en?x=-996&y=-71&KeyWords=811-2196-5-ND

I think that'll do much better.
Thanks again for your help!

 

[BobK]

and i already have a very large heatsink.

But there is something wrong if you already have a large heatsink, and it is getting that hot at 83mA. There should be only about 1.5W to dissapate and anything I would call a very large heatsink should run pretty cool at that. Can you measure the current?

Bob