Need help with calculations

[resistcircuitresist]

Working on a project with a 12v power supply that is dropped to 5v by a lm317 breakout board.  I am current drawing around 160mA but need it to be okay drawing up to 250mA max.

Problem is with heat dissipation, this so223 lm317 is getting really hot.  12v-5v ->      7v*.250mA = 1.75 Watts of power  or 7V*.166 = 1.162


How can I drop the the input voltage closer to 5v so I wont have such a big difference in input/output.  I was planning on using a large resistor, but I'm concerned with  fluctuating voltages into the lm317 board when i vary the load.

[IanB]

Size the resistor for the maximum current expected at the load.

For instance if the maximum load current is 250 mA and you need 8 V on the supply side of the LM317, then you need to drop 12 - 8 = 4 V over the resistor. That would give you a resistance of 4 / 0.25 = 16 ohms. The resistor will dissipate 4 x 0.25 = 1 W, so choose a 2 W resistor or better to give it a safety margin.

You don't need to worry about fluctuating voltages on the supply side of the LM317 as that is what voltage regulators are designed to take care of.

[alm]

As long as you give the LM317 enough head room so it doesn't drop out at max current, I don't see the issue with a series resistor. Line regulation should be quite good, and you're not striving for the ultimate accuracy. A more elegant solution would be to get an LM317 in a larger case and use some PCB copper area as heat sink. You're dissipating the energy anyhow, not much point in having a largish hot resistor on your board instead of a hot regulator. Better still would be a buck regulator.

[Jeff1946]

Not elegant but should work fine, just use a few 1N400x (x=1 to 7) diodes, they are cheap and rated for 1A.   They will drop about 0.7 V per diode.

[resistcircuitresist]

So assuming I want 7v input to regulator.    12-7= 5v across resistor
5v* .25A(max load) = 20ohms

5v*.25 = 1.25W Over Resistor

Would a wire wound resistor be ideal for this application?

[alm]

Any 2W+ resistor should be fine in this application, I don't see anything wrong with wire wound. You don't want to run the resistor to close to their rated power, they can get very hot (>= 150°C) at their maximum power. You should also mount them a little away from the PCB to improve cooling. Don't put them near heat sensitive components like electrolytic caps.

[mariush]

You're saying "Working on a project with a 12v power supply that is dropped to 5v by a lm317 breakout board."

Are you forced to stick to that breakout board or you just prefer to use it out of convenience? Do you have to keep the area of this regulator very small?

My suggestion would be to make your own "breakout board" using a TO-220 or a D-Pak/to-252 or a TO-263 package 7805 or LM317. 

Such package will have already a larger metal surface so it will transfer heat better than the so223 package, and in addition you can connect a cheap heatsink to it making it dissipate heat even faster.

For example:

0.5$  7805 TO-220 : http://www.digikey.com/product-detail/en/MC7805CT-BP/MC7805CT-BPMS-ND/804682

0.9$ 7805 to-263 : http://www.digikey.com/product-detail/en/MC7805ACD2TR4G/MC7805ACD2TR4GOSCT-ND/1139740

heatsinks to-220

0.23$ http://www.digikey.com/product-detail/en/274-2AB/345-1087-ND/655555
0.3$  http://www.digikey.com/product-detail/en/507302B00000G/HS115-ND/5849

heatsinks to-263

0.66$ http://www.digikey.com/product-detail/en/V-1100-SMD%2FA-L/A10760-ND/3476156
1.12$ http://www.digikey.com/product-detail/en/V-1100-SMD%2FA/A10752CT-ND/3476162

You should be able to find all these on eBay or other stores, they're plain parts with lots of companies making them and kinda fool-proof.

Being fixed 5v output, there's no need for adjusting resistors (which will also get hot on the pcb as the regulator heats up so their resistance might change a bit). You only need 2 capacitors (only one really recommended, the second can be ignored) besides these and you have the complete solution.

The resistor will still get hot and will conduct heat through its leads onto the board, where you'll get more heat from the pins of the so223 chip .. which I'm surprised it can do that much current - most in such small packages are rated for 100mA current.

Unless you're really space restricted, it seems to make more sense to have the heat dissipated in the air further away from the board.


If you insist on using a resistor, I'd suggest using at least a 3w one. A 2w one will be quite hot, and as it gets hot the resistance value will change, but as you're using a regulator after it, it won't matter much.

[resistcircuitresist]

Using the regulator board because we are trying to only us off the shelf products.  Ended up buying a chassis mount 25W resistor. We shall see how well it works.

[c4757p]

Wait - how is a discrete chassis mount resistor "off the shelf" and a 7805 and heatsink are not?

[resistcircuitresist]

The boss only wants prebuilt solutions, and doesn't want to do any modifications.


Also, having an issue where the voltage out of the resistor into the lm317 module is fluctuating from 6.5 to 8.4v, the output is also fluctuating a bit..... I assume this is because the lm317 module w/ caps doesn't have enough supply voltage and is temporarily turning off?  Also the reason that the output voltage isn't seeing the same sag is that the caps are smoothing.

[IanB]

Also, having an issue where the voltage out of the resistor into the lm317 module is fluctuating from 6.5 to 8.4v, the output is also fluctuating a bit..... I assume this is because the lm317 module w/ caps doesn't have enough supply voltage and is temporarily turning off?  Also the reason that the output voltage isn't seeing the same sag is that the caps are smoothing.

It doesn't turn off, it just drops out of regulation (it tries to regulate, but does not do a perfect job).

If you want the output regulated at 5 V, you must not let the input voltage drop below 7.5 V (8 V even, at load currents above 1 A). 6.5 V is way too low for regulation. Either your load is drawing more current than you think, or your resistor value is too high.

The reason the output voltage isn't seeing the same sag is that the regulator is regulating it--that's what it does. It tries to stop the output voltage from changing.

[mariush]

A 7805 is no different than a capacitor or a resistor, it's an off the shelf part, from a reputable manufacturer, comes with datasheets, tests, it's an established components (years if not decades old, classic design) and so on.  It has three pins: input, ground, output ... it's a basic dc-dc converter. Can't be more simple than that.

Just like you buy the tl317 pcb from a store, you buy the 7805 from a store... what difference does it make that you plug the tl317 pcb on your design, or you plug a 7805 on your pcb... it's simple pins.

And like I said, this 7805 can provide up to 1A needing only about 1-1.5v above the 5v to be stable - so 6.5-20v would be no problem.

His rule makes sense only if you'd make a switching converter or something you plug in the mains - customs designs would have to be EMI certified and all that, which costs money. But this is not a part or design you make from zero, you're not making it out of transistors, diodes, resistors etc so it doesn't have to be tested or certified.

Anyway, if you insist on getting something manufactured buy a company, buy one of these then, it's off the shelf, commercially made, with all the guarantees and stuff your boss wants:

http://www.digikey.com/product-detail/en/RI-1205S/945-1191-ND/2256371

Unlike linear regulators which must have a bit over the output voltage to work right ( in your case about 2-2.5v), these are switching regulators and in addition the one above it's isolated. It takes anything between 10.8 and 13v and outputs 5v without warming up.

The downsides is that there will be a slight bigger ripple on the output (being switching kind) and the minimum input voltage is higher at 10.8-11v.