LT3080 and reducing power dissipation

[chaser]

Hello,

I am trying to build a small backpack for a video transmitter and an integral part of schematic is a voltage regulator which gives me 5v 1A from a 12.6v source. As I have a few of those LT3080 at hand in SOT-223 package I thought I could use those. Looking at the datasheet I can understand that how to get the output voltage down to 5V but the heat dissipation with a resistor gives me a bit trouble. I am choosing the resistor in parallel solution but can't seem to calculate the correct value. With the following circuit parameters:

Vin = Vcontrol = 11.1V (SOT-223 has those pins connected)
Vin(max) = 12.6V (3S LiPo battery)
Vout=5V
Iout(max)=1A
Iout(min)=450mA (assuming Rp carries no more than 90% of Iout(min) = 405mA) <- don't exactly understand what this part means

I get:

Rr = (12.6-5) / 0.405 = 18.77 ohm (so with a standard 5% its 19 ohm)

Max Power Dissipation calculations:
(12.6 - 5) x 1A = 7.6W

1A - (12.6-5) / 19 = 0.6A
Pdis = (12.6 - 5) x 0.6 = 4,56W
Rp= 7.6-4.56 = 3,04W

So I need a 19ohm resistor capable of handling 3W?

I attached the eagle screenshot of my circuit and the LT3080 datasheet is here

[c4757p]

5V 1A from 12.6V will, as you have calculated, need to dissipate 7.6W. That's a ton of power; you'll need a big heatsink. Not really good for a "small backpack". Use a buck converter instead. Watch this video:

[Tac Eht Xilef]

Iout(min)=450mA (assuming Rp carries no more than 90% of Iout(min) = 405mA) <- don't exactly understand what this part means

It means that, for the resistor in parallel configuration, the rule-of-thumb is that you choose the resistor to carry no more than 90% of the minimum load current at minimum load (the regulator supplies the remaining 10%).

That said, 450mA looks like a suspiciously round number - did you get it from the specs of the transmitter? I'd be measuring the actual standby current myself; you might find it's lower than that.

The other thing is that, from the rest of your calculations, at maximum load (1A) you've got the resistor dissipating ~3W and the regulator dissipating ~4.6W. Good luck getting a SOT-223 package to dissipate that much power (hint: you won't; not without some real fancy heatsinking). You'd be much better off with a regulator in at least a DDPAK, or preferably TO-220.

Unless there's real good reason not to, a switching buck regulator would seem to be a better solution than a linear regulator like the 3080.

[trackman44]

Since you already have a few LT3080s lying around, use one to step down to 9V say, and the second one to 5V, like that you spread out the heat desipation.

Will

[minime72706]

You can parallel them using either small ballast resistors (10's of milliohms), but really, this is not a great application for a linear regulator.

[chaser]

Thanks for all your replies. Despite the wall of math envolved the MC34063 looks to be a good solution. Ill try it over the weekend. The LT3080 will just have to wait for their turn.

[minime72706]

Thanks for all your replies. Despite the wall of math envolved the MC34063 looks to be a good solution. Ill try it over the weekend. The LT3080 will just have to wait for their turn.

Uhh ... do not use the MC34063. Many people on this forum have reasons to say likewise. It's unreliable and old - and sometimes dangerous. DO NOT USE IT.