LT3080 and reducing power dissipation - Page 1
Forum section: electronics / beginners

author: chaser
author_rank: Newbie
timestamp: December 09, 2013, 12:43:18 pm
content: 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=5VIout(max)=1AIout(min)=450mA (assuming Rp carries no more than 90% of Iout(min) = 405mA) <- don't exactly understand what this part meansI 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 = 7.6W1A - (12.6-5) / 19 = 0.6APdis = (12.6 - 5) x 0.6 = 4,56WRp= 7.6-4.56 = 3,04WSo I need a 19ohm resistor capable of handling 3W?I attached the eagle screenshot of my circuit and the LT3080 datasheet ishere

author: c4757p
author_rank: Super Contributor
timestamp: December 09, 2013, 01:39:31 pm
content: 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:

author: Unknown
author_rank: Unknown
timestamp: December 09, 2013, 01:53:36 pm
content: Quote from: chaser on December 09, 2013, 12:43:18 pmIt 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.

author: trackman44
author_rank: Regular Contributor
timestamp: December 09, 2013, 03:19:15 pm
content: 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

author: minime72706
author_rank: Frequent Contributor
timestamp: December 09, 2013, 04:33:03 pm
content: 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.

author: chaser
author_rank: Newbie
timestamp: December 10, 2013, 06:51:26 am
content: 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.

author: minime72706
author_rank: Frequent Contributor
timestamp: December 10, 2013, 06:10:12 pm
content: Quote from: chaser on December 10, 2013, 06:51:26 amUhh ... 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.