Heatsink recurrent problem on linear power supply project - Page 1
Forum section: electronics / projects

author: efron
author_rank: Contributor
timestamp: January 28, 2012, 11:55:19 pm
content: Dear Dave, dear all,Thank you all for this blog stuff.It's taking me now long time before I'll go for final schematic/pcb with my personal linear power supply project.Really nothing from the other world:* 15Vdc max (variable output)* 1.5A max should be enoughWell, the project also includes additional outputs (fixed ones) but this is not big issue for the moment.Even if it isn't really part of the spec, I really want to use the well known LM317T (in TO220 format).According to its datasheet, it really well fits my needs (and I have two of them in my lab).However, I've been fighting against a recurrent problem; heatsink stuff.To start with, the LM317T package has a Rth_jc of about 5°C/W (worst case), which requires a heatsink on it.Generally speaking I can find heatsinks in the market of 5°C/W or greater values without big problems and atreasonable prices. However, for better performances (let's say 3°C/W or less), the price raises andstock availability goes down.Because of this constraint, I would like to limit my design so that I can use a heatsink of 5°C/W.Without going into details for maths, this forces to limit the maximum powerto about 10W if 25°C ambient temperature is considered (and 125°C max junction temperature).And now comes my recurrent problem. If I want 15V output, I need about 18V a the input pin of theLM317 (drop out max of 3V). With the input pin at 18V and in case of 1.5A max and minimum possible output voltage (1.25V),this leads to a power consumption on the LM317T of 25W, which means that the LM317 would not support the heat.In other words, the current through the LM317 should be limited to 600mA.A first solution was to use a by pass power transistor in parallel, activated in such a way thetotal max current throug the LM317 is 600mA, the transistor supporting the additional current up to 900mA (600+900=1500).For information, this design is made of standard bipolar NPN transistor in Darlintong configuration with a power bipolar NPN transistor.I skip the reason why of this design - although you can ask for it if you want).This said, the DC input for this configuration must be 18V + 0.7V (Vbe).But having 18.7V at the input implies at least 15Vac at the output of the transformer (graetz bridge considered = 1.2V loss).However, 15Vac seems not to be very standard value, and I had to choose a 18Vac then (against 12Vac that was not enough).This implies an input dc voltage of about 24Vdc on the LM317 (instead of 18V desired), which obliges to several possibilities:* to reduce even more the maximum current on LM317, but increasing it through the power transistor, implying heatskink considerations again.* to introduce a power resistance to cope with the voltage excedent (24-18=6V >> 6x1.5=9W)* to add another LM317 in serial, with Vin-Vout = 6V (but 9W would also require heatskink)* Obviously there are more suitable solutions.But at the end I don't really know what would be the best one. Maybe by adding additional LM317 the final output will be even better(regulated twice) but it seems to become too much complex for my needs. Or maybe a simple power resistance could be enough ...I would appreciate any suggestions,(hope it wasn't too much abstract as problem)Bye

author: Rerouter
author_rank: Super Contributor
timestamp: January 29, 2012, 12:54:29 am
content: another approach is to run 2 in parrellel and use 1 ohm ballast resistors on the output, or similar, havent read over the datasheet in a while, (yes you loose a volt and a half at full output, but the regulators will be slightly offset from one another, that means for 15V out, its +1.25V for the regulator, + 1.5V for the maximum drop of the resistors, which is only 18V DC, that gets you to above 1.2A at minimum voltage, (using your math) as the resistors drop means the regulator has a smalled voltage drop across it,

author: IanB
author_rank: Super Contributor
timestamp: January 29, 2012, 01:05:26 am
content: You can address the problem in part by using a series drop resistor ahead of the LM317. If you want 18 V on the input of the regulator at 1.5 A, measure the smoothed output voltage of the transformer at the same current load. Suppose it is 21 V, then you have 3 V too much. To drop 3 V at 1.5 A you need a resistor of 2 ohms. So place a 10 W power resistor in series and reduce some of the load on the regulator.This won't solve your whole problem, but it will help a bit if your transformer output voltage is higher than you really need.

author: Rufus
author_rank: Super Contributor
timestamp: January 29, 2012, 02:30:52 am
content: There is no way to make numbers that don't add up add up.You have a lot of power to dissipate so need a big heatsink or multiple smaller ones.Don't forget your input voltage to the regulator includes ripple and you need to accommodate variation in the mains supply voltage so it is a bit worse than you seem to have been calculating.If you put a transistor across an LM317 to take some current you loose the current limit protection in the LM317 and thermal shutdown protection (although mounting the LM317 and transistor on the same heatsink gives some delayed thermal protection).The only solution I am seeing is a tracking pre-regulator to limit the voltage across the LM317. The National LM317 datasheet has an example using another LM317 as a tracking pre-regulator and another using PNP and NPN transistors for a high current regulator. If you combine those in a tracking pre-regulator you get most of the power dissipated in an NPN transistor which you can get with better thetaJC. You will still need a big heatsink (or a fan).

author: amspire
author_rank: Super Contributor
timestamp: January 29, 2012, 03:07:07 am
content: You could try something like this.The maximum voltage across either LM317 is 10V, but only one of 317s can be running hot at any time. For voltages less then 7V out, all the current comes from the 10V source and the left 317 is not in the circuit.Over 7V output, all the current comes from the 20V source via the left LM317, and the right LM317 has a fixed 2.4V drop across it.Richard.

author: Bored@Work
author_rank: Super Contributor
timestamp: January 29, 2012, 08:23:34 am
content: If you do a linear power supply, or any other piece of electronics, the rule is simple:The power that goes in, minus the power that goes out, is the power converted into heat.If you manage to beat that rule you are a prime candidate for a Nobel Price.For a linear power supply, in the worst case the maximum input power is all converted into heat. This is not specific to the LM317. Every linear regulator is subject to that problem.And there are just two principle solutions1) Prepare to dissipate all the heat that can be generated in the worst case2) Limit the input power when the difference between the output power and the input power becomes too large.You can limit the input power by2.1) limiting (reducing) the current, or by2.2) reducing the input voltage.2.3) Or of course both.Reducing the current typically affects the output performance of the power supply. So your options are usually just 1), or 2.2).You can reduce the input voltage when needed by using a transformer with multiple tabs, i.e. multiple output voltages, and automatically switching between the tabs when needed. This switching can introduce transients in the output.Another way is to use a switching (not a linear) pre-regulator. A switching regulator introduces ripple and high-frequency noise in the power supply output, depending on how good the linear part of the power supply manages to suppress them or not.A "dropper resistor" is not a really good solution. Guess what happens? It gets hot. And you have to get rid of that heat, too. You are just back to solution 1), just that you have distributed the heat generation.

author: efron
author_rank: Contributor
timestamp: January 29, 2012, 09:06:17 am
content: Thans all for your quick and dedicated answers,@AmspireYour proposal seems good and tempting, however, as I have a transformer of 18Vac giving 24Vdc after rectification, you need somehow to reduce voltage from 24 to 10V.I understand your circuit aims at limiting the maximum voltage across the LM317s, but still needs a power device (power resistor or power transistor) before getting 10Vdc.At the end, effectively, there will be most probably serveral heatsinks to cope with the whole power disipation, like the other bloggers noticed.efron.