Power supply issue - Page 1
Forum section: electronics / beginners

author: Anson
author_rank: Regular Contributor
timestamp: April 12, 2013, 02:33:49 am
content: I recently finished my first power supply. It's a simple linear dual polarity adjustable power supply. It works fine however today I decided to see how many amps it could put out. So I stuffed my breadboard full of LED's enough o pull about 360mA I was pretty sure it could do more but I didn't have anything to load it anymore. Then after a few seconds i noticed the reading on my meter lowering steadily down around 100mA I realized the LM317 was extremly hot. I have a heatsink on it not a large one but it has one and I read on the datasheet its good to 1.5A why is mine crapping out at 350?

author: c4757p
author_rank: Super Contributor
timestamp: April 12, 2013, 02:38:01 am
content: Simple answer: Because it also has a temperature limit and you've exceeded that. Use a bigger heat sink.

author: BravoV
author_rank: Super Contributor
timestamp: April 12, 2013, 02:41:23 am
content: And don't expect your breadboard to carry full current of that regulator, move them to pcb and solder.

author: David_AVD
author_rank: Super Contributor
timestamp: April 12, 2013, 02:42:16 am
content: Remember that a linear regulator has to dissipate heat to regulate. That's determined by the current multiplied by the difference between the regulator input and output.For example, if the input to the regulator is 20V and the regulator output is 5V, that's a 15V drop. Multiply that by 0.36A and suddenly you've 5.4 Watts of heat the LM317 need to get rid of.

author: c4757p
author_rank: Super Contributor
timestamp: April 12, 2013, 02:42:37 am
content: Longer answer: The temperature rise above room temperature is roughly equal to the power dissipated multiplied by the total thermal resistance (including the 317, the heat sink, and whatever thermal compound, if any, you have between them). Find the thermal resistances in the various datasheets. The power dissipated is equal to Iout(Vin-Vout).Quote from: BravoV on April 12, 2013, 02:41:23 amWhat breadboard can't do 350mA? I've used mine at multiple amps - sure, the contacts get a bit toasty, but it hasn't turned into a pool of molten plastic yet!

author: eevblogfan
author_rank: Frequent Contributor
timestamp: April 12, 2013, 07:21:31 am
content: heythe lm317 can heat up if the filtering cap ( the one right next to the lm317 from the side of the P.S ) is longer then 10C'M . or you've exceeded the power can be dissipated by the lm317 , say he can dissipate 750~1.5W ( 60~100+C ) then the OTP kicked into play and limited the current to 100mA ,

author: Anson
author_rank: Regular Contributor
timestamp: April 12, 2013, 05:20:16 pm
content: I have it soldered onto a pcb. When I put it together I didn't really take into account heat dissipation. I'm new to this so yeah thats a dumb mistake. I have a 48v center tapped transformer powering it so yeah thats an issue. I have some beefier heat sinks I can try. I was running it at 12v.

author: lgbeno
author_rank: Frequent Contributor
timestamp: April 12, 2013, 06:01:16 pm
content: Yeah you are dropping 4W with this device. If you are using a TO220 package the thermal resistance is 4deg C per watt however that is junction to case and expects that you have a perfect heat sink.Unfortunately you do not so add the deg rise per watt of your heatsink to 16 and that is your junction temp. Airflow can help power dissipation a lot too.

author: MasterOfNone
author_rank: Regular Contributor
timestamp: April 12, 2013, 07:35:47 pm
content: Sorry “Anson” I don’t understand, think I’m being a bit thick. You said you “stuff your breadboard with LEDs, enough to pull 360mA“, then later you said “I was running it at 12V”. Did you have enough 12V LEDs (with the internal resistor) or did you place a resister in series with each LED or did you have enough LEDs to arrange them so that groups of 3 or more where in series and each group was connected in parallel. Or did you just connect some standard LEDs to the breadboard in parallel and then connect them to your PSU when it is set to 12V rather than around 3V? Doing the last option would draw a lot more than 360mA, it would be like doing a short circuit test on the PSU.

author: eevblogfan
author_rank: Frequent Contributor
timestamp: April 12, 2013, 08:08:58 pm
content: heymaster please look at the needed part , he say he draw 360mA and that's the only part we should be looking at ,Hence , 360mA ( don't care how ! ) at 12V from 48V , so I can guesstimate the voltage DC is higher then 48 , am I wrong ?then . say the 48V is RMS and with full load , then the voltage DC should be 1.15*48V = 55.2V so 55V-12 = 43V so the Vdrop across the LM317 is 43V . then the current is 360mA ( by quoting OP ) so 0.36*43= 15.48W so probably quiet hot , I'd recommend you to tell us the full detailed circuit in order to help you solving the over heating/abusing of the lm317 ,Cheers

author: mariush
author_rank: Super Contributor
timestamp: April 12, 2013, 08:20:52 pm
content: He has 48v CENTER TAP , and he says he has a "dual polarity" power supply.I assume from that he has +/- around 24 v , somewhat like this design:There's only one forward voltage drop as only one diode works on each side so let's just neglect it in this situation.Also going to assume the capacitor is big enough so that we won't have much ripple (assume the voltage regulator has 24v at the input).Then at 12v 500mA of current, your linear regulator will dissipate as heat the following(Input Voltage - Output Voltage ) x Currentand that's (24-12 )x0.5 = 6 watts.At 0.36A , you have about 2.1 watts. A small heatsink should be enough to dissipate the heat away from the regulator.... not sure though, it's not clear exactly what the input voltage each regulator receives.

author: eevblogfan
author_rank: Frequent Contributor
timestamp: April 12, 2013, 08:28:22 pm
content: Quotesilly meany way .he need something like 3CM wide 4CM high and 2CM deep in order to dissipate it properly ( perhaps smaller size but I am pretty sure that's okay size too )if size and power handling is a problem , go for the switching reg method ( assuming ripple is not a big of a deal !)Cheers

author: MasterOfNone
author_rank: Regular Contributor
timestamp: April 12, 2013, 09:27:23 pm
content: Quote from: eevblogfan on April 12, 2013, 08:08:58 pmI’ll try to explain my statement in a different way.As you know LEDs are Diodes so connecting them across the PSU is effectively shorting the output of the PSU if there isn’t any series resistors (internal or External). You can place multiple LEDs in series and connect them to a higher voltage source since each LED will have it’s own voltage drop.So I was thinking he could have been measuring the overload protection current because the output would be almost shorted if he place normal LEDs across a 12V PSU.Hope that makes it clearer

author: Anson
author_rank: Regular Contributor
timestamp: April 12, 2013, 11:22:57 pm
content: Yes that is my circuit. I have 50v rated 1000uf caps for c1 and c2. I have no way to measure ripple but I am pretty sure those should take care of it. I used 470 ohm resistors to set my LED's to each pull around 25mA they are rated up to 30ma. It took me an hour to set them all up in my breadboard. I used 12v Because I am thinking of using these ultra bright LED's for boat lights for a friend for when he goes frog gigging and I didn't want to have to use any more parts than possible. I am afraid they may fry the frogs before he gets them though. So the test was two fold one to see how many amps the lights pulled and two to see ho many amps my power supply could handle. As I said I did not fully take into account heat dissipation. I have a pair of aluminum heat sinks I can use that are 5mm thick 18mm wide and 50mm tall with fins portruding from the top 20mm that should take care of it. Just need to drill holes in the right place.

author: eevblogfan
author_rank: Frequent Contributor
timestamp: April 13, 2013, 05:17:12 am
content: sounds okay , assembole the whole thing and report us

author: stratogazer
author_rank: Contributor
timestamp: April 15, 2013, 01:52:03 am
content: Anson,didn't see you mention a fan for the heat sinks in your PSU. Even though the temperature stability of the LM-317 is 1% up to 125C, keep the regulator cool as you can.