[Solved] Problem charging 10F caps - Page 1
Forum section: electronics / beginners

author: carcanhol
author_rank: Regular Contributor
timestamp: November 04, 2014, 04:18:58 am
content: Hi, i just made this circuit with LM317T voltage regulator but it charges very slow, like 1mV per second, did i fried the lm317?I can even use a green led on the output and its getting 2.2V(pot adjusted) but very low current.Is there a way to increase current or my circuit it's not the apropriate or is badly built?Circuit used in attachments.Thanks.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 04, 2014, 07:05:37 am
content: Charging the caps shorts the regulator output, which puts it into current limit, and then thermal limit. They don't last long in thermal limit...You need a current limited circuit.Tim

author: Odysseus
author_rank: Regular Contributor
timestamp: November 04, 2014, 07:15:36 am
content: It should be obvious if the 317 is in thermal overload, as it will become quite hot.For a input-output differential of 12V, the 317 is limited to around 2A, which would charge 60F at a rate of 33mV/s. That's about 2 min to reach 4V, during which the regulator will likely enter thermal shutdown, reducing the charging current to something much lower.

author: Andreas
author_rank: Super Contributor
timestamp: November 04, 2014, 11:38:59 am
content: Hello,I think the problem is that you are using Gold-Caps which have a too high internal resistance for your application.Do you have a datasheet for the caps.With best regardsAndreas

author: ManCave
author_rank: Regular Contributor
timestamp: November 04, 2014, 01:30:43 pm
content: Quote from: carcanhol on November 04, 2014, 04:18:58 amHi, you should add another LM317 circuit after your voltage regulator to limit the current which will prevent the voltage regulator going into the thermal shut-down due to over-current. Circuit is simple - like the one below. Google is your friend...It's a very useful circuit to know so I highly recommend you get yourself familiar with it. Use the resistor R1 to set the current limit and make sure that the resistor has sufficient power rating (e.g. the power rating is higher than power dissipated) using formulaPower dissipated by resistor = I2RGood Luck!Filip @ ManCave

author: carcanhol
author_rank: Regular Contributor
timestamp: November 05, 2014, 12:22:48 am
content: Quote from: Odysseus on November 04, 2014, 07:15:36 amIt didn't got hot, so i was so confused, today i saw a bit of solder on the pcb tracks and removed it it worked! but i was using a 1k resistor, so it was very slow charging, i changed that one with a 2.2Ohm 5W ceramic resistor and it was consuming ~0.800A while charging the caps so i guess everthing is ok now(at least they got charged and light up the led on the Joule thief).Sorry for the noobish problem, i never thought of something so small would make a circuit unusable.Thanks for the help.

author: carcanhol
author_rank: Regular Contributor
timestamp: November 05, 2014, 12:26:12 am
content: Quote from: ManCave on November 04, 2014, 01:30:43 pmFilip, you are the best, im going to make that for sure, i got today 5x LM317T ST brand so tomorow i will make 2 circuits, one for the capacitor charger and one for the power supply.Thanks.

author: carcanhol
author_rank: Regular Contributor
timestamp: November 05, 2014, 12:30:43 am
content: Quote from: Andreas on November 04, 2014, 11:38:59 amDont know, i just bought them from ebay, here is the linkhttp://www.ebay.com/itm/10F-2-7V-farad-Supercapacitor-x-6-pcs-100-000-cycles-EDLC-high-reliability-/261619792435?pt=LH_DefaultDomain_0&hash=item3ce9c13a33

author: Paul Moir
author_rank: Frequent Contributor
timestamp: November 05, 2014, 01:58:57 am
content: You might find that heat sink a little inadequate. Just guessing but I'm thinking you're looking at something like 17 degrees C per watt die temperature rise. Given that you can only have (max) a die temperature of 125 and ambient is say 25, that means you can only have 100 degree temperature rise. 100/17 = 5.9W is the max you can dissipate in the LM317. The LM317 will be dissipating the most when the voltage in the caps is low. Say with 3v on the caps and 12 on the supply, 5.9 / (12-3) = .66A is the max you can be allowing through the LM317 while keeping it out of shutdown.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 05, 2014, 02:17:13 am
content: Just be careful not to overcharge the caps. A current limiter circuit alone won't stop at 2.5V, it'll just keep on going until it runs out of supply.Tim

author: carcanhol
author_rank: Regular Contributor
timestamp: November 05, 2014, 04:53:04 pm
content: Quote from: ManCave on November 04, 2014, 01:30:43 pmJust a small question, i just connect the output of the circuit i have made to the input of the circuit you posted so i get the limited voltage and current power supply?I saw one circuit made on the ON SEMI. lm317 datasheet but it uses negative voltages and that is confusing for me now because i use an atx psu from a pc to power everything up.

author: carcanhol
author_rank: Regular Contributor
timestamp: November 05, 2014, 04:57:48 pm
content: Quote from: Paul Moir on November 05, 2014, 01:58:57 amif i use a big heatsink from a lga775 pentium with a fan it could do for 1A or more output or the rate the lm317 temp rises faster than the heatsink capability to conduct heat?thanks

author: ManCave
author_rank: Regular Contributor
timestamp: November 05, 2014, 09:32:27 pm
content: Perhaps I should have asked this in the first place, can you share the purpose of the circuit? What are you trying to accomplish? There might better/simpler ways of doing it...

author: Paul Moir
author_rank: Frequent Contributor
timestamp: November 05, 2014, 10:10:16 pm
content: A fan makes a world of difference. I would bet that if you blew a fan on your current heat sink, you would be able to do 1A. I'd be shooting to keep the thing under 50 deg C, which is conveniently just above the threshold of pain for most people.Using a heatsink that was meant to use a fan without one is a bit questionable, or I should say experimental. Sometimes the fins are too close together for efficient convection (main way heat gets out of the heat sink without a fan).

author: carcanhol
author_rank: Regular Contributor
timestamp: November 06, 2014, 02:41:31 pm
content: Quote from: ManCave on November 05, 2014, 09:32:27 pmThe purpose is to charge 6x 10F 2.7v capacitors in parallel to 2.5V to power a joule thief and then a White Led.Later i will be using a hand crank generator.

author: carcanhol
author_rank: Regular Contributor
timestamp: November 06, 2014, 08:41:24 pm
content: The schematic right now.And the final board.Some parts were replaced or changed.

author: carcanhol
author_rank: Regular Contributor
timestamp: November 07, 2014, 06:23:15 am
content: Demo video:

author: IanB
author_rank: Super Contributor
timestamp: November 07, 2014, 08:51:15 am
content: Quote from: carcanhol on November 05, 2014, 04:53:04 pmGenerally you would put the current limiting circuit before the voltage regulating circuit rather than after it so that the current limiting circuit doesn't affect the final output voltage. But otherwise yes, you can cascade the circuits.

author: ManCave
author_rank: Regular Contributor
timestamp: November 07, 2014, 01:45:54 pm
content: Well done!Quote from: IanB on November 07, 2014, 08:51:15 amIan is right! There will be a voltage drop through the current limiting circuit!

author: carcanhol
author_rank: Regular Contributor
timestamp: November 07, 2014, 03:37:26 pm
content: Quote from: ManCave on November 07, 2014, 01:45:54 pmYes, i ended up using a 1k pot on the adj on the voltage side because i forgot there was a voltage drop on the current limiter and then another drop on the diode near de caps. so now its fine. Thanks for the help.