lm317cc cv.  - Page 1
Forum section: electronics / beginners

author: guygibbs
author_rank: Newbie
timestamp: April 02, 2015, 05:56:25 pm
content: I am a pretty new hobbist and I've managed to use the lm317 datasheet page 7-8 minus the transistors I just run to ground rather than -10v. Well my question is the circuit surprisingly works well it seems to me and I do get current and voltage adjustment and it will even go to 0. 0001 volts with both pots turned down, but here is what I've done lol (newbie) I run a 1k to the input of cv lm317 and a pot from the adjust with one side to ground and the other to v out on the cv side of the 1k resistor as usual which this is NOT how the schematic says to do and if I correct it the way the datasheet says it does not work half as well and I only get max 1 amp out . This way I get up to 2 amp out . But does anyone believe this will last OK ? Also the pot does not get hot but of course the lm317 doesI have big heat sink and will add a fan . I'm sorry if this isn't enough info I can add more if needed . Thanks in advance!

author: Rick Law
author_rank: Super Contributor
timestamp: April 02, 2015, 06:59:30 pm
content: Quote from: guygibbs on April 02, 2015, 05:56:25 pmGuyGibbs,You may want to provide a link or more info on which datasheet you are referring. I looked at the TI's datasheet (OCTOBER 2014), STMicroelectronics' datasheet, and Fairchild's datasheet. Non of them show a circuit on page 7/8.It will help get to a better answers if you can include aphoto of a simple sketchon exactly how you did it.Try as I might, I could picture what your words are describing. I think you mean 1K to the ADJ pin then from ADJ is trimpot to ground, or perhaps you had a 1K trim with pin 1 to Vout, pin 2 to ADJ, and pin 3 to ground...Also - what load is used testing is important.- If you just connect a resistor, Vout and Iout are dependent on each other.Vout is always equal to Iout times Resistor, so in this case, once you set Vout, Iout is also fixed.- If you just connect an LED+resistor, then Vout will only vary within a specific range based on the LED (and until you pump too much in a burn out the LED).Rick(a relatively newbie as well)

author: guygibbs
author_rank: Newbie
timestamp: April 02, 2015, 11:43:16 pm
content: OK this is the way it is supposed to be ...

author: guygibbs
author_rank: Newbie
timestamp: April 02, 2015, 11:55:04 pm
content: And thus us what I've got haha.. test load is a 4r70jsmt 5 watt load resistor. And also tested with dead short since they are protected and I get over 2 amps out and that's is no matter what voltage adj. Pot is set at. And down to .001 I out when set current pot down . And yes the voltage pot does drop current as well when limit is reached .

author: guygibbs
author_rank: Newbie
timestamp: April 02, 2015, 11:57:06 pm
content: I'm sorry it won't let me post my hand drawn schematic of how I have everything .. any ideas?... it is virtually identical only instead of the current pot going to a transistor and to negative rail I have that side going straight to ground .

author: guygibbs
author_rank: Newbie
timestamp: April 06, 2015, 04:35:31 pm
content: I've figured it out , I basically just run the two regulators in series and it allows for the double current and both are adjustable . Also made another newbie mistake while mounting on heat sink (big) I had to isolate one but I did both because of the tab connections. It may not be perfect but I'm happy with it since it does go to 0.01v . Thanks for the reply

author: Rick Law
author_rank: Super Contributor
timestamp: April 07, 2015, 12:01:40 am
content: Goodie, happy for you that you got it working.