What adjustable linear regulator to use? - Page 1
Forum section: electronics / projects

author: LukeW
author_rank: Frequent Contributor
timestamp: March 17, 2019, 07:30:45 am
content: Are there any good candidates for a "jellybean" stock part as a more modern replacement for the 317?In particular, I'd like an adjustable linear regulator to meet the following requirements - any recommendations? Thanks!500 mA max load0-5V out"Rail-to-rail" output voltage, all the way from 0 to 5V.Adjustment with a manual 10k pot, ideally output voltage should be linear with respect to pot position.

author: awallin
author_rank: Frequent Contributor
timestamp: March 17, 2019, 07:55:47 am
content: you'll get as many part-suggestions as replies to this onemy go-to LDO lately has been LT1963.. but I'm sure someone will quickly inform us about a cheaper and better part!?https://www.analog.com/media/en/technical-documentation/data-sheets/1963aff.pdf

author: coromonadalix
author_rank: Super Contributor
timestamp: March 17, 2019, 11:42:22 am
content: The lm317 has 2.5v dropoutLD1086 LT1083 LT1086 they have 1.5v dropoutLT1085 has 1vLM1117 has 1.2vBut many of them (3 legged to220) wont go at zero untill you null out the adjustment pot with a negative voltage, like the lm723 ....Ultra Low Dropout LDO regulatorshttps://www.st.com/en/power-management/ultra-low-dropout-ldo-regulators.html?querycriteria=productId=SS1733

author: bd139
author_rank: Super Contributor
timestamp: March 17, 2019, 01:36:49 pm
content: LT3080. The thing is the bee's knees!https://www.analog.com/media/en/technical-documentation/data-sheets/3080fc.pdf1. Max 1A out2. Current referenced so goes down to 0V3. 350mV dropout4. Single resistor programming5. Stable with small output capLook at the application notes under "lab power supply". That sounds like what you need.They cost 10x what an LM317 does however in low quantities.

author: Zero999
author_rank: Super Contributor
timestamp: March 17, 2019, 05:09:01 pm
content: Most regulators don't work down to 0V, without a negative power supply. The LT3080 has a minimum load current requirement which means a current sink is required for it to work down to 0V with light loads. Unfortunately such a current sink would require a negative supply to get a true 0V out, although a current mirror does a reasonably good job of getting close without a negative supply.

author: bd139
author_rank: Super Contributor
timestamp: March 17, 2019, 05:18:21 pm
content: Good point. I've never run my LT3080 board without a load.

author: LukeW
author_rank: Frequent Contributor
timestamp: March 18, 2019, 02:04:56 am
content: The LT3080 does seem a little expensive (well duh it's LT) but it seems really simple and easy, quality regulation - better than setting up extra parts and a negative rail converter on an LM317 to make it go down to zero, which will be nearly as expensive and not as good.Here's a challenge.The program current on the LT3080 is 10 uA.Therefore, to go from 0-5V you would have a 500kvariable resistor.But what if the pot was already committed, already glued to the devices, absolutely can't be changed and it's a 10kpot?Is there any nice way that we could set 0-5V using a pot which must be 10k?

author: soldar
author_rank: Super Contributor
timestamp: March 18, 2019, 07:34:35 am
content: Quote from: LukeW on March 18, 2019, 02:04:56 amThe answer to your question is right there in the application notes.

author: Bassman59
author_rank: Super Contributor
timestamp: March 18, 2019, 07:19:41 pm
content: LT3080 is pretty great. Just mind the Vc control voltage, which has to be something like 1.5 V above the output. So unless you have a separate higher-voltage supply it can’t really be used as an LDO, even though Vin to Vout min is 300 mV or thereabouts. (Numbers from memory, I don’t have the data sheet right here.)

author: LukeW
author_rank: Frequent Contributor
timestamp: March 19, 2019, 02:25:44 pm
content: Thanks for pointing out that datasheet schematic I overlooked.R1 and R2 need to be quite small in order to get as close to 0 as possible when Rset is 0.Let's say R1 is 3.3k and R2 is 68 ohms. RSet is 0-10k.That should give us Vout = (3.3k*10uA) + (10k*10uA)*(1+(3300/68)) = 4.99 volts at the 10k pot maximum.If I've calculated that correctly. Can you confirm that is right?