LM2577 step-up voltage regulator design problem - Page 1
Forum section: electronics / beginners

author: BishopB7
author_rank: Newbie
timestamp: May 03, 2018, 07:36:37 pm
content: Hi everyone,I have the LM2577 step-up step-down voltage regulator. R1 resistor is replaced with a potentiometer. Conversion ranges are 1.2V - 34V. Thing is I need to limit the output voltage up to 12V (in step-up mode) even if potentiometer is turned over 12V level. Putting a 7812 regulator on the output stage - doesn't sound right. I guess something has to be done with the feedback circuit, but I can't think of anything, that would work. Any ideas ?P.S. Sorry for my English, it's not my native language.

author: Rigolon
author_rank: Regular Contributor
timestamp: May 03, 2018, 08:57:11 pm
content: A power regulator, as you thought, will not work since they usually have some drops in them, usually around 2V. So if you have 5V on the input of the 7812, your output will get down to 3V and your step-up won't work as intended.What you want is a voltage limiter a.k.a Zener Diode. I believe that it will do the job.

author: BishopB7
author_rank: Newbie
timestamp: May 04, 2018, 05:30:57 am
content: Yes I know the part about linear regulator that it requires 2V more at the input than its regulating voltage. Zener diode? Where to add it ? I forgot to mention that minimum input voltage of step up regulator is 3V.

author: ramonest
author_rank: Contributor
timestamp: May 04, 2018, 08:16:20 pm
content: Maybe playing with the value of the potentiometer? So that when you have 12 V at the outut the potentiometer is already maxed? I do believe 12(R2/(R1+R2))=1.23 is the expression you need to look at.Regarding the zenner thing I guess he ment to put a zenner pararell to the potentiomenter.

author: johnkenyon
author_rank: Regular Contributor
timestamp: May 05, 2018, 12:59:06 pm
content: Quote from: BishopB7 on May 03, 2018, 07:36:37 pmMake life easier - choose the values for the resistors and potentiometer so that in the worst case scenario, the maximum output voltage is 12v.If I were doing this...Select a potentiometer value to replace R1 - 10kWork out the value for R2 to give 12v out at full 10k resistance.Vout=1.23* (1+R1/R2)Rearrange:R2=R1*(1.23/(Vout-1.23))For R1=10k, and Vout=12vR2 this gives me 1.14k for R2.1.14k isn't standard, so we choose 1.2k and see how this affects the voltage:Vout=1.23*(1+10/1.2) = 11.48vA little on the low side, so we use the Vout equation to work out a new value for R1 at 12v with 1.2kR1=((Vout-1.23)/1.23)*R2This gives me R1=10.50kSo for R1 lets put a 510R resistor in series with the potentiometer.This gives us R1=0.51k to 10.51k, and R2=1.2k and we see if these values work.Putting these values intoVout=1.23* (1+R1/R2)gives us a range of 1.75v to 12.00vNote that this assumes that a 10k potentiometer will give exactly zero ohms at one end and exactly 10k at the other, and that the fixed resistors are exactly 1.2k and 510R.In the real world, this won't happen so (for example) your 510R resistor might have to be 560R or 470R, and/or you'll be happy with something other than exactly 12v max./John

author: BishopB7
author_rank: Newbie
timestamp: May 10, 2018, 04:25:42 pm
content: Thanks John, that really seems like simplest solution(I always start from difficult ones)