9V wall wart showing 15V at low load - okay to hook it to <15V input? - Page 1
Forum section: electronics / beginners

author: Kold
author_rank: Newbie
timestamp: February 19, 2015, 06:07:55 pm
content: If I have a regulator that has a maximum input voltage of 12V, and I provide a 9V adapter input that temporarily puts out 15V at start-up before dropping to 9V, will I still destroy the IC?

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 19, 2015, 06:11:47 pm
content: I wouldn't risk it, but it all depends on the IC, input caps, etc. What does the IC list as the absolute maximum input?

author: Richard Crowley
author_rank: Super Contributor
timestamp: February 19, 2015, 06:18:52 pm
content: What does "temporarily" mean? Now did you perform this experiment?What is the nature of this "9V adapter"? Does it have some sort of internal active circuit? Is it just a "dumb" supply with a linear transformer and rectifier (and maybe a capacitor or maybe not)?Virtually ALL un-regulated (raw) external DC supplies have higher open-circuit voltages than they do under the designed load.It seems quite probable (IME) that this is completely normal and nothing to worry about.

author: f5r5e5d
author_rank: Frequent Contributor
timestamp: February 19, 2015, 06:31:10 pm
content: yes you can destroy a IC by exceeding datasheet absolute max specs, often some will survive small overvoltage but you are taking a real risk of failure - maybe over time, after hundreds of startsengineers really like to strictly meet abs max rating specs under all operating conditions - usually with some margin for circuit tolerances and rare conditionshobbyists may take bigger chances with less cost than a company risks in customer returns, loss of reputation

author: Kold
author_rank: Newbie
timestamp: February 19, 2015, 06:37:07 pm
content: Quote from: suicidaleggroll on February 19, 2015, 06:11:47 pmAh, sigh. So it maybe does matter. The 12V I'm referring to is the absolute maximum input voltage. There would be input caps as per a typical regulator application.Quote from: Richard Crowley on February 19, 2015, 06:18:52 pmTemporarily means that the quiescent current of the IC in question is enough to bring the wall wart AC adapter's voltage down to its rated value, but that during start-up the input voltage may exceed the IC's absolute maximum input voltage on the order of hundreds of microseconds to a couple milliseconds.The 9V adapter would be a worst case transformer + rectifier + peak detector.Quote from: f5r5e5d on February 19, 2015, 06:31:10 pmThank you, f5r5e5d. That's very helpful.

author: Richard Crowley
author_rank: Super Contributor
timestamp: February 19, 2015, 06:51:42 pm
content: Is there no filter capacitor here anywhere? Why doesn't charging up the filter capacitor absorb this turn-on transient?Do we have a schematic for this circuit? There is a limit to the practical discussion of theoretical circuits.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 19, 2015, 07:05:42 pm
content: Is it possible this spike you're measuring is simply overshoot caused by the inductance in the wire overcharging a low ESR cap at the input of your circuit? Depending on the L, C, and R at play, you can get up to ~1.5x overshoot when you first connect the power supply. A soft turn-on circuit could fix it, higher ESR cap could fix it, etc.

author: Jeroi
author_rank: Newbie
timestamp: February 19, 2015, 08:40:46 pm
content: If your max input voltage is 12v and charger is 9v then you are ok. Most linear power bricks tend to have higher voltage while off load. When you connect it to load, the regulator will maintain the 9V like the spec says. Also have to mention the absolutely max is just for reference where should be stayed at normal operation. Usually the components can handle several seconds higher voltage. The absolutely max value is just the manufacturers garantie to chip work. So you are all set with that power brick.