Two regulators, same line issue. - Page 1
Forum section: electronics / beginners

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 06:45:22 pm
content: Hi All, I have a small problem I cannot figure out why exactly this is happening due to lack of knowledge and experience in things like this:Here is what's going in.I have one battery source (3AAs in series) and two voltage regulators connected to this source. One regulator is supplying 3.3V voltage for microprocessor (top one) and another regulator (bottom one) supplying variable voltage to peripherals, you can see I placed red resistor and switch in parallel to one of the feedback resistors so I can chose one of two output voltages 2V or 7V.The problem is that if I apply power to this setup it stars up just fine, but once I attempt to switch from 2V to 7V by closing that red switch, - microcontroller dies. I tried to study why and it seems that this is because once I enable that switch, battery voltage drops by 2V for about 500uS and I think it sucks charge from first regulator's input capacitors (I think) so it temporarily fails to supply power to microcontrolles and it dies.What would you recommend to fix this issue?Thank you in advance.

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 06:55:53 pm
content: You are doing this without soft start functioning. Likely internal switch opens for prolonged time and high current starts flowing through it. Also 2.4 MHz buck-boost converter is not a thing to tinker with. Do suboptimal PCB design and it won't work correctly even without doing something like this. And it's not a microcontroller. If by microcontroller you mean something that is powered by this circuit, then why 7V would be there in the first place. In any case you are attempting to do something strange.

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 06:58:56 pm
content: QuoteDon't use this IC at all. Without understanding what you are doing and properly designing PCB, you won't get good results from it.

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 07:03:02 pm
content: Datasheet says that this has built in soft start (page #8http://www.ti.com/lit/ds/symlink/tps63060.pdf, it limits current below 2A. Anything I can do to limit it even lower?Quote from: wraper on October 12, 2017, 06:55:53 pm

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 07:05:38 pm
content: Not an option, I am already using it and trying to figure out ways to fix this particular issue, I just do not have enough theoretical knowledge to know what solution to try before I can experiment with all of them and what works better, etc.Quote from: wraper on October 12, 2017, 06:58:56 pm

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 07:07:02 pm
content: Quote from: alexg on October 12, 2017, 07:03:02 pmAnd why soft start is supposed to function when circuit already started? Read the words "soft start", not soft something in the middle of doing something stupid. In any case switching feedback loop in the middle of operation is not something you should do in the first place.

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 07:16:37 pm
content: This chip has short circuit protection. However with suboptimal PCB design it may fail functioning properly. Most likely inductor gets saturated, and IC dies. BTW what inductor you are using?

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 07:17:11 pm
content: What about this sequence then: set EN to low to shot it down, change feedback loop to different voltage, set EN high?Quote from: wraper on October 12, 2017, 07:07:02 pm

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 07:19:11 pm
content: I will look it up later once at home, off the memory, 1.2uH, about 2A saturation current, 30mOhm.Quote from: wraper on October 12, 2017, 07:16:37 pm

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 07:21:28 pm
content: Quote from: alexg on October 12, 2017, 07:19:11 pmSo you basically make a short circuit on the output while switching feedback loop and use inductor which saturates at the short circuit protection current.

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 07:26:26 pm
content: Quote from: alexg on October 12, 2017, 07:17:11 pmThat would be the best. If this is not allowed, you should implement soft voltage ramp in the feedback loop.

author: alexg
author_rank: Regular Contributor
timestamp: October 12, 2017, 07:27:35 pm
content: Great, thanks for explaining, now I can look and study this issue and learn about it. Also a follow up question: if you look at 63070 regulator which is practically the same characteristics as 63060 (i posted its schematic picture below) they have FB2 pin, which serves same purpose, it is essentially a FET enabling and disabling extra resistor in the loop, kind of the same thing I was doing, right?Quote from: wraper on October 12, 2017, 07:21:28 pm

author: wraper
author_rank: Unknown
timestamp: October 12, 2017, 07:36:45 pm
content: Apparently, just a MOSFET within IC on above circuit is pretty simplified. Page 10 of datasheet shows it differently. In any case if your circuit may expierience high current through the inductor during voltage change, inductor must be beefy enough to not saturate.