My first KiCAD PCB: Boost converter - Page 1
Forum section: electronics / beginners

author: kevin1024
author_rank: Newbie
timestamp: August 27, 2014, 08:24:42 am
content: Hi! I did the "getting to blinky" KiCAD tutorial and got overconfident and decided I'd like to create my first PCB. I'd like to build a switching power supply that will let me use a 3.5V lipoly battery to power a 12V 120mm computer fan.I looked around quite a bit at different datasheets and selected the LT1308B:http://cds.linear.com/docs/en/datasheet/1308abfb.pdfPage 20 of the datasheet has pretty much exactly the circuit I'd like to build, so I'm pretty much ripping it off, but adding a couple things to the example application circuit:1. A switch to let me shut down the boost converter (using the SHDN pin)2. A low-battery indicator LEDHere is the schematic I came up with:You can also see the full KiCAD files here:https://github.com/kevin1024/fanboostI'm looking for general advice / review, and have a couple specific questions:1. In the example application on page 20 of the LT1308B datasheet, it says the output is 12v 300mA. I was hoping for output more around 400mA, how do they get this figure?2. I'm using the LBO pin for the low battery detector. It's an "open collector", and I think the way I have it wired up, the LED will stay on until the battery is low. I think I'd rather have this work the opposite way (the LED lights up when the battery is low). Or maybe even have 2 LEDs, one red and one green. I think I have to use a transistor for this, does that seem right?3. Should I just wire the battery in with a SPST switch instead of using the SHDN pin? (this just occurred to me)

author: ConnorGames
author_rank: Regular Contributor
timestamp: August 27, 2014, 09:00:18 pm
content: 1) Probably a combination of capacitor size and diode and inductor inductor current ratings, but I'm not really sure. Looks like the chip can probably do 400mA with the right components2) Sounds correct3) Switches that can handle >1A are a lot more expensive than switches that only need to switch signal current. I would use the SHDN pin if you are not aiming for ultra-low power consumption when the converter is off.

author: kevin1024
author_rank: Newbie
timestamp: August 27, 2014, 10:36:41 pm
content: > Probably a combination of capacitor size and diode and inductor inductor current ratingsIs there a guide I could use to select these? I have to admit I'm a bit lost here. They list some suggested components in the datasheet, but I don't see any mention of current ratings other than "The inductor must be able to handle current of 2A steady-steate, as well as support transient and start-up current over 3A without inductance decreasing by more than 50% to 60%. Does this have some impact on the actual output current?> 2) Sounds correctI think I might design something like this:> 3) Switches that can handle >1A are a lot more expensive than switches that only need to switch signal current.Cool, that makes sense. One problem I might run into though is if I use the SHDN pin, the LBO output will be low-impedance when the IC is off. That means my LBO LED will be on, even if the battery isn't low. Hmmm....Also, should the VCC for the LBO LED come directly from the battery or should I use the output from my boost converter to power the LED?

author: kevin1024
author_rank: Newbie
timestamp: August 28, 2014, 07:43:11 am
content: Well, I just spent some time messing around in LTspice and I seem to have created a simulation of my power supply. As far as I can tell this looks pretty good:The current and voltage are stabilizing at 12V and 400mA. I put a 30 ohm load on there, which makes it draw the 400mA.It looks like it's drawing between 1 and 1.9 amps from the battery, which doesn't seem too bad. That means (3.75V)(((1.9A + 1 A) / 2)) = 5.4375 watts drawn, and (12V)(.4A) = 4.8 watts output. So that's ((4.8W)/(5.4375))(100) =~ 88% efficiency.... right? Is that good?Is there something I could be doing to optimize this?