Adjustable Buck Boost Converter - Page 1
Forum section: electronics / beginners

author: Menxiu
author_rank: Contributor
timestamp: November 09, 2017, 12:46:26 pm
content: I'm trying to design adjustable buck boost converted powered by USB-C, the output voltage i'm looking for is 12V.So far I've decided on using the LT1370.Questions:Is it possible to use USB-C to power this supply and how would i do that?How can i design it so it is adjustable? (I've read into some parts of this and altering the feedback resistors is an option)Is it possible to have LCDs to display the current and voltage output?

author: mariush
author_rank: Super Contributor
timestamp: November 09, 2017, 01:36:51 pm
content: If you want it to be step-up only (from 5v up to 12v), you can use the example circuit on the first page of the datasheet.If you want to be able to set up a voltage lower than 5v (for example from 1.25v up to 12v), then you need to use the regulator in SEPIC mode, see example circuit on page 13 but you may need to adjust the inductor value for the narrower input voltage range and also note that it's a special inductor with two coils required, not a regulator inductor.The datasheet contains the instructions about how to choose various components.The output voltage is configured as explained in the datasheet, through two resistors. You can replace one of the resistors with a potentiometer, or you can put a potentiometer in series with the resistor to change the total resistance on-the-fly.One of the resistors (R2) should be a value below 7 kohm, they recommend 6.19 kOhm, but you can probably go with a more standard value like 4.7 kOhm or 5.6 kOhmThe output voltage is basically defined by the formulas (page 7) : Vout = Vreference x ( 1 + R1 / R2) where Vreference is 1.245v OR R1 = R2 x (Vout / 1.245 -1)So let's say you want to be able to adjust voltage between 1.5v and 12v and you pick 5.6 kohm for the R2 resistor ... then your R1 must be minimum R1 = 5600 x ( 1.5 / 1.245 - 1) = 1147 ohm and for 12v your resistor would have to be R1 = 5600 x ( 12/1.245 -1) = 48376 ohmSo the best solution would be to replace R1 with a 1000 ohm resistor in series with a 47k or 50k potentiometer, this way you'd have R1 between 1000 ohm (which sets output voltage to 1.47v) and 48-51kOhm (which sets output voltage to 12.58v)You can buy one of those cheap seven segment display boards from eBay if you want to see the output voltage.As for usb type c cable ... you can simply google for the pinout and connect your board to +5v and ground wires of that cable, nothing special.Here you go :https://en.wikipedia.org/wiki/USB-C#Connector_pinoutsVbus is your +5v , GND is your ground... all the other pins (wires, whatever) you can leave unconnected.ps. you don't have to use exactly the same parts as the ones in the example circuits. If you can't find one just look for its datasheet and determine the most important "parameters" for that part and search at distributors like Digikey, Mouser, Farnell etc components with parameters very close to those of that part.Or, read the notes in the datasheet where they explain why they chose that part, or what properties of that part are important, what they looked for when they made the selection.

author: Menxiu
author_rank: Contributor
timestamp: November 20, 2017, 01:46:22 pm
content: So I've decided to go with the SEPIC mode, I'm wondering if this coupled inductor would do the job?http://uk.rs-online.com/web/p/wire-wound-surface-mount-inductors/7445182/Also, is USB-C the right choice for this or would an older gen. of USB will do?Current reasoning for USB-C = Max voltage and current rating of 20V and 5A .Thank you for the previous answer.

author: mariush
author_rank: Super Contributor
timestamp: November 20, 2017, 08:48:14 pm
content: I don't know if that inductor will work, you'll have to read the datasheet for the chip you decide to use and see if it's technical parameters match or exceed what the regulator chip requires from an inductor.I think USB Type C 's default voltage is 5v at 0.9A or 4.5w. At least that's the voltage for plain usb 3.0 . Computer motherboards may give you up to 2A without complaining (because there's usually 2 or 4 connectors grouped together behind a 2A resettable fuse or something like that).Anything higher than 5v (12v or 20v) must be negotiated - you must have a chip connected to the usb data wires, and the chip must "talk" to the other end of the cable and say "i want 12v" or "i want 20v" ... and there's no guarantee the other side will give you (for example majority of computer motherboards don't have usb chips smart enough or the circuitry to switch from 5v to 12v on the fly, and 12v is readily available on motherboards.Also understand that this is different than what phone chargers do. Some phone chargers have Quickcharge, which is a different protocol and again, some chip needs to talk to the charger to enable voltages higher than 5v ... it's not the same as usb does with voltages. Seehttps://en.wikipedia.org/wiki/Quick_Charge( Quickcharge 4 is supposed to be compatible with usb power deliver as wikipedia says but still that's far away)

author: David Hess
author_rank: Super Contributor
timestamp: November 21, 2017, 06:52:02 pm
content: A coupled inductor has advantages however two separate inductors may be used.http://www.ti.com/pwr-contrib-006-mc2-en

author: ogden
author_rank: Super Contributor
timestamp: November 21, 2017, 07:08:11 pm
content: I would pick buck-boost regulator. There's many:http://www.linear.com/products/Internal_Power_Switch_Buck-BoostOutput voltage can adjusted by altering feedback voltage circuit using DAC or just potentiometer. First hit that shows idea:http://www.microchip.com/forums/m688260.aspxIn case of microcontroller you better use regulator with enable pin because you want to set proper feedback correctionbeforeregulator is enabled.