Boost Converter IC Dilemma  - Page 1
Forum section: electronics / beginners

author: Danny_El90
author_rank: Newbie
timestamp: June 26, 2025, 06:06:57 pm
content: Hello everyone, I am working on a project that I want to boost from 2.00V – 3.6V DC (from different lithium batteries), to 12V, with at least a250mA60mA - 80mAoutput current (load will not be drawing a constant250mA60-80mA, but could spike up to200mA80mA).I have been looking around at different boost converter IC's but can't seem to find one that hits all my requirements: low input *startup* voltage (2.00V - 2.3V min), low quiescent current (ideally less than 200uA), and an output current capability of at least250mA, 200mA60-80mA, would be nice to have capability for the higher end.I get my hopes up when I see IC's with "Burst Mode" capabilities, but cry upon looking at the output current @ 12V with my Vin.I was looking at chips like the LTC3122, LT3124, etc. But from the dozen I've seen so far, one requirement is always missing. I am not even sure if a boost converter IC is the right call any more given the voltage input/output ratio and output current requirement. Quite unsure what to do, and what the best move is, has anyone been in a similar situation?

author: Benta
author_rank: Super Contributor
timestamp: June 26, 2025, 06:27:30 pm
content: Check ti.com

author: langwadt
author_rank: Super Contributor
timestamp: June 26, 2025, 06:38:56 pm
content: https://jlcpcb.com/partdetail/HX_hengjiaxing-HX3608/C296278https://jlcpcb.com/partdetail/Everanalog_ICdesign-EA2208LT6R/C22462489?

author: Danny_El90
author_rank: Newbie
timestamp: June 26, 2025, 07:01:18 pm
content: Quote from: langwadt on June 26, 2025, 06:38:56 pmThese look really interesting, especially EA2208L... I would love to use these however, I am restricted to only main stream component manufacturers unfortunately.Thanks for recommendation!

author: langwadt
author_rank: Super Contributor
timestamp: June 26, 2025, 08:38:33 pm
content: Quote from: Danny_El90 on June 26, 2025, 07:01:18 pmclose enough?https://ww1.microchip.com/downloads/en/DeviceDoc/MIC2288-Data-Sheet-DS20006034C.pdf

author: MariuszD
author_rank: Regular Contributor
timestamp: June 27, 2025, 07:21:37 am
content: You need an integrated circuit with a 2A switch current. MIC2288 have 1A switch. Also battery cannot have significant voltage drop at 2A

author: Danny_El90
author_rank: Newbie
timestamp: June 27, 2025, 09:26:32 am
content: Quote from: MariuszD on June 27, 2025, 07:21:37 amMy battery only has a maximum current draw of 700mA. The load will only draw current spikes up 50mA - 80mA from the 12V boost output periodically, maybe every few minutes - and maybe draw a steady 2-5mA constantly (MCU, clocks, etc).I was thinking, would if be possible to daisy chain two boost IC's? Like do it in two stages, maybe go from 2V - 5V, and 5V to 12V. Wouldn't this decrease my input current requirement? Although, I am bit concerned about efficiency using this approach.Another main concern is battery life, I aim to have it last as long as possible (it's a 12Ah battery). I suspect the average current draw might be at most 10mA.

author: mikerj
author_rank: Super Contributor
timestamp: June 27, 2025, 10:10:24 am
content: Quote from: Danny_El90 on June 27, 2025, 09:26:32 amWhat happened to the 200-250mA requirement you mentioned a couple of times? Dropping to 80mA is a big change.Quote from: Danny_El90 on June 27, 2025, 09:26:32 amIf you are planning (as implied) to power the MCU and other low voltage parts from the 12v rail then you already need a second regulator. Cascading two boost regulators (i.e. 2v->5v->12v instead of 2v>12V->5v) may well be beneficial in this case, but you'd need to work the numbers. Be aware that burst mode on switching regulators can cause electrical noise that is difficult to filter.Quote from: Danny_El90 on June 27, 2025, 09:26:32 amDoes the 12v rail need to be continuously powered or could it be powered only when required?

author: Danny_El90
author_rank: Newbie
timestamp: June 27, 2025, 12:39:59 pm
content: I have two very similar main boards. One is designed to be powered by a 3.6V battery, and the other by a 12V battery. Right now, I only have access to 3.2–3.6V batteries, so I want to design a small, separate PCB that will boost the voltage from ~3.2 - 3.6V up to 12V. This boost board will act as a voltage adapter, allowing me to use my 12V main board with the lower-voltage batteries I have now.Previously, I mistakenly assumed the current draw would be the same as when powering my 3.6V board directly from the same battery. I didn’t account for the fact that to maintain the same power output (~800mW), stepping up the voltage to 12V means the current draw on the battery side must decrease. That’s why I’ve now adjusted the expected input current draw to around 80 mA.The new boost board will only handle the voltage step-up and provide connector ports to interface with my 12V main board (which includes the MCU and other components). The boost board will always need to be active while the main board is powered, so battery life is a concern—I want to ensure the adapter is as efficient as possible and doesn't drain the battery unnecessarily, hence the low quiescent current requirement as well.Also, the reason why I have an even lower input voltage requirement (2.00V to 2.4V) is because of the reverse protection diodes I intend to put at the battery input.

author: Danny_El90
author_rank: Newbie
timestamp: June 27, 2025, 12:44:43 pm
content: On my main boards, I already have other regulators (5V, 3.3V, 1.8V) rails to supply the circuitry local to that board.The 12V on the booster board would be the main rail if that makes sense (for my 12V variant main board), and fed into my main board via connectors.

author: langwadt
author_rank: Super Contributor
timestamp: June 27, 2025, 01:18:24 pm
content: Quote from: Danny_El90 on June 27, 2025, 12:39:59 pmthat is written in very confusing way. if you need 800mW you either need ~66mA from 12V or 222mA from 3.6V,power in = power out ...

author: Danny_El90
author_rank: Newbie
timestamp: June 27, 2025, 01:53:37 pm
content: Quote from: langwadt on June 27, 2025, 01:18:24 pmYes sorry, I should have articulated myself more clearly. But this is the gist of what I am trying to achieve.

author: mariush
author_rank: Super Contributor
timestamp: June 27, 2025, 08:33:11 pm
content: Not quite ... if you want 12v at 80mA (12 x 80 = 960 mW), assume the step-up regulator is gonna be around 80% efficient converting 2-4v to 12v, so in reality you're gonna have probably closer to around 1200 mW of power draw from battery. At 2v, you're looking at around 600mA pull from battery, so the regulator's switching current should be quite high, maybe close to 2-3A or even higher.TLV61046 should work, it should be able to produce more than 100mA ,,, figure 6.1 shows it being close to 70mA output with 1.8v input.link here :https://www.digikey.com/en/products/detail/texas-instruments/TLV61046ADBVR/8133008if you want higher efficiency, have a look at something like tps61288 which will work with minimum 2.4v but will be over 90% efficienthttps://www.digikey.com/en/products/detail/texas-instruments/TPS61288LRQQR/15856656Quoteit would make more sense to use a buck-boost regulator that supports 2v ... 14v input voltage, configured to output 5vexample of buck boost .. tps630701 (2v to 16v input, fixed 5v up to 2A output) :https://www.digikey.com/en/products/detail/texas-instruments/TPS630701RNMR/6175215This way you can feed the 5v regulator with either that battery's input or the 12v battery input, without involving an extra conversion, and more losses due to conversion inefficiency. Of course a buck-boost regulator is a bit more expensive but not by much.For the 3.3v and 1.8v outputs ... you can either power them from the 5v regulator. The 1.8v regulator could be powered directly from either input (2...4v or 12v), but the 3.3v if you use a cheap step-down regulator you would need at least 3.3v input voltage to output 3.3v so you either power from regulated 5v or you accept that when the battery voltage goes below 3.3v your output will no longer be 3.3v. A buck-boost would be more expensive but accept the wider input voltage range without any complaints.