Lithium Ion battery and microcontroller - Page 1
Forum section: electronics / general

author: chinoPR
author_rank: Newbie
timestamp: September 06, 2021, 04:00:38 pm
content: Hello community, I am writing about a problem that I have in a new project.I have to power a microcontroller with Lithium Ion battery, I must lower the voltage to 3.3V without increasing the current consumption. Unfortunately in Argentina it is difficult to get certain components, it would be very helpful if someone with experience has some recommendations ... thank you very much

author: agehall
author_rank: Frequent Contributor
timestamp: September 06, 2021, 05:15:19 pm
content: You might want to provide some more details. What components are you looking to use? What kind of current consumption are you looking for?The answer to your current question is pretty much “how long is a piece of string”…

author: nctnico
author_rank: Super Contributor
timestamp: September 06, 2021, 06:01:44 pm
content: Quote from: chinoPR on September 06, 2021, 04:00:38 pmFor such a small drop you better use an LDO; don't bother with a switching regulator. The trick is to find an LDO which is stable with ceramic capacitors and has a low supply current.

author: chinoPR
author_rank: Newbie
timestamp: September 06, 2021, 07:06:12 pm
content: I am using an XMEGA microcontroller with a SigFox transmitter. I transmit only once a day (I read digital inputs and put the uc to sleep) and the client asks me to use a lithium ion battery. And I thought of an LDO unfortunately there is no low current in the local market. Beyond current calculations, I am interested in finding a method to lower the voltage without having to increase consumption significantly.

author: SiliconWizard
author_rank: Super Contributor
timestamp: September 06, 2021, 07:15:15 pm
content: Quote from: chinoPR on September 06, 2021, 04:00:38 pmFirst thing before picking a solution is to get your requirements right. For starters:* How much current your complete circuit is going to draw (min, max, average) from the 3.3V power rail?* Do you really need to power your circuit at exactly 3.3V?For the second question, it's an important one. Many people think they absolutely have to use 3.3V, but this is rarely the case. Many ICs rated for 3.3V operation have a relatively wide operating range. Allowing your design to operate at a different voltage opens a number of new possibilities.

author: olkipukki
author_rank: Frequent Contributor
timestamp: September 07, 2021, 03:30:28 am
content: Quote from: chinoPR on September 06, 2021, 04:00:38 pmJust try to evaluate some designs using TI WEBENCH and find out what you will able to source...https://webench.ti.com/power-designer/switching-regulator?update=1&powerSupply=0

author: asmi
author_rank: Super Contributor
timestamp: September 07, 2021, 01:52:52 pm
content: Quote from: nctnico on September 06, 2021, 06:01:44 pmThis is a bad idea, not only does LDO waste power for no good reason, it will also not utilize a full battery capacity if you want 3.3 V output. Even the best LDO require 0.2-0.3 V minimum dropout, so you will have a cutoff voltage of 3.5 - 3.6 V, which is about half of battery's capacity.The most efficient way to take advantage of full battery capacity while getting a stable system voltage is to use a buck-boost converter(hereis the one example, but there are a ton of others of course). These things tend to have very high efficiency (typically 90+%) so you minimize power losses due to heat.

author: kripton2035
author_rank: Super Contributor
timestamp: September 07, 2021, 02:39:50 pm
content: HT7333 LDO has a mean 90mV dropout...

author: Siwastaja
author_rank: Super Contributor
timestamp: September 07, 2021, 03:53:05 pm
content: For very low-power (peak power is important! Capacitors can supply some sub millisecond only; bear this in mind if the device uses a radio modem or other high current peaky loads), LDO linear regulator isn't necessarily a bad choice at all. With low output current, dropouts down to around 100mV are specified, and the open-circuit voltage of an empty cell is some 3.4V, so there is little or no wasted capacity at all. The reason why cell datasheets specify discharge cutoff like 2.5V or 2.8V is that cell DC resistance rises sharply when the cell is getting empty, causing increasing voltage drop when there is significant load current (like C/2 or C/5 as typically specified in cell datasheets). But for low current projects that need only maybe some C/20 peak current, 3.4V cutoff is acceptable and maybe loses only some 5-10% of capacity, definitely not "half".In low power devices, efficiency is dominated by quiescent current draw anyway and it's hard to beat some 70% real-world efficiency of an LDO linear regulator with a switcher.

author: mariush
author_rank: Super Contributor
timestamp: September 07, 2021, 04:08:55 pm
content: Quote from: chinoPR on September 06, 2021, 07:06:12 pmIf the XMEGA microcontroller can work up to 4.25v or whatever, run it directly from the battery.Use a linear regulator with an ENABLE pin or SHUTDOWN pin to keep the transmitter off until it's needed - if necessary use some voltage translators on your data wires between the transmitter and the micro (but a lot of devices have data inputs that are 5v tolerant so check datasheet)If you only need to transmit once a day then you don't really care about how efficient the voltage conversion is for those few seconds, just get a linear regulator with low dropout voltage.You could go a step further and monitor the battery voltage and once it goes below let's say 3.5v then your micro could turn on a mosfet that overrides your regulator and powers the transmitter directly (any 3.3v device should have an absolute maximum, like 3.6v for example an since the battery is already at 3.5v and most likely dropping a bit with the current load as you transmit... )

author: nctnico
author_rank: Super Contributor
timestamp: September 07, 2021, 05:47:17 pm
content: Very simple solution could be a good old fashioned parallel regulator using a 3.6V zener diode with a reasonably low leakage. There is not much energy in a Li-ion battery when it is over 4.2V so burning that away doesn't matter much. Once the voltage drops enough, the zener stops drawing current.

author: thm_w
author_rank: Super Contributor
timestamp: September 07, 2021, 11:35:59 pm
content: Quote from: asmi on September 07, 2021, 01:52:52 pmIts a $5 chip, chances are OP cannot acquire it easily. Most likely OP could run off of 3.0V as well, and you'll easily find a 3V reg with say 0.1V drop across it, using 95%+ capacity, and average efficiency something around 80%. Not far off from that IC (which is 80-95% depending on voltage + load).Of course with the complete lack of details they have provided, no conclusion can be made here.

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 08, 2021, 02:58:18 am
content: Use a LiFePO4 battery and its operating voltage range is pretty much exactly right for 3.3V.

author: nctnico
author_rank: Super Contributor
timestamp: September 08, 2021, 07:36:42 am
content: Quote from: asmi on September 07, 2021, 01:52:52 pmBetter do the math first! That buck-boost chip is using at least 30uA (up to 60uA and not including losses in other parts) versus a low power LDO which can easely go sub 2uA worst case (30 times better than your solution!). Also drop-out voltage is not a big problem at low currents like the few mA needed to power a microcontroller. Since most Li-ion batteries spend most of their time at around 3.6V an LDO dissipates next to nothing.

author: asmi
author_rank: Super Contributor
timestamp: September 08, 2021, 02:41:21 pm
content: Quote from: nctnico on September 08, 2021, 07:36:42 amISL91110IR's shutdown current is only 1uA max, 0.05 uA typical, which is much better than what you say LDO have. There are better LDOs of course, but the same can be said about buck-boost ICs, the one I've mentioned is just an example I personally used, and it provides 2 Amps of sustained current with bursts of 3 Amps. If you need less current, you can pick other cheaper devices.Quote from: nctnico on September 08, 2021, 07:36:42 amA cutoff voltage for typical Lipo battery is about 3 V, so at 3.6 V you leave about half of it's capacity on the table.

author: olkipukki
author_rank: Frequent Contributor
timestamp: September 08, 2021, 07:40:06 pm
content: The buck-boosts are getting better and more attention last few years for OP's related applications, for example, TPS63900 is<100nA quiescent.I doubt more than 50-100mA burst is requires, but all these relative to what OP's can source, buy and handle...

author: thm_w
author_rank: Super Contributor
timestamp: September 09, 2021, 10:51:09 pm
content: Quote from: asmi on September 08, 2021, 02:41:21 pmHow is using shutdown current valid when the microcontroller should be running?If that is acceptable then you can use an LDO with shutdown pin that would have similarly low currents.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: September 09, 2021, 11:21:12 pm
content: Plenty of LDOs that are suitable, e.g. MIC5504 160mV dropout @ 300mA , 38uA quiescent.Also don't forget that a lot of nominally 3.3v devices will run quite happily at 3.0 or even 2.7V, so using a lower voltage LDO is also an option.

author: bhave
author_rank: Contributor
timestamp: September 10, 2021, 01:24:08 pm
content: I run an MCP1700T LDO at 2.8V off a 3.7V nominal lithium cell with great success. Just remember the leakage of every component on the system can make an impact. My setup with the LDO, a SAMD10 MCU, various sensors, and bluetooth radio lasts over a year on a single 18650 cell.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 22, 2021, 03:45:35 pm
content: An alternative method with very simple components is to just put a few diodes in series with your battery, and combine that with an P-channel MOSfet to short the diodes one your battery is half empty.You can also do this with an N-channel MOSfet if you put the diodes in the GND lead.Be aware though that you have to use a fet with the right UGS. Some are over 4V which obviously will not work, but if it's too low then it may open the FET and short the diodes before it should.For some battery powered operation a 30uA quiescent current of an LDO is no big deal, for other applications even 1uA can be too much. There are quite a lot of microcontrollers that can have deep sleep modes of 100nA or less, and some can even run on 100nA