Why do i get better battery life than my calculation? - Page 1
Forum section: electronics / beginners

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 09:58:42 am
content: I've been doing electronics for a while now, but I can not get my calculations to agree with what happens in practice regarding battery life.I have made a temperature gauge (wireless) that sends temperature to a server that I can read again in an app. This is controlled with a Wemos D1 Mini (ESP8266). It is programmed to send data once a minute and the process takes approx. 5 seconds before going into "deep sleep" mode again.I have measured that it uses 200uA (micro amperes) in deep sleep and 80mA in the five seconds it is alive per. minute. I use 1 pc. 18650 battery with 2000mA capacity. (brand new battery taken from a newly purchased but cheap battery bank)Can anyone help me to calculate approx. how long will such a battery last with the consumption figures I have mentioned above?The reason I ask is because the battery has lasted as long now as I have assumed it should last, but when I measure the battery it still has 4.1V in it ...Either I have calculated completely wrong, or I have a multimeter that is lying. Thelast, however, is very unlikely ....

author: kaz911
author_rank: Super Contributor
timestamp: December 12, 2021, 10:54:20 am
content: What did your own calculation spit out ?If it is 80mA for 5 - that is then 400mA(seconds) - is that not about 0.11 mAh?<- sorry just me being absolutely idiotic.Just "reaching" here - have not had time calculate it.It runs for 80mA for 1/12th of the time. So 6.66mA "all the time"

author: NickEdison
author_rank: Contributor
timestamp: December 12, 2021, 10:57:09 am
content: It is important to realize that the capacity in your case is likely expressed in mAh; meaning how much mA it can deliver for 1 hour.The average current drawn each minute:200 * 55/60 + 80 * 5/60 = 0.2 * 55/60 + 80 * 5/60 = 6.85 The average current each minute is the same as that per hour, hence your consumption remains 6.85  (average for 1 hour). The battery life can now be estimated as follows:2000/6.85  = 291.97...  (291.97.../24 gives an expected life of approximately 12 days)The main parameters that impact the consumption above is the time it is alive and what it consumes then:Maybe it is only alive for 3 seconds or lessNot sure what else happens before it goes into "deep sleep" mode, but maybe it is in another state where it is consuming less current

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 11:25:00 am
content: Thanks for good explanationMy calculation was almost the same, but since we now know the average consumption (6.85mAh) and the battery should last around 12 days, i can tell that the battery has now last 14 days, and when i measure the voltage on the battery i get 4.11V.I do not understand why. I'm absolutely sure that my measurement is correct, but it is a chance that the units on time is +- 1 or 2 sec.I thought the battery should be almost empty now, but 4.11V is just 0.6ish volt usage....

author: RoGeorge
author_rank: Super Contributor
timestamp: December 12, 2021, 11:50:05 am
content: Are you sure about the 5 seconds activity to transmit data?That, seems a huge time to send a temperature. A single data packet would be enough, so more like 0.5s, maybe way less than that.Wire a small series resistor in series with the battery (1 ohm or smaller), and capture the voltage drop on that resistor during a full minute. If you don't have a digital oscilloscope to capture the voltage drop, maybe use a soundcard to record the voltage, then look at the results with a program like Audacity.

author: BeBuLamar
author_rank: Super Contributor
timestamp: December 12, 2021, 12:59:49 pm
content: I would expect you to get better battery life than the rated 2000mAh because your discharge rate is too low. Most Li-Ion battery would have higher mAh rating if discharged at lower rate.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 01:01:52 pm
content: Quote from: BeBuLamar on December 12, 2021, 12:59:49 pmThat can be an good explanation yes, if it's true.

author: ledtester
author_rank: Super Contributor
timestamp: December 12, 2021, 01:21:08 pm
content: Quote from: ElectricPower on December 12, 2021, 09:58:42 amBattery capacity depends on the current draw and the 2000mA rating is usually measured when drawing 100's of milliamps (like 500mA or 1A). A battery will last longer if you draw less current.To reduce transmission time you might look at this video:An Efficient WiFi Button For Home Automation Using ESP-NOW | ~68 Years From 2xAA Batterieshttps://youtu.be/iEkGRI8_txEThey got the transmission time down to 280ms.Edit: Here's an example showing how the discharge rate affects the battery capacity measurement:https://www.powerstream.com/9V-Alkaline-tests.htmIn the table note how the capacity measurement depends on the discharge rate -- e.g. 310 mAH @ 100 mA discharge rate vs. 90 mAH @ 1A discharge rate. This example uses alkaline batteries, but it is generally true for all batteries.

author: richard.cs
author_rank: Super Contributor
timestamp: December 12, 2021, 02:33:51 pm
content: For what it's worth it's possible to measure and report temperature way quicker than that, and to achieve lower sleep currents. For example at work we have a sensor measuring and transmitting every 30 seconds with a 7 year life on an AA. I've not looked at all at the components you're using but in general I would expect you to be able to get the sleep current down to a couple of microamps, and the measure and TX to a few tens of milliseconds. 80 mA during TX sounds kinda high too. Getting to the lowest power sleep modes often takes some careful datasheet reading and quite a bit of care.Of course if 12 days meets your requirements that's cool, well done for making something that works. :-) If you'd like to improve it or shrink the battery then there are probably options to do so.When working out battery lifetimes I prefer to work entirely in charge units (Coulmbs or mAh) but some people like to calculate it as an average current which is equivalent.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 04:23:26 pm
content: I see.So all my calculation regards to battery life do not give me any good data at all. That was new for me. Measured voltage to day (17 days in use) and the voltage shows 3.98V.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 12, 2021, 05:48:36 pm
content: Many wrong answers, real reason is simple: multimeter was never supposed to measure spiky current correctly. There's a reason why it says DC.For measuring sleep time leakage (which is pretty much DC), it's OK, but that 80mA peak could be actually anything. It's a complex waveform, you need the integral (or average).You really need a scope, and a device such as Dave's µCurrent, or your own similar shunt amplifier circuit. You may be just barely be able to it with just scope, with relatively high-value shunt resistor.As others have said, 5 seconds just to send temperature is massive waste of time. Even if the current actually isn't 80mA, it's still something quite high, so fix this and it'll last even longer. You should be able to do this in 50ms easily.

author: bdunham7
author_rank: Super Contributor
timestamp: December 12, 2021, 06:21:38 pm
content: Quote from: Siwastaja on December 12, 2021, 05:48:36 pmAnd among those wrong answers--Peukert's effect doesn't apply to Li-Ion cells.Given his results, I'd guess that his device isn't actually staying on for 5 seconds. I've actually done something like this to look at battery draw where I used a 1-ohm resistor as a shunt (which is probably what his DMM has anyway) and then measured across it with two different channels of a DSO so that I could set them at different scales to see the baseline and the high points. This is probably marginal at 200uA depending on the scope, but should capture the power-on interval fairly well and allow him to piece together a pretty good idea of total draw without using an SMU.

author: bdunham7
author_rank: Super Contributor
timestamp: December 12, 2021, 06:25:29 pm
content: Quote from: ElectricPower on December 12, 2021, 04:23:26 pmAs they say, Garbage IN, garbage OUT. Your measurements of the current draw are pretty clearly higher than actual, likely by a huge amount. So yes, your DMM is 'lying'. What exactly are you using?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 06:32:37 pm
content: Quote from: bdunham7 on December 12, 2021, 06:25:29 pmI will rather say in this case, "garbage in, refinement out"I measure input voltage (DC) and current with Brymen 869S. I know how to read, and when i measure the current it says between 60-90mA. In about 5 seconds.

author: NickEdison
author_rank: Contributor
timestamp: December 12, 2021, 06:37:19 pm
content: Quote from: Siwastaja on December 12, 2021, 05:48:36 pmThis is a good point and to do accurate measurement good hardware is definitely required. However if you would like to get a better estimate of the behavior while transmitting data, you could measure the current over a shunt with a 2nd micro controller. This will not be accurate (although you might be able to improve it significantly with an amplifier circuit) but might give you a better idea of what is really happening.If you are interested in how to, let us know and either me, or someone else will be able to help you with that.Quote from: bdunham7 on December 12, 2021, 06:25:29 pmAlthough possible I think it's more likely that a DMM is not the right tool for this task because of what Siwastaja stated.

author: bdunham7
author_rank: Super Contributor
timestamp: December 12, 2021, 06:47:12 pm
content: Quote from: ElectricPower on December 12, 2021, 06:32:37 pm60mA for 5s per minute is 5mA average. Your battery should be stone dead in 400 hours from that alone.Do you have a DSO? If so, use a 1-ohm resistor as a shunt as I posted above and have a close look at that power on cycle.. If not, try these:Look at the bargraph display--does it show a steady(ish) reading for five seconds?Use the AC+DC current mode, see if anything changes.Put a capacitor, maybe 10uF, across the multimeter.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 12, 2021, 06:49:08 pm
content: Quote from: bdunham7 on December 12, 2021, 06:21:38 pmExactly, although really high currents accompanied with not shifting low-voltage cutoff limit result in effective loss of capacity, which can be gained back by lowering LVC limit or CV discharging until current drops to datasheet conditions. In any case, this could account for maybe 10% capacity loss and is completely insignificant in such low-power gadget.Capacitor over the multimeter is a good low-cost, simple idea, while it won't make the readings accurate, it will significantly reduce the measurement error and give the right ballpark.

author: james_s
author_rank: Super Contributor
timestamp: December 12, 2021, 06:57:28 pm
content: I think the question has been pretty well answered, but only 12 days on a battery for something that is meant to be used continuously sounds grossly inadequate. For a simple wireless temperature measurement each minute you ought to be able to get *years* out of a battery. The ESP8266 is probably not the best choice here. No matter what you use, you will want to avoid linear voltage regulators, especially the cheap ones that are used on most of the various modules from China. The quiescent draw can easily be more than what the load is drawing for micropower stuff.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 12, 2021, 07:08:56 pm
content: Linear regulators are OK if you use low-Iq types. Low-Iq yet high efficiency switchers are a relatively new thing. That's OK too.The key here is to minimize that 5 seconds into something sensible. It could go down to 5ms with proper software design. 50ms should be very easy to achieve regardless of what bloat libraries are used or not.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 12, 2021, 07:15:06 pm
content: I using an HT7333 regulator with an 1000uF and an 100nF capacitor.

author: james_s
author_rank: Super Contributor
timestamp: December 12, 2021, 07:17:13 pm
content: In the small number of battery powered devices I've made I didn't use a regulator at all, I used a 3V lithium battery powering the microcontroller directly. Low quiescent regulators designed for battery powered devices do exist, but all of the cheap Arduino-like devices I've seen have cheap regulators with a high quiescent draw.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 16, 2021, 05:50:59 am
content: 16 days now, and still going.Voltage is 3.88V

author: james_s
author_rank: Super Contributor
timestamp: December 16, 2021, 08:00:15 am
content: I think you will find that the voltage will drop rather suddenly at the end. Lithium ion batteries have a very flat discharge curve with a sudden steep slope when they are depleted.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 16, 2021, 08:17:52 am
content: Quote from: james_s on December 16, 2021, 08:00:15 amMaybe. Time will show

author: SteveyG
author_rank: Unknown
timestamp: December 16, 2021, 11:05:36 am
content: How are you measuring battery SoC? Battery voltage on Lithium batteries is not a reliable measure. What load are you applying whilst trying to infer SoC?

author: Siwastaja
author_rank: Super Contributor
timestamp: December 16, 2021, 01:58:36 pm
content: Quote from: SteveyG on December 16, 2021, 11:05:36 amMaybe, but OP does not use lithium battery but a lithium ion battery, where voltage is a good enough SoC indicator. If they measure 4.1V, that means SoC is at least 90%. If they measure 4.1V under load, that just means it's even closer to 100%.For low-power devices, voltage is very reliable indicator. You can always filter out high-current peaks if you so wish.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 19, 2021, 07:18:10 am
content: 19 days and still going. Battery voltage is now 3.85V

author: cemelec
author_rank: Contributor
timestamp: December 19, 2021, 09:04:26 am
content: Battery capacity is rated at a given discharge rate, e.g. a 10Ah cell may be rated as 1Afor 10 hours.That same cell will not give 2A for 5 hours, or 10A for 1 hour. Indeed at 10A discharge you might only get 5-6 Ah.The converse is true, at a very low discharge rate (as in your case) you will get significantly more than the rated capacity, as evidenced by your experience.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 19, 2021, 11:16:10 am
content: Quote from: cemelec on December 19, 2021, 09:04:26 amComplete BS. Does not apply to li-ion cells. You are describing the Peukert law that models the behavior of lead acid cells. No one uses lead-acid cells on IoT gadgets. This was already discussed, so it would be nice if you read the thread before wasting our time, thank you in advance.

author: cemelec
author_rank: Contributor
timestamp: December 19, 2021, 05:16:18 pm
content: Dear SiwatajaMy apologies, I didn't realise you owned this thread, or indeed this forum, otherwise I would have shown more servility.You're still wrong mind.

author: Nusa
author_rank: Super Contributor
timestamp: December 19, 2021, 05:39:16 pm
content: Quote from: Siwastaja on December 19, 2021, 11:16:10 amAs usual, the truth is somewhere in the middle:https://iopscience.iop.org/article/10.1149/1945-7111/abad69

author: bdunham7
author_rank: Super Contributor
timestamp: December 19, 2021, 06:12:54 pm
content: Quote from: Nusa on December 19, 2021, 05:39:16 pmThat paper contradicts every model, both theoretical and demonstrated, of Li-Ion battery operation that I know of. I didn't scour every detail, but it seems to me that the differences in capacity that they are trying to fit to a Peukert-like equation are simply attributable to the remaining charge left in the cell when it reaches some arbitrary cutoff voltage at some rate of discharge. Peukert's Law and the Peukert effect are a description of an identifiable effect that occurs in electrochemical batteries using a redox reaction like the common lead/acid/lead oxide variety. Li-Ion batteries work on a much different principle and can't really tolerate 'slip', or mismatch in the charge-discharge amounts measured in coulombs. For much the same reason they cannot tolerate float charges.Here's a telling excerpt from the paper:In conclusion, we note one important fact. For any batteries, a high enough current always exists, at which already at the battery discharge start, on its terminals, the voltage will be lower of the lower cutoff value, i.e. for such high current, the battery released capacity will be equal to zero. However at the present time, this fact is taken into consideration by neither the classical Peukert's equation nor any of its generalizations.With a lead-acid battery you can directly and fairly definitively measure it's charge with a hydrometer, or by other means. With an Li-Ion battery you do that by letting its temperature stabilize and them measure the OCV and cell temp. I can pretty much guarantee you that all of the cells that they tested with different charge rates had different charges remaining when measured this way, and thus more current could have been recovered at a lower rate--IOW the cell wasn't really discharged. And that could be proven by taking two cells subject to the two different rates and then comparing them--they won't be in the same state. You do get a bit of a similar effect with lead-acid batteries, but with those, you can have two cells arrive at the same absolute state of discharge, measured in the same way, after providing different amounts of ampere-hours at different rates. AFAIK, that won't happen to any degree of significance with Li-Ion cells unless you damage them.That whole paper looks like a PhD-level attempt to hammer a square peg into a round hole. I'd like to see what Jeff Dahn would say about it. Perhaps I'm wrong.

author: wizard69
author_rank: Super Contributor
timestamp: December 19, 2021, 08:00:40 pm
content: Quote from: ElectricPower on December 12, 2021, 04:23:26 pmYou should be careful here because many Lithium chemistries have very flat voltage curve while discharging. You will want to know what the specific voltage is to consider a battery fully discharged. Beyond that temperature itself is a factor. Also it is not impossible for a battery to be better than speced out of the box. Also there is a drop off in performance after a few charge/discharge cycles before the battery starts to age at a more linear rate. I really wouldn't assume that the battery will be doing this well after some usage.edit:I'm likely bumping heads with people that know a lot more than I do about batteries but on the other hand it is pretty clear that not all chemistries are the same. In fact with everybody and their brother going after the EV market, batteries are becoming "interesting" to say the least. If you want to know what is going on with a battery it is a good idea to have the data sheet and manufactures application notes, for a specific model.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 19, 2021, 08:06:27 pm
content: Quote from: Nusa on December 19, 2021, 05:39:16 pmNo, the truth is "not in the middle", the truth is the truth, you know, the actual physical reality. Truth does not follow social games, we don't "make deals" in order to be nice to each other. Giving correct information is my only priority, everything that works against it, I ignore or fight against.Yes, it's very common mistake to mis-apply the Peukert formula due to small observable capacity difference caused by the combination of Low-Voltage Cutoff (which is always arbitrary, you have to usesomething) and equivalent DC resistance of the cell. In any case, the effect on capacity is small, a few %. The question is only about whether the exponential Peukert law is a good model for li-ion, and the answer is no.If you choose LVC poorly, or the application demands high currents and can't work down to lower voltages (which is of poor design, similar to crappy gadgets which refuse to work with alkaline cells at 1.2V, causing unused charge to remain), then yes, you can force-fit the observations into the Peukert law andcalculatea non-unity Peukert exponent. This does not make the Peukert law a suitable tool for analyzing li-ion cells, which is why it is never normally used. Fitting an exponential formula into a fairly linear system makes little sense.Nothing prevents you from:* Using low enough LVC, or* Using dynamic LVC, or* CC-CV discharging the cellto fully utilize the charge in the li-ion cell. Done that way, calculated Peukert exponent will be very very close to 1.Finally, li-ion cells are rated by the manufacturer typically at 0.5C or 0.2C discharge down to typically 2.5V or 2.8V. There is very very little extra to be gained, no matter how low current you are using. Maybe a few %, unless you take a risk of overdischarging the cell.Yes, some of us actually have a scientist background, some of us have published papers in peer reviewed journals, or reviewed papers by others so know the process, and some of us have actually made some career on analyzing and testing li-ion cells. I'm really not the only one here who qualifies on all these. Those who have an ego issue and need to try to shine on something they have not the slightest clue about, maybe think about what damage it causes to the forum and the projects of others? The options for us really are stay quiet and let the wrong information shine, or correct the errors with the risk of the clueless egoists starting to fight back, derailing the thread. What would you choose if I spewed some BS about somethingyouknow very well, and you think is important? Think about this for a while and forget about your pride, it's worthless, mine at least is and I love to be shown wrong. Thank you in advance.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 19, 2021, 08:08:00 pm
content: Quote from: wizard69 on December 19, 2021, 08:00:40 pmYou can guess from the 4.2V voltage mentioned that this is a typical consumer-type LCO, NCA or NMC cell. Their curves are not that flat, voltage gives fairly good estimate of SoC. If it reads 4.1V, it has to be around 90%, it just simply can't be something like 70% for example.

author: wizard69
author_rank: Super Contributor
timestamp: December 19, 2021, 08:24:13 pm
content: Quote from: Siwastaja on December 19, 2021, 08:08:00 pmWell it is flat compared to most alternative batteries. My concern is that we really would be better off knowing specifically what that battery is. That is its manufacture, type and other information. The reality is this is a rapidly changing field with fairly constant tweaking to chemistries. It seems like almost yearly that we have higher energy batteries in the same form factor.

author: bdunham7
author_rank: Super Contributor
timestamp: December 19, 2021, 08:24:20 pm
content: Quote from: Siwastaja on December 19, 2021, 08:06:27 pmThe paper he cited was published in the Journal of the Electrochemical Society (which I know nothing about) and the primary author is a professor at Don State Polytechnic, Rostov-on-Don, Russia. So I could certainly see how someone might find it credible and I'm still wondering if perhaps I missed the point as opposed to them being wrong. I didn't comb through it for details because I think they are obscuring their experimental data a bit, or at least I couldn't find it easily. I'm inferring that they used the manufacturer's listed LCV in each case and I don't see any of the very careful measurement of total charge and discharge currents and energy as I have in other studies.https://orcid.org/0000-0002-1613-8659

author: bdunham7
author_rank: Super Contributor
timestamp: December 19, 2021, 08:36:32 pm
content: Quote from: Siwastaja on December 19, 2021, 08:08:00 pmCertainly not flat like a Lithium-Iron or Thionyl primary cell, but to be fair, a lot of Li-ion cells (especially NMC in my limited experience) have a fairly wide area in the middle of the discharge curve where you need to know the temperature of the cell quite accurately to determine the SOC from the OCV. My car's BMS uses quite a few tricks to accurately indicate charge in this area, it certainly doesn't rely on voltage alone. At the ends, no doubt-->4.1 is pretty full and <3.5 is pretty dead.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 20, 2021, 08:49:28 am
content: Quote from: wizard69 on December 19, 2021, 08:24:13 pmLead acid, NiCD, NiMH are all flatter. LFP is also flatter. So are lithium primary cells.Alkaline primary cells are pretty much the only battery tech that comes in my mind which is less flat (or quite similar) to consumer gadget LCO/NCA li-ion.Yes, some NMCs are flatter in the middle than classic consumer LCOs, making it hard to make distinction between say 45% and 55% SoC. But rarely a typical IoT device needs this level of accuracy. Even laptop computers with sophisticated Coulomb counters (which finally kind of work) tend to sacrifice the 1% resolution and show some 3- or 4-level bar graph in the UI, because people are fine with this.In general, I suggest people actually measure the curves, or look them up because someone else has done that already. lygte-info.dk has tested a lot of li-ion cells.For example, look at the classic Samsung INR18650-29E (thousands and thousands have gone through my hands into battery systems):https://lygte-info.dk/review/batteries2012/Samsung%20INR18650-29E%202900mAh%20%28Blue%29%20UK.htmlAs you can see, first the capacity seems to drop a bit as a function of discharge current, but then as you expect a total collapse per the exponential Peukert Law, the opposite happens, the cells seem to gainmorecapacity at highest discharge currents! How is that possible?It's possible because the apparent loss of capacity is actually about the increasing voltage drop dV = dI * R_esr with increasing I; this causes the test to be terminated early, as the cell hits the arbitrary 2.8V stop condition (datasheet condition is given as 2.5V, mind). In highest current test, however, the cell heats up considerably because of I^2*R losses (where the square I^2 term swamps the linear dV = dI*R). High temperature then lowers the heavily temperature-dependent R_esr of the cell, so the voltage drop is actually lower at the highest discharge currents, allowing the test to run for longer.All of this can be avoided by using dynamic (current-dependent) low-voltage cutoff point. Or, just accepting small (in this lygte-info test, 7%) differences in capacity as a function of discharge current.

author: RoGeorge
author_rank: Super Contributor
timestamp: December 20, 2021, 01:40:50 pm
content: Quote from: Siwastaja on December 20, 2021, 08:49:28 amVery useful measurements and charts, thank you.I'm trying to sort out my pile of recheargeables collected with time. Apart from a few LiIon, most of them are 1.2V NiMH AA/AAA, a few NiCd, and a few battery packs from the handheld tools I have (either LiIon or NiCd). In total probably less than 50-100 batteries, but still too many to measure them one by one. I'm building a multichannel voltage datalogger as we speak, so to discharge them in batches of 10-20 or so, and any good data would help with the design ant the software.Do you happen to have any link with similar measurements/info, but for 1.2V NiMH and/or NiCd cells, please?

author: tunk
author_rank: Super Contributor
timestamp: December 20, 2021, 03:27:36 pm
content: HKJ has some tests of NiMh AAs:https://lygte-info.dk/review/batteries2012/Eneloop%20AA%20BK-3MCCE%201900mAh%20(White)%20UK.htmlhttps://lygte-info.dk/review/batteries2012/Eneloop%20AA%20BK-3MCCE%201900mAh%20(White)%202019%20UK.htmlhttps://lygte-info.dk/review/batteries2012/CommonAAcomparator.php

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 23, 2021, 01:28:01 am
content: 23 days:

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 26, 2021, 08:57:15 am
content: 26 days has gone, and the battery is now dead. 2.72V and the TP4056 has shut the battery off.26 days is still impressive, but yes, from 3.7 ish volt and down it falled wery quick. From 3.7V to 2.72 in 2 days.

author: bdunham7
author_rank: Super Contributor
timestamp: December 26, 2021, 02:59:48 pm
content: Quote from: ElectricPower on December 26, 2021, 08:57:15 amWhat is the ambient temperature in your lab?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: December 26, 2021, 05:27:17 pm
content: Quote from: bdunham7 on December 26, 2021, 02:59:48 pmAbout 15 degrees celcius.

author: bdunham7
author_rank: Super Contributor
timestamp: December 26, 2021, 11:38:40 pm
content: Quote from: ElectricPower on December 26, 2021, 05:27:17 pmThat's interesting, I would expect the 'edge of the cliff' to be a bit under 3.6V at very small draws unless you have an unusual cell chemistry, like the ones marketed as 3.8 or 3.85 nominal voltage. But I've only seen those in pouch cells. But as you've noticed, wherever the 'cliff' is, you don't recover much additional energy going over it.

author: mayor
author_rank: Regular Contributor
timestamp: December 27, 2021, 12:33:16 pm
content: Hi OP,this might be an interesting read (shameless plug!):https://hackaday.io/project/182323-long-range-thermocouple-temp-sensor

author: Siwastaja
author_rank: Super Contributor
timestamp: December 27, 2021, 01:39:18 pm
content: Yeah, I'm a bit surprised it fell off so quickly below 3.60V - 3.60V is usually around 40% SoC -, but that said, curves can be quite different between products.Is it reading 2.72V even with load disconnected, or just during peaks?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: January 30, 2022, 10:03:52 pm
content: I tried with an another 18650, and i get 29 days lifetime. It was an 2.500mA noname battery from an bran new but sheap power bank.