EEVblog #772 - How To Calculate Wasted Battery Capacity - Page 1
Forum section: electronics / general

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 04:43:45 am
content: In this tutorial Dave explains how to precisely measure and calculate the remaining (or wasted) energy capacity in a battery using the graphical analysis technique using a spreadsheet.The differences between constant power and constant current loads and when to use them is discussed.Also, the importance of ESR and measuring the battery voltage under load is demonstrated.This is a particularly relevant to the Batteriser product and proof is provided that shows that even over the entire current range, the wasted energy in a battery can be no better than a few 10's of percent for a 1.1V cutout voltage. Blowing the batteriser claim of 80% wasted energy out of the water.

author: ornea
author_rank: Unknown
timestamp: July 28, 2015, 05:16:41 am
content: Considering the importance (as Dave pointed out) of measuring the battery when loaded, how could this technique be adapted to characterise a real device with transient power demands.Would the battery measurements need to be somehow synced to the peak power demands.

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 05:51:57 am
content: So ... Batteriser can get 10-20% extra out of the battery? That soundsgoodto a lot of people! Those 10-20% extra will add up if you use a lot of batteries.You forgot to mention that Batteriser can't be 100% efficient. Itwillwaste some power. Even the best converters are only about 90% efficient (10% wasted as heat). Your real gain is only 0-10%.Does that still sound worthwhile? We didn't factor in Ohm's law yet. Ohm's law tells us that if you raise the voltage at the input of a resistive or constant current device it will use more power.eg. If the battery is putting out 1.1-1.3V under load then raising it to 1.5V will cause the device to use 15-30% more power.Add in the 10% waste due to the voltage conversion and Batteriser gives you a netlossof 25-40% in those devices.Yes, you got 10-20% from the battery but you had to throw away 25-40% to achieve that.What about the other type of device - constant power? To achieve constant power the device needs a built-in DC converter - ie.it already has a batteriser built-in! Adding another Batteriser in series just means 10% loss (assuming 90% efficiency of conversion).No matter how you look at it, Batteriser means aLOSSin overall power.The only devices it could possibly be useful for are devices with a stupidly high cutoff voltage, eg. 1.3V under load. These might exist but nobody's come up with one yet and even Batterizer has changed its claim of cutoff voltage to 1.1V (they initially started out at 1.3V). In reality most modern devices will go down to 1.0V and Batteriser will be a LOSS in those devices, ie. your batteries will last less time with a Batteriser than they would without one.nb. I've assumed 90% efficiency of the Batterizer here. In reality it's probably worse than that simply Because they're forced to use a microscopic inductor to fit it inside the product. Only time will tell how efficient it is (actual measurements) but I'm betting 90% is a generous number.

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 06:00:18 am
content: Quote from: ornea on July 28, 2015, 05:16:41 amYou basically have to connect up your device and measure it during typical operation.Quote from: ornea on July 28, 2015, 05:16:41 amYou can get a good estimation by measuring at very small time intervals. IIRC Dave started with 100,000 datapoints for the curve in the video.(And that's why they manufacture those big devices Dave was using ... to log/store/analyse all that data)If you want it 100% perfect instead of 99.5%? That's a lot more difficult.

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 06:35:27 am
content: Quote from: Fungus on July 28, 2015, 05:51:57 amI thought I did mention that? But could have forgot....In any case I've mentioned in other videos.This video not meant as a Batteriser debunking video actually, it's a tutorial on how I go that capacity graph I showed in my blog post on the Batteriser a while back.

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 06:38:07 am
content: Quote from: Fungus on July 28, 2015, 05:51:57 amThey are welcome to send me one for testing, but I doubt they'll do that because according to them I don't know enough about the subject

author: VingTor
author_rank: Contributor
timestamp: July 28, 2015, 07:02:49 am
content: Quote from: Fungus on July 28, 2015, 05:51:57 amHe did briefly mention it, but not as elaborate as you did. You had a lot of good points that should have been mentioned.I think Dave need to get a batterizer to test and teardown, then throw it away at the endAnother reason to not use this product, not sure if it has been mentioned, lets say you are using it in a product with a flash memory/eeprom or something. Such products will if they are well designed ensure that batteries are OK before writing to the memories to ensure that the memory is not corrupted by sudden power loss. I would hate if the memory card in my camera got corrupted by the use of this product. There will be little warning to low battery (if any), as the batterizer will push the voltage up until it self get a low power cutoff, at which time the voltage will drop from ok to nothing in no time... I prefer some warning and some wasted power instead of no warning and being stuck without replacement batteries in time.Most good products will also allow specifying which battery type is used, both my camera and GPS allows to specify alkaline, lithium, and NiMH battery types in its firmware. If you fail to set these settings correct, battery performance will of course be bad (actually, I think the cutoff settings in my camera is quite bad, nikon e4600, never tested to see what cutoff it has).

author: VingTor
author_rank: Contributor
timestamp: July 28, 2015, 07:08:00 am
content: Quote from: EEVblog on July 28, 2015, 06:38:07 amSomeone else could send it to you in mailbag

author: BravoV
author_rank: Super Contributor
timestamp: July 28, 2015, 07:21:49 am
content: As there are many nuts in this forum here(volt,time,resistance etc), to complement Dave's explanation video at 27:30 when the voltage at 1 Volt cut off, using graph as he did approximate the remaining capacity is approx 5%.To be exact, its at 4.487101838% capacity, straight from the table.Another interesting fact is reading straight at the table, with 10% of the remaining capacity, the voltage is rather high at between 1.08 to 1.09 Volt.

author: robbak
author_rank: Contributor
timestamp: July 28, 2015, 08:05:57 am
content: Um, Dave, your formulae to invert the 'capacity remaining' graph was nice - but it all simplifies to 1-(Eused/Etotal). Take the previous formula and prefix it with 1-().And in your graphical 'area under the curve' drawings, you need to include the area between 0.8V and 0V - it is all voltage that needs to be multiplied by the current. It does make that area bigger - but it would also make the remaining area bigger, leaving the ratios largely unchanged.

author: Unknown
author_rank: Unknown
timestamp: July 28, 2015, 08:11:12 am
content: Just wanted to make a couple of comments on this tutorial:First, Dave's calculation assumes that there is no energy available once the voltage under load reaches 0.8 volts. Presumably, if we had a load with a dropout voltage lower than 0.8, we could draw a small amount of extra energy beyond what the calculation considers to be 100% --- it's pretty clear that the discharge curve becomes close to vertical, however, so this extra energy is pretty much negligable.In the tutorial, Dave scales the graph so that the voltage axis runs from 0.8 to about 1.6, and with the corresponding energy remaining axis running from 0% to 100% --- the rescaling certainly makes for a more readable graph, but certain things that he says about this are a little misleading, I think:1. The graphical method given (horizontal line from voltage to discharge curve, vertical from discharge to percent remaining curve etc) is correct (subject to the caveat about a small amount of energy remaining even at 0.8 V), but Dave claims that the axes need to be scaled in this way to make the graphical method work. Actually, the method would work fine in any case.2. (At constant current) the energy is the area under the voltage curve, but this area should be taken from the zero volt line, not the 0.8 volt line. Because of the way the voltage axis was scaled, the "wasted energy" area that Dave drew towards the end of the video looks misleadingly small, since it neglected the rectangular part of the area between 0 and 0.8 volts --- the wasted energy is around 5%, but that "triangular" region was actually in area quite a lot smaller.

author: silvas
author_rank: Newbie
timestamp: July 28, 2015, 08:21:31 am
content: Great video. I loved the description of when to use the constant current vs. constant power curves - makes perfect sense.This post actually mentioned both things that dk27 just posted, so I'll spare the discussion. I'll still attach the drawing I made. The graph @24:02 is the clearest IMO since it is obvious why the discharge is nearly linear (the derivative varies only sligntly).

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 08:23:23 am
content: Quote from: dk27 on July 28, 2015, 08:11:12 amThat's the standard way to draw graphs, and that's why most people's initial guess at how much energy is left is skewed. How many people would have guessed 10% instead of the 4.5% that actually came out of the calculation?So Dave isn't misleading people, he's clearing things up. He's doing the math and providing a definite numerical result instead of looking at a chart and guessing.

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 09:38:42 am
content: Quote from: VingTor on July 28, 2015, 07:02:49 amI've mentioned this at length in the previous video and blog post on the topic.This was not meant to be a Batteriserdebunkingvideo, I just mentioned it a few times.

author: Unknown
author_rank: Unknown
timestamp: July 28, 2015, 09:43:24 am
content: Quote from: Fungus on July 28, 2015, 08:23:23 amMaybe people are terrible at estimating areas, but a comparison of the areas between 0.8 V and the discharge curve instead of between 0 V and the discharge curve actuallyunderestimatesthe wastage:Let V(t) be the discharge curve at constant current, monotonically descending to 0.8 at time t=t2. Suppose the dropout voltage is reached at t1<t2.The ratio Integral[V-0.8,{t,t1,t2}]/Integral[V-0.8,{t,0,t2}] is less than the ratio Integral[V,{t,t1,t2]/Integral[V,{t,0,t2}].As a really simple example, imagine t1=1 and t2=2 and that the discharge curve is linearly decreasing from 1.6 (at t=0) to 0.8 (at t=t2=2), and that we have a dropout voltage of 1.2 (ie half-way) --- these are chosen not for realism, but because it makes the areas easy to calculate since everything is just triangles and rectangles. If we measure the areas above 0.8 we find the "total" area to be 0.8 and the "wasted" area to be 0.2, which would lead us to estimate a wastage of 25%. If, instead, we measure the areas from 0, we have a total area of 2.4 and a wasted area of 1, giving an actual wastage of nearly 42%.QuoteI certainly didn't intend to suggest that Dave meant to mislead and in the greatest part the math presented in the tutorial was correct and informative --- just that the way the area under the voltage curve was depicted had the potential to be misleading.

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 09:44:27 am
content: Quote from: dk27 on July 28, 2015, 08:11:12 amThis is why the manufactures usually don't provide curves below 0.8, it is the defacto industry standard dropout voltage at which point any extra energy in the battery is consider negligible. As I mentioned in the video, there is some energy left below 0.8V for really small currents, and this is why a joule thief can flash a LED down to bugger all. Very small amount of energy, very niche applications where it can be used.I didn't want people thinking they should go to the ends of the earth to design products with a 0.5V cutout voltage, that's rarely done in the industry.

author: silvas
author_rank: Newbie
timestamp: July 28, 2015, 09:57:54 am
content: Quote from: Fungus on July 28, 2015, 08:23:23 amI think this is what dk27 was suggesting would have been nice to point out in the video.

author: EEVblog
author_rank: Unknown
timestamp: July 28, 2015, 10:36:35 am
content: Quote from: dk27 on July 28, 2015, 08:11:12 amYes actually, this is correct, it will work either way.QuoteIf there is no significant energy below 0.8V then you don't have to go to 0V. What I did was correct when you assume (as is industry standard practice to do) there is zero usable energy below 0.8V.The numbers come out exactly the same regardless of whether you include all the way down to 0V or just down to 0.8V. Those who are unsure about this need only try it.If you did have a niche application like a joule thief then you'd want to include all the way down to 0V or wherever the last gasp of usable energy is.

author: lapm
author_rank: Frequent Contributor
timestamp: July 28, 2015, 11:06:20 am
content: Quote from: EEVblog on July 28, 2015, 06:38:07 amSo can they name someone that does and what that persons qualifications are to be expert in field?

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 11:08:31 am
content: Quote from: EEVblog on July 28, 2015, 09:44:27 amIt would actually be bad for rechargeables and increase the risk of leakage in alkalines.Areallywell designed device should probably power itself down at about 1.05V.

author: stuner
author_rank: Newbie
timestamp: July 28, 2015, 11:14:41 am
content: Quote from: EEVblog on July 28, 2015, 10:36:35 amI think you're not talking about the same thing. The problem with the graphical method shown at the beginning of the video is not that you ignore all the stored energy once the battery is discharged down to 0.8V. The problem is that in your energy calculation you are neglecting the area below 0.8V while the battery voltage is still above 0.8V. The graphic silvas posted shows this very nicely:If you don't count this area, both the integrated energy that can be used as well as the remaining energy in the battery are incorrect, but especially the remaining energy is reduced to almost nothing. If you then compare the two areas you get a percentage for the remaining energy that issignificantly lowerthan the actual figure. Your spreadsheet calculations in the rest of the video actually avoid this issue because they use the total battery voltage and are still correct.

author: Unknown
author_rank: Unknown
timestamp: July 28, 2015, 11:40:33 am
content: Quote from: stuner on July 28, 2015, 11:14:41 amThanks stuner (and silvas), I think you've got this exactly right, and maybe explained it better than I did.Dave, thanks for making your videos, I've been watching the channel for a while now and have been both entertained and informed.

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 11:50:42 am
content: Quote from: silvas on July 28, 2015, 11:14:41 amEstimated time until that image appears on the Batteriser web site...?Quote from: stuner on July 28, 2015, 11:14:41 amThe point is that we'renotcomparing the areas. Comparing the areas would be wrong.Quote from: stuner on July 28, 2015, 11:14:41 amWe're doing the calculations in the time domain, and the time domain isn't linear.The spreadsheet shows a transformation to a domain where the discharge is withinhalf a bee's dickof a straight line going down to zero. This is a domain where calculations are valid.

author: Unknown
author_rank: Unknown
timestamp: July 28, 2015, 12:46:56 pm
content: Quote from: Fungus on July 28, 2015, 11:50:42 amComparing thewrongareas would be wrong. But (in the constant current case) comparing the areas is completely sensible.Power at time t is P(t)=I(t)V(t), if current is constant, then P(t) is proportional to V(t). Energy is power integrated over time, so then energy is then proportional to voltage, V(t) integrated over time. Graphically, this integral is the area under the curve V(t). So then the energy in a given time is proportional to the area under V(t) over that time. (When I say "area under the curve" I mean the area between the x-axis, the appropriate limits of integration, and the curve given by V(t) --- I'm assuming that V(t) is positive.)QuoteI'm pretty sure we didn't do any transformation out of the time domain. (Normally when someone has an alternative to the "time domain" they might mean the "frequency domain", but what do you mean?)As I understood the spreadsheet, it was simply numerically integrating the quantity P(t)=I(t)V(t) with respect to time, perhaps up to some constant factor depending on the time step Dave was using. That integration was done in the time domain.What do you mean by the time domain being not "linear"?

author: Fungus
author_rank: Super Contributor
timestamp: July 28, 2015, 12:50:58 pm
content: Quote from: dk27 on July 28, 2015, 12:46:56 pmI mean it needs integrating before you can make any sense of it.

author: Pentium100
author_rank: Frequent Contributor
timestamp: July 28, 2015, 02:59:33 pm
content: I'd like to see constant current and constant power discharge curves down to zero volts. It would be interesting to see how much more energy a germanium (for example) based device can use. Or a device that uses two batteries in series, but cuts out at 0.8V.However, I do not have the necessary equipment to do it.

author: apis
author_rank: Super Contributor
timestamp: July 28, 2015, 03:09:30 pm
content: Would be interesting to see the temperature response, but that's not so easy to test I suppose.

author: apis
author_rank: Super Contributor
timestamp: July 28, 2015, 03:13:18 pm
content: Quote from: Fungus on July 28, 2015, 05:51:57 amYeah, that's my conclusion as well. The truth will come out eventually.

author: Wytnucls
author_rank: Super Contributor
timestamp: July 28, 2015, 06:15:53 pm
content: Quote from: EEVblog on July 28, 2015, 09:44:27 amThey claim their DC to DC converter works down to 0.6V, so perhaps it is worthwhile to investigate how much energy is actually available in that uncharted region.My rough discharge test points to 60mW down to 30mW over a 5 hour period between 0.8V and 0.6V for a AA battery.

author: HKJ
author_rank: Super Contributor
timestamp: July 28, 2015, 06:17:03 pm
content: Quote from: Pentium100 on July 28, 2015, 02:59:33 pmI cannot do it to zero, but to 0.3 volt with 0.1 watt constant power:All the spikes is because I pauses the discharge 1 minute every 10 minutes to see the unloaded voltage. I.e. my time scale is 10% to long.If I had held a longer pause the cell would have recovered more voltage.Here is a zoom of the last part of the above curve:

author: HKJ
author_rank: Super Contributor
timestamp: July 28, 2015, 06:44:14 pm
content: I wonder why Dave want to do all that drawing on his curves to get remaining energy, why not directly make a curve showing it.A normal discharge curve with 0.1A discharge:Using energy scale:Showing remaining energy (It is the wrong way, but my charting system want increasing scales):At some other currents (0.2A, 0.5A and 1A):

author: Dr. Frank
author_rank: Super Contributor
timestamp: July 28, 2015, 07:17:55 pm
content: Dave,that could have been a nice educational video, but you really flunked it, sorry!As others already mentioned, you did not show the power integral area correctly, i.e. from 0.8 to the discharge line only, instead of zero volt to the discharge line.That is really annoying, and may disturb students, for whom this video is intended for, obviously.It could also have been very easy to bust the rest out of the residue energy myth, by discharging the battery to its very limits, below 0.8V, as these energy saver gadgets claim to operate well below that voltage.It's a pity you missed that chance.Frank

author: Pentium100
author_rank: Frequent Contributor
timestamp: July 28, 2015, 08:16:31 pm
content: Quote from: Fungus on July 28, 2015, 05:51:57 amBut what if you stick it on only after the device turns off because of low battery voltage?

author: nitro2k01
author_rank: Frequent Contributor
timestamp: July 28, 2015, 08:20:20 pm
content: Quote from: EEVblog on July 28, 2015, 10:36:35 amWell, you have a nice setup of a logging electronic load, so this would have been easy to prove empirically. Just keep loading the battery until at can no longer provide your x mA (or if you were doing a constant power measurement x mW.) If this point is 0.75 V instead of 0.8 V, then the numbers wouldn't come outexactlythe same for a 0.8 V or a 0 V cutoff. I'm not saying this because it would necessarily be viable to use this last amount of energy, but because it would have been so easy to extract this data and get a true "0%" point for a certain load, and for a particular battery.

author: jimon
author_rank: Regular Contributor
timestamp: July 28, 2015, 09:00:00 pm
content: Quote from: HKJ on July 28, 2015, 06:17:03 pmThank you ! Now I can rest because I know that after 0.8v point there are 1-2% of charge left

author: dcac
author_rank: Frequent Contributor
timestamp: July 28, 2015, 09:12:03 pm
content: Quote from: Pentium100 on July 28, 2015, 08:16:31 pmYeah there's no question as long as the battery by it self has enough voltage to power the device Batteriser will simply be drawing/wasting extra energy.But BR claims: "Works on all new and most "used" batteries." so this suggest it can/should be used with new (unused) batteries.But then again the BR Apple keyboard video shows it being used after battery status indicates Low level.

author: RobEE
author_rank: Contributor
timestamp: July 28, 2015, 10:10:07 pm
content: Dave,This video was very informative, as always, but the videos where you talk while doing screen capture of your computer screen can be a little boring. Have you ever considered setting up a camera and recording yourself while you talk? Then you can do either a picture in picture shot where your talking head is in the corner of the screen, or you can cut back and forth between you and your computer screen. Doing either of those, or a combination, would make your screen capture videos more entertaining. Sorry if this has already been brought up before, but I don't spend much time on the forums. Keep up the great work.-Robert

author: alter Ratz
author_rank: Contributor
timestamp: July 28, 2015, 10:34:53 pm
content: Hello,I think one point that is missing is using some Batterizer-Type switching pre-regulator for products using a linear regulator.If you start from 1.5V and yor device drops out at 0.8V then you could pre-regulate the 1.5V to lets say 0.85V. Assuming a current consumption of 1A this would drop the battery load from 1A to 1*0.8/1.5/0.9 = 0.59A (assuming a generous efficiency of 90%), thus reducing less power loss in the linear voltage regulator of the product and (theoretically) increasing battery lifetime.However, I did not take the time to calculate if reduced losses in the linear voltage regulator make up for the efficiency losses in the switching converter.But however, this is not how the batterizer works (rather it does the opposite of converting the voltage up!). So the batterizer is still bullshit.Best regards,Bernhard

author: LabSpokane
author_rank: Super Contributor
timestamp: July 28, 2015, 11:02:29 pm
content: Quote from: Dr. Frank on July 28, 2015, 07:17:55 pm? So all the battery companies do it wrong as well? If there's a STEM student who can't add or subtract a rectangular area from an integral... Oh who cares? Somebody has to drag down class averages I suppose.I thought the presentation was perfectly clear. Maybe those unhappy souls who don't "get" it should ask for refunds.And I thought the 0.8V cutoff was also to prevent battery leakage.

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 12:08:50 am
content: Quote from: Dr. Frank on July 28, 2015, 07:17:55 pmThe numbers come out exactly the same.Too bad if they are disturbed about it, I wanted to explain how I got the graph I showed in my Batteriser blog post.QuoteThere is bugger-all energy left under 0.8V at any reasonable product current. Why do you think it's the de-factco industry standard cutout voltage and is where the graphs stops in all the datasheets?As I said in the video, you can extract a little bit more energy out under that depending upon your applications. In most cases it's not worth bothering about.To get characteristic plots of low current drains to see how much energy is left under 0.8V is many hundreds of hours of data logging.QuoteRubbish, I can always do another video. Every one of my video doesn't have to be a tour-de-force, holy grail, be-all-end-all, the mother of all tutorials

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 12:29:44 am
content: Quote from: RobEE on July 28, 2015, 10:10:07 pmI don't particularly like those. I don't think my small face in corner would add any value, as you should be watching what's happening on-screen, not my head.Also, the boom mic and pop filter is big and covers my face, and I wouldn't be looking at the camera.

author: c4757p
author_rank: Super Contributor
timestamp: July 29, 2015, 12:47:08 am
content: Quote from: RobEE on July 28, 2015, 10:10:07 pmFor those who find information boring, there is always Mailbag.

author: pmj
author_rank: Newbie
timestamp: July 29, 2015, 01:59:25 am
content: If i can take the data to plot the graph until 0.8v using my product as load, then I will use my product until the voltage would be 0.8v. So by this video, my product would use 100% of the capacity of the battery, what it isn't 100% true. But it is the conclusion that i would take form this method.

author: Unknown
author_rank: Unknown
timestamp: July 29, 2015, 03:55:51 am
content: Quote from: EEVblog on July 29, 2015, 12:08:50 amI assume you mean something other than that the apparent capacity remaining that one would get from integrating from 0.8V to the discharge line only would not differ from that if you instead integrated from 0 volts to the discharge line. In other words, the difference between measuring the "area under the curve" with the area's base at 0.8 V instead of the area's base at 0 V. Because, there is actually a substantial difference.In a.png, I've sketched a hypothetical discharge curve at constant current. (I know that no cell actually behaves as depicted, but my purpose is only to illustrate the point.) Note that this curve drops vertically after it reaches 0.8 V, so in this case there isnoenergy at all after this point.In b.png, I've changed the vertical scale to only go from 0.8 V to 1.6V, and I've asked myself the question, "The cell has been discharged to 1.2 V, what fraction of the original energy remains?". Naively, I measure "areas under the curve" of this plot, and obtain the answer 25%.In c.png, I keep the scale that starts from 0 V, and ask exactly the same question, now, when I correctly find the areas under the curve, find an answer which is more like 42%. So, where we measure our areas from does make a difference.I believe that 42% is the correct answer and that this would agree with the spreadsheet method that you spent most of the time in the video explaining. The point though is that during the video, you did sketch over a discharge plot an "area under the curve" with a base at 0.8 V. If we forget about the need to add the rectangles down to the base at 0 V, we are liable to be mislead, often substantially. Even if we remember that the extra areas need to be added in, speaking for myself at least, it is very difficult to imagine the right amounts, and how this should change the interpretation. All this is just another example of the importance of using the right scale on a graph for a given purpose.We can always, of course, compute and get the right answer, but the importance of the graphical method (in this case sketching the area remaining under the curve) is that done correctly, it gives us an intuition for what computed result we should expect --- that's good engineering.So, my point is that using the graphs with the given scales is likely to substantially mislead our intuition about this problem --- and it happens to do so in the direction of suggesting less unused capacity than is in fact the case.I've been getting the feeling that in this group, saying that there is more energy left than looking at the (wrong) area suggests, makes me look like I am trying to support Batterizers (sp?) claims --- I'm not. The numerical calculations that Dave made (with the spreadsheet), are (I think) correct, and the unbiased visual depiction of the areas under the discharge curve would support them.

author: LabSpokane
author_rank: Super Contributor
timestamp: July 29, 2015, 04:11:31 am
content: Quote from: wilfred on July 29, 2015, 02:48:14 amI'm a EE, but I don't understand why a freshman college student with a 101 physics class behind them could not readily grasp Dave's presentation. I was actually thinking Dave was going to show how to fit the curve to a polynomial. (For embedded, I suppose they use a lookup table of sorts.) The graphical solution technique I thought was a great compromise. That's how every pilot's operating handbook works and teenagers are able to work through those with no issues. So, I'm confused why everyone's confused. Maybe the rest of the world is so far past graphical solution techniques that they now seem strange.

author: BobC
author_rank: Unknown
timestamp: July 29, 2015, 05:23:56 am
content: Sorry for posting so late to this thread, but I only now just stopped giggling.Am I the only one who parsed the video title as: "How To Calculate `Wasted Battery' Capacity"?My mind immediately saw a cartoon battery, laid-back and holding a joint.So, of course, if the battery is as wasted as it looks, I doubt there's enough capacity to worry about.Oh, come on. "Wasted Battery". Get it?

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 07:08:51 am
content: Quote from: dk27 on July 29, 2015, 03:55:51 amSee attached.As I said it makes no difference with my technique if you extend down to zero or not.I think I now understand what you are getting at, and the confusion seems to be that you have to integrate the data under the curve. I suspect I didn't make that clear enough. The integral gives you the area under the curve.

author: Smokey
author_rank: Super Contributor
timestamp: July 29, 2015, 07:09:57 am
content: Just think of all the extra battery sales that are going on now from engineers discharging perfectly good brand new batteries just to record the discharge curves. Time to buy some Duracell stock!

author: stuner
author_rank: Newbie
timestamp: July 29, 2015, 07:26:39 am
content: Quote from: EEVblog on July 29, 2015, 07:08:51 amOnce again you're talking about entirely different things. Nobody in this thread is arguing that there is a significant ammount of energy left in the battery once it's discharged below 0.8V. We're also not saying that the calculations you did you the spreadsheets is wrong, but the graphical method you used throughout the video does not yield the same result, because you ignored a part of the energy area.dk27andsilvasposted some nice graphics to illustrate your mistake.Edit: It seems that you realized the difference in the time it took to write this post.Quote from: dk27 on July 29, 2015, 03:55:51 amExactly. And just because there are some trolls on the internet trying to sell people a product that has practically no use does not mean that we shouldn't point out mistakes in Dave's videos. In fact I think that it is benefitial to correct them, becaus it prevents Batteriser from using the errors to weaken Dave's argument. In the end even with the correct calculations it turns out that their product is essentially useless.

author: Wytnucls
author_rank: Super Contributor
timestamp: July 29, 2015, 07:28:13 am
content: The 0.8V industry cut-off is an arbitrary figure, deemed necessary to avoid malfunctions in a majority of battery-operated equipment.Since their company claims a lower 0.6V cut-off for their converter, that portion of extra energy should be measured to avoid accusations of 'not knowing enough'.A discharge test from 0.8V to 0.6V at a constant 50mW for instance, shouldn't take more than 3 hours. That might not be sufficient energy to drive the monkey for any length of time, but should be substantial enough for a low consumption device, like a multimeter.

author: Unknown
author_rank: Unknown
timestamp: July 29, 2015, 08:00:49 am
content: Quote from: EEVblog on July 29, 2015, 07:08:51 amThanks for taking the time to understand my point. I didn't notice when I first watched your Batteriser video, but take a look at EEVblog #751 between about 12:31 and 13:07 here you draw on a plot of the discharge curve with time, and highlight the comparison between areas, but you only extended those areas down to 0.8 V rather than all the way down to 0 V --- I think you do this simply because the scale on the plot you had only went down to 0.8 V but I think you'll agree that people could be easily be left with the impression that only the areas down to the horizontal line at 0.8 V need to be considered.In this example, you estimate the wasted capacity (the area you coloured green) to be about 20%, which is a very good estimate --- by counting squares (I hope reasonably carefully) I got about 17%. However, if we only compare the areas actually depicted in the plot, I count the green area to be much closer to 10% --- the green area you shaded in certainly doesn't come close to a fifth of the total area.So I do think (unusually for you) that there's been some confusion caused by this.

author: silvas
author_rank: Newbie
timestamp: July 29, 2015, 08:39:03 am
content: Quote from: EEVblog on July 29, 2015, 07:08:51 amThis makes sense. The two curves are in different units, so the vertical scaling relationship must be arbitrary.Quote from: EEVblog on July 29, 2015, 07:08:51 amI don't think that's the confusion. Your spreadsheet and graphical analysis with the line drawing in IrfanView was bang-on and I don't think anybody is debating that.What people are pointing out is that the spreadsheet's integration procedure can be viewed graphically as summing up a bunch of skinny rectangles, whose height corresponds to the voltage measurements (assuming a constant current). But a skinny rectangle of height 1V goes from 0V on the vertical axis to 1V on the vertical axis, and cannot be drawn on a graph whose vertical axis starts at 0.8V (without going outside the graph). Since the areas you were shading with the sharpie were starting at 0.8V on the vertical axis they gave a qualitatively different picture of the situation, which missed out a bit on some good intuition to be had such as why the %remaining curve was so close to linear.You didn't focus too much on interpreting the areas you were drawing, so I don't think it is a big issue. You explained a different approach (the line drawing approach) which is interesting because it gives more quantitative results since the tick marks where the axes are intersected give you the quantitative information. Actually it is interesting because the line drawing approach is in some sense a graphical way of doing BravoV's "look at the values of columns C and E in the same row" approach mentioned earlier in this thread.

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 09:27:18 am
content: The problem here is that it is a common technique in the industry to compare areas under the curve on 0.8V cutout graphs to get a ballpark figure, and it's actually not that bad, you can get reasonable close for most purposes, and this is what I was showing in the video. But yes, this just happens to work on the 0.8V cutout graphs the manufacturers publish.I should have made this clear in the video.When people (including me) talk about the area under the curve representing energy, they are of course talking about the fact that when you integrate the data you get the area. I showed this in an old video on battery consumption I think it was. Once again, I should have pointed this out.

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 09:30:49 am
content: Quote from: silvas on July 29, 2015, 08:39:03 amYes, it's exactly the same thing, but tables of data are boring, and the whole point of the video was to explain how I got the graph I published on my Batteriser blog post last month, because many people had not seen that approach before:http://www.eevblog.com/2015/06/07/the-batteriser-explained/

author: Unknown
author_rank: Unknown
timestamp: July 29, 2015, 09:40:32 am
content: Quote from: EEVblog on July 29, 2015, 09:27:18 amI hate to suggest that common practice in the industry is wrong, but, under the assumption that the load draws constant current:1. Your integration technique gives the correct answer.2. Using the 0.8 V cutout graph by comparing areas of regions who's base is at the horizontal 0.8V line often produces results substantially different from what you get by extending those areas down to 0 V.3. The answers you get by extending the areas down to 0 V (ie the non-"industry standard" method) agree with the answers you would get from your integration technique.Your integration technique involves finding the function:f(t_1) = the integral of I(t)V(t) between t=0 and t=t_1Then if t=t_max is the time at which V(t) reaches 0.8 V you compute the capacity remaining at time t by:[1 - f(t)/f(t_max)] * 100%(The method of numerical integration used in your spreadsheet actually introduces a constant factor inversely related to the time-step you use, but this doesn't matter as it just cancels out in the ratio, in the same way that my constant will cancel out below.)Now if I(t) is constant and positive, theng(t_1) = the integral of V(t) between t=0 and t=t_1is proportional to f(t_1) simply because V(t) is proportional to I(t)V(t). So g(t)=a*f(t) for some a(=I(t), the constant current).So, g(t)/g(t_max) = [a*f(t)]/[a*f(t_max)] = f(t)/f(t_max) since the constants of proportionality just cancel off.So I can just as well use:[1 - g(t)/g(t_max)] * 100%as my capacity remaining.Note thatg(t_1) = integral V(t) between t=0 and t=t_1is exactly the same as the area between the t-axis (at 0 V), the curve given by the graph V(t), and the vertical lines at t=0 and t=t_1.The capacity remaining at time t_cutoff is:[1 - g(t_cutoff)/g(t_max)] * 100% = [ [g(t_max) - g(t_cutoff)] / g(t_cutoff)] * 100%g(t_max) is the area under the whole curve from time zero until the time t_max where the curve V(t) reaches 0.8V.g(t_max)-g(t_cutoff) is the area under the curve from time t_cutoff until t_max.As noted above, all these areas are taken from the t-axis which is the 0 V line, and as claimed, the capacity remaining is the ratio of the two areas in question.

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 10:13:35 am
content: I've added an annotation to clear this up.

author: EEVblog
author_rank: Unknown
timestamp: July 29, 2015, 10:43:36 am
content: Quote from: Wytnucls on July 29, 2015, 07:28:13 amYou can't just do that, because the pedants will come out and crap on about how it wasn't a proper test, no idea how the battery was drained before hand, doesn't give you the true % of the capacity at that drain, was allowed to recoverIf I'm going to do that I'd do it properly and do the full discharge at several different (low) constant power settings.50mW would take over 70 hours.Do that couple of times for different power values and that's a fair bit of work.Although granted it is a set and forget kind of thing, so I might do it for fun.

author: j.a.mcguire
author_rank: Contributor
timestamp: July 29, 2015, 02:57:42 pm
content: If the batteriser can save you 10-20% and it is typically only 90% efficient (I think its efficiency should be measured empirically before casting aspersions here). Then the best use-case is to only insert the batteriser into your device when the device stops working due to low battery cut off. In that scenario, you're looking at 90% efficiency of 20% remaining capacity which should give you an extended life of 18%. So it's potentially still a worth-while product.

author: Bud
author_rank: Super Contributor
timestamp: July 29, 2015, 08:05:44 pm
content: Those theorizing on capacity bean counting below 0.8v and even below 1v should sit down, build a boost DC to DC converter (or use an off the shelf one) and try it for themselves. I had a real product and in testing stage i quickly abandoned that idea of boosting the battery voltage and had to redesign the power supply because power just dropped like a brick after reaching a certain point in battery discharge.

author: Fungus
author_rank: Super Contributor
timestamp: July 29, 2015, 08:42:54 pm
content: Quote from: j.a.mcguire on July 29, 2015, 02:57:42 pmNope.The current required to keep the booster going is proportional to the battery voltage.As the voltage hits 1.0V the current demands will start going upwards and easily overwhelm the battery's ability to keep up. Result: The voltage will drop, the booster will ask foreven morecurrent, this causes the voltage to drop even further... etc.Short version: It will go into an exponential death spiral.

author: BobC
author_rank: Unknown
timestamp: July 29, 2015, 10:29:07 pm
content: The relatively undocumented portions of alkaline battery consumption curves below 0.8V looks worthy of some special attention.Much of the history of battery use (and thus analysis) should be reexamined in the light of today's world of micro-power circuits and efficient boost converters.Why should we avoid discharging an alkaline battery below 0.8V, and do those reasons still apply to all of today's loads? If the limit doesn't apply, are any battery manufacturers providing curves below 0.8V? If they do apply, then why?I can think of several points worthy of investigation:1. Does deep discharge of an alkaline battery create new hazards, such as a significantly increased risk of leakage or fire?2. If we ignore the arbitrary 0.8V lower limit, how low can we go for small loads?3. If we assume a 95% boost converter efficiency, how much additional operating time can we obtain for various loads at a nominal 1.0V output? (The argument being that it may not be worth going below 0.8V no matter what technology is available.)4. What are some typical alkaline battery self-discharge currents? Does the self-discharge current dependent on the charge state of the battery? Is there a point in the life of a battery where the self-discharge current will consume the entire current production capability?I'm certain there are additional factors to consider, and I hope others will mention the ones I've missed.Eventually, I'd like to place the above discussion within the larger topic of how best to power systems that are disconnected from mains AC power, based on all applicable considerations: Initial cost, lifetime cost, peak wattage, standby wattage, minimum required operating time (between energy system recharge/replacement), environmental considerations (mainly temperature, but perhaps also altitude). The discussion should include all commercially available technologies, such as various battery technologies and ultra capacitors. And perhaps a separate discussion about how adding a solar cell changes things.

author: EEVblog
author_rank: Unknown
timestamp: July 30, 2015, 01:32:38 am
content: Quote from: wilfred on July 29, 2015, 10:57:11 pmRelax. I simply explained a practical reality what I'm pretty sure would happen if I did the quick 3 hour test.Yes they might be right, and I agree they would be to a certain extent, but I also think there is valuable data to be had from such a test. But trying to convince others of that may not be easy. It's likely I'd cop more flack than not for doing such a test, regardless of how valuable the data actually is.I may end up doing that if I need some data on that and don't have the time to do the full test. But I'd much rather get the full proper test, and multiple characteristic curves.

author: LabSpokane
author_rank: Super Contributor
timestamp: July 30, 2015, 02:25:21 am
content: Why is the region below 0.8V so interesting to people? Can someone explain this to me? Compare the relative areas. Instead of trying to utilize some tiny sliver of what is left, why would you not put your efforts in to minimizing consumption of the overwhelming majority of the energy available?

author: EEVblog
author_rank: Unknown
timestamp: July 30, 2015, 02:29:57 am
content: Quote from: LabSpokane on July 30, 2015, 02:25:21 amIt's because they think there is something there, and there is, it's just not much.They see joule thief circuits flash a LED down to bugger all voltage and think maybe they can do more than that.But the reality is as you said, for most products that draw any significant current (i.e. don't get the shelf life of the battery), then it's usually not worth worrying about.I think there is very little real hard data on it though, and it would be good to get, just for those cases where it can be made use of.Ironically the Batteriser claim they had to design their own ASIC converter chip because there was nothing on the market that went down to 0.5V and 1A. And of course for the vast majority of the usage scenarios they have just wasted that time and money, especially with the aforementioned snowballing effect of lower volatge=higher current, they could have used an off the shelf one that went down to 0.8V and have been done with it. But that doesn't market nearly as well

author: BravoV
author_rank: Super Contributor
timestamp: July 30, 2015, 02:30:44 am
content: Quote from: LabSpokane on July 30, 2015, 02:25:21 amI guess certain people love to drive their car until the "last drop" of the fuel used ?

author: BravoV
author_rank: Super Contributor
timestamp: July 30, 2015, 02:38:45 am
content: Quote from: EEVblog on July 30, 2015, 02:29:57 amAssuming the wafer is made from ordinary silicon based semiconductor, is it possible for the chip to start kicking at 0.5 volt ? Again, at 0.5 volt, not the magic number at 0.6 to 0.7 volt.We're talking the "almost" depleted battery's voltage that is at 0.5 volt even its "unloaded".

author: LabSpokane
author_rank: Super Contributor
timestamp: July 30, 2015, 04:01:05 am
content: Quote from: wilfred on July 30, 2015, 03:35:27 amI was throwing you a bone. But to make you happy and fulfilled I have filled in the dreaded, missing rectangle. I'm trying to point out a region of interest and illustrate that the region over 0.8V is what matters. A good engineer focuses on conserving the power in the large, blue region. Bad engineers focus on tiny sliver of at the tail end. There's more variation from battery to battery than there is to be recovered in that cyan region.QuoteThis quote is from just a few posts up. That's what I'm addressing. Not disingenuous all. But to the point. It may be an undocumented region, but that's because it's hardly worth the effort.

author: bitwelder
author_rank: Super Contributor
timestamp: July 30, 2015, 05:57:08 am
content: Quote from: Fungus on July 29, 2015, 08:42:54 pmPlus usability would be horrible:- put fresh batteries without the gadget- use the product- battery low: first of all, find where you put the Batterisers. Once found and inspected they are still OK, open the product, remove batteries, install the magic sleeves, remount the batteries- after a short time lapse: out of juice for good.Was it really worthy?

author: amyk
author_rank: Super Contributor
timestamp: July 30, 2015, 06:15:17 am
content: Quote from: BravoV on July 30, 2015, 02:38:45 amLook atTPS6120xQuoteThese can boost all the way to 5.5V too, although they can't output 1A with a 0.5V input and the efficiency is going to be horrible down there.TI and Linear have some ultra-low input voltage ICs for energy harvesting that will go down to the tens of mV range, so it's not a silicon limitation. There's alsothis one, intended for solar cells, which starts up at 0.4V.That brings to mind the question, if experienced companies like TI and Linear can't make 0.5V input 1A boost converters with 90%+ efficiency, what are the chances these Batteriser guys can...?

author: Unknown
author_rank: Unknown
timestamp: July 30, 2015, 07:10:17 am
content: Quote from: LabSpokane on July 30, 2015, 04:01:05 amHere's why the area underneath thehorizontalline at 0.8 V is important. Have a look at a.png. Here we are considering discharging a cell at a constant current of 100 mA until we reach 1 V. What is the green area in this plot? Many people on this forum (including you) have made similar plots and say that the area painted green is the unused energy in the cell, right?Now take a look at b.png, If we stop discharging at 1.6 V, we essentially still have a fresh cell, so the green area in this plot should be the total energy in a fresh cell, right?Ok, the green area in b.png is less than the green area in c.png, right?And the green area in c.png is 3/4 of the green area in d.png, okay?So what is the green area in d.png? We have to be a bit careful with our units, but the vertical distance is easy, it's 0.8 V. Now the horizontal axis is in hours, but we have to remember that these are 100 mA hours, so in total, we have 30*100 mA hours = 3000 mA hours = 3 A hours. Multiplying these together I get the area in d.png as 2.4 watt hours.So the green area in c.png is 3/4 * 2.4 watt hours which is 1.8 watt hours.So the total capacity of my cell must be less than 1.8 watt hours. Substantially less, in fact.Thing is, I reckon that I can take a 1V at 100 mA load and appropriately regulated, use this cell to drive that load for at least 20 hours. (Since after 20 hours at 100 mA, I should still have more than 1.1 V from the cell.) Fair enough? So, since my load draws 1 V * 100 mA = 0.1 watts, and I can run this for 20 hours, I've somehow managed to extract 0.1 watts * 20 hours = 2 watt hours, from this cell. But we worked out that the capacity of the cell was somewhat less than 1.8 watt hours.Now, I don't believe in free energy, so there must be something wrong here.What is wrong is that to quote wilfred, "for a pictorial representation of the area under the curve to represent energy it should be shown from 0 V". If you extend the areas down to 0 V properly, then everything works out fine.Many people have been showing plots like a.png, and claiming that the "wasted energy" is only the green region. It just isn't, and in earlier posts, I've demonstrated that the depiction can be substantially misleading.BTW, I don't care at all about the little cyan sliver showing the energy available after the cell has discharged to below 0.8 V --- I think I have been careful in each post I've made to explain that I'm happy to consider it to be negligible, and in all calculations I have done I've set it tozero. It's just not the point.

author: silvas
author_rank: Newbie
timestamp: July 30, 2015, 08:36:33 am
content: Quote from: LabSpokane on July 30, 2015, 02:25:21 amYour image is the perfect way to explain the misunderstanding that is going on here. When we are saying "the area below 0.8V", we mean the large white area below the large blue area, *not* the tiny cyan sliver. In fact, as far as the "area below 0.8V" people are concerned, the tiny cyan sliver is assumed to be zero area. Does that make sense? All this ruckus is about remembering the large white area. It's nothing more complicated than that.

author: silvas
author_rank: Newbie
timestamp: July 30, 2015, 09:01:30 am
content: Quote from: EEVblog on July 29, 2015, 09:27:18 amThanks for adding the annotation.Quick question: why is the area-down-to-0.8V used as a ballpark figure? Since the discharge curve is so close to linear, you can get a very good approximation very easily without even looking at areas. For example, if 90% across the graph the battery is down to 1.0V, then you estimate that it has used up 90% of the energy when it is down to 1.0V. Or to explain the approximation graphically in terms of your IrfanView line technique, draw the %remaining tick marks on the horizontal axis, then after drawing your first line, directly draw a vertical line down to the horizontal axis. Since the discharge curve is so close to linear, this is a very good approximation.This seems both easier and more accurate than ballparking an area-down-to-0.8V. Is there a historical reason that area-down-to-0.8V is in common use?

author: LabSpokane
author_rank: Super Contributor
timestamp: July 30, 2015, 01:01:20 pm
content: Quote from: silvas on July 30, 2015, 08:36:33 amOK, what Dave did is a very common graphical technique. He modified the graph so that only the time variant portion was there. The calculations in his spreadsheet account for that energy in that area already. There is some perception that if the battery is run below 0.8v that the energy in that region in white will be accessed. It's already gone. The only energy that can be recovered running the battery below 0.8v is in the cyan triangle.

author: nessatse
author_rank: Regular Contributor
timestamp: July 30, 2015, 02:16:29 pm
content: In my mind the idea of calculating remaining energy, ie integrating the discharge curve, etc, is unnecessary for purposes of analysing/debunking the batteriser. All that matters is how much battery life remains. For example, say you have a device that requires 100mA and the discharge curve crosses the x axis at 10 hours (we can all agree that there will be very little difference between 0.8V and 0V, seeing that the discharge curve basically drops straight down at the end). Now, if your device cuts out at say 1V and that point is reached at 9 hours, then there is an absolute maximum of 1 hour or 10% life left in the battery, FOR THIS DEVICE/APPLICATION, regardless of what magic doodad you clip on to the battery.

author: Wytnucls
author_rank: Super Contributor
timestamp: July 30, 2015, 03:02:45 pm
content: The energy doesn't drop right off below 08V. There is still some left below that voltage level. Whether it can be harvested successfully is another matter. They claim they can do it, so let's wait and see.In the meantime, try to plot a AA battery discharge curve with a constant 10 Ohm load below 0.8V and you should get about 60mW decreasing to about 35mW at 0.6V over close to 4 hours (~60mA).Of course, most of the recovered energy will be between 1.1V and 0.8V, but it won't hurt to pad that up with anything left over.We know already that their converter works at about 1.1V and that is commendable in itself, for the miniaturization required. Its usefulness is still debatable though.

author: silvas
author_rank: Newbie
timestamp: July 31, 2015, 05:56:23 am
content: Quote from: LabSpokane on July 30, 2015, 01:01:20 pmI don't think there's that perception at all (or there has been a misunderstanding). When the battery is at 1.1V, the electrons drop across a 0V-to-1.1V potential, not a 0.8V-to-1.1V potential. Shading in an area only down to 0.8V means you shade in an energy area equivalent to the electrons dropping across a 0.8V-to-X potential, which gives a misleading picture of how much energy the electrons have given up. This has nothing to do with what the battery does when it is discharged to 0.8V or below; when that has happened, the energy in the large white rectangle has already been delivered.We aren't saying that there is more or less energy available. Just that you have to draw it right for the picture to actually be an accurate representation of the energy. Nobody is claiming that the large white rectangle represents some sort of "untapped energy". The large blue area plus the large white rectangle's area is an accurate graphical representation proportional to the energy. Just the blue area is not.There are other people in this thread discussing what happens when the battery has been discharged to 0.8V or below, but they are not the people worrying about the large white rectangle (as you would expect, since the large white rectangle has absolutely nothing to do with what happens when the battery is discharged to 0.8V or below).

author: LabSpokane
author_rank: Super Contributor
timestamp: July 31, 2015, 06:17:20 am
content: QuoteNo shit.I was *trying* illustrate the region of *interest* which is the time variant portion of the curve.Over and out.

author: miguelvp
author_rank: Super Contributor
timestamp: July 31, 2015, 06:26:41 am
content: Quote from: silvas on July 31, 2015, 05:56:23 amDave already explained that:https://www.eevblog.com/forum/blog/eevblog-772-how-to-calculate-wasted-battery-capacity/msg719534/#msg719534It makes no difference if you extend it to 0V or cut it at 0.8V because after 0.8V it just discharges way too fast, maybe you get half a percent extra if you are lucky.

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 07:02:34 am
content: Quote from: LabSpokane on July 31, 2015, 06:17:20 amBut LabSpokane, in an earlier posting on this forum you attached the following plot, where you've plotted a green area, and labeled it "Green area is what energy remains -->".Did you actually mean to make the caption say "Green area is what energy remains provided we only count the portion of energy obtained by letting the electrons drop to the 0.8V potential --- but don't worry because the rest of the energy remaining isn't very interesting"?PS: I know that LabSpokane probably didn't plot the area he meant to --- I think he meant to have the left hand boundary of the green region intersect where the red 0.925V line and the blue discharge lines also intersect, unless LabSpokane disagrees, lets just assume that's what he intended to draw.

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 12:42:52 pm
content: Quote from: miguelvp on July 31, 2015, 06:26:41 amPlease oh please at leasttryto comprehend what you read before you respond. There areother peoplearguing over how much energy remains in a cell once it discharges to 0.8V, silvas and I are making no claims at all about this. For the record, my personal opinion is that it isnegligible, I'll happily call itzeroorhalf a percent, whatever you like. I'd be extremely surprised if silvas disagreed on this point.The actual issue is whether when someone tells us that the "area under the (discharge) curve is energy" we are meant to take it seriously or not. And people do draw pictures of areas and say this.If we're not meant to take the phrase seriously, then why would people say it? My best guess would be that it's just part of the ritual passed down from electrical engineer to electrical engineer, and you have to say it, otherwise the blue magic smoke might escape or something. In other words the whole thing is just "cargo-cult calculus".I happen to think that we are meant to take the phrase seriously --- we ought to be able to measure areas, compare areas, and come up with the right results. Our claim is that to get sensible answers out of such procedures, those areas have to be taken down to the 0V line. If you don't do this, if you instead only extend the area down to the 0.8 V line, you can't take the notion that area corresponds to energy at all seriously. Look for an earlier post where I show for example that you end up concluding that you can draw 0.1 watts for over 20 hours from a cell with a capacity of less than 1.8 watt hours --- nonsense.Now, there are a few possibilities:1) It turns out we're wrong. So far everyone who says this simply attacks thestraw-manabout energy left in the cell after it's been discharged past 0.8V. How about instead showing me the "right" way to compute the "area under the curve", without that area extending below 0.8 V, so we don't get a violation of the law of conservation of energy.2) We're not meant to take "areas under curves" seriously, it's just some meaningless words we say. Then wouldn't it be better to stop saying those words, stop drawing pictures of the energy being an area under the curve and save everyone a lot of time and confusion?3) We're right, but it doesn't matter, because using the "incorrect" areas either results in the same answers for the problems we've been talking about since whilst the area representing the unused capacity of the cell is smaller than it should be, so is the area representing the total capacity of the cell, and these two effects cancel out either exactly or approximately. The problem is that they don't cancel exactly, and they generally don't even cancel approximately most of the time --- I earlier gave an example which when calculated using only the areas above 0.8 V gave an unused capacity of 25%, when calculated using what I claim were the correct areas gave an unused capacity of approximately 42%. As an approximation, being nearly a factor of two out is pretty bad for this application.4) We're right, but so what? "We like to draw the area only down to 0.8 V, but everyone knows you have to extend right down to 0 V." Well, many of the cases in which the argument has been used has been to refute (for example) the claims of Batteriser. I think its fairly safe to guess that much of the intended audience for these arguments can't necessarily be expected to know that the areas drawn on the graph are not really the areas that need to be compared --- wouldn't it be better not to mislead that audience? Even worse however, the effect of comparing the wrong areas substantiallyunderestimatesthe capacity left in the cell. It will look like mendacity on your part when your opponent points this out --- straight out ofHow to Lie with Statistics.A much better strategy is to use the depiction that more straightforwardly tells the truth, so that you don't give the likes of Batteriser ammunition to use against you.5) We're right, and it turns out that some engineers have been wrong about this point. Well, great, because once they learn they have been wrong, they will stop being wrong in the future and as a result be slightly better engineers. I would like there to be better engineers, because a lot of the stuff that I use in my life gets made by engineers. If engineers are better then I get to have nicer stuff.

author: rs20
author_rank: Super Contributor
timestamp: July 31, 2015, 01:30:16 pm
content: With a constant current discharge, the x axis is both time and charge (since at a constant current, time and charge are proportional). The y axis is voltage. So you take the integral under the curve ("all the way down to zero" is the correct way to do it, but that doesn't need to be said because that's just what integration is). And you get voltage * charge = energy. So dk27, this translates to your options 3/4/5, I don't particularly care to pick one of those in particular.With constant resistance or constant power discharge however, the x axis still measures time, but no longer measures charge. So taking the area under the curve is meaningless in this sense (unless there's some way to ascribe meaning to it that's beyond me). So in that case, dk27, corresponds to option 2.In particular, with constant power discharge, the x axis is directly proportional to energy, no need to take any areas under anything.This raises a larger question, which is why we care so deeply about theenergyin the battery. I mean, in this case, the batteriser people are making claims centered around energy, so kind of fair enough I guess. But the x axis in all these graphs is "how long the device runs for". Isn't that more relevant to the end user than how much energy is in the battery? I mean, with a constant current discharge, there's more area under the curve for the first minute than later minutes, but that's only because the linear regulator in my device is burning off more power as heat. Why do I care that more energy disappeared from my battery that first minute than any subsequent minute? Don't I only care that I've got one less minute of battery life left? So taking the area under the curve is, I dare say, always a useless endeavour, and if Dave is suggesting to do that as a part of product design, I boldly claim that's just wrong -- just get the discharge curve corresponding to your device, and read off how long it'll last off the x axis. Job done.This is basic dimensional consistency stuff.

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 01:59:56 pm
content: Quote from: rs20 on July 31, 2015, 01:30:16 pmAnd that's quite right. Admittedly I didn't mention constant current in my last post, but I think that in all my preceding posts, I've been careful to state that we're talking about the constant current case. The interpretations that I've been critiquing have all involved discharge curves at constant current.QuoteAnd I think you make a very good point here. For almost all aspects of product design, you care about battery life, not energy dissipation. I'd disagree mildly with your statement that it's always a useless endeavour, for example I can imagine applications where you might possibly care about how much heat your regulator is dissipating with time. But most of the time engineers want to know about service life under different loads, and the plots provided in the data-sheets answer those questions very well without any need to think about areas under the curve at all.So since actually needing to compute the energy is needed so rarely, I suppose I shouldn't be surprised that there appears to be substantial confusion about how to go about doing it. And we shouldn't be surprised that the plots provided in the data-sheets aren't optimal for doing this calculation. (Well, at least not in the constant current case.)The discussion in this thread, however, was about energy in the cell, not about service life, so in that context, the right way to depict that energy (in this case as area under a curve) is a valid topic for discussion.QuoteYes, that's what I thought.

author: Fungus
author_rank: Super Contributor
timestamp: July 31, 2015, 02:11:44 pm
content: Quote from: rs20 on July 31, 2015, 01:30:16 pm...and completely wrong with a constant-power device like the Batteriser. Constant power devices draw more current as the voltage drops, completely skewing that discharge curve.

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 02:14:44 pm
content: Quote from: rs20 on July 31, 2015, 01:30:16 pmIf my ability to do maths is to be trusted, option 3 is ruled out. When I made my first post, I assumed it was just a case of option 4 and that I could contribute towards strengthening the arguments that people were making against the Batteriser. At this point, I'm pretty sure that option 5 is the case, but hey, I like nice stuff

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 02:51:09 pm
content: Quote from: Fungus on July 31, 2015, 02:11:44 pmAgain, it would be nice if people would take the time to comprehend what they are replying to before they post --- it's not hard, and really just basic courtesy.I'm pretty sure that rs20 explained how you can deal with the constant-power case. Hint: Read the bit that mentions "constant resistance" and "constant power discharge". After I read that it becomes quite natural to me that when rs20 says "just get the discharge curve corresponding to your device", he is suggesting that if you do happen to be in a constant power situation that you use the plot showing a constant power discharge curve. And then youcan"just read off how long it will last from the x axis".If the device you're considering powering using the Batteriser is normally (say) constant current, you can do the proper calculation for that, as well, and then compare the resulting times.

author: LabSpokane
author_rank: Super Contributor
timestamp: July 31, 2015, 03:03:19 pm
content: Quote from: dk27 on July 31, 2015, 07:02:34 amOh FFS, of course I didn't plot that. That was a capture from Duracell's datasheet - the same datasheet Dave used - and the same datasheet anyone could download if they Googled "duracell AA datasheet." And of course the energy remaining is calculated by referencing zero potential!

author: miguelvp
author_rank: Super Contributor
timestamp: July 31, 2015, 03:55:01 pm
content: Quote from: dk27 on July 31, 2015, 12:42:52 pmI don't think the problem is me comprehendingBut I guess I would have to explain a bit instead of giving you the link to the graph and assume that you would be able to see what I meant, my bad.The area under the curve is energy, but it doesn't matter if you use Dave's graph to 0.0V or 0.8V because the percentage of used energy takes that into account and it would move from 0.8V to 0.0V if you include it in the graph.So for the energy below 0.8V this is what you'll get:If you look at the blue line at 0.8V let it intersect with the constant current curve (dark blue), then project it vertically until it intersects the energy capacity curve (red) then finally project it to the percentage axis (right in red) and you'll see that it's pretty much at 0% (other than my paint fu is not that great and the line I used was too thick) The area in blue (triangle) is the left over energy past the 0.8V at 100mA constant current discharge.Yes there is some, but it's negligible.

author: stuner
author_rank: Newbie
timestamp: July 31, 2015, 04:08:02 pm
content: Quote from: miguelvp on July 31, 2015, 03:55:01 pmBut it is. Nobody is talking about the graphs Dave made with Calc (becuase they're perfectly fine). We were talking about theareas Dave marked in the datasheet graphs. He has since added an annotation, which should have conculded this discussion imo...

author: miguelvp
author_rank: Super Contributor
timestamp: July 31, 2015, 04:21:05 pm
content: Quote from: stuner on July 31, 2015, 04:08:02 pmAnd I get it, but the only missing power is that little sliver. If you extend the graph to 0.0V you are not just extending the unused area but the used area as well so it's a good approximation.

author: stuner
author_rank: Newbie
timestamp: July 31, 2015, 04:28:55 pm
content: Quote from: miguelvp on July 31, 2015, 04:21:05 pmThe effect is not the same on both areas (the remaining energy is reduced much more significantly). Whether or not it's a useful approximation obviously depends on the exact scenario and requirements, but I wouldn't consider it a "good" approximation.

author: Fungus
author_rank: Super Contributor
timestamp: July 31, 2015, 05:32:44 pm
content: Quote from: stuner on July 31, 2015, 04:28:55 pmYes,WE GET IT!!!No need to repeat it endlessly. Either do the math, do the experiment, or dosomethingto back up the loud noises.

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 10:25:34 pm
content: Quote from: LabSpokane on July 31, 2015, 03:03:19 pmYou didn't colour in the green region and add the label? Anyway, I'm glad you understand that the energy remaining is calculated by referencing zero potential --- which means the green area isn't literally "what energy remains". So we're in option 4:Quote

author: Unknown
author_rank: Unknown
timestamp: July 31, 2015, 10:36:05 pm
content: Quote from: Fungus on July 31, 2015, 05:32:44 pmI posted some time back some support for what stuner is saying here, ie that the approximation isn't good:https://www.eevblog.com/forum/blog/eevblog-772-how-to-calculate-wasted-battery-capacity/msg719467/#msg719467It's a bit rich to accuse us of making the "loud noises" when you're the one YELLING!

author: EEVblog
author_rank: Unknown
timestamp: July 31, 2015, 10:45:10 pm
content: Quote from: miguelvp on July 31, 2015, 03:55:01 pmJust to be clear, that data under 0.8V on my graph is fake, I added an extra data point juts to get that 0V graph for visual explanation.My latest video (in two parts) looks at energy under 0.8V

author: LabSpokane
author_rank: Super Contributor
timestamp: August 01, 2015, 03:59:50 am
content: And so ends the tale of Encyclopedia Brown and The Case of the Missing Microwatt.

author: Fungus
author_rank: Super Contributor
timestamp: August 01, 2015, 07:17:18 am
content: Quote from: dk27 on July 31, 2015, 10:36:05 pmYeah, but five pages in and there's still more noise than signal...Nobodyis arguing that the graph's perfect. The math is complex, so...debate is over. Time for experiment.And thanks to Dave the results are in: 30 seconds of extra life in 10.5 hours.Less than 0.2% of the power is left below 0.8V.

author: Unknown
author_rank: Unknown
timestamp: August 01, 2015, 08:13:32 am
content: Quote from: Fungus on August 01, 2015, 07:17:18 amIt has been explained over and over again that the argument we are making hasnothingat all to do with whether or not there is any energy (not power) available after the cell has been discharged below 0.8 V. I don't know whether the most charitable interpretation of your post is that you are disingenuously trying to distract from the point, or simply that you are thick. (Actually Hanlon's razor,http://rationalwiki.org/wiki/Hanlon%27s_razor, tells us we should presume stupidity before malice, but I'm beginning to find it very hard not to favour the latter explanation.) Knowingly attempting to win an argument by intentionally misrepresenting your opponent's claims really would be malicious.Now, back to the point. I admit that no one expects a graph to be perfect, but we can at least attempt to keep it from being outrageously misleading. In this case, we are talking about a distortion due to truncating the y-axis of the graph. Marketers and other people who would like to dishonestly persuade often use the trick of y-axis truncation, for some nice examples of this seehttp://data.heapanalytics.com/how-to-lie-with-data-visualization/. People who care about the truth should probably try to avoid both using such tricks and being perceived to be using such tricks.Now, when we say that "the approximation" isn't very good we are trying to call attention to such a distortion, hopefully so we can avoid making plots that unintentionally mislead --- read earlier posts to see exactly what the distortion is, but basically the problem is that when the graph is truncated the area representing the energy remaining at a given discharged voltage always looks smaller (in comparison to the apparent whole) than it in truth is. The distortion is not small, I gave an example where the 42% actual energy remaining ends up looking like 25%. (EDIT: fixing a mistake I just realised I made in the algebra --- long day) Even worse, the effect scalesquadraticallyinverse linearly --- halve the remaining energy and the distortion effect increases byroughly 4roughly 2 times --- the actual remaining energy remaining halves, but the amount apparently remaining reduces by a factor of 4. That's bad, really bad.

author: Fungus
author_rank: Super Contributor
timestamp: August 01, 2015, 08:52:33 am
content: Quote from: dk27 on August 01, 2015, 08:13:32 amAre you about to do it again...?Quote from: dk27 on August 01, 2015, 08:13:32 amYes,we get it!Quote from: dk27 on August 01, 2015, 08:13:32 amArea calculations where the area is incomplete: Bad!!!!Finger-in-the-air calculations of remaining battery life by looking at the graphs: Surprisingly good! As the next video shows it really WAS half a bee's dick.

author: Unknown
author_rank: Unknown
timestamp: August 01, 2015, 09:06:12 am
content: Quote from: Fungus on August 01, 2015, 08:52:33 amI never claimed that any of those predictions were wrong. Again you are trying to distract, by suggesting that I did. That is disingenuous. I now no longer believe that you ever intended honest argument.

author: nitro2k01
author_rank: Frequent Contributor
timestamp: August 01, 2015, 10:14:08 am
content: Quote from: dk27 on August 01, 2015, 09:06:12 amSo what exactly are you claiming? Yes, Dave does draw the area under the area of that graph which is incorrect, but he also names the percentage from the graph, which is numerically generated has a Y axis from 0-100%, and is accurate. (To within the bee's dick's margin, as the calculation assumes the battery is dry at 0.8 V, but that was not the issue you complained about to begin with, right?) So if you listen to Dave when he says that 4% of the battery's capacity is left for a 1V cutout voltage, thisisaccurate no matter what he's filling in under the voltage graph.

author: Unknown
author_rank: Unknown
timestamp: August 01, 2015, 10:49:42 am
content: Quote from: nitro2k01 on August 01, 2015, 10:14:08 amOk, that's a reasonable question, I think I covered it in the following post:https://www.eevblog.com/forum/blog/eevblog-772-how-to-calculate-wasted-battery-capacity/msg721176/#msg721176, especially possibility 2.If you accept that the ratio of the areas depicted gives the wrong answer, and you simultaneously use some other means of finding the right answer, then why bother talking about those areas in the first place? The exercise becomes a meaningless ritual --- we talk about "areas under the curve", but we're not really taking the idea seriously. I think I'm justified to be concerned about that ritual, because at the same time he's saying that 4% of the battery capacity is left, he's pointing to an area that has a size more like 2% of the total area, so if someone were to mistake the ritual of pointing to areas as something to be taken seriously, although they will hear 4%, they will see about 2%.I'm sure that for Dave, pointing to areas is not meant to be merely a meaningless ritual, and, to his credit, he later added an annotation to the videos that acknowledged that the areas in question actually need to be integrated down to 0 V to make sense. However there was also the suggestion that eyeballing the "wrong" areas in order to get a reasonable estimate was a standard practice "in the industry". This is covered by my possibility 3 in the referenced post. The point being that this is generallynota very sensible practice, since as an approximation, it is really bad.

author: miguelvp
author_rank: Super Contributor
timestamp: August 01, 2015, 04:09:33 pm
content: Dave said in the video that the area under the curve "represents" the energy in the battery. No one claimed that the area "is" the energy in the battery.Also he mentions that is a "common technique in the industryto compareareas under the curve on 0.8V cutout graphs"The area under the graph without the integral part of the constant 0.8V added does change the percentage of remaining capacity(which is what we are talking about, hence the title of the thread) but the 0.8V cutout graph is used to "compare" not "compute" the areas, semantics are important here.What does that mean? well if the battery is considered to be fully discharged at 0.8V using the graph you provided, the ballpark comparison gives you 50% of energy remaining, sure there is an error from the actual 41.(6)% but that is good enough for government work and far away from your "computed" absolute area capacity of 25%.Edit: the discharge curve on your graph, happens to be the same as the percentage of energy used. So a straight line will do.If you apply that to your other graph that it doesn't have the cutout, the percentage will be more accurate and closer to the actual 41.(6)%Engineering even if it's very precise it needs tolerances and constrains to apply the science and math to a "practical" solution, it's not about absolutes because absolutes don't work in engineering, there are way too many external factors.For example, you can't build a bridge to absolute specifications because external factors out of your control will affect it, so you need to compute what are the best tolerances you are willing to live with to get a practical solution done.

author: Fungus
author_rank: Super Contributor
timestamp: August 01, 2015, 06:17:45 pm
content: Quote from: miguelvp on August 01, 2015, 04:09:33 pmThis.

author: Unknown
author_rank: Unknown
timestamp: August 01, 2015, 10:22:17 pm
content: Quote from: miguelvp on August 01, 2015, 04:09:33 pmReally?We're going to get all philosophical about this? Well, okay...In your mind, do you have a precise notion of how your area under the curve is meant to "represent" an energy? For example, is there some quantitative relationship between the area and the energy. Alternatively do you have a qualitative notion in mind? Or do you mean that the area bears no particular relationship to the energy at all, and we, by convention, are simply using it as a symbol of the energy?QuoteUntil we know precisely what you mean by an area "representing" an energy, there is no basis to decide whether it is reasonable to compare the areas, or even whether such a comparison is "good enough" for your application.For example, if you claim that areas are merely symbolic of energies then I can learn nothing by comparing the areas. In the same way, I learn nothing about the differences between cats and dogs by comparing the symbols "cat" and "dog".Remove enough meaning from the areas and it's time to ask whether it's worth thinking about them at all.QuoteOk, if you are happy with that approximation, that's fine. However, that approximation has nothing to do with any areas, so why do anything with the areas at all? Talking about the areas would just be a meaningless ritual.(Of course, I believe that by treating the areas seriously, and working with the right ones, you can get answers that are even better than those good enough for government work. You might even be able to get a job in private industry!)QuoteThe percentage of energy dissipated is quadratic in time (recall that my graph was representing a constant current draw), the discharge curve on my graph is not quadratic in time. The service time remaining is a straight line with respect to time, but that's obvious.

author: miguelvp
author_rank: Super Contributor
timestamp: August 02, 2015, 07:27:16 am
content: Well, the graph clearly shows the 0.8 cutoff so you just have to add the integral of that constant if you really want to calculate the area of that representation. It's pretty trivial, all the information is at hand.If people don't know how to read graphs, well, that's their problem.Yeah, 8% error is not ideal by any means at 50% of the time, but a good approximation and not as much as a 25% error as suggested. If some people don't know how to read graphs, then send them back to elementary school.The thing is that quadratic or not it doesn't matter if you have the discharge curve, and if you don't, I recon a linear approximation will do the job as well for an off the cuff approximation. At the end of the day we are talking about under a minute of battery life after it reaches the 0.8V.

author: Unknown
author_rank: Unknown
timestamp: August 02, 2015, 07:55:54 am
content: Quote from: miguelvp on August 02, 2015, 07:27:16 amAll very macho, but perhaps people who don't know how to present graphs so that they are as clear as possible should also be sent back to school. At the very least, if you're going to present a graph, point to an area you've coloured in, say that it represents the energy remaining, and then also emphasise how small it is, you should remind your audience that they need to imagine adding in the extra rectangles in order to get a true impression of that energy. Not adding that reminder is a sign of either ignorance, forgetfulness, or dishonesty --- if you're arguing against the claims of people like Batteriser, you don't want to give them the opportunity to accuse you of dishonesty. People being misled by the graph that you present can very quickly become your problem. (I realise that you personally might not be trying to refute Batteriser's claims, but that is an activity of a lot of the readers of this forum.)QuoteAnd again the attempt to distract by talking about the battery life after it reaches the 0.8 V --- this is a dishonest tactic, you know full well that my position has nothing to do with this question.

author: Fungus
author_rank: Super Contributor
timestamp: August 02, 2015, 08:13:18 am
content: Quote from: dk27 on August 02, 2015, 07:55:54 amThe graph shows voltage against time. There's nothing unclear about it.Very few people are probably interested in the exact number of joules remaining (and probably even fewer are trying to calculate it using that graph).Battery lifetime and voltage are FAR more important than the remaining Joules. Reducing the Y axis would be more damaging than chopping it off.If you really want the rest you can just draw a box underneath. It still won't get you any hard numbers though - you'll need to do the math if you want real numbers.

author: EEVblog
author_rank: Unknown
timestamp: August 02, 2015, 09:34:46 am
content: Quote from: dk27 on August 02, 2015, 07:55:54 amYou guys can nerd fight all you want in here over the technical details, just leave me out of it thanks.

author: Don Hills
author_rank: Regular Contributor
timestamp: August 03, 2015, 12:46:48 am
content: (Moved to a different thread.)

author: EEVblog
author_rank: Unknown
timestamp: August 03, 2015, 12:47:10 am
content: For those playing along at home, dk27 decided to take his bat and ball and go home.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: April 03, 2016, 11:41:59 am
content: Quote from: EEVblog on July 28, 2015, 09:44:27 amWould you develop Integral[V-0.8,{t,t1,t2}] here, for us?JPD