Spreadsheet "pro tip": Instead of dragging a formula down, you can double-click the little black square that marks the lower right corner of the selected cell, i.e. the one you'd normally drag down. This expands the formula to all the data rows. And with many rows it saves a crapload of time. :-)
EEVblog "So if your still hanging in there, thank you for that..." No, thank YOU Dave! That was super helpful and a topic I have never seen discussed elsewhere on the internet. Top notch!
Ben Bogard Thanks, that was the idea. I hadn't seen the capacity done graphically with line intersections like this before. And just putting it up on the Batteriser blog post without explanation didn't seem enough, so thought I'd explain how I got that graph and why.
Whenever you have a formula that gives you what you want, but is inverted, you can just subtract the max value by it and it'll inverse it... in your case just do =(1-D5/MAX(D))*100 .... Also, a faster/easier way to get a more accurate number is to use the hard numbers itself to get the actual percentage... Just find the shutoff voltage (1.1v) on the C column then when you have the formula plugged in go to column D and it'll tell you a very good estimate (assuming you have enough data points) of your remaining percentage.... It's more accurate than doing it in the graph.
Hey Dave, nice Video! But the graphical analysis with the areas you did in the beginning was actually wrong. The area you marked is _not_ the remaining capacity of the battery. You have to expand the graph all the way down to 0V to get the correct areas/proportions between the areas. (You are also mixing up capacity (Wh) with volthours, but considering the constant current that's fine).
Stuner Agreed, but it's nitpicky. The voltage-time plot is not energy but proportional to the energy in this case (Dave may have been sloppy with his words a couple times but he knows this of course...). Also the energy past 0.8 volts is probably negligible and iirc it's the cutout voltage of the batteriser.
wanswers I didn't write that as clearly as I should have. The problem is not with stopping at 0.8V (this is probably negligable), but rather not counting the voltage area below 0.8V which _significantly_ changes the proportions of the voltage areas.
Stuner Ah, I think I see what you mean. Despite the area below 0.8v not being on the graph (which can be visually misleading), the method Dave used to calculate the area, i.e. Riemann sum style integral approximation, did in fact account for that extra area. The graphical description sort of ignored this, but the numerical analysis is still correct.
Stuner It should be clear that the "area under the curve" is the area to the x-axis. Otherwise you're claiming that just zooming into a curve changes its integral.
Instead of dragging down for excel to complete the calculations of all the cells you can double click and it will auto fill down to the adjacent line, saved me a lot of time for large amounts of data. I am using MS excel 2013, not sure of this function on other versions. Dave, There is always energy remaining E=MC^2 - Thank you for channel!!!
I've never even seen this kind of graph for measuring remaining battery capacity... I learned something today! :D Also, testing batteries under load... Never something I had ever considered, guess I'll be retraining myself on that aspect. I knew they regained some voltage, but I was told for years that you just hook the trusty old multimeter to the battery to see if it's good. Curse my know-nothing teachers of old~
Zeigwolf Just using a meter on the battery is good enough to determine if you should just toss it out or not, but you shouldn't use it to determine capacity. This is why the manufacturers give you under load voltage discharge graphs.
This is good to know if you are trying to program a battery capacity calculator in your electronics project, it is not linear, it must match the voltage curve of your battery.
I can't tell you how many years I would tell this to people in the video production industry using wireless mics. They would buy a battery tester, and it would only measure voltage(radio shack?). It it showed close to 9 volts, they would put it back into the wireless pack, and put it back on the talent. That was back in the late 90's - early 2000's. By 2005's everyone got burned too many times and would only use a 9 volt battery for 3-4 hours at the most, then throw them away. Urrr.
GeorgeGraves I used to measure batteries with no load with my multimeter . Someone gave me a battery tester that was from Radio Shack and it even says on the front that it uses a 150mA load for the normal setting (also has 1mA load settings for button-type and 3V lithium button cells, as those are generally in low-drain applications).
Elektronik He does say that he's not really using watt hours as a unit, and that the particular unit doesn't matter for the purpose of what he's doing. The "Wh" figure was just an example as he was talking.
EEVblog Let's say that expected capacity is 3 Wh, then 12 / 3 = 4 hours. Duracell graph says that it takes ~27 hours to discharge battery to 0.8v under 100mA load. What's going on ? :)
Elektronik I think that technically, you need to multiply that I*V by the (delta t) i.e. the time-step size, after adding the previous energy value. So, E = E(n-1) + V(n)*I(n)*{t(n)-t(n-1)} Since your time resolution was 1 second, you don't get any errors (apart from being in Joules rather than Wh). But if you had any other value, your energy value will be wrong I think.
Hi EEVblog, I have not seen all then videos in your sequence but will watch them as you also show how to use that spreadsheet. Thank you for your upload. Take care mrbluenun
Good video. However you have some errors in it. When you say that the capacity is the area under the graph that is correct but when you mark this area its not. You must include the area from 0v to 0.8v aswell.
Constant resistance is used for life calculations in flashlights, among other things. And there are occasionally units sold that take batteries, thus the spec. ;-)
Thanks for the Excel details mate:Not boring at all; and just the kind of gory detail most people gloss over, which is a shame, because it's the tricky bit!!!
If you want to find the remaining battery percentage all you have to do is look at your table (21:39). At 1V cut of you have used 95.51289816% of your capacity.
Nice video, good to see some facts for once in this discussion. I loved the way they used the Apple keyboard power graph as an example. It's tricked into showing 100% as it's presumably based on voltage and is obviously not a valid measure if using a DC-DC converter. So the power gauge will always show 100% then drop sharply to 0% when the DC DC drops out. The keyboard with batteries alone will slowly drop towards 0% as the voltage (under load) decreases. Whether that is before or after the one with DC-DC probably depends on load I guess, the efficiency loss through the DC-DC, vs the small gain under the cutoff. Be interesting to see if a 90% DC-DC converter would be more efficient at any load, at what load would it break even, and where it would be a lot worse to have the DC-DC. Of course, they should have done that and presented those results, but I can't seem to locate them........
***** Yep, the battery life gauge is calibrated for an undoctored battery. Of course it'll show a falsely-high battery remaining capacity when you trick it with a DC-DC converter, as it hides the variation in voltage that's necessary to be able to measure battery capacity. A DC-DC converter in line is like having a bank that always says you have thousands in your account, but then suddenly goes to zero when you make the final $10 withdrawal.
gblargg Exactly, maybe they could remarket it as a device with introduces some exciting jeopardy into you life, you'll never quite known when your keyboard is going to die.
Dc-dc converters vary in efficiency a lot with respect to load and duty ratio - you'd have to define type of dc-dc and when it should kick in to be able to retrieve some information from the experiment.
Instead of doing the graph stuff, one could just look at the table at the cutout-voltage and get the wasted capacity in the most right column in an instand. but hey, it looks nice and even a manager can understand this visual approach :P
I wish they showed the short-circuit current on datasheets. Knowing that, you would be able to infer the worst case cut off for a device with a boost converter. At some point, the current would not be sufficient to charge the inductor no matter how wide the PWM control signal.
There is an error in the spreadsheet work. The first energy consumption number is 0, which is an apriori figure. But the second entry shows you taking previous energy consumed (t=previous) + V (t=current) * I (t=current) when in fact it should be energy consumed (t=previous) + voltage (t=previous) * current (t=previous). This isn't a grievous since energy consumed at t=previous and t=current are nearly equal but it does mean that you should have one additional entry, so you're total will be off by a little more than approximately 1/# of entries.
EEVblog Dave, couldn't you just look at the spreadsheet data to find the remaining capacity? Find the row where you drop below 1 volt and look at the number you generate for remaining battery capacity at that point. That seems simpler, quicker, and more accurate than looking it graphically and estimating where the line intersects the right Y-axis.
Or you could just do =100-D5/D$99 You don't need the $D since you are not dragging the cell horizontally... although technically $D$99 will work fine also. Great video Dave!
Very interesting. I was kind of hoping for an old school area under the curve measurement where you make very large print on a strip chart recorder then cut out and weigh the relevant sections with a triple beam balance.
Nice vid Dave, always a learning experience! Its a pretty good technique to show those with little understanding of batteries how much energy your wasting.
Odd, it seems as though one of my canon point-and-shoot cameras rejects the batteries at a very high voltage, like 1.2 or so. The thing eats batteries like nothing else.
***** cameras have a relatively high current demand. And I think you have measured the voltage after removing the battery from the camera which is not what Dave is talking about. Dave was speaking for the voltage under load. But yeah: Old consumer devices eat a set of batteries for breakfast, lunch and dinner. But with modern DCDC converts and micro-controllers it got better.
***** Is that by any chance when using the flash option and charging the capacitors for that? When using decent NiMH batteries with a much lower internal resistance than alkalines I find that I usually get more flashes out of the batteries and as a bonus the capacitors tend to charge faster too meaning less time between photos. Give it a shot sometime, especially since the energy capacity left in the battery isn't entirely wasted you'll just top it off faster when putting them back in the charger if it does leave some in there. The only real problem would be if its not optimised for the lower nominal voltage of NiMH/NiCd but I find this to be a rare thing in any photography gear with flash.
As right as you are abotu the Batteriser marketing I still have one argument to defend the usability of this thing. I just made a test with an alkaline battery that had 1.35v with no load in the beginning. I intended to find out how much energy in Wh I can squeeze out of it when draining it from 0.8v to 0v. To first bring it down to 0.8 I put a load of 200mA on it which caused it to drop down to 0.85v under load. After draining 500mAh it had 0.64v with the 200mA load on it and 1.1v with no load. As you can see that were 500mAh and it's still going in the range where everything apart from solely resistive devices like flashlights or some dumb fans or so would've shut off. I can't think of many devices with microcontrollers and low voltage cut off that would put such high loads on the battery but that would be a situation where the batteriser could really help you.
Honey BooBoo So you're saying that the battery was 0.85V under a 200mA load, and you ran it for 2.5 hours and it only dropped to 0.64V? Look at the battery discharge curve for a 200mA load at 10:35 in the video; it's almost vertical at 0.8V. Once you hit 0.8V you don't even have an hour left. Your test doesn't make sense.
gblargg But that's exactly the point. Under load it drops dramatically. It's an exponential behaviour. The difference in the voltage drop between 100mA and 200mA is very high and the reason it could sustain the voltage under load so well is that even after the 500mAh discharge if you had put a 100mA load on it it would have been well above 0.8v maybe like 0.9v-1.0v Edit: To understand it you also have to know that the reason why the 200mA load graph from the vid differs so much from the results I had is that the graph is for a fresh Duracell Coppertop and I used some years old Philips alkaline that I found lying around somewhere. Age and quality makes a big difference in voltage drop under load but alkaline batteries don't lose that much capacity from just lying around. A nimh would've been completely flat after lying around for so long.
Do you read the graph differently than me? My reading is that once the cell is discharged to the point where it outputs 0.8V when 200mA current is being drawn, it has less than half an hour left (the slope is nearly vertical). How could you draw 200mA for 2.5 hours at all when the cell was already only 0.8V when you start, let alone have the voltage still be 0.65V under load after that time?
gblargg Because it's an old cell that has higher internal resistance etc. than a new one. Because of that it drops dramatically more under load than a fresh cell without having remarkably less energy in it. Just a slight bit is lost to time and more internal resistance ----> heat.
I really wish we had the current/voltage measurements down to zero volts. While I'm pretty sure the falloff is so quick that it isn't significant, this experiment does not actually measure the total (theoretically) possible battery capacity; rather, it measures the capacity only as far as the designated cutoff voltage of 0.8V. I reckon there is another 5-10% in there. Not that you could get at it easily without a bunch of cell in series, and without battery leakage to ruin your day, etc. etc. :)
You can determine if a AA battery is good or not with multimeter. Just stick the meter on amps and do a 1 second reading. If it will not deliver enough current it is bad.
I have no idea what you are talking about mate. It does make me want to learn about it though. I think people from candlepower forums(where all flashaholics hang out) will enjoy this.
Power Max Yeah that was going to be my next comment. Even if the curve drops off really fast, you still have a huge rectangle hidden from this graph since 0.8 -> 0 is even bigger than the 1.5 -> 0.8 voltage range that we can see on the current graph
Power Max yes that is true that there is much more "graph" to be seen, but notice that the area we are talking about is not the entire area under the curve up until the cut off voltage we've chosen (ie 1.5V -> 1.1V for example), that area is the energy we have used. The wasted energy is the area under the curve from the cutoff voltage to 0.8V (ie 1.1V -> 0.8V). Because these graphs basically become vertical at 0.8V the area under the curve from 0.8V -> 0V is negligible.
I think that technically, you need to multiply that I*V by the (delta t) i.e. the time-step size, after adding the previous energy value. So, E = E(n-1) + V(n)*I(n)*{t(n)-t(n-1)} Since your time resolution was 1 second, you don't get any errors (apart from being in Joules rather than Wh). But if you had any other value, your energy value will be wrong I think.
Awesome, I wouldn't thought of doing this that way. I would use least square method to approximate the function and then calculate the area below by integration. That's my computer science degree in action ;)
I remember getting a Sinclair programmable calculator (24 step) RPN and put in a fresh standard 9v battery and it did not even turn on. It only ran with an alkaline battery, and that didn't last long. C1978
You don't actually need to align the graphs like that. At least not for this purpose. You'd get the same values if you left the characteristic discharge curve going from 0. Nice video.
Very interesting. My first digital camera (a toy really at 640x480) only worked with non-rechargeable AA batteries and even those didn't last long in the camera, but still worked for ages in other devices after removing them from the camera. Obviously a poor design with a high voltage cutoff!
Does the wasted percentage of energy have a dependancy on the current drawn? Looking at those curves on the datasheet, it looks like you may waste more and more energy at higher current levels for a given cutoff voltage. It does make sense since higher current levels will drop more voltage across the ESR so therefore it will prematurely reach cutoff level.
It seems to me like hours of usage would be more relevant and simpler to calculate rather than using remaining energy. You would not have even needed to compile the second curve using just time.
Of course that would be convenient if it worked, but it doesn't. Depending on how what you plug the battery to you get very much different usage time from it.
superdau Yes, remaining energy vs. time is not a linear relationship. But you know what else changes over time? The power requirement of the device does. From a user's standpoint the remaining energy isn't as relevant as remaining usage time. If you establish a plot for the specific device you are using that extends to the full .8 volts and measure the shut down voltage of the device, then the theoretical remaining usage time could be calculated easily if the device could actually function all the way down to .8 volts.
friendofCHAKA Is it just me, or do some of the Amprobe products on Amazon look fake? In some of the pictures you can see that "contractor" is spelled "contrator" on the front cover of the multimeters. It's the sort of Engrish you'd expect to find on counterfeit products.
Your video is just about the right length, so don´t worry :) When i´m looking at your discharge curve, i can almost hear my professor again sayin: ok now we need to derive the equation for that, using some integral math and derive where we have our desired 10% for *every* current. Homework guys, see you all next week....that was the usual....didn´t like it back then and don´t like it today...i came up with these gfx as well....he didn´t like that, but hey, i was way faster, as i used to do this in exams as well... He didn´t like it anyway and, as usual, i failed...hahaa. I don´t need that math thing throughout, i only need to know the result for my specific problem! Hey Prof. do you hear me...;) We haven´t got that amount of time in the industry to sit back and derive formulars/equations....we have mathlab/mathcad/maple do let it work that out for us, if we need that.....we have to solve problems, that´s why we been hired for...;)
Well, aligning both y-axis was not really required. As long as they share the same timescale everything is fine. Also you should be careful with area under the curve because displaying data only down to 0.8V skews the real ratio!
Just to get this right, when looking at your curves I must come to the conclusion that after these kind of batteries are discharged to 0.8V (under load of course) there is only a very very small amount of energy left in that cell? Otherwise it would not be very engineering like to just put the zero point of energy left right there, or would it? I'm sorry for bothering you, but showing the complete curve (if even possible) would be a better way of debunking that silly product. I'm really keen to see the results of your long term battery test. I hope that it will show the complete curve way down to zero volts (if, as I said before, this is even possible). Thanks for your great videos, both educational and entertaining, I love it! :)
Great video Dave, as a process engineer I live by the motto, "Question Everything" and your in depth analysis of measuring batter efficiency was well done. Was this in response to the new video that Batteriser posted?
Is that a line graph or a scatter plot? (More important if the X data is not at regular intervals...) For constant power discharge, instead of constant current, the line would actually end up linear, since the capacity is also a percentage of energy... With the constant current example. the remaining energy is a power percentage, and the power of the load is decreasing as the voltage decrease... (A normal (mostly) constant resistance load is a non-LED flashlight....)
Useful video, been aware of it for ages and I usually test the old alkaline battery pile of my parents every year by plugging some banana plug on one end and alligator clips on the other end leads then hook a resistor up between the two probes that way and measure voltage. If it drops quickly I discard them if they're still decent enough for use in low power applications I'll put them in the "half dead but still useful box". An interesting fact to be aware of though is that there are some battery chemistries where the open circuit voltage is a sound indication of state of charge. With my 18650 lithium ion batteries its a rather good indication and while I'm no expert at lead acid batteries, if I'm not mistaken it applies to them as well to an extent?
Would the mid-point intersection be the minimum cutoff voltage that the Batterizer has a chance of being useful? As in, above that point might be useful, below that point useless to detrimental.
In general, you explain how to compute an integral and basically, what you say is somewhat correct and it will certainly give you a result which is reasonably close to the precise data. But you made some systematic errors (which luckily won't change the result very much). 1. The capacity is the area under the full curve which goes right down to zero, not 0.8. That is why your first guess of 10 to 20% remaining capacity was a bit on the exaggerated side. If you see the full area, you would - right from the start - have gotten a better estimate. 2. When doing an integral with empirical data points, you should consider taking the previous data points into your sum. You can simply use a linear interpolation by just taking the average of the current and the previous point. Would be a tiny bit more accurat, but of course still based on assumptions. Does not make a huge difference, but just relying the current data points is not the best assumption. 3. You Do Not Divide by the last data point. That is wrong. You divide by the maximal value. Which you do not have measured. It will of course not be far off of what is writte in your last row but remember that the curve goes down to zero. There is a tiny amount of capacity. This does not show in your graph. You can guess it to be only about 1 or 2% of the full capacity, but nontheless, your graph is (strictly speaking) having systematic errors. Overall, again, nothing new under the sun. Of course, for simple calculations, that is more than good enough. But I just like to be a smart-ass. :)
Very informative. It would be interesting to see what happens below 0.8V. It won't change your general conclusions much, but it would be nice to close that 'gap' because sure as bob's your uncle the 800% BSers will try to claim that there is a 'ton'of energy in that gap. Keep it up man!
mart fart Just looked that gem up, I think the battery voltage was the least if it's problems! RPN, need I say more. I'm so glad we don't have technology like that now.
Wobblycogs Workshop Haha it was the scientific programmable in a black case and they were selling them off cheap, I was still at school and it probably hindered me in math more than helping. I also bought a Science of Cambridge MK14 micro kit @39.95 also Sinclair c1978 I used to boast I built a computer as a kid at school...the truth is more prosaic.
Also I hardly believe your graphical analysis method is the best method out there when compared to actual integration using Simpson's or trapezoidal rule. Which you could've just as easily done with the data at hand.
Sorry Dave but if you want to talk about ratios of used energy capacity vs remaining energy capacity and make that visible as an area under a graph, you *do* want to make that area go down to 0V (what you almost did at 17:48). Otherwise it's actually not correct. Btw: There is a very neat built-in Excel function called VLOOKUP which does some magic for you. You should include that in your next spreadsheet tutorial ;)
Dave if you have any super capacitors it would be a interesting test to see the comparison with a Cap on its own and with a battery connected in parallel. I have 18 x 3000F Maxwell caps in series and parallel configuration for jumpstarting cars and trucks. 3 sets in parallel of 6 in series 2.7V Caps Max each 16.2v max for the setup. its a beast ! and made 2 of them. often wondered what energy capacity they actual have stored in them.
I have a question about battery discharge safety with a PWM application. Maybe someone can answer this or point me to the right direction to where can I read or learn or ask about this? Thanks! I have a 8.4V rechargeable battery (dual 18650 in series for vaping) and a heating element with 0.1 ohm and lets say a 10% duty cycle at maybe 100hz (arduino PWM with a mosfet). The average amp draw would be 8.4 amp but the amp draw during the duty on would be 84 amps! I plan to use Samsung 25R "safe" chemistry batteries (not lipo) that are rated for 20 amps continuous discharge. In the datasheet it actually mentions high amp pulse discharge tests. So basically I'm wondering if this is safe to average? What makes batteries actually explode? Temperature? In any case thanks for the awesome video!
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.
Dave, I use the AA, AAA and "AAAA" batteries ! The "AAAA" fit the Streamlight Stylus and a Laser Pointer that fit next to a couple pens that fit the Fisher Space Pen Refills ! Also use the CR123A in a LED flashlight and an UV Water Pureifier. Have used the 1.5 VDC "A" battery and even took one apart ! Have you done a teardown of an "A" Battery ? Like the Alkaline, Li and High Tech Rechargeable Batteries ! (Say NO to Cadium Batteries) Eyes UP and lights down, tjl Sent by Win7Pro64 w/ADSL
EEVblog Why only go down to 0.8V? In case you're using a DC-DC converter, it would put out a "useable" voltage anyways, so while looking at the discharge graph on the data sheet tells me that there isn't much left in the cell from 0.8V to 0.0V (or, let's say 0.1V), why not go ahead and see/measure the graph all the way down to zero? OK, in case of rechargeable batteries, this would obviously be fatal for the battery, but if you're using disposable alkaline batteries anyways, it wouldn't hurt a thing. Any hey, maybe the curve behaves somehow unexpectedly below 0.8V? OK, to get useable power out of it, it would have to draw a much higher current, which means a constant power curve is the only thing where this area could be relevant, but still, I'd be totally interested how a battery behaves when it's operated so much outside its specs (but lack the means of easily measuring the thing...writing a value down every minute from the multimeter just isn't an option if it takes several hours for the battery to get down to 0.1V).
Don't you have to take the first timestep into account? If I understand it correct, there was 0.105A at 1.65V for 1 timestep, but it wasn't used in the calculations. You started with the second timeslot. So don't you have to calculate E(t)=E(t-1)+I(t-1)*U(t-1) (with E for Energy)?
dhufi How does your formula change anything. With it you're just shifting the error to the end of the data, because now the last entry of the table isn't used. That's the error you get from discrete math. The voltage/current isn't constant for the whole timestep anyway and still it's used in the calculation. So for a better calculation you should use the voltage/current average of two consecutive points (this assumes a linear function within each timestep, which is more accurate, while the first version assumes a constant function). But the best way around: make the timesteps smaller, reduce the error.
hey Dave I am an electronics noob so don't mind me asking is the cut off voltage a design feature or a product of efficiency of the components used in the circuit i.e something that's an after thought
hyronov A good designer would try try and make the cutoff voltage as low as possible in order to use the most energy from the battery. Sometimes however that's not easy or cost effective for various reasons. It should not be an afterthought, if it is, you're a bad designer.
EEVblog well thanks for the reply adding to that question. Say two manufactures using the same components is it possible that one really well engineered will have a lower cut off voltage. what rough of order of magnitude lower could it be .1V~.5V for AA cell.
hyronov Under 0.8V cutoff is pretty pointless, there is very little energy left under 0.8 for any sort of real load. This is why the industry uses 0.8V as the defacto standard minimum cutoff voltage.
HI dave, I don't know if you could help me but I have a LCD screen that I want to use with my arduino. The problem is that I can find the datasheet anywhere (I even send an email to the company that sells them) do you know how I could get the pinout ?
I hate batteries period. I'd never use anything but NiMh or better though. I don't see this as a problem except the unit may cut out earlier than the batteriser installed though...
EEVblog No. You say: "and if you want watt-hours and stuff you'd put like divide by 3600 .." The problem is that you are doing a newbies guide to calculate energy, but you completely skip the proper way to calculate watt-hours, yet you label your column "Wh". You're showing 11+ Wh for a single AA cell. Next time someone new to this tries to calculate battery energi, they will do as you and do current * voltage. Why not show proper engineering practice? (T2-T1)*((I2+I1)/2)*V
SirJMDDK First he specifically said it's not really Wh. Secondly it's totally irrelevant for getting the remaining percentage. Thirdly everyone doing real energy calculations should know this basic math anyway and how to convert units like Wh/Ws/Joules whatever. Since he collapsed multiple measurements into a single data point the first column isn't any usual timeunit anyway. It might be in units of "27.2seconds" for example (that's why he said "put *like* divide by 3600"). Imagine the confusion when doig this division to getto the correct Wh.
Not quite right. If you're comparing a DC-DC converter to constant current, you have to account for BOTH the cutoff voltage and the constant current vs constant power. If your constant current product needs 1v, you're wasting all the area between the horizontal 1v line and the battery voltage curve above it. If you're going to bother doing it right, you should also continue the curve beyond .8v, I'd continue it until the battery no longer delivers enough power to run the device, whatever that might be(at any voltage/current combination), then compare actual active and standby time using your DC-DC converter vs linear regulator.
sleepib If you carry on past 0.8V the voltage drops off a cliff and has very little remaining capacity. You might as well treat the cell as being dead. Good point about the wasted power and this leads to another nail in the coffin of Batteriser. If you add a Batteriser to boost the voltage to a constant current load it will waste more energy than using the cell on its own. The regulator has to drop more voltage and waste more power!
sleepib A decent DC-DC-converter is a constant power load. So why would it waste the energy above the output voltage? That's how a linear regulator works.
superdau the constant current example uses a linear regulator, but doesn't account for the energy wasted by the voltage drop across the linear regulator, it only accounts for the energy left in the battery when your device no longer functions. If you're trying to decide whether it's worth switching to a DC-DC converter, you need to account for both.
EEVblog Thanks for all the steps. I hate when people, books or Internet Tutorials skip steps, even when I know how to do it, because if I did something wrong I can compare both methods.
Sasha Whitefur "unusable" is inaccurate as the power could be used, but it is not economical. And it is not even power, it is energy. It should be "unused" or "remaining" energy. But either way: It´s just changing words around and adding half sentences. For a PhD thesis, this can make all the difference, but here it is about the general statement/concept. Off course, general statements can be significantly off (like when driving with the last gallon to the gas station and either to make or not), but in those cases more care has to be applied anyway.
your calculation in the spreadsheet seems to be wrong to me. First: why does is the first row have 0 WH at 1min? you should use the mean value between two rows - but sure that only makes a little difference. Second: you should take the measurements down to 0V to get the real capacity if you want to show that there is not much energy left at .9V. Randomly setting the zero reference to some Voltage level gives you wrong result. imagine that you only took measurements to 1V and took that as 0-Energy reference. The results would look similar while still a bit of energyis present. this is an attack point in your proof
I saw that Yahoo posted a news article today about the Batteriser that actually called you out, EEVblog. www.yahoo.com/makers/breakthrough-battery-gadget-answers-critics-125063020800.html
Under load, OK, I got that the first time it was stated but wholly hell, stop repeating yourself please. It's so dam annoying when you tubers constantly repeat things.. I lost count after the 3rd time you stated under load......
EEVblog Really? I'd like to thank my mother, my father, the dead rat I saw coming home, and the refresh button.... Without those, I would have never been first. Still really interesting stuff.
![[MV] A leap of faith...ความเชื่อใจครั้งสุดท้าย (From "Petrichor the series")](http://i.ytimg.com/vi/U1piZH2CNXA/mqdefault.jpg)
Spreadsheet "pro tip": Instead of dragging a formula down, you can double-click the little black square that marks the lower right corner of the selected cell, i.e. the one you'd normally drag down.
This expands the formula to all the data rows. And with many rows it saves a crapload of time. :-)
EEVblog "So if your still hanging in there, thank you for that..." No, thank YOU Dave! That was super helpful and a topic I have never seen discussed elsewhere on the internet. Top notch!
Ben Bogard Thanks, that was the idea. I hadn't seen the capacity done graphically with line intersections like this before. And just putting it up on the Batteriser blog post without explanation didn't seem enough, so thought I'd explain how I got that graph and why.
Whenever you have a formula that gives you what you want, but is inverted, you can just subtract the max value by it and it'll inverse it... in your case just do =(1-D5/MAX(D))*100 .... Also, a faster/easier way to get a more accurate number is to use the hard numbers itself to get the actual percentage... Just find the shutoff voltage (1.1v) on the C column then when you have the formula plugged in go to column D and it'll tell you a very good estimate (assuming you have enough data points) of your remaining percentage.... It's more accurate than doing it in the graph.
Hey Dave, nice Video! But the graphical analysis with the areas you did in the beginning was actually wrong. The area you marked is _not_ the remaining capacity of the battery. You have to expand the graph all the way down to 0V to get the correct areas/proportions between the areas. (You are also mixing up capacity (Wh) with volthours, but considering the constant current that's fine).
Stuner Agreed, but it's nitpicky. The voltage-time plot is not energy but proportional to the energy in this case (Dave may have been sloppy with his words a couple times but he knows this of course...). Also the energy past 0.8 volts is probably negligible and iirc it's the cutout voltage of the batteriser.
wanswers I didn't write that as clearly as I should have. The problem is not with stopping at 0.8V (this is probably negligable), but rather not counting the voltage area below 0.8V which _significantly_ changes the proportions of the voltage areas.
Indeed
Stuner Ah, I think I see what you mean. Despite the area below 0.8v not being on the graph (which can be visually misleading), the method Dave used to calculate the area, i.e. Riemann sum style integral approximation, did in fact account for that extra area. The graphical description sort of ignored this, but the numerical analysis is still correct.
Stuner
It should be clear that the "area under the curve" is the area to the x-axis. Otherwise you're claiming that just zooming into a curve changes its integral.
Instead of dragging down for excel to complete the calculations of all the cells you can double click and it will auto fill down to the adjacent line, saved me a lot of time for large amounts of data. I am using MS excel 2013, not sure of this function on other versions.
Dave, There is always energy remaining E=MC^2 - Thank you for channel!!!
I've never even seen this kind of graph for measuring remaining battery capacity... I learned something today! :D
Also, testing batteries under load... Never something I had ever considered, guess I'll be retraining myself on that aspect. I knew they regained some voltage, but I was told for years that you just hook the trusty old multimeter to the battery to see if it's good. Curse my know-nothing teachers of old~
Zeigwolf Just using a meter on the battery is good enough to determine if you should just toss it out or not, but you shouldn't use it to determine capacity. This is why the manufacturers give you under load voltage discharge graphs.
Zeigwolf The current range on multi meters is good for testing batteries, but it's not too good for the batteries though :-)
This is good to know if you are trying to program a battery capacity calculator in your electronics project, it is not linear, it must match the voltage curve of your battery.
I can't tell you how many years I would tell this to people in the video production industry using wireless mics. They would buy a battery tester, and it would only measure voltage(radio shack?). It it showed close to 9 volts, they would put it back into the wireless pack, and put it back on the talent. That was back in the late 90's - early 2000's. By 2005's everyone got burned too many times and would only use a 9 volt battery for 3-4 hours at the most, then throw them away. Urrr.
GeorgeGraves I used to measure batteries with no load with my multimeter . Someone gave me a battery tester that was from Radio Shack and it even says on the front that it uses a 150mA load for the normal setting (also has 1mA load settings for button-type and 3V lithium button cells, as those are generally in low-drain applications).
At 17:00: Fully Battery Capacity of your AA-Battery is calculated to 12 Wh. The correct Capacity is about 3 Wh (@100mA)
Elektronik He does say that he's not really using watt hours as a unit, and that the particular unit doesn't matter for the purpose of what he's doing. The "Wh" figure was just an example as he was talking.
Mythricia that means he tested for 4 hours?
if 1 hour = 3Wh then 4 hours = 12Wh ? maybe? just curious
Elektronik No, I mentioned the actual units are not corrected to actual Wh.
EEVblog Let's say that expected capacity is 3 Wh, then 12 / 3 = 4 hours. Duracell graph says that it takes ~27 hours to discharge battery to 0.8v under 100mA load. What's going on ? :)
Elektronik I think that technically, you need to multiply that I*V by the (delta t) i.e. the time-step size, after adding the previous energy value.
So,
E = E(n-1) + V(n)*I(n)*{t(n)-t(n-1)}
Since your time resolution was 1 second, you don't get any errors (apart from being in Joules rather than Wh). But if you had any other value, your energy value will be wrong I think.
Hi EEVblog,
I have not seen all then videos in your sequence but will watch them as you also show how to use that spreadsheet. Thank you for your upload.
Take care
mrbluenun
Wow, yesterday I was actually wondering "how much space in my batteries are wasted....". What a coincidence.
Good video. However you have some errors in it. When you say that the capacity is the area under the graph that is correct but when you mark this area its not. You must include the area from 0v to 0.8v aswell.
Magnus Arvidsson
It should be clear that the "area under the curve" is the area down to the axis.
Magnus Arvidsson Dave must make your comment an _integral_ part of this video, _derived_ from your observation.
superdau Well it should be but at 28:05 Dave clearly shows that this is not clear at all.
Constant resistance is used for life calculations in flashlights, among other things. And there are occasionally units sold that take batteries, thus the spec. ;-)
Thanks for the Excel details mate:Not boring at all; and just the kind of gory detail most people gloss over, which is a shame, because it's the tricky bit!!!
If you want to find the remaining battery percentage all you have to do is look at your table (21:39). At 1V cut of you have used 95.51289816% of your capacity.
Nice video, good to see some facts for once in this discussion. I loved the way they used the Apple keyboard power graph as an example. It's tricked into showing 100% as it's presumably based on voltage and is obviously not a valid measure if using a DC-DC converter. So the power gauge will always show 100% then drop sharply to 0% when the DC DC drops out. The keyboard with batteries alone will slowly drop towards 0% as the voltage (under load) decreases. Whether that is before or after the one with DC-DC probably depends on load I guess, the efficiency loss through the DC-DC, vs the small gain under the cutoff. Be interesting to see if a 90% DC-DC converter would be more efficient at any load, at what load would it break even, and where it would be a lot worse to have the DC-DC. Of course, they should have done that and presented those results, but I can't seem to locate them........
***** Yep, the battery life gauge is calibrated for an undoctored battery. Of course it'll show a falsely-high battery remaining capacity when you trick it with a DC-DC converter, as it hides the variation in voltage that's necessary to be able to measure battery capacity. A DC-DC converter in line is like having a bank that always says you have thousands in your account, but then suddenly goes to zero when you make the final $10 withdrawal.
gblargg Exactly, maybe they could remarket it as a device with introduces some exciting jeopardy into you life, you'll never quite known when your keyboard is going to die.
Dave, thanks man. I was blind but now I can see. Thanks a million.
Now redo this video with a ~ 80-90% efficient DC-DC converter on it to simulate the batteriser.
Dc-dc converters vary in efficiency a lot with respect to load and duty ratio - you'd have to define type of dc-dc and when it should kick in to be able to retrieve some information from the experiment.
Instead of doing the graph stuff, one could just look at the table at the cutout-voltage and get the wasted capacity in the most right column in an instand.
but hey, it looks nice and even a manager can understand this visual approach :P
hassiaschbi Correct. And it's nicer to publish a graph than a table of numbers.
I wish they showed the short-circuit current on datasheets. Knowing that, you would be able to infer the worst case cut off for a device with a boost converter. At some point, the current would not be sufficient to charge the inductor no matter how wide the PWM control signal.
There is an error in the spreadsheet work. The first energy consumption number is 0, which is an apriori figure. But the second entry shows you taking previous energy consumed (t=previous) + V (t=current) * I (t=current) when in fact it should be energy consumed (t=previous) + voltage (t=previous) * current (t=previous). This isn't a grievous since energy consumed at t=previous and t=current are nearly equal but it does mean that you should have one additional entry, so you're total will be off by a little more than approximately 1/# of entries.
EEVblog Dave, couldn't you just look at the spreadsheet data to find the remaining capacity? Find the row where you drop below 1 volt and look at the number you generate for remaining battery capacity at that point. That seems simpler, quicker, and more accurate than looking it graphically and estimating where the line intersects the right Y-axis.
Aaron Hurd Yeah there is a function for that called VLOOKUP so you don't even have to do the search yourself.
Aaron Hurd As soon as I saw the hand-drawn lines come out, I went back and looked at the table.. "4.48%" :D
Or you could just do =100-D5/D$99
You don't need the $D since you are not dragging the cell horizontally... although technically $D$99 will work fine also.
Great video Dave!
Very interesting. I was kind of hoping for an old school area under the curve measurement where you make very large print on a strip chart recorder then cut out and weigh the relevant sections with a triple beam balance.
Nice vid Dave, always a learning experience! Its a pretty good technique to show those with little understanding of batteries how much energy your wasting.
Thanks Dave!
Just what i needed to know for the project I am working on.
Odd, it seems as though one of my canon point-and-shoot cameras rejects the batteries at a very high voltage, like 1.2 or so. The thing eats batteries like nothing else.
***** cameras have a relatively high current demand. And I think you have measured the voltage after removing the battery from the camera which is not what Dave is talking about. Dave was speaking for the voltage under load.
But yeah: Old consumer devices eat a set of batteries for breakfast, lunch and dinner. But with modern DCDC converts and micro-controllers it got better.
***** Is that by any chance when using the flash option and charging the capacitors for that?
When using decent NiMH batteries with a much lower internal resistance than alkalines I find that I usually get more flashes out of the batteries and as a bonus the capacitors tend to charge faster too meaning less time between photos.
Give it a shot sometime, especially since the energy capacity left in the battery isn't entirely wasted you'll just top it off faster when putting them back in the charger if it does leave some in there. The only real problem would be if its not optimised for the lower nominal voltage of NiMH/NiCd but I find this to be a rare thing in any photography gear with flash.
As right as you are abotu the Batteriser marketing I still have one argument to defend the usability of this thing.
I just made a test with an alkaline battery that had 1.35v with no load in the beginning. I intended to find out how much energy in Wh I can squeeze out of it when draining it from 0.8v to 0v. To first bring it down to 0.8 I put a load of 200mA on it which caused it to drop down to 0.85v under load. After draining 500mAh it had 0.64v with the 200mA load on it and 1.1v with no load. As you can see that were 500mAh and it's still going in the range where everything apart from solely resistive devices like flashlights or some dumb fans or so would've shut off. I can't think of many devices with microcontrollers and low voltage cut off that would put such high loads on the battery but that would be a situation where the batteriser could really help you.
Honey BooBoo So you're saying that the battery was 0.85V under a 200mA load, and you ran it for 2.5 hours and it only dropped to 0.64V? Look at the battery discharge curve for a 200mA load at 10:35 in the video; it's almost vertical at 0.8V. Once you hit 0.8V you don't even have an hour left. Your test doesn't make sense.
gblargg But that's exactly the point. Under load it drops dramatically. It's an exponential behaviour. The difference in the voltage drop between 100mA and 200mA is very high and the reason it could sustain the voltage under load so well is that even after the 500mAh discharge if you had put a 100mA load on it it would have been well above 0.8v maybe like 0.9v-1.0v
Edit:
To understand it you also have to know that the reason why the 200mA load graph from the vid differs so much from the results I had is that the graph is for a fresh Duracell Coppertop and I used some years old Philips alkaline that I found lying around somewhere. Age and quality makes a big difference in voltage drop under load but alkaline batteries don't lose that much capacity from just lying around. A nimh would've been completely flat after lying around for so long.
Do you read the graph differently than me? My reading is that once the cell is discharged to the point where it outputs 0.8V when 200mA current is being drawn, it has less than half an hour left (the slope is nearly vertical).
How could you draw 200mA for 2.5 hours at all when the cell was already only 0.8V when you start, let alone have the voltage still be 0.65V under load after that time?
gblargg Because it's an old cell that has higher internal resistance etc. than a new one. Because of that it drops dramatically more under load than a fresh cell without having remarkably less energy in it. Just a slight bit is lost to time and more internal resistance ----> heat.
+EEVblog I know we're engineers not math nerds but you could just use integral calculus to calculate the area under the curve.
I really wish we had the current/voltage measurements down to zero volts.
While I'm pretty sure the falloff is so quick that it isn't significant, this experiment does not actually measure the total (theoretically) possible battery capacity; rather, it measures the capacity only as far as the designated cutoff voltage of 0.8V. I reckon there is another 5-10% in there. Not that you could get at it easily without a bunch of cell in series, and without battery leakage to ruin your day, etc. etc. :)
Rock solid engineering as usual.. Thanks Dave..
You can determine if a AA battery is good or not with multimeter. Just stick the meter on amps and do a 1 second reading. If it will not deliver enough current it is bad.
Just take the integral?
MogDog66 I've done a video on that: th-cam.com/video/Dh0xYu8YvaE/w-d-xo.html
EEVblog What's taking so long to make the LM741 video for video 741?????
MogDog66 Actually he is calculating the integral using the rectangle method.
***** that would be approximating the riemann integral by a sum - not actually taking the integral.
How would you go about taking the integral though?
I have no idea what you are talking about mate. It does make me want to learn about it though. I think people from candlepower forums(where all flashaholics hang out) will enjoy this.
When you say to take the "area under the curve", do you mean down to 0 as well? The graphs all seem to end at 0.8v.
Power Max Yeah that was going to be my next comment. Even if the curve drops off really fast, you still have a huge rectangle hidden from this graph since 0.8 -> 0 is even bigger than the 1.5 -> 0.8 voltage range that we can see on the current graph
Power Max yes that is true that there is much more "graph" to be seen, but notice that the area we are talking about is not the entire area under the curve up until the cut off voltage we've chosen (ie 1.5V -> 1.1V for example), that area is the energy we have used. The wasted energy is the area under the curve from the cutoff voltage to 0.8V (ie 1.1V -> 0.8V). Because these graphs basically become vertical at 0.8V the area under the curve from 0.8V -> 0V is negligible.
I really appreciated the tutorial.
Pure gold ! something extremely usefull for the young players ;)
I think that technically, you need to multiply that I*V by the (delta t) i.e. the time-step size, after adding the previous energy value.
So,
E = E(n-1) + V(n)*I(n)*{t(n)-t(n-1)}
Since your time resolution was 1 second, you don't get any errors (apart from being in Joules rather than Wh). But if you had any other value, your energy value will be wrong I think.
Awesome, I wouldn't thought of doing this that way.
I would use least square method to approximate the function and then calculate the area below by integration.
That's my computer science degree in action ;)
And of course, as you've mentioned before, at best would the extra 10% be lost because of inefficiencies in the in the dc-dc converter.
I remember getting a Sinclair programmable calculator (24 step) RPN and put in a fresh standard 9v battery and it did not even turn on. It only ran with an alkaline battery, and that didn't last long. C1978
You don't actually need to align the graphs like that. At least not for this purpose. You'd get the same values if you left the characteristic discharge curve going from 0.
Nice video.
Very interesting. My first digital camera (a toy really at 640x480) only worked with non-rechargeable AA batteries and even those didn't last long in the camera, but still worked for ages in other devices after removing them from the camera. Obviously a poor design with a high voltage cutoff!
Does the wasted percentage of energy have a dependancy on the current drawn? Looking at those curves on the datasheet, it looks like you may waste more and more energy at higher current levels for a given cutoff voltage. It does make sense since higher current levels will drop more voltage across the ESR so therefore it will prematurely reach cutoff level.
It seems to me like hours of usage would be more relevant and simpler to calculate rather than using remaining energy. You would not have even needed to compile the second curve using just time.
Of course that would be convenient if it worked, but it doesn't. Depending on how what you plug the battery to you get very much different usage time from it.
pocoapoco2
Then you didn't understand the part where he showed the straight percentage line. See the deviation to the real values?
superdau Yes, remaining energy vs. time is not a linear relationship. But you know what else changes over time? The power requirement of the device does. From a user's standpoint the remaining energy isn't as relevant as remaining usage time. If you establish a plot for the specific device you are using that extends to the full .8 volts and measure the shut down voltage of the device, then the theoretical remaining usage time could be calculated easily if the device could actually function all the way down to .8 volts.
WOOOWW!!! Thanks Dave!!!
Proyectos LED You're welcome!
After watching this I got an Amazon ad for Amprobe battery testers. haha that was quick.
friendofCHAKA Is it just me, or do some of the Amprobe products on Amazon look fake? In some of the pictures you can see that "contractor" is spelled "contrator" on the front cover of the multimeters. It's the sort of Engrish you'd expect to find on counterfeit products.
haha right its just an artist's rendition of a battery tester.
Can you do a teardown of the "Bagdad Battery".
Like for using LibreOffice :3
Your video is just about the right length, so don´t worry :)
When i´m looking at your discharge curve, i can almost hear my professor again sayin: ok now we need to derive the equation for that, using some integral math and derive where we have our desired 10% for *every* current. Homework guys, see you all next week....that was the usual....didn´t like it back then and don´t like it today...i came up with these gfx as well....he didn´t like that, but hey, i was way faster, as i used to do this in exams as well...
He didn´t like it anyway and, as usual, i failed...hahaa.
I don´t need that math thing throughout, i only need to know the result for my specific problem! Hey Prof. do you hear me...;)
We haven´t got that amount of time in the industry to sit back and derive formulars/equations....we have mathlab/mathcad/maple do let it work that out for us, if we need that.....we have to solve problems, that´s why we been hired for...;)
Well, aligning both y-axis was not really required. As long as they share the same timescale everything is fine. Also you should be careful with area under the curve because displaying data only down to 0.8V skews the real ratio!
Just to get this right, when looking at your curves I must come to the conclusion that after these kind of batteries are discharged to 0.8V (under load of course) there is only a very very small amount of energy left in that cell?
Otherwise it would not be very engineering like to just put the zero point of energy left right there, or would it? I'm sorry for bothering you, but showing the complete curve (if even possible) would be a better way of debunking that silly product. I'm really keen to see the results of your long term battery test. I hope that it will show the complete curve way down to zero volts (if, as I said before, this is even possible).
Thanks for your great videos, both educational and entertaining, I love it! :)
Great video Dave, as a process engineer I live by the motto, "Question Everything" and your in depth analysis of measuring batter efficiency was well done. Was this in response to the new video that Batteriser posted?
Is that a line graph or a scatter plot? (More important if the X data is not at regular intervals...)
For constant power discharge, instead of constant current, the line would actually end up linear, since the capacity is also a percentage of energy... With the constant current example. the remaining energy is a power percentage, and the power of the load is decreasing as the voltage decrease...
(A normal (mostly) constant resistance load is a non-LED flashlight....)
Useful video, been aware of it for ages and I usually test the old alkaline battery pile of my parents every year by plugging some banana plug on one end and alligator clips on the other end leads then hook a resistor up between the two probes that way and measure voltage. If it drops quickly I discard them if they're still decent enough for use in low power applications I'll put them in the "half dead but still useful box".
An interesting fact to be aware of though is that there are some battery chemistries where the open circuit voltage is a sound indication of state of charge. With my 18650 lithium ion batteries its a rather good indication and while I'm no expert at lead acid batteries, if I'm not mistaken it applies to them as well to an extent?
Dave -- a singular axis is AXIS, not AXES!
********** Sorry, didn't mean to sound rude! I just am a bit nuts, is all.
***** No, because I've heard him say Axis, singular, and I think he even said it at some point in this video.
Would the mid-point intersection be the minimum cutoff voltage that the Batterizer has a chance of being useful? As in, above that point might be useful, below that point useless to detrimental.
In general, you explain how to compute an integral and basically, what you say is somewhat correct and it will certainly give you a result which is reasonably close to the precise data. But you made some systematic errors (which luckily won't change the result very much).
1. The capacity is the area under the full curve which goes right down to zero, not 0.8. That is why your first guess of 10 to 20% remaining capacity was a bit on the exaggerated side. If you see the full area, you would - right from the start - have gotten a better estimate.
2. When doing an integral with empirical data points, you should consider taking the previous data points into your sum. You can simply use a linear interpolation by just taking the average of the current and the previous point. Would be a tiny bit more accurat, but of course still based on assumptions. Does not make a huge difference, but just relying the current data points is not the best assumption.
3. You Do Not Divide by the last data point. That is wrong. You divide by the maximal value. Which you do not have measured. It will of course not be far off of what is writte in your last row but remember that the curve goes down to zero. There is a tiny amount of capacity. This does not show in your graph. You can guess it to be only about 1 or 2% of the full capacity, but nontheless, your graph is (strictly speaking) having systematic errors.
Overall, again, nothing new under the sun. Of course, for simple calculations, that is more than good enough. But I just like to be a smart-ass. :)
Very informative. It would be interesting to see what happens below 0.8V. It won't change your general conclusions much, but it would be nice to close that 'gap' because sure as bob's your uncle the 800% BSers will try to claim that there is a 'ton'of energy in that gap. Keep it up man!
Last! :D Awesome video as always Dave!! All the best from Argentina!
Very interesting video, thanks. I wonder if anyone can find a device that cuts out at 1.4 volts? I'd bet that they are very very few and far between.
Wobblycogs Workshop Yep, they'd be very rare. Possible though.
Wobblycogs Workshop Sinclair programmable calculator c1976 would not even turn on with standard battery.
mart fart Just looked that gem up, I think the battery voltage was the least if it's problems! RPN, need I say more. I'm so glad we don't have technology like that now.
Wobblycogs Workshop
Haha it was the scientific programmable in a black case and they were selling them off cheap, I was still at school and it probably hindered me in math more than helping. I also bought a Science of Cambridge MK14 micro kit @39.95 also Sinclair c1978 I used to boast I built a computer as a kid at school...the truth is more prosaic.
Use Ni-Zn 1.6V nominal rechargeable for that product?
Also I hardly believe your graphical analysis method is the best method out there when compared to actual integration using Simpson's or trapezoidal rule. Which you could've just as easily done with the data at hand.
Sorry Dave but if you want to talk about ratios of used energy capacity vs remaining energy capacity and make that visible as an area under a graph, you *do* want to make that area go down to 0V (what you almost did at 17:48). Otherwise it's actually not correct.
Btw: There is a very neat built-in Excel function called VLOOKUP which does some magic for you. You should include that in your next spreadsheet tutorial ;)
Yey! DaveCalc!
Dave if you have any super capacitors it would be a interesting test to see the comparison with a Cap on its own and with a battery connected in parallel.
I have 18 x 3000F Maxwell caps in series and parallel configuration for jumpstarting cars and trucks. 3 sets in parallel of 6 in series 2.7V Caps Max each 16.2v max for the setup. its a beast ! and made 2 of them. often wondered what energy capacity they actual have stored in them.
Hello. Anyone got link to DaveCAD? ;)
Elmer T You can buy it at Amazon or Staples
Elmer T Sry no warez on youtube.
Dr. Lecter I take my DaveCAD from the work supply cupboard. Doesn't everyone?
Dan Fishlock Only when no one is looking.
Could you use this to fine-tune battery indicators as well? Like the remaining battery icon in cellphones
I have a question about battery discharge safety with a PWM application. Maybe someone can answer this or point me to the right direction to where can I read or learn or ask about this? Thanks!
I have a 8.4V rechargeable battery (dual 18650 in series for vaping) and a heating element with 0.1 ohm and lets say a 10% duty cycle at maybe 100hz (arduino PWM with a mosfet). The average amp draw would be 8.4 amp but the amp draw during the duty on would be 84 amps! I plan to use Samsung 25R "safe" chemistry batteries (not lipo) that are rated for 20 amps continuous discharge. In the datasheet it actually mentions high amp pulse discharge tests.
So basically I'm wondering if this is safe to average? What makes batteries actually explode? Temperature?
In any case thanks for the awesome video!
***** But wouldn't the heat average out?
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.
Dave, I use the AA, AAA and "AAAA" batteries ! The "AAAA" fit the Streamlight Stylus and a Laser Pointer that fit next to a couple pens that fit the Fisher Space Pen Refills ! Also use the CR123A in a LED flashlight and an UV Water Pureifier. Have used the 1.5 VDC "A" battery and even took one apart ! Have you done a teardown of an "A" Battery ? Like the Alkaline, Li and High Tech Rechargeable Batteries ! (Say NO to Cadium Batteries)
Eyes UP and lights down, tjl Sent by Win7Pro64 w/ADSL
Did you know that battery is not the only one that came with size C and D?
Rizki Pratama Yes boobs also comes with C and D sizes, I prefer D.
EEVblog Why only go down to 0.8V? In case you're using a DC-DC converter, it would put out a "useable" voltage anyways, so while looking at the discharge graph on the data sheet tells me that there isn't much left in the cell from 0.8V to 0.0V (or, let's say 0.1V), why not go ahead and see/measure the graph all the way down to zero? OK, in case of rechargeable batteries, this would obviously be fatal for the battery, but if you're using disposable alkaline batteries anyways, it wouldn't hurt a thing. Any hey, maybe the curve behaves somehow unexpectedly below 0.8V? OK, to get useable power out of it, it would have to draw a much higher current, which means a constant power curve is the only thing where this area could be relevant, but still, I'd be totally interested how a battery behaves when it's operated so much outside its specs (but lack the means of easily measuring the thing...writing a value down every minute from the multimeter just isn't an option if it takes several hours for the battery to get down to 0.1V).
Just great, thanks
Don't you have to take the first timestep into account? If I understand it correct, there was 0.105A at 1.65V for 1 timestep, but it wasn't used in the calculations. You started with the second timeslot. So don't you have to calculate E(t)=E(t-1)+I(t-1)*U(t-1) (with E for Energy)?
dhufi
How does your formula change anything. With it you're just shifting the error to the end of the data, because now the last entry of the table isn't used. That's the error you get from discrete math. The voltage/current isn't constant for the whole timestep anyway and still it's used in the calculation. So for a better calculation you should use the voltage/current average of two consecutive points (this assumes a linear function within each timestep, which is more accurate, while the first version assumes a constant function).
But the best way around: make the timesteps smaller, reduce the error.
Thumbs up!!
You don't need to do $D$19, you're just wanting to keep the row the same so you only need D$19.
hey Dave I am an electronics noob so don't mind me asking is the cut off voltage a design feature or a product of efficiency of the components used in the circuit i.e something that's an after thought
hyronov A good designer would try try and make the cutoff voltage as low as possible in order to use the most energy from the battery. Sometimes however that's not easy or cost effective for various reasons. It should not be an afterthought, if it is, you're a bad designer.
EEVblog well thanks for the reply adding to that question. Say two manufactures using the same components is it possible that one really well engineered will have a lower cut off voltage. what rough of order of magnitude lower could it be .1V~.5V for AA cell.
hyronov Under 0.8V cutoff is pretty pointless, there is very little energy left under 0.8 for any sort of real load. This is why the industry uses 0.8V as the defacto standard minimum cutoff voltage.
HI dave, I don't know if you could help me but I have a LCD screen that I want to use with my arduino. The problem is that I can find the datasheet anywhere (I even send an email to the company that sells them) do you know how I could get the pinout ?
Alexandre Rouma
Totally wrong place to ask. Search one of the many arduino forums.
superdau
already done that.
Alexandre Rouma Do you know what driver IC it's using?
urdnal No, but it came from a TOMTOM GPS
Why is an AC impedance at 1kHz specified for a DC, AA battery?
Eagerly awaiting the tutorial on how to suck eggs, especially with Easter coming up.
If their claim of 1.3v cutoff from before was right then you get >45% which is probably what they based this on and then just exaggerated things...
Does the size of the load matter? if i put a 1k load or a 1M load what i measure?
***** chanage what i measure :)
I hate batteries period. I'd never use anything but NiMh or better though. I don't see this as a problem except the unit may cut out earlier than the batteriser installed though...
Why didn't you sampled the data down to zero volts? That 0.8V aligned to 0% capacity bothers me.
Unless your time is in hours you can't just do current * voltage. It is wrong on so many levels when you label it as Wh.
SirJMDDK I mentioned that.
EEVblog No. You say: "and if you want watt-hours and stuff you'd put like divide by 3600 .."
The problem is that you are doing a newbies guide to calculate energy, but you completely skip the proper way to calculate watt-hours, yet you label your column "Wh". You're showing 11+ Wh for a single AA cell.
Next time someone new to this tries to calculate battery energi, they will do as you and do current * voltage.
Why not show proper engineering practice? (T2-T1)*((I2+I1)/2)*V
SirJMDDK
First he specifically said it's not really Wh. Secondly it's totally irrelevant for getting the remaining percentage. Thirdly everyone doing real energy calculations should know this basic math anyway and how to convert units like Wh/Ws/Joules whatever.
Since he collapsed multiple measurements into a single data point the first column isn't any usual timeunit anyway. It might be in units of "27.2seconds" for example (that's why he said "put *like* divide by 3600"). Imagine the confusion when doig this division to getto the correct Wh.
superdau Correct. The whole point of this video is the percentage remaining calculation using the graph.
How dare you not start time at 0...
i did not waste any energy on my 12 V lead acid it had 2 V left :(
Not quite right. If you're comparing a DC-DC converter to constant current, you have to account for BOTH the cutoff voltage and the constant current vs constant power. If your constant current product needs 1v, you're wasting all the area between the horizontal 1v line and the battery voltage curve above it.
If you're going to bother doing it right, you should also continue the curve beyond .8v, I'd continue it until the battery no longer delivers enough power to run the device, whatever that might be(at any voltage/current combination), then compare actual active and standby time using your DC-DC converter vs linear regulator.
sleepib If you carry on past 0.8V the voltage drops off a cliff and has very little remaining capacity. You might as well treat the cell as being dead. Good point about the wasted power and this leads to another nail in the coffin of Batteriser. If you add a Batteriser to boost the voltage to a constant current load it will waste more energy than using the cell on its own. The regulator has to drop more voltage and waste more power!
sleepib
A decent DC-DC-converter is a constant power load. So why would it waste the energy above the output voltage? That's how a linear regulator works.
superdau the constant current example uses a linear regulator, but doesn't account for the energy wasted by the voltage drop across the linear regulator, it only accounts for the energy left in the battery when your device no longer functions. If you're trying to decide whether it's worth switching to a DC-DC converter, you need to account for both.
Is this video a freaking spreadsheet tutorial ?
lacombar Kinda.
EEVblog Thanks for all the steps. I hate when people, books or Internet Tutorials skip steps, even when I know how to do it, because if I did something wrong I can compare both methods.
EEVblog Excel for dummies!
Kostadin IV Kostadinov It's LibreOffice (OpenOffice), much better piece of software
Brendan Wilding it's a lot of hassle for something which could be done trivially with, say, gnuplot...
Instead of saying wasted, it would be more accurate to call it Unusable power.
Sasha Whitefur "unusable" is inaccurate as the power could be used, but it is not economical. And it is not even power, it is energy. It should be "unused" or "remaining" energy. But either way: It´s just changing words around and adding half sentences. For a PhD thesis, this can make all the difference, but here it is about the general statement/concept. Off course, general statements can be significantly off (like when driving with the last gallon to the gas station and either to make or not), but in those cases more care has to be applied anyway.
It's still power, it can be used more by only drawing .8V.
Real science, real engineering. Opposite of this would be buying 1k dislikes from Vietnam...
1172 people are Batteriser backers lol
but these are cells not batteries.....
Anas Malas The math scales.
watersne oh. Not the video itself... I was talking about the duracell datasheet....
Anas Malas Haha yes
your calculation in the spreadsheet seems to be wrong to me. First: why does is the first row have 0 WH at 1min? you should use the mean value between two rows - but sure that only makes a little difference. Second: you should take the measurements down to 0V to get the real capacity if you want to show that there is not much energy left at .9V. Randomly setting the zero reference to some Voltage level gives you wrong result. imagine that you only took measurements to 1V and took that as 0-Energy reference. The results would look similar while still a bit of energyis present.
this is an attack point in your proof
Holy crap, so many bot dislikes.
I saw that Yahoo posted a news article today about the Batteriser that actually called you out, EEVblog. www.yahoo.com/makers/breakthrough-battery-gadget-answers-critics-125063020800.html
Under load, OK, I got that the first time it was stated but wholly hell, stop repeating yourself please. It's so dam annoying when you tubers constantly repeat things.. I lost count after the 3rd time you stated under load......
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joblessalex You win the internet.
EEVblog Really? I'd like to thank my mother, my father, the dead rat I saw coming home, and the refresh button.... Without those, I would have never been first. Still really interesting stuff.
joblessalex Of course, there's some of us that watched this yesterday already! ;)
Aadil Shah Very true. Patreon and forum backers saw it yesterday.
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