EEVblog #772 - How To Calculate Wasted Battery Capacity

[EEVblog]

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.

[ornea]

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.

[Fungus]

So ... Batteriser can get 10-20% extra out of the battery? That sounds good to 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. It will waste 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 net loss of 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 a LOSS in 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.

[Fungus]

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.

You basically have to connect up your device and measure it during typical operation.

Would the battery measurements need to be somehow synced to the peak power demands.

You 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.

[EEVblog]

So ... Batteriser can get 10-20% extra out of the battery? That sounds good to 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.

I 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.

[EEVblog]

Only time will tell how efficient it is (actual measurements) but I'm betting 90% is a generous number.

They 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 

[VingTor]

You forgot to mention that Batteriser can't be 100% efficient. It will waste some power. Even the best converters are only about 90% efficient (10% wasted as heat). Your real gain is only 0-10%.

He 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 end 

Another 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).

[VingTor]

Only time will tell how efficient it is (actual measurements) but I'm betting 90% is a generous number.

They 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 

Someone else could send it to you in mailbag 

[BravoV]

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.

[robbak]

Um, Dave, your formulae to invert the 'capacity remaining' graph was nice - but it all simplifies to 1-(E_used/E_total). 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.

[dk27]

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.

[silvas]

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).

[Fungus]

certain things that he says about this are a little misleading, I think:

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.

That'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.

[EEVblog]

He did briefly mention it, but not as elaborate as you did. You had a lot of good points that should have been mentioned.

I've mentioned this at length in the previous video and blog post on the topic.
This was not meant to be a Batteriser debunking video, I just mentioned it a few times.

[dk27]

That'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?

Maybe 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 actually underestimates the 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[t]-0.8,{t,t1,t2}]/Integral[V[t]-0.8,{t,0,t2}] is less than the ratio Integral[V[t],{t,t1,t2]/Integral[V[t],{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%.

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.

I 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.

[EEVblog]

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.

This 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.

[silvas]

certain things that he says about this are a little misleading, I think:

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.

That's the standard way to draw graphs, and that's why most people's initial guess at how much energy is left is skewed.

I think this is what dk27 was suggesting would have been nice to point out in the video.

[EEVblog]

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.

Yes actually, this is correct, it will work either way.

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.

If 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.

[lapm]

Only time will tell how efficient it is (actual measurements) but I'm betting 90% is a generous number.

They 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 

  So can they name someone that does and what that persons qualifications are to be expert in field?

[Fungus]

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.

It would actually be bad for rechargeables and increase the risk of leakage in alkalines.

A really well designed device should probably power itself down at about 1.05V.

[stuner]

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.

If 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.

I 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 is significantly lower than 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.

[dk27]

I think you're not talking about the same thing.

Thanks 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.

[Fungus]

(image of discharge graph extended down to 0.0V)

Estimated time until that image appears on the Batteriser web site...?

If you then compare the two areas you get a percentage for the remaining energy that is significantly lower than the actual figure.

The point is that we're not comparing the areas. Comparing the areas would be wrong.

Your spreadsheet calculations in the rest of the video actually avoid this issue because they use the total battery voltage and are still correct.

We'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 within half a bee's dick of a straight line going down to zero. This is a domain where calculations are valid.

[dk27]

If you then compare the two areas you get a percentage for the remaining energy that is significantly lower than the actual figure.

The point is that we're not comparing the areas. Comparing the areas would be wrong.

Comparing the wrong areas 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.)

Your spreadsheet calculations in the rest of the video actually avoid this issue because they use the total battery voltage and are still correct.

We'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 within half a bee's dick of a straight line going down to zero. This is a domain where calculations are valid.

I'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"?

[Fungus]

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"?

I mean it needs integrating before you can make any sense of it.