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.
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.
Only time will tell how efficient it is (actual measurements) but I'm betting 90% is a generous number.
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%.
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
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.
He did briefly mention it, but not as elaborate as you did. You had a lot of good points that should have been mentioned.
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.
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.
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.
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.
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
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.
Quote2. (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.
(image of discharge graph extended down to 0.0V)
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.
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.
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"?