Boost circuit math
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Are there different formulas based on efficiency? Or is efficiency just a coefficient?
Why not just measure it? Bust out the multimeter and put it in-line with the + wire.
Is that really how you apply the 93% efficiency coefficient?
Perhaps with another mod more conducive to testing and repeated disassembly
But was looking for the conceptual calculations, just trying to make sense of the results I am seeing.
Anecdotally I have seen better battery life from regulated devices running high resistance/high voltage vs running low resistance/low voltage given the same wattage. I had always thought it was the result of the lower amp draw on the load side. However by these calculations, when a circuit is regulated (specifically boosted) the battery should see the same amp draw regardless of the load. So ... *headscratch*
It's usually difficult to know the exact converter efficiency at any given portion of it's operational range. Some of them can vary quite a lot in efficiency at different output levels.
That could easily explain the phenomenon that you describe. For instance, some buck converters that run from two batteries in series lose efficiency drastically as you lower the voltage, basically wasting the excess voltage as heat at the converter.
For a VV running from one battery, running 5 volts into 2.5 ohms, it's 2 amps and 10 watts at the load. The battery will be supplying more than that wattage, due to losses in the converter.
Using the max charge and min charge voltage levels of the battery and ohm's law, 10 watts from 4.2 volts (fully charged) is 2.38 amps, and 10 watts from 3.2 volts (needs to be charged) is 3.125 amps. That's not counting efficiency loss.
If you know the efficiency at those points, or at least can estimate it, then I believe that you can divide that result by the efficiency. For 85%: 3.125/.85 = 3.676 amps pulled from the battery needing a recharge. Not entirely sure about that efficiency calc. I'm tired.
That could easily explain the phenomenon that you describe. For instance, some buck converters that run from two batteries in series lose efficiency drastically as you lower the voltage, basically wasting the excess voltage as heat at the converter.
For a VV running from one battery, running 5 volts into 2.5 ohms, it's 2 amps and 10 watts at the load. The battery will be supplying more than that wattage, due to losses in the converter.
Using the max charge and min charge voltage levels of the battery and ohm's law, 10 watts from 4.2 volts (fully charged) is 2.38 amps, and 10 watts from 3.2 volts (needs to be charged) is 3.125 amps. That's not counting efficiency loss.
If you know the efficiency at those points, or at least can estimate it, then I believe that you can divide that result by the efficiency. For 85%: 3.125/.85 = 3.676 amps pulled from the battery needing a recharge. Not entirely sure about that efficiency calc. I'm tired.
That does make some sense. Though it is interesting that the original notion I had, which I believe is somewhat conventional wisdom, doesn't actually pan out, despite anecdotal appearances.
Yup I believe this is how I saw it applied in my research. Or at least it wasn't adding the inefficiency percentage straight.
Without knowing to much about the guts of the DNA20 design, I will say, that what you are seeing is exactly what should happen, with a typical "Boost" circuit design, with a regulated output.
Take a look at this link about the theory on why a boost circuit works, but don't get to involved with the engineering math.
http://en.wikipedia.org/wiki/Boost_converter
That circuit that they show would be the front end to your "Regulator" output.
Here is the key. Under light loads, that little (or sometimes big) inductor, does not have to completely discharge. Therefore to maintain a specific voltage at the Capacitor on the right of this wiki boost circuit, the battery does not have to work as hard, or in other words it doesn't have to discharge high amperage.
But, when the load increases (the regulator side now), then the inductor discharges completely, and the battery now has to supply more current, thus reducing the battery life faster.
This output regulator is now dissipating more energy in the form of heat, when you see, low resistance/low voltage, on the output, because the input to this regulator is probably held at a higher voltage then the output.
So high resistance/high voltage, should give you longer battery life.
If it didn't then you would be violating the Second Law of Thermodynamics. Now you don't want to break any laws do you???
If all you need is a ball park number. Convert the output into watt's, then apply the output watt's to the battery voltage to come up with the current (amps), V x Amps = Watts, and then add in the in-efficiency of the circuit. I would use +10% for the in-efficiency.
so lets say you are pulling 12 watts and the battery is reading 4 volts. Then your pulling 3 amps of current, before accounting for the in-efficiency. After using 10% in-efficiency you are pulling 3.3 amps.
so lets say you are pulling 12 watts and the battery is reading 4 volts. Then your pulling 3 amps of current, before accounting for the in-efficiency. After using 10% in-efficiency you are pulling 3.3 amps.
This is not a 2nd Law of Thermodynamics problem (entropy), but rather a Conservation of Energy problem. This means the energy used at the output (Pout=Vout x Iout) + inefficiency loss must equal the energy drawn from the battery at the input (Pin=Vin x Iin).
So, if
Pout + loss = Pin,
then
Vout x Iout + (inefficiency loss) = Vin x Iin
If one observes a difference in battery life between high resistance/high voltage and low resistance/low voltage at the same nominal output power, then it means the regulator has significant differences in operating efficiency between the two operating regimes.
For switching power supplies, it is generally true that they operate most efficiently in the region close to 100% duty cycle. The losses occur mostly from the FETs during switching on/off. The Power MOSFETs used in PWM converters have extremely low resistance when they're fully on (minimal loss) and essentially infinite R when they're fully off (zero loss); however, when they're actively transitioning between the ON and OFF states, there is a brief period of time (microseconds) when they conduct current AND have some finite and non-trivial resistance, generating heat in the amount of I²R (that's why FETs heat up when in use). At low voltage, the FETs switch more often, causing them to transition thru the heat-generating region more often, resulting in more loss/lower efficiency compared to operating at a higher output voltage.
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Electrical energy calculations aside, I have often wondered just how much effect the physical properties of the coil can have on the power requirements of atomizers of differing resistances. I'm not at all learned enough in physics to make anything but guesses on the subject, but I was curious enough to look into it a bit, today.
There was a thread not too long ago that presented an interesting argument: 'Watts do not matter. Its all about wire temp.' Thinking of that thread, I looked up some data on my preferred resistance wire:
_NiChromeData
One of the pieces of data there is the amount of current needed for the wire to reach a certain temperature. Using that and other data there, I came up with some interesting numbers:
Now, assuming that most atomizer manufacturers use a lower gauge wire to make a lower resistance coil (seems to be the case for ones that I have disassembled,) the lower resistance coil seems to require about 18.4% more power to reach 1400 degrees than the higher resistance coil does. While that's not enough to convince me of anything, it makes me wonder about the efficiency of the resistance wire itself.
Anecdotal evidence from the experiences of people on these forums seems to indicate that the lower gauge wires also take longer to heat up than the higher gauge wires. I have noted that, myself. This could be considered to be power lost due to slower response time. This would seem to further support any perceived run-time benefit from running higher resistance with higher voltage.
There are many other things that I'm unsure of, such as the efficiency of heat transfer from the wire to the e-liquid, based upon the mass and surface area of the wire. All I can say for sure is that this subject is too complicated for most to understand to a high degree.
There was a thread not too long ago that presented an interesting argument: 'Watts do not matter. Its all about wire temp.' Thinking of that thread, I looked up some data on my preferred resistance wire:
_NiChromeData
One of the pieces of data there is the amount of current needed for the wire to reach a certain temperature. Using that and other data there, I came up with some interesting numbers:
Code:
Gauge Ω/mm A@1400F
----- ------ -------
31 .02692 2.17
34 .05374 1.41
31g - 55.72mm - 1.5Ω - 2.17A - 3.255V - 7.063W
34g - 55.82mm - 3.0Ω - 1.41A - 4.230V - 5.964WNow, assuming that most atomizer manufacturers use a lower gauge wire to make a lower resistance coil (seems to be the case for ones that I have disassembled,) the lower resistance coil seems to require about 18.4% more power to reach 1400 degrees than the higher resistance coil does. While that's not enough to convince me of anything, it makes me wonder about the efficiency of the resistance wire itself.
Anecdotal evidence from the experiences of people on these forums seems to indicate that the lower gauge wires also take longer to heat up than the higher gauge wires. I have noted that, myself. This could be considered to be power lost due to slower response time. This would seem to further support any perceived run-time benefit from running higher resistance with higher voltage.
There are many other things that I'm unsure of, such as the efficiency of heat transfer from the wire to the e-liquid, based upon the mass and surface area of the wire. All I can say for sure is that this subject is too complicated for most to understand to a high degree.
Electrical energy calculations aside, I have often wondered just how much effect the physical properties of the coil can have on the power requirements of atomizers of differing resistances. I'm not at all learned enough in physics to make anything but guesses on the subject, but I was curious enough to look into it a bit, today.
There was a thread not too long ago that presented an interesting argument: 'Watts do not matter. Its all about wire temp.' Thinking of that thread, I looked up some data on my preferred resistance wire:
_NiChromeData
One of the pieces of data there is the amount of current needed for the wire to reach a certain temperature. Using that and other data there, I came up with some interesting numbers:
Code:Gauge Ω/mm A@1400F ----- ------ ------- 31 .02692 2.17 34 .05374 1.41 31g - 55.72mm - 1.5Ω - 2.17A - 3.255V - 7.063W 34g - 55.82mm - 3.0Ω - 1.41A - 4.230V - 5.964W
Now, assuming that most atomizer manufacturers use a lower gauge wire to make a lower resistance coil (seems to be the case for ones that I have disassembled,) the lower resistance coil seems to require about 18.4% more power to reach 1400 degrees than the higher resistance coil does. While that's not enough to convince me of anything, it makes me wonder about the efficiency of the resistance wire itself.
Anecdotal evidence from the experiences of people on these forums seems to indicate that the lower gauge wires also take longer to heat up than the higher gauge wires. I have noted that, myself. This could be considered to be power lost due to slower response time. This would seem to further support any perceived run-time benefit from running higher resistance with higher voltage.
There are many other things that I'm unsure of, such as the efficiency of heat transfer from the wire to the e-liquid, based upon the mass and surface area of the wire. All I can say for sure is that this subject is too complicated for most to understand to a high degree.
Oddly enough a lot of people credit low resistance coils with a *faster* response time. Perhaps they are not comparing apples to apples there. I would say the discussion in that thread is not just interesting, it is definitive. But it's a little obvious -- wire temp is the objective of the whole enterprise. Watts matter for what watts matter for, but I think the general idea expressed in that thread -- that comparing watts for watts' sake is misleading -- is completely valid.
I wrote a couple of blog posts just for this very topic...
[h=4]Calculating Battery Drain Current[/h]
[h=4]Battery Life - Low Resistance, High Resistance, and Efficiency.[/h]
A lot of the material is already in this thread, but they might help put the puzzle together, or they might not. Either way, getting all of these brains together in one place is my kind of thread.
[h=4]Calculating Battery Drain Current[/h]
[h=4]Battery Life - Low Resistance, High Resistance, and Efficiency.[/h]
A lot of the material is already in this thread, but they might help put the puzzle together, or they might not. Either way, getting all of these brains together in one place is my kind of thread.
Perhaps what needs be said these days is resistance doesn't matter in the same way. There's nothing magical about "sub-ohm" per se ...
I wrote a couple of blog posts just for this very topic...
Calculating Battery Drain Current
Battery Life - Low Resistance, High Resistance, and Efficiency.
A lot of the material is already in this thread, but they might help put the puzzle together, or they might not. Either way, getting all of these brains together in one place is my kind of thread.
Thanks, I had one of those bookmarked, the other is exactly what we're examining in this thread. I wonder if manufacturers would provide an efficiency curve for their converters?
Some manufacturers have complete Data Sheets, but I don't think we can expect something like that from a Zmax or DNA module.
For example, look up the okr-t/6 Data Sheet on digikey or mouser and you'll find a bunch efficiency curves for it.
For example, look up the okr-t/6 Data Sheet on digikey or mouser and you'll find a bunch efficiency curves for it.
I would guess that the DNA and some other boards might be designed around an existing converter chip, with no info printed on it to protect their IP. If so, they would probably not want to divulge more details than what is commonly given for a consumer product.
The chips usually do have some efficiency curves listed in their datasheets, as mentioned. I know that the PTN04050C does, for instance.
The chips usually do have some efficiency curves listed in their datasheets, as mentioned. I know that the PTN04050C does, for instance.
I doubt it. They generally would try to obfuscate the components on boards such as those.
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