So which APV are you adjusting down to 1.5 - 2.0 volts to fit those figures?
Not until they repeal Ohm's law.
High Resistance Vaping
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So which APV are you adjusting down to 1.5 - 2.0 volts to fit those figures?
Not until they repeal Ohm's law.
Not sure what the first question means?
Efficiency of the circuit is going to vary depending on which one you are using, but it's just a solid fact that a boost circuit in an APV isn't going to be 100%. Which means additional current draw over the calculated requirement for the voltage or power setting you have.
Ohms law has nothing to do with why APVs draw more current per watt, the efficiency of the circuit does. If it's not 100% it's drawing additional current to meet the required output, meaning more current than a mechanical at the same level.
There is slight loss in any circuit, a mechanical mod is no different, but you are only dealing with small losses due to conductivity, not losses due to conversions.
To give you something with solid figures, consider the dna 30, which Zen says draws 12amps at highest setting (30w)
That means, in order to produce 30 watts, the DNA30 draws 12 amps.
In order to produce 30 watts on a mechanical, you only draw around 7 amps.
On a mechanical, you can put out over 40 watts of power drawing the same current as the DNA 30 at only 30 watts.
That's some pretty serious inefficiency, and the DNA30 is considered a high-end chip.
I suppose 58% efficiency (12 A vs. 7 A) is possible, but that's terribly low. Even a linear chip can do better than that. Heck, I can manage something somewhat reasonably close to that with a rusty nail, two wires, and two cheap signal transistors (probably 35%).
I'd need to ask if the 12A is the maximum pulsed current. In said above example with the rusty nail, at its peak the inductor might be drawing 5 A from the battery. At its low point, it's actually going to be trying to recharge the battery. That is not a highly tuned circuit, however, and has no protection against current reversal when the magnetic field on the nail collapses (fortunately, something like an NiMH will put up with that even if it doesn't particularly like it that much).
With a bit better design, the inductor that charges to raise the voltage won't be reversing itself into the battery, but it'll still be pulling down the amperage like mad when empty and just starting to charge. Just at the moment when it fills and gets ready to discharge into the rest of the circuit, it'll be drawing exactly zero. That's typical of buck/boost controllers, and we tend to limit their draw to avoid nasty things happening with our batteries.
Still and all, the average may be 8 but you report the maximum draw of 12 particularly when using a lithium battery as you really do need to have one where the amp limit is above peak.
That's a possibility there.
I'd need to ask if the 12A is the maximum pulsed current. In said above example with the rusty nail, at its peak the inductor might be drawing 5 A from the battery. At its low point, it's actually going to be trying to recharge the battery. That is not a highly tuned circuit, however, and has no protection against current reversal when the magnetic field on the nail collapses (fortunately, something like an NiMH will put up with that even if it doesn't particularly like it that much).
With a bit better design, the inductor that charges to raise the voltage won't be reversing itself into the battery, but it'll still be pulling down the amperage like mad when empty and just starting to charge. Just at the moment when it fills and gets ready to discharge into the rest of the circuit, it'll be drawing exactly zero. That's typical of buck/boost controllers, and we tend to limit their draw to avoid nasty things happening with our batteries.
Still and all, the average may be 8 but you report the maximum draw of 12 particularly when using a lithium battery as you really do need to have one where the amp limit is above peak.
That's a possibility there.
I'll throw a factor in to muddy the waters for ya: I've watched reviews that show that some of the regulated devices don't perform too well when the resistance load on the output is relatively low. Have a look at this video, and you will see that an MVP2 would work the best toward the higher end of common coil resistances; the load tests start at about 13 minutes in:
A PBusardo Review & Contest - iTaste MVP V2 - YouTube
A PBusardo Review & Contest - iTaste MVP V2 - YouTube
Don't make me turn this vape around. Because I will.
You're actually both kind of right here.
Any chip will have built-in inefficiencies. Good switching regulators may have efficiencies near 90%, though, particularly if the difference between voltage in and voltage out isn't too extreme. Jumping from a nearly-drained lithium's 3.4 to a setting of 6.0 is not going to be that efficient in most cases (but more expensive chips with higher efficiencies certainly exist).
By the time you add the OLED screen (minor draw at, I'm guessing [sight unseen], around 10-15 ma), button press detection, and the little machine that goes "bing"...er, sorry, the microchip that monitors the input/output and adjusts accordingly, the amount of power used can become relatively appreciable.
The mechanicals have none of that and only have losses inherent to the (low) resistance of the case and wires.
Since both share the losses inherent in the battery's internal resistance, I'm not going to count it for either.
I'm simply unsure of whether the numbers given are sustained or peak draw. Given that I design power converters for funsies, I'm going to guess peak as it's the standard I was given--never give the user the average sustained as they might choose a battery that's only up to the average. That's a recipe for slight disappointment with my designs as the voltage won't peak high enough to run the lights I build at full brightness (putting an almost drained AA alkaline into one of my lamps results in rather dim, but still usable, light--hardly a major issue).
With a lithium battery, that's potentially a recipe for disaster as the battery can enter thermal runaway and vent.
That makes sense. 3.3V at 1 ohm results in a sustained 3.3 A, or just under 11 watts. While under the screen-available setting of 15 W...well, it's 1 ohm, or well outside the availability of most stock attys.
Most devices limit the sustained draw, there's a limit on how much amperage can go through the device.
That power curve is amazingly flat, so I commend the designers on...gee, actually using either a REALLY high frequency pulse that doesn't show on his scope, or using adequate (and expensive) capacitance.
My coils almost always come out around 4 ohms using 32 awg kanthal wrapped around 3 strands of 2mm silica. I use a vivi nova tank (3ml) with a modified head (to accept more wick) and a Vamo with stacked 18350's.
I vape around the 6.5-7 watt setting normally in I believe No. 2 mode (RMS). I go through several tanks a day and the batteries usually last me all day if not part of the next.
I've been using the same hardware (minus the wick and wire of course) for the last 8 months. Vaping has been very consistent with what I've settled into above, except when I mess up a coil lol
Just from my experiences and what others are saying you probably don't have anything to worry about.
I vape around the 6.5-7 watt setting normally in I believe No. 2 mode (RMS). I go through several tanks a day and the batteries usually last me all day if not part of the next.
I've been using the same hardware (minus the wick and wire of course) for the last 8 months. Vaping has been very consistent with what I've settled into above, except when I mess up a coil lol
Just from my experiences and what others are saying you probably don't have anything to worry about.
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Yeah it seems unreasonably low to me too... I knew there was some loss but that seemed really significant. You could very well be right about that, that it's pulsed load. I guess the Evolv people would be the best to ask.
Either way I think we can both agree it's not going to be 100%, which means increased drain for a given power.
Breadboard testing done over in the Modding forum showed efficiency for switching regulators running at 90% - 95% It was a long term discussion when they were first replacing older style regulators.
So long as P = E*E / R - varying the voltage will have an effect of a square, versus varying the resistance in any circuit. And it is the wattage that heats a coil.
So long as P = E*E / R - varying the voltage will have an effect of a square, versus varying the resistance in any circuit. And it is the wattage that heats a coil.
So, a 5-10% loss, that's still going to require additional current to make up the difference, no?
I can run the figures all I want, they seem to support what I'm saying as far as I can tell. but people keep disagreeing on this, can someone post some v/r/a numbers that show where a APV will draw less current than a mechanical at the same wattage? (taking into account standard mech batt voltage, as well as voltage/amperage/wattage limits of the APV, and taking into account 5-10% loss)
What matters is the input and output. The input is always the same, it doesn't matter mech or APV, input is at max 4.2v (lets not get into stacked batts)
Lets aim for 15 watts. For the same output in watts, the calculation is exactly the same, it doesn't matter if it's an APV or a mech. The only difference is whether you build a 1.2 coil straight on the battery, which introduces a very tiny loss, or your chip puts a 1.2 ohm load, and then converts that additional current into higher voltage.
If you are trying to get 15 watts, from a 4.2v battery, there is only one way, putting a 1.2ohm load on it. Whether it's a coil, or a chip, it's still a 1.2 ohm load, same current. Except the chip will have loss, which either means your power isn't as high as you set it, or the chip draws additional current to compensate. And if the power isn't as high as you set it, then we are no longer comparing watt-for-watt. If the chip draws additional current, it's stressing batteries more, watt for watt, than a mechanical will.
I've heard that at lower power levels the DNA 20 boards totally stop regulating and just dump straight battery voltage into the coil. Assuming the method used to accomplish that has a similar, and very low loss comparable to a simple straight 510 center pin, load on the battery could be about the same as a mech, but then you still got a display and some sort of software running, very very tiny amounts of power consumption, but still there.
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So if a regulated mod increased the draw from 3 amps to 4 amps it's stressing the battery more than a mech which needs to increase from 10 amps to 15 amps.....
and the mech mod does it without any loss to resistance other than the coil in the circuit.
No. Where are you getting those figures?
Ill try to explain again, the battery inside the apv is only operating at 4.2 v max.
Go use an ohms law calculator, and try to get a given wattage, at the same 4.2v, with different current levels. You can't, its impossible.
When calculating amp draw on a battery, you use the battery voltage, not the set output voltage of the device.
Ill try to explain again, the battery inside the apv is only operating at 4.2 v max.
Go use an ohms law calculator, and try to get a given wattage, at the same 4.2v, with different current levels. You can't, its impossible.
When calculating amp draw on a battery, you use the battery voltage, not the set output voltage of the device.
So if a regulated mod increased the draw from 3 amps to 4 amps it's stressing the battery more than a mech which needs to increase from 10 amps to 15 amps.....
No, but it would be stressing the battery more than a mech would that only needed 2.8 and 3.7 A to produce the same wattage.
Within reasonable efficiency range, I'm not going to quibble about 10% loss in the system. While not trivial, it's not large enough to be particularly significant, and people will need to counterbalance the value of the APV's abilities against a 10% loss in effective battery life.
In my case, there's no contest. I'll stick with my basic regulated battery as it produces a nice consistent vape through the battery's charge....except those durned last 10 puffs or so.
Neither regulated or mechs are 100% efficient, not a major factor.
If you take a simple example, run a 6 volt mod with a 2 ohm atty, 3 amps, 18 watts
a mech mod 3.7 volts 18 watts = 4.86 amps and needs a sub ohm 0.76Ω atty to do it.
Even if the efficiency of the chipset is horrible, 50% would only bring the regulated to 4.5 amps vs. mech at 4.86 amps
The efficiency of the mod is
(output power of the mod) / (power expended by the battery)
If the efficiency of the regulated mod is 50% then the battery in it is expending 36 watts...
Looks to me like you've tied you efficiency number to the voltage instead of the engineering meaning of efficiency which is power or more correctly - energy.
For an APV to output 18 watts, it's going to draw about 4.28amps on a fully charged battery at the very very least assuming 100% efficiency, and amperage draw still increases as the battery discharges.
The 3 amp figure you have isn't accurate at the battery, because it uses the output voltage of the device, not the output voltage of the battery. As was mentioned the battery inside the mod is never above 4.2v, the only way to get additional voltage is to put a lower resistance load on the battery and convert that additional current into higher voltage. 6v and 2ohm is only 3 amps, but you can't use those figures. You must use 18watts and 4.2v, which gives you the 4.28 amps. Any additional loss from the chip means lower power or higher current draw, there is just no way around that.
I'm not trying to beat a dead horse here, but it seems like the common assumption is that APVs are easier on your batteries because you don't actually see the low resistance load being placed on the battery. Out of sight, out of mind it seems. It doesn't matter if it's an APV or a mech, high power always equals high current draw if you stick with a single cell battery, and APV's are going to draw a little more than mechs to accomplish the same power.
I'm not into beating dead horses, either, but I'm also one that's been trying to dispel this myth for quite some time.
Comparing a single-cell buck/boost model to a single-cell simple circuit model, both lithium-based cells:
The battery can never output more than 4.2v.
For the device to output more than 4.2v, the device requires more current from the battery.
Disregarding converter efficiency, the minimum current required from the battery is the output wattage divided by the battery voltage. Then the efficiency factor can be applied to that number by dividing the current by the efficiency.
The 'Overview' section of this Wikipedia article describes the concept simply, with links to further information:
http://en.wikipedia.org/wiki/Boost_converter
So, boosted circuits will draw more power from the battery than a simple circuit would, because of converter inefficiency.
In your simple example, the 6v boost circuit output into 2 ohms draws 3 amps from the boost circuit output, making the total power 18w. The law of conservation of energy applies here, in that you can't just make a 4.2v cell somehow create 18w of power at 3 amps of draw. That would only be 12.6w.
In order to get 18w out of the device, at least 18w must be drawn from the battery, which cannot change it's voltage. Therefore, the boost circuit input requires a minimum of 4.28 amps from the battery to convert the battery voltage to the 6v device output. If the converter is very efficient, say 95%, then the draw is 4.28/.95 = 4.5 amps from the battery.
Compare that to the simple circuit, which requires 4.28 amps to make 18w at 4.2v (0.98 ohms,) and the boost circuit does require more power. It does offer better adaptability and consistency, though.
Note that I'm using fully-charged battery voltage in both cases, not nominal. The numbers are in proportion, either way.
And I just wanted to make sure that a coil I wrapped at 4 ohms was ok on my equipment and whether there were any problems. I opened up a large can of worms...LOL. Thanks to all for the Info.
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