I have to disagree, in order for the board to supply 30 watts to the coil, the battery would have to supply the board with 30 watts, the only difference from resistance might be a slight difference in copper losses due to the board drawing higher current, but it shouldn't make much of a difference
Don't forget about Watts Law
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Yes they are completely seperate, ohms law is relationships with resistance, watts law is relationship with power. It's a rifle the mod is resting on
Lol, the board needs to know. You don't. I'm sorry if I can't explain it so you can understand it.
I won't argue. We're saying the same thing again. You're perfectly correct the batteries must supply applied wattage plus board losses.
Haha my apologies, this is why I became an electrician and not something to do with English, wording things is not my strong suit.
Nor mine, apparently. You just saw me fail to explain it
I was an electrician some time ago myself. Betting your life on that tagout makes you think about things...I'm mostly residential, mostly just work with 120/240 so not much worry about my life, not a huge fan of 347/600 or higher but luckily it's not something I run into very oftenNor mine, apparently. You just saw me fail to explain it
I will admit that I learned most of this when variable voltage was just hitting the scene, and then I chose to use mechs so resistance is paramount to me. But now I'm intrigued.
Let's say you want to run 100 w, and you have two coils for your tank, a 0.2 and a 1.0. If I understand correctly, the 1 ohm coil is pulling 10 amps(I know that's 10v, but this is hypothetical), and the 0.2 is pulling roughly 22 amps. I know the circuit handles some of this, but isn't that important to know? Shouldn't you be choosing a battery that can handle the amp draw?
Let's say you want to run 100 w, and you have two coils for your tank, a 0.2 and a 1.0. If I understand correctly, the 1 ohm coil is pulling 10 amps(I know that's 10v, but this is hypothetical), and the 0.2 is pulling roughly 22 amps. I know the circuit handles some of this, but isn't that important to know? Shouldn't you be choosing a battery that can handle the amp draw?
You'll have to worry about whether or not your device itself can handle the current draw, for example certain gauges of wire are rated to handle certain amps, 22 amps might be too high for your devices wiring to handle, when it comes down to battery you need to use watts law by dividing 100w by the nominal voltage of your batteries (in my Sigelei it would be 8.4 on a fresh charge or 6.4 when the mod says my batteries are depleted) whether you have a .2 ohm coil or a 1 ohm coil. I guess to get technical let's say you draw 22 amps and your device is wired with 18 AWG wire. That wire isn't designed for that high of current and if it were to draw that too long it could burn the wire off, which would then cause a short and that's when your batteries can fail. But as far as sizing your batteries to what is safe resistance won't come in to play in a regulated mod
Ohh yes most mods do now offer Buck voltage regulation. The phase of not dropping the voltage has pretty much passed. But many still do PWM output when the desired output wattage requires lower voltage than battery voltage.But more and more are incorporating per customer demand.
Might check new releases.
High power Mod demand pushed Buck aside.......................until Low power Vapers decided they needed the High power mods.
Made since to me....., not, however it is good to see Manufacturers do listen.
Yes I think it is output current not battery current as well. It would measure/calculate that to figure wattage anyway. Still not a bad reference.
Ok, I think I'm starting to get it. Not a side of the equation I've payed attention to before, since I use single battery mechs.
yep, I'm surprised someone hasn't come out with a mod yet that will tell you the nominal voltage on your battery at whatever state of charge it is at as well as the current being drawn on it, would sure be nice and alleviate a lot of questions people have
The VTC mini will show battery voltage now. with power off press and hold all 3 buttons.
That is with V3 software.
It shouldn't take much for them to incorporate current drawn then, although with the cu losses on the board the current drawn won't be exact if they just have it calculate based on the set wattage, no doubt with the way this technology seems to be advancing it won't take long before these things are completely fool proof.
I suspect that additional circuitry would be required to display the battery current. But they already monitor the battery voltage to have low battery shutdown and the battery graphical representation of charge.
With a software change they should be able to show the actual voltage of the battery instead of the graph.
With a software change they should be able to show the actual voltage of the battery instead of the graph.
It would just have to do a basic P/E calculation to find current draw, so if it can accurately measure voltage it shouldn't take much, maybe even just a software upgrade, but yes if it is actually measuring straight current draw then it would need added circuitry, I was meaning just calculating it based on known values.
Measuring voltage vs current requires different circuitry and also causes a bit of resistance called a shunt to be added to the lead between the battery and the circuit board so they can measure the voltage drop across it and calculate the current using ohms law. So they would also have to measure voltage on each side of the shunt for this calculation.
Yes they could calculate a close guess of current draw using wattage and such and would have to figure in a variable chip efficiency as the chip efficiency varies according to load/wattage.
You are perfectly correct you must always concern yourself with the amp draw on your batteries. The thing about the resistance is the batteries can't "see" it. It's downstream of the board. The board sees resistance, yes, and adjusts it's output voltage to make the set watts. But that doesn't mean anything to the batteries. Upstream of the board is the batteries, and they don't care what the board does on the other side. Upstream, voltage is fixed (sort of, it varies with battery charge) and watts are always watts. If the board is outputting X watts, then it must draw X watts plus board losses from the batteries and it must take it at whatever voltage the batteries are supplying at that charge state. Thus, to hold watts constant, if voltage is fixed, it is current that must vary. Nothing else can as resistance is also fixed by the wire size connecting the board to the batteries. Varying the resistance on the other side of the board is not meaningful. Watts is Watts, and they stay the same.
Yes but if the device can calculate the nominal voltage, and the power, it should be capable of displaying the solution of one divided by the other, the same way it calculates voltage needed to supply a set wattage to different ohm coils. It won't be as accurate as the method you're describing but it will be accurate enough to know if you're in the safe parameters of your battery. (Assuming you're not maxing out your current rating of your battery as high as you can)
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