battery and coils

[namelessss]

If I use 30 watts on a 0.9 single coil will I get the same battery life off 30 watts on a 2.0 single coil or more battery life?

Plus for more vapour should I have my coils right or loose?

 

[Swordsbeer]

Not sure you want to run 30 watts on 2.0 or could. That would take like 7.2 volts and 3.87 amps. using around 6 watts or so looks good, so you would get a lot longer battery life.


Edit
Sorry, here is where I got that information: http://www.steam-engine.org/

 

[Completely Average]

If you ran both at the same wattage you would get less battery life with the higher ohm coil. The more resistance, the more voltage is required to reach the same wattage.

However, you wouldn't have to run at the same wattage to produce the same amount of vapor. The larger surface area of the 2.0ohm coil would mean that more juice can be converted to vapor at once rather than having to run a really hot coil and feed a lot of juice to it really fast. Think of it like boiling water. A low ohm coil is like boiling water in a small diameter but really tall pan. To boil it all quickly you'll need a lot of heat. The higher ohm coil would be like using a really large diameter pan so the water would boil with less heat because more of the water is touching the bottom of the pan where the heat is.

In the end you would find that battery life is similar either way. You'll turn down the wattage for the larger ohm coil to keep from burning the juice or getting dry hits, but the voltage would end up close to the same either way.

 

[wesson]

Nicely explained. Even I understand that

 

[Kloudz]

Run a 2.0 coil at 8 watts and you'll have a awesome vape and great battery life.

 

[Baditude]

Plus for more vapour should I have my coils right or loose?

Seems like no one has addressed this yet. I'm assuming you made a typo, and meant to say tight and not right. Right?

I was taught that a microcoil (compressed so the wire loops touch each other) is better than an open spiral coil. A spiral coil has a higher chance to have "hot spots" or shorts. A microcoil should also have a more even spread of heat throughout the coil part of the wire.

microcoil

 

[RamShot Rowdy]

If I use 30 watts on a 0.9 single coil will I get the same battery life off 30 watts on a 2.0 single coil or more battery life?

Since we're talking about running two different resistance coils at the same wattage, I assume we're talking about a regulated variable wattage device. In that case, the current draw from the battery should be the same with either resistance atomizer. So the battery would drain at approximately the same rate with either atomizer assuming the device was set to 30 watts in both cases. There could be slight differences depending on the devices efficiency at powering one resistance vs the other.

 

[bwh79]

If you ran both at the same wattage you would get less battery life with the higher ohm coil. The more resistance, the more voltage is required to reach the same wattage.

Nicely explained. Even I understand that

Except it's wrong. With a mechanical/unregulated device, you simply do not get the same wattage with different resistances, and with a regulated/variable-wattage device, the resistance doesn't matter. As far as the battery is concerned, 30 watts is 30 watts, and it doesn't matter if you're pumping them into one ohm or ten, you will get about the same amount of battery life either way.

The key thing to remember here, is that watts in = watts out. The battery itself can only output a single voltage, which is whatever its native charge happens to be at the time (generally around 4.2 volts at full charge, and less as it begins to drain.) The battery puts out whatever voltage it has, and the chip adjusts its internal resistance to achieve the desired wattage. So when you draw 30 watts from a battery at, say, 4 volts, then it doesn't matter what the resistance of the atomizer is, or what voltage gets applied to it on the other side of the chip. The drain on the battery is still the same 7.5 amps, any way you slice it.

 

[DingerCPA]

I'm going to step out on a limb here, but my understanding is that the resistance DOES play a factor because it's going to determine what current is required by the coil.

Ohm's Law - Volts = Current * Resistance
Watt's Law - Watts = Volts^2 / Resistance (or Watts = Current^2 * Resistance)

Resistance is the constant, whereas either volts or watts gets set on a regulated device, or the remaining charge on the battery determines the available volts for a mech.

On a regulated device, if I have a 1ohm coil and I throw 30W on that, the device will apply 5.5V (even though the charge of the battery may only be 4.2V max - buck/boost circuity will push the higher voltage.) Therefore, I'd be running 5.5A through the coil. If I now do the same thing with a 2ohm coil, I'd be applying 7.8V across the coil, but only drawing 3.9A through the coil. Drawing lower Amps should mean that the battery would last a little longer.

On regulated mods, you set either Watts or Volts, and because the resistance is a constant, the opposite (Volts or Watts) is derived accordingly. Mech mods don't have the buck/boost to change the voltage applied to the coil, so the power output effectively changes as the charge on the battery drains. In order to "generate" higher wattage, the resistance of the coils has to decrease, but that can only decrease to not overdrive the discharge current of the battery. If I have a freshly charged battery (presume 4.2V) and a 0.5 ohm coil, I'm going to yield just over 35W (4.2*4.2/0.5), and I need a battery that can continuously deliver 8.4A current (4.2/0.5) As the charge on the battery decreases, let's say 3.8V, you'll be achieving about 29W, but you still need to deliver 7.6A from the battery.

 

[bwh79]

I'm going to step out on a limb here, but my understanding is that the resistance DOES play a factor because it's going to determine what current is required by the coil.

You might want to re-read the second paragraph of my post right above yours. On a regulated device, it's only the wattage setting that matters. So long as the watts remain constant, then the resistance of the atomizer does not matter, because the battery is outputting the same voltage no matter what, which, since we've already established that the watts are a constant, means that the amp draw on the battery remains constant as well.

A change in the resistance means that the voltage applied on the output side of the chip will have to be adjusted in order to compensate, but on the input side, where the battery is, it's still X-number of watts set by the user, divided by Y-number of volts output by the battery, which makes for a consistent amp draw from the battery, any way you work it.

With a mechanical/unregulated device, you're correct in that the coil resistance does affect the amp draw, but in that case, the watts do not (and can not) remain constant when you change the resistance, since they are a factor of the other values combined.

 

[RamShot Rowdy]

I'm going to step out on a limb here, but my understanding is that the resistance DOES play a factor because it's going to determine what current is required by the coil.

Ohm's Law - Volts = Current * Resistance
Watt's Law - Watts = Volts^2 / Resistance (or Watts = Current^2 * Resistance)

Resistance is the constant, whereas either volts or watts gets set on a regulated device, or the remaining charge on the battery determines the available volts for a mech.

On a regulated device, if I have a 1ohm coil and I throw 30W on that, the device will apply 5.5V (even though the charge of the battery may only be 4.2V max - buck/boost circuity will push the higher voltage.) Therefore, I'd be running 5.5A through the coil. If I now do the same thing with a 2ohm coil, I'd be applying 7.8V across the coil, but only drawing 3.9A through the coil. Drawing lower Amps should mean that the battery would last a little longer.

On regulated mods, you set either Watts or Volts, and because the resistance is a constant, the opposite (Volts or Watts) is derived accordingly. Mech mods don't have the buck/boost to change the voltage applied to the coil, so the power output effectively changes as the charge on the battery drains. In order to "generate" higher wattage, the resistance of the coils has to decrease, but that can only decrease to not overdrive the discharge current of the battery. If I have a freshly charged battery (presume 4.2V) and a 0.5 ohm coil, I'm going to yield just over 35W (4.2*4.2/0.5), and I need a battery that can continuously deliver 8.4A current (4.2/0.5) As the charge on the battery decreases, let's say 3.8V, you'll be achieving about 29W, but you still need to deliver 7.6A from the battery.

With a mech mod or variable voltage mod, you are correct. Increasing resistance will result in less current draw (Assuming everything else remains the same.) and cause the battery to drain more slowly than with a lower resistance atomizer.

On a regulated variable wattage mod set to X watts, the battery still has to produce ~X watts of power, so the battery drain should be close to identical regardless of the atomizer resistance.

You can verify this with Steam Engine.

 

[edyle]

I'm going to step out on a limb here, but my understanding is that the resistance DOES play a factor because it's going to determine what current is required by the coil.

Ohm's Law - Volts = Current * Resistance
Watt's Law - Watts = Volts^2 / Resistance (or Watts = Current^2 * Resistance)

Resistance is the constant, whereas either volts or watts gets set on a regulated device, or the remaining charge on the battery determines the available volts for a mech.

On a regulated device, if I have a 1ohm coil and I throw 30W on that, the device will apply 5.5V (even though the charge of the battery may only be 4.2V max - buck/boost circuity will push the higher voltage.) Therefore, I'd be running 5.5A through the coil. If I now do the same thing with a 2ohm coil, I'd be applying 7.8V across the coil, but only drawing 3.9A through the coil. Drawing lower Amps should mean that the battery would last a little longer.

On regulated mods, you set either Watts or Volts, and because the resistance is a constant, the opposite (Volts or Watts) is derived accordingly. Mech mods don't have the buck/boost to change the voltage applied to the coil, so the power output effectively changes as the charge on the battery drains. In order to "generate" higher wattage, the resistance of the coils has to decrease, but that can only decrease to not overdrive the discharge current of the battery. If I have a freshly charged battery (presume 4.2V) and a 0.5 ohm coil, I'm going to yield just over 35W (4.2*4.2/0.5), and I need a battery that can continuously deliver 8.4A current (4.2/0.5) As the charge on the battery decreases, let's say 3.8V, you'll be achieving about 29W, but you still need to deliver 7.6A from the battery.

You have compared the 5.5Amp output to the 1 ohm coil with the 3.9Amp output to the 2 ohm coil.
You are correct that lower amp draw should mean longer lasting battery, but the amp draw from the battery is the same in both cases and is approx 7.5 amps.

 

[DingerCPA]

Yes. Thanks all who cleared that up for me. Regulated device power out must = power in. 10w with 3.7v input doesn't care about output resistance. Still need 10w......

Sent from.... ooh look! Something shiny!!!