Currently im running a Sigelie 75w with dual 2.5mm 26g kanthal showing .6 ohms and my voltage is showing 4.7 at 45 watts. Im using a single samsung 25r. Is this ok and what voltage range should i stay in?
Voltage question
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Agreed with the above.
On a regulated mod, the voltage or wattage setting you use should be what most satisfies you. Start low, gradually work you way up higher until it begins to taste burnt, and then drop the setting down a smidge. Different flavors will taste best at different voltages/wattages.
On a regulated mod, the voltage or wattage setting you use should be what most satisfies you. Start low, gradually work you way up higher until it begins to taste burnt, and then drop the setting down a smidge. Different flavors will taste best at different voltages/wattages.
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Ok great. Just wanted to make sure the battery would be okay. How is the batteries rated at 4.2 volts and the thing reads 4.7?
Pay attention more to amps for the battery part.
0.6 ohms and 4.7v you're at about 8 amps. If you're using legit 25Rs then you're well within the safe limit. You're at what 36w or so now? I was running my 75w at 55w before I sold it.
Even at 75w you're at 11 amps with the 0.6 ohm coil. You're fine with it.
0.6 ohms and 4.7v you're at about 8 amps. If you're using legit 25Rs then you're well within the safe limit. You're at what 36w or so now? I was running my 75w at 55w before I sold it.
Even at 75w you're at 11 amps with the 0.6 ohm coil. You're fine with it.
Yeah I seen the amps as i use steam engine and ohm reader before i fire anything!!
Newer mods use buck to up the power. The chip acts like a capacitor, storing voltage until it's needed, and applying more under the right settings than what the battery is rated for.
Myself, though I lean at the safer side of doing things with batteries, up to 50 to 60watts the Samsung 25r is fine, 55 watts and above on a single battery mod I'd plug in a 30amp battery into the mod like an LG HB6, Sony VTC3 or VTC4.
Working out amps with a regulated mod like you have is not totally straightforward.
Say you have 35W at 4.7V, Watts=Volts x Amps, so Amps = Watts / Volts. You are drawing 35/4.7 = 7.4 amps from the mod.
However the mod is drawing more than that from the battery. If the battery is fully discharged, it might be supplying only 3.2V, but still 35W. So the amp draw from the battery is 35/3.2 = 11 amps. It will be a little more than that, since the voltage conversion is not 100% efficient, but you don't want to get too close to the rated current anyway.
The thing to remember is that power is conserved. No matter what happens to volts or amps, the power taken from the battery is the same as the power delivered to the coil.
Say you have 35W at 4.7V, Watts=Volts x Amps, so Amps = Watts / Volts. You are drawing 35/4.7 = 7.4 amps from the mod.
However the mod is drawing more than that from the battery. If the battery is fully discharged, it might be supplying only 3.2V, but still 35W. So the amp draw from the battery is 35/3.2 = 11 amps. It will be a little more than that, since the voltage conversion is not 100% efficient, but you don't want to get too close to the rated current anyway.
The thing to remember is that power is conserved. No matter what happens to volts or amps, the power taken from the battery is the same as the power delivered to the coil.
Um... No
It's a boost circuit that raises voltage. A buck converter lowers voltage. The chip is not acting like a capacitor.
No, the boost converter draws current (Amps) to raise the output voltage above the battery voltage.
The batteries voltage is independent from what the power supply is putting out. Fully charged the battery voltage is 4.2V it slowly drops as you use the device until the battery voltage is around 3.2V. Then you recharge it
To expand on this, with a regulated variable power device. Lithium battery at Nominal Rated Voltage is a platue of 3.6 to 3.8volts. Device like example above 35w at 4.7v, at full charge of 4.2v the mod will pull an extra amp of current out of the battery to pass into the boost circuit that converts that extra amp into an extra 0.5v, then slip streams that 0.5 in with the 4.2v to make 4.7v. As the battery drains the boost circuit pulls more and more amps out of the battery to slipstream extra volts into the output, so by the time you are at 3.6v on the battery charge you could be pulling upwards of an extra 10amps current, so 11amps in example above, then add in another 9amps for voltage boosting conversion, total amp draw on the battery could be 20a, the max continuous discharge rating of the Samsung 25R. Thus if I am over 50watts, I'll use a 30amp CDR battery that can handle that much stress without damage, over 80watts on a regulated device I'll use a dual battery mod, not a single battery mod.
I'm confused by your description of a regulated mod needing up to 20A when firing at 35W. Especially when you say another 9 amps is needed for the voltage conversion. @rurwin was right, at 3.2V at 35W you ideally draw 10.94A. Assuming 90% efficiency in the regulator circuit that means you would actually be drawing 12.16A from the battery, not 20A. You only need another 1.2A from the battery to handle voltage conversion inefficiencies.
I said could potentially ask up near there, especially at higher wattage, as stated above also what the conversion is not exact as a 1amp to 1volt conversion, knowing how people like to skirt the envelope and boundaries around here anymore, was merely trying to empart they should add in an extra amp draw buffer, which could be max 5amps or so, Ohms law Calculators work real well when talking what to expect as potential max output of a battery directly connected to a coil, in a regulated with a boost circuit a lot of people do not take the extra amp draw the mod does into account or do not know that it exists.
I agree with your last point. Always best to use the lowest voltage the mod will operate at and divide the max wattage by that. Then increase it by 10% to handle regulator inefficiency. This will give you the worst-case current draw from the battery. If the low voltage cutoff for the mod is not known, use 3.2V. I haven't seen a regulated mod operate below that.
[edit] @edyle's equation below is a great way to simplify things for most setups.
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Minor point here, and practically useless, but I know of at least one mod that does allow the batteries to drain past 3.2V. The gi2 clone off fasttech will definitely do that.... not that anyone would buy it now, but mine was bought in November or December of last year, and is now in my box of spare parts. Turns out it stays on and drains the batteries when you're not using it for a weekend and kills the batteries.
the 4.7 volts is the output voltage of the mod.
the battery itself has a voltage of 3.2 to 4.2 volts.
To estimate the amp draw on your vw mod battery, use:
amp draw = watts/3
Hmmm...I guess you would use 100W divided by 0.0V then.
Lousy "feature" that mod has! It doesn't have a cutoff while you're vaping?
Well, the cutoff is around 5 to 5.5 volts (series) if it's been used recently. If it hasn't been used, it'll just slowly drain the batteries until they're real dead lol. As in 0.0 volts dead. It was really only used for builds that required more power than I could use with my kbox or my lil smok mod, so I'm stuck with 40 ish watts for now. It had quite a few other "features" that didn't endear me to it, but the whole eating batteries thing is probably the worst.Hmmm...I guess you would use 100W divided by 0.0V then.
Lousy "feature" that mod has! It doesn't have a cutoff while you're vaping?
All the voltage and current goes into the boost convertor; it doesn't add a bit to the battery voltage. Boost convertors work like the ignition coil in a car. You have a switch across a coil. Opening and closing that switch causes a large voltage across the coil because that's what coils do. A diode and a capacitor (in the simplistic ideal circuit) then smooth the voltage to its average value. By adjusting the on and off periods of the switch you can adjust the average voltage. Other components measure the output voltage and adjust the switch to give the required output voltage. By the conservation of energy, if the voltage goes up then the current goes down. Power is lost in the switch but it is not a large amount. If you consider that any energy lost in the circuit can only cause the circuit to heat up, (conservation of energy again,) and the board is well insulated, then it is obvious that anything over a few watts will cause the mod to overheat quite quickly.
I've just run the maths: A 5 gram circuit (Sp. Heat cap = 400J/KgC) will heat up by 0.5 degrees per watt second. So a five second draw with an inefficiency of 10 watts would heat it by 25 degrees. Three or four draws in succession would make it too hot to operate. In your approximation, 9 amps at 3.2V is 28.8W, so a temperature rise of 14 degrees per second, or 140 degrees before the ten second mod cut-out. It would be worse than this because the heat would take time to conduct out of the chip and onto the PCB. In practise I doubt that the temperature rise is over ten degrees for a ten second draw, so the inefficiency must be not more than 2W, which is 0.6A in the worst case. The usual advice is to stay 20-30% below the battery rated current, which is at least 5A for a 20A battery.
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