I built a dual coil 26g kanthal reading about .8ohms. I have it set to 55 watts giving me 6.6 volts and 8.33 amps. I am used to .2 to .3 builds. My first question is this: what drains the battery faster higher volts or higher amps? I know that batteries discharge (at full capacity) 4.2 volts. Does setting the higher volts affect the batter performance and life?
What drains a battery faster, Volts or Amps?
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The question doesn't make sense. The two are inextricably linked to each other, by the constant of your atomizer resistance. That's like asking "which uses more gas, RPM or MPH?" As long as the relationship between the two remains constant (say, you keep the car in third gear), then as one factor goes up, so does the other. Likewise with volts and amps. As long as the atomizer resistance remains constant, increasing voltage will automatically increase the amps, and vice versa. So neither drains the battery any "faster" than the other, you're just looking at the same picture from two different angles.
Watts over time, actually. A 1 volt, 10 amp current is the same thing as a 10 volt, 1 amp current. They both make 10 watts over the time the circuit is energized. batteries are rated in mAh, milliamphours, meaning the battery will provide a current at the rated mAh for one hour. This gives you an idea what the battery will do in practice. But the true measure of capacity is watt hours, how many watts of power the battery will provide for what time period.
Higher watts drains battery faster.
watts = volts x amps
the 6.6 volts and 8.33 amps that you quote are the volts and amps *output* of the electronics.
the battery itself has a voltage (at full charge) or 4.2 volts which is the *input* to the electronics.
Current or amps is the ONLY thing that drains a battery. The more amps drained the faster it runs down.
This is all about the watts. The same watts with any resistance will drain the battery at the same rate. Change the resistance of your coil(s) and the volt and amp output from the mod will change but if you keep the watts the same the only difference on battery drain will be the difference in regulator efficiency at different mod volt outputs, which is minor.
Noob here, now you have me curious . Is it watts or is it amps. In theory 1ohm at 4v=4A=16W or .5ohm at 2v=4A=8W. Both at 4A. Would they drain battery same even though one at higher watts?
With a regulated mod if you are putting out voltage higher than the battery voltage it draws a bit more amps per volt out out of the battery as the efficiency decreases a bit to make the higher voltage.
but as far as the battery itself goes only amps drains the battery.
but as far as the battery itself goes only amps drains the battery.
If you energized both circuits for the same time, 16w drains the battery faster than 8. Power is measured in watts over time, just like you pay the power company for kilowatt hours. Energizing both for 1 second yields 8 watt seconds vs. 16 watt seconds. The latter is twice as much power. The 8 watt circuit for two seconds = the 16 watt circuit for 1; both are 16 watt seconds.
Let me simplify the question.. If I build a .8 ohm coil and run it in my IPV4 at 50 watts, and then tomorrow build a .2 ohm coil in the same IPV4 and run it at 50 watts, Batteries at full charge when I first build, which set up will I be able to vape longer on before my batteries need to recharge?
Neither one. 50 watts is 50 watts. Each time you push the fire button for 1 second, it uses 50 watt seconds of power, and the battery only holds so many watt seconds. There are some minor considerations-- higher amp loadings increase heat latency, which reduces efficiency, for example-- but in practical terms, the coil resistance doesn't matter. The mod compensates for it. If you tell it 50 watts (and that lies within its capacity) then it does 50 watts. The rest of it is how long you hold the button down vs. how many watt seconds the battery holds.
Due to inefficiencies in the mod DC/DC Buck Boost converter circuitry output voltages higher than your actual battery voltage drains the battery just a bit faster, but for all intents and purposes no real difference in battery life.
Are there any threads/good reads /links about this watts/sec. Ma/hr battery life?
The answer to his question is amps. The answer to your question ,what uses more gas rpm or mph, is rpm. Its actually pretty black and white for both situations, there shouldn't be any confusion.
Not as far as I know. The math is hairy for an exact answer because it involves considerations not contemplated in the current conversation, such as cell temperature and design optimization of the cell, etc.; and must further be integrated over time by some factor as cell output voltage varies directly with charge state. However, a quick and dirty estimate is easily derived. Divide battery mAh by 1000 to yield amps, viz. an 1100 mAh batt is also a 1.1 amp hour battery. Multiply battery amp hours by nominal battery voltage; 3.7v in an IMR cell. In our example, your 1100mAh, 3.7 volt battery holds *about* 4.07 watt hours. Multiply this by 3600-- example: 14,652-- for watt seconds or Joules. Clearer, I hope? Batteries are rated in mAh or Ah to give a better indication of how their design is optimized; maximum current output capacity or maximum total power storage. This is the difference between "starting" batteries for cars and "hotel" batteries for boats or RVs. The former will tolerate very high currents, but don't have huge total storage. The latter hold more power, but can't put it out as quickly.
Good replies here. Tony16 here's a short read and you can find more in the web-site if your search 'watt hour' Calculating the Battery Runtime - Battery University
One joule = one watt second. Batteries produce energy. The joule is a handy measure of energy.
It is.
Because it takes things out of a Static, Non-Deferential, view and puts it in a Time Dependent frame of Reference.
1: It is watts.
But if you compare amps you have to compare the amps output of the battery.
2: The example of 0.5 ohm at 2v=4A=8W is using 2 volts which would be the regulated mod output voltage, and so the 4 amps in that case is the amps output of the regulated mod
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