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Thread: Ohms/Amps/Volts -???

  1. #1
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    Default Ohms/Amps/Volts -???

    After all these changes and developments in the e-cig technology, I'm constantly upgrading to new and better stuff-- and it's great!!

    I've been really happy with my Ego-T battery (1000) and a 510-lr atomizer. I tried the Ego tank for a while but came back to dripping. It's the perfect/strong throat hit and NO burning.

    I wanted to upgrade (for more battery power basically) so I bought a 5 volt E power with the new tank. (2.0 cartomizer) No throat hit at all... I'm now trying a 1.5 ohm cartomizer and the burn I keep receiving trying to break it in is killing me. The cartomizers work on my Ego-t battery and it's good--- But did I just waste my money on the 5volt battery thing? Can I lower it down somehow with different batterys to make it have the same output as my Ego-T?

    On a pie chart for instance..

    How would you relate ohms/mAh/Volts to battery longevity and throat hits with no burn?

    Sorry for so many questions..this gets complicated the deeper you get

    Please discuss!

    btw - I'm brand new to cartomizers..why are they so much cheaper than atomizers?

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    Lower ohms = higher current

    Higher ohms = lower current

    For simplicity accept that current is what heats things up and that is called Power.

    Power = Current x Current x Resistance (you will notice current is exponential)

    so at say 4 Volts and 2 ohm carto/atty

    P = 2 amps x 2 amps x 2 ohms = 8W

    The higher the wattage, the hotter the coil, the more juice gets vaporized, the higher the TH

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    Cartomizers are designed to be disposable. A feature that I use.

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    Now onto second part ... maH

    This is the capacity of the battery (how much energy it can store)

    1W of heating is equal to 1 joule of energy for 1 second. You can only store X number of joules into a battery. Once the 'tank' is empty ...its empty.

    The higher the wattage you vaping at, the more energy per second you are consuming (foot heavy on the gas pedal), the quicker you will drain the battery.

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    Here's an Ohms/Voltage Chart. You need more Ohms, not less, SR cartos, not LR, at least a 3.0, probably higher to get the same taste/feel of the Ego which was only 3.4 volt.
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    Last edited by dearme; 02-04-2012 at 10:21 AM.

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    The more amps you are drawing on your battery, the faster it will drain.
    The "sweet spot" vapers sometimes refer to, when determining a particular voltage/resistance combination boils down to wattage.
    For example, I tend to like to vape a certain juice at 6.5 watts. If I were using a 3.0 ohm cartomizer, I would need to set my device to about 4.4 volts, to achieve 6.5 watts. In doing so, I would be drawing 1.47 amps.
    Now, if I were using a 2.0 ohm cartomizer, I would only need about 3.6 volts to achieve this 6.5 watts. However, the amperage draw would be 1.8 amps.
    In both examples, I have achieved my desired 6.5 watt "sweet spot", however, in the second example, the amperage draw is more, which will mean that the battery will drain faster.

    You can determine these values by using an Ohms Law calculator. Here is the one I used for the above examples:
    Ohm's Law Calculations With Power

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    Quote Originally Posted by Rule62 View Post
    .
    In both examples, I have achieved my desired 6.5 watt "sweet spot", however, in the second example, the amperage draw is more, which will mean that the battery will drain faster.

    [/url]
    Please explain more...?

    So if I set my PV to 6V and use a 5.54ohm carto (6.5W)

    and compare it to

    Setting the PV to 3.3V and use a 1.68ohm carto (6.5W)

    The battery is going to last longer on 6V example ???

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    Quote Originally Posted by Sicarius View Post
    Please explain more...?

    So if I set my PV to 6V and use a 5.54ohm carto (6.5W)

    and compare it to

    Setting the PV to 3.3V and use a 1.68ohm carto (6.5W)

    The battery is going to last longer on 6V example ???
    Yes, because in your second example, you are drawing approx. .9 amps more.

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    watt = amp * volt
    watt hour capacity = amp hour capacity * voltage
    Wh capacity = (mAh capacity / 1000 ) * voltage

    Since the PV is boosting the voltage from nominal voltage (say 3.7V) to 6V and power is constant (ignore efficiency) the battery itself is supplying 1.76A to the boost circuit at 3.7V.

    Since the PV is bucking the voltage from nominal voltage (say 3.7V) to 3.3V and power is constant (ignore efficiency) the battery itself is supplying 1.76A to the boost circuit at 3.7V.

    There is no difference in apparent load to the battery and the only questionable extra life would be from the efficiency in either the buck of boost circuits themselves.

    Since the battery can only store a finite number of Wh there should be very little difference it battery time irrespective of voltage.

    My

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    How many amps is too many (i.e. dangerous or battery killing)?

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