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Dual 3.7s???

Discussion in 'Modding Forum' started by MidnighToker, Jul 21, 2009.

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  1. MidnighToker

    MidnighToker Super Member Verified Member ECF Veteran

    Jun 27, 2009
    NC
    Is anyone running a mod with 2 3.7 batteries? Just curious if the 7.2v was too much for the atty or not. Made sense to ask this BEFORE I built my box mod...then again, I'll prolly build it anyway.
     
  2. Sun Vaporer

    Sun Vaporer Moved On ECF Veteran

    Jan 2, 2009
    Florida
    Yes--doing some experimenting I ran one with 2 3.7 volts---it lasted less then a hour---others may have had other experiences but that was enough for me.

    Sun
     
  3. jeffakamax

    jeffakamax Unregistered Supplier ECF Veteran

    Jun 4, 2009
    USA
    Only in parallel, not series.
     
  4. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    That's me, as well - I'm waiting on my switch and batteries to complete my twin 14500 nicostick.
     
  5. jonnyb

    jonnyb Unregistered Supplier

    Jun 12, 2009
    redford,michigan
    If you run a resistor in series it will limit the voltage to 5v,which will give you a killer hit and not burn out your atomizer.
     
  6. SmokinScott

    SmokinScott Super Member ECF Veteran

    Apr 21, 2009
    Acton MA, USA
  7. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    Sort of, but he's talking about series and not parallel, so he'd need a larger resistor to cut it down to 5v.
     
  8. MidnighToker

    MidnighToker Super Member Verified Member ECF Veteran

    Jun 27, 2009
    NC
    For parallel I would run 2 "separate circuits" to the atty? (Sorry...been a LONG time since electronics in HS) One for each battery?

    P.S. How much resistance would I need to drop that to 5v? I was actually looking to hit that "sweet spot"
     
  9. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    Parallel vs. serial battery setup, which is dependent on how your battery buses are in your mod (i can't see your pics). Since you say you're pushing 7v, you're obviously running your batteries in serial.

    You'll need to know the resistance of the atomizer - that and a little calculation will give you your amperage through the circuit, which from there lets you figure what resistor you'll need to get down to 5v.

    I showed the calcs in another thread... lemme find it again and I'll post it in here.
     
  10. MidnighToker

    MidnighToker Super Member Verified Member ECF Veteran

    Jun 27, 2009
    NC
    Cool. I appreciate it. I do have meter and am not a total ..... when it comes to electricity and circuits....just can not for the life of me remember the calculations and methods.
     
  11. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    Well, first you have to measure your total resistance. From that, you divide the voltage (7v) by the resistance to get the amperage

    I (amperage) = E (voltage) / R (resistance)

    Then , you solve for the voltage drop that you will need - you want to drop 2 volts across the resistor you're going to put into the circuit.

    I (amperage) * R (resistance) = E (voltage)


    Let's do a hypothetical, using common values I've seen tossed about in the forum:

    3 ohms seems to be a common value given for the resistance of an atomizer. 1.5a seems to be the max current they'll handle, long-term.

    I = E / R
    I = 7v / 3ohms
    I= 2.333 amps (sorta high)

    Now, let's figure out how much resistance we need to add to drop the voltage to 5v at the atomizer.

    I * R = E
    2.333 * R = 2
    R= .86 ohm - that's gonna be hard to find, I think.

    Let's look at it the other way around, and try to limit the amperage to no more than 1.5a

    R = E / I
    R = 7 / 1.5
    R = 4.666 ohm total resistance (5 ohm would be closest) That means adding 2 ohms to that 3 ohm atomizer.

    Let's work with that, and see what we get for voltage and amperage at the atomizer, shall we?

    I = E / R
    I = 7v / 5ohm
    I = 1.4 amp - pretty close!

    Now, let's figure the voltage drop:
    I * R = E
    1.4a * 2ohm = 2.8v dropped across the resistor, leaving 4.2v at the atomizer - a bit low.

    Now you have to figure POWER, so we don't burn up the resistor:
    P (power) = I * E
    P = 1.4a * 7v
    P = 9.8 watts - you're going to need at least a 10watt resistor, and 15-20w would be even better.
     
  12. MidnighToker

    MidnighToker Super Member Verified Member ECF Veteran

    Jun 27, 2009
    NC
    That just made my head hurt....... is that codeine?!?!

    Thanks a lot man. That help a LOT for the "dream mod"....someday. Then again, what do I have to lose by doing it now? A resistor?
     
  13. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    Google "Ohm's law calculations" :) It'll help.

    Yeah, you'd pretty much only lose a resistor if it was too light. 4 ohms total resistance at 7v, with 1 ohm before the atomizer would give you 1.75a current draw and 1.75v dropped, leaving 5.25v at the atomizer. Wattage would go up, though - you'd need a minimum of 15w dissipation, and 20w would be better.
     
  14. Lucacri

    Lucacri Senior Member Verified Member ECF Veteran

    May 24, 2009
    Manhattan, New York
    In this case, though, when the batteries will start to drain (but still above 5v combined), you'll get less voltage to the atomizer, right?
     
  15. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    Yes, but not as much as you'd think - 3 ohm atomizer, 1 ohm drop resistor:

    7v / 4ohm = 1.75a

    1.75a * 1ohm = 1.75v dropped

    7v - 1.75v = 5.25v at atomizer
    ------------------------------
    6v / 4ohm = 1.5a

    1.5a * 1ohm = 1.5v dropped

    6v - 1.5v = 4.5v at atomizer
    ------------------------------
    5v / 4ohm = 1.25a

    1.25a * 1ohm = 1.25v dropped

    5v - 1.25v = 3.75v at atomizer

    So, by the time the battery reaches 5v, you're still at the level of a 'normal' lithium battery....and with the larger batteries, the time it takes to drop to that level will be MUCH longer.
     
  16. Lucacri

    Lucacri Senior Member Verified Member ECF Veteran

    May 24, 2009
    Manhattan, New York
    That's great.... umh me thinking about creating new mod... :)
     
  17. mnealtx

    mnealtx Super Member ECF Veteran

    Jun 16, 2009
    Camp Bondsteel, Kosovo
    You'd still have to do the resistance measurements on YOUR atty, though - the numbers above are for a hypothetical atty with 3ohm resistance - I don't know if that's the actual case or not, I just remembered people saying that attys seem to run about 3ohms.
     
  18. MidnighToker

    MidnighToker Super Member Verified Member ECF Veteran

    Jun 27, 2009
    NC
    Just wanted to update this thread with the results.

    Batteries FINALLY showed up today from DX. After patiently waiting a few hours for them to charge up, I popped them into my box mod. Screwed on an atty and fired it up...

    Yup...like a dumbass I didn't check the voltage BEFORE firing up the atty. POOF!!:mad:

    Two things came of this. 1 ohm is NOT enough for this, and apparently 801 atomizers don't like 8.5v
     
  19. jimmieu

    jimmieu Senior Member ECF Veteran

    Apr 4, 2009
    Thanks for the great info mnealtx. I always wanted to know how That worked.
     
  20. SmokinScott

    SmokinScott Super Member ECF Veteran

    Apr 21, 2009
    Acton MA, USA
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