Dual 3.7s???
- Thread starter MidnighToker
- Start date
- th_trl_thread_readers 0
- Status
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
Sun
That's me, as well - I'm waiting on my switch and batteries to complete my twin 14500 nicostick.
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.
Like this one?
http://www.e-cigarette-forum.com/forum/modders-forum/27494-just-box-mod-my-first-one.html by Muldrick
http://www.e-cigarette-forum.com/forum/modders-forum/27494-just-box-mod-my-first-one.html by Muldrick
Last edited:
Sort of, but he's talking about series and not parallel, so he'd need a larger resistor to cut it down to 5v.
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"
P.S. How much resistance would I need to drop that to 5v? I was actually looking to hit that "sweet spot"
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.
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.
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.
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.
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.
Last edited:
That just made my head hurt....WTF 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?
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?
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.
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?
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.
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.
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!!
Two things came of this. 1 ohm is NOT enough for this, and apparently 801 atomizers don't like 8.5v
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!!
Two things came of this. 1 ohm is NOT enough for this, and apparently 801 atomizers don't like 8.5v
Sorry to hear about the POOF. Might I suggest using a 2ohm 2watt resistor, which should bring it down to about 4.75 with fully charged batteries. Credit to Mars_mcc:
http://www.e-cigarette-forum.com/forum/battery-mods/26110-simple-box-mod-swappable-resistors.html
http://www.e-cigarette-forum.com/forum/battery-mods/26110-simple-box-mod-swappable-resistors.html
- Status
Similar threads
