agreed, they wont vape at 3.7v alone...meant for HV onlyFor the regular 510 setup I wouldn't get them. IMO
agreed, they wont vape at 3.7v alone...meant for HV onlyFor the regular 510 setup I wouldn't get them. IMO
just curious but wouldnt they outlast reg atty's even for vaping at 3.7? I am going to put this atty on my 905 and see how it fairs in comparison to my old broken in atty and let you know what difference I am getting.For the regular 510 setup I wouldn't get them. IMO
It's in the ohms highping...the atomizers that drewsworld (nhaler) carries are 5.2 ohms. Apply the math and it reduces voltage load to atty by ~2 volts.
Check out this thread for the detailed info:
http://www.e-cigarette-forum.com/fo...terest-check-custom-resistance-atomizers.html
Ed
Well explain why my brand new HV atty with 5.2 ohms on the dot runs at higher voltage under load then my current 510 atty i've been using the past week at 3 ohms.
I just tested it and i could not believe it and tested it again and again swaping between the 2 attys.
Brand new HV 510 atty @ 5.2 ohms on the Xhaler (tight cap):
Unloaded = 6.60v
Loaded = 6 VOLTS! 8-o
Regular 510 atty @ 3ohms
Unloaded = 6.6v
Loaded = 5.40-60ish volts
I guess you could say its defective but umm.. it reads 5.2ohms like its supposed to. I guess its not a high enough ohms to lower the voltage. maybe needs to be 10 ohms?
P.S. This is a brand new tube and switch i just got yesterday from Drew. 5.40v is the normal average vape under load that i've found on the Xhaler and also tested it yesterday when i got the new tube and switch.
It's in the ohms highping...the atomizers that drewsworld (nhaler) carries are 5.2 ohms. Apply the math and it reduces voltage load to atty by ~2 volts.
Check out this thread for the detailed info:
http://www.e-cigarette-forum.com/fo...terest-check-custom-resistance-atomizers.html
Ed
I am expecting mine in a day or two, so I haven't gotten a chance to try for myself yet. The link I posted was for reference as to the theory applied in the concept of these custom HV attys. A lot of people in there have been loving their HV 510's, so I'm thinking it's possibly a bad atty you got. I am also under the impression that these attys are meant for 2 x 16340 3.6v Li-Ion batts for optimal performance. I have both sets of batts to run 2 x 3.0 or 2 x 3.6, but again, my HV attys are still in route. I'll give em a ride at both to see.
eclypse, have you tried running them with 2 3.6 batts yet?
Ed
Yeah, I've read that. There's all kinds of squirrely math going on in that thread.
Bottom line is that it's very basic DC electronics laws at play in a PV. And it doesn't matter what the ohm rating is, the voltage dropped across a resistor (heating coil in this case) equals the voltage applied to that resistor. You put 1 volt on a 10000 ohm resistor and there is going to be a 1 volt drop across it. You put 1 zillion volts across a (gynormous) 1 ohm resistor and the drop across it is going to be a zillion volts. (and before anybody tells me, I know a zillion isn't a number)
The only thing that will change this is if there is a second resistor in the circuit wired in series. Then the math comes into play and it would depend on the ohm rating of the two resistors as to what percentage of the applied voltage is being dropped across each. I don't think that is the case here, unless there is a second resistor cramed into the atty somewhere.
Now, if you are talking about the amount of voltage that the battery drops when a load is placed on it, it's a completely different thing. But if that is what is being refered to in the claim, it's still (even more) wrong. The batt is going to drop the supplied voltage more if the resistance is less (ie load is higher).
Regardless, I just wanted to know where they came up with the "drops 2V" claim.
For further explanation...E(volts) over I(amps) R(resistance) is ohms law...P(power or watts ) over I(amps) E(volts) is the power factor...So you realy not lowering the voltage but you are lowering the watts to decrease the heat of the atomizer and drawing less current , therefore making the batteries last longer as well...By raising the resistance the ams are iversely proportional to the amps and or power... Now you know why I described it the other way
E Divide by R equals I
P Divided by E equals I
For further explanation...E(volts) over I(amps) R(resistance) is ohms law...P(power or watts ) over I(amps) E(volts) is the power factor...So you realy not lowering the voltage but you are lowering the watts to decrease the heat of the atomizer and drawing less current , therefore making the batteries last longer as well...By raising the resistance the ams are iversely proportional to the amps and or power... Now you know why I described it the other way
E Divide by R equals I
P Divided by E equals I