For those using LR atomizers: How much time is everyone getting out of the 10440's when using a LR attomizer?
LR 510 Atties on the Precise and S6
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You don't have to. I don't. LR atty's are the bees knees, though!
I put a fully charged 10440 on my precise with a LR 510 atty today and vaped it like mad to see how long it would last.
I got 1 hour and 45 minutes off of it. When the vapor died down I checked the battery with my multimeter and it was reading 3.75 volts.
I can usually take these batteries down to about 3.55 volts with a normal Joye atty, but the LR atty only wanted to go to 3.75 YMMV
Regards
Jim
Just remembered to try the LR atty's on my Precise. I've had some 1.8 ohm attys from Altsmoke for about a week and love them. Man, what a difference on the Precise, the Dealextreme batts I got were giving me headaches, but this is a dream. Can't wait to get even better 10440's in my possession. Might be too close to bed time to find out how long they'll last, but I'll report back. Wasn't getting much 'satisfactory' time out the TF 10440's to begin with.
Just remember these LR's are drawing like 2 -2.5 Amps so battery life will be less than a regular 510, drawing 1 amp ,,,,,ICR hi drain Batts.
yeah like 2.47A for a 1.5Ω LR versus 1.61A for a standard joye 510 (2.3Ω)
Got about an hour, the TF battery was at 3.5v when it started losing power. That's pretty much the same deal as when I use a 2.1ohm 510. The big difference is inside the atty, the 2.1's were getting gummed up easily when voltage started to drop, if I pushed it I would eventually get a nasty, cooked taste from it. The 1.8ohm atty cuts through it easier, vaporizes more easily and doesn't allow it to just sit and cook onto the coil, even down around 3.6v. No nasty taste if I push it either, just less and less vapor.
yep same deal with me... once voltage dropped to almost recharging time I would get an almost burnt but not burnt taste.... Not like a dry burnt taste, but that gunk tasting vapor but almost burnt like. Simply put its not tasty.
I compared it to simmering sauce in a pan and forgetting about it. Have fun cleaning that pan. I think scorched is the word.
I just did my first "under load" test, fairly ghetto style. I couldn't figure out a way to do it on this tight device, so I just touched a wire straight to the atty coil, and the other end was on the button as I pushed. Tried it multiple times and my meter was jumping frantically between 1.8 and 2.4v. I assume the jumpiness was due to the lousy connection to the coil, I was unable to get it to settle for even a second.
If I connect the battery and atty outside of the precise and test voltage under load, can I just factor in the Precise's resistance or something to get a more accurate answer? I'm not up on all these power equations floating around here but I'm trying to read as much as I can to understand it all.
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OK, I found the how to thread for testing under load. Going to make an adapter now.
If these are protected batts, then the protection circuit must of kicked in because of over current draw. 10440 are not even recommended for LRs, they are not a high drain batts.
I'm not up on it either but I don't think you can factor in the precise later because you don't know the capability of the battery. It might not be linear. but maybe.
V=IR
Resistance is going to be fixed. Can measure the atomizer and the resistance through the precise/switch should be known.
If you measure the unloaded voltage of the battery you can calculate the value for I.
If the battery is not capable of delivering that current then you'll see a drop in voltage to make the equation work.
So far so good. Now the question is if you can take your voltage measurement without the precise's resistance and just factor that into the equation later. Since R is a constant you would expect V and I to remain proportional but I don't know if the battery works that way. I don't know if maybe the batteries internal resistance changes non-linearly with voltage. You'd probably be pretty close anyway.
So, what's R for the Precise, switch and spring? I'm *guessing* something between 0.5 ohms to maybe something over 1 ohm?
See what you get without the Precise. That will be interesting enough
V=IR
Resistance is going to be fixed. Can measure the atomizer and the resistance through the precise/switch should be known.
If you measure the unloaded voltage of the battery you can calculate the value for I.
If the battery is not capable of delivering that current then you'll see a drop in voltage to make the equation work.
So far so good. Now the question is if you can take your voltage measurement without the precise's resistance and just factor that into the equation later. Since R is a constant you would expect V and I to remain proportional but I don't know if the battery works that way. I don't know if maybe the batteries internal resistance changes non-linearly with voltage. You'd probably be pretty close anyway.
So, what's R for the Precise, switch and spring? I'm *guessing* something between 0.5 ohms to maybe something over 1 ohm?
See what you get without the Precise. That will be interesting enough
I think I remember that. The guy never did give any figures for resistance of the circuit he was measuring did he? It's fine to see the voltage vary but kind of meaningless without knowing the circuit resistance, battery max discharge rate, etc...
Well, my soldering iron crapped out on me. I spent the last hour watching law and order while trying to rig up alligator clips and old atty adapters. Wasn't having much luck. Guess I gotta wait till I pick up a new iron to play more with this, maybe try to get the old battery iron to work later.
ETA: Here's that how-to page. Doesn't say much of anything, but I didn't read into the thread.
http://www.e-cigarette-forum.com/fo...-device-under-load-prodigy-other-devices.html
ETA: Here's that how-to page. Doesn't say much of anything, but I didn't read into the thread.
http://www.e-cigarette-forum.com/fo...-device-under-load-prodigy-other-devices.html
Ok, got off the tube and managed to piece together something that held together long enough. 4.01v TF 10440 in Precise w/ 1.8ohm altsmoke atty under load came down to 2.99v. So about a 1v drop.
Sooo, that's about 1.66 amps. Any idea on the discharge rate on those? Or the capacity? Max discharge rate can typically be about 2C so maybe those are like 550mAh or 750mAh? I have no idea. Maybe a lot less?
oops, disregard the above. Measure the resistance of your test circuit.
oops, disregard the above. Measure the resistance of your test circuit.
I am not sure how to tally the actual resistance of it, it was 0.1ohm on the hot wire and 0.9ohm on the ground side. My meter's internal resistance is 0.1. Would it just be .9+.1-.1=.9???? Should I be closing one side and measure it as a whole?
ETA: ok, I crossed one end and got 0.2, minus my meter that's 0.1ohm resistance in my little load tester.
ETA: ok, I crossed one end and got 0.2, minus my meter that's 0.1ohm resistance in my little load tester.
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Err, ummm, can you measure the whole circuit? like atomizer and everything. Ideally you would remove the battery and put one test lead on the positive and the other on the negative where they would connect to the battery without the battery there. That would give you the resistance of the circuit that you measured voltage on.
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