xtar 2100mAh 30A wrapper removal

[Mikenekro]

So I had an xtar 30A laying around and I wanted to see if it actually was just a rewrapped VTC4 because I have noticed quite poor performance from this battery before. I was quite surprised that when I unwrapped it, it was a genuine VTC4 cell (from what I can see), due to the poor performance I had out of the battery.




Just as I thought. They soldered the positive onto the top. I could see the 3 posts of the VTC4 underneath the bulky positive of the xtar.
What I am wondering is if this soldered positive could have affected the overall performance of the battery?

I'm not the only one who noticed the performance drop of this battery. If you search for xtar 30A tests there is at least 1 test I found that confirms this.



All of the features seem to match up with a bare VTC4 cell.


So I am not sure if the positive on the Xtar was cutting out performance or if I just got a bad cell or if Xtar is using old cells. So I am going to re-wrap it with the soldered on positive off and see if it performs any better without it.

 

[inswva]

Please do update the thread. I'm interested in your findings. I use four of the XTAR VTC4 re-wraps in my regulated mods. I've not noticed any profound performance difference between the XTARs and my OEM VTC4s. I actually prefer the XTARs due to the beefy wrap and large positive contact. But, if it is determined the addition of the XTAR positive is hampering performance, I'll likely remove it and re-wrap the ones I have.

 

[Fictitious Character]

I too am interested in your findings on this matter. I also am pleased to see the reason why the batt is a tad longer than the normal vtc4.

Out of curiosity what device are you using them with that you notice such a voltage drop? I am assuming a mech device?

 

[Ryedan]

Good stuff Mike! I'm wondering what performance drop you noticed, mAh drop or voltage drop and how did you measure it? Also, I would have thought they would have spot welded the positive plate on, not soldered it. Are you sure it was soldered?

It will be interesting to see how it goes without the positive plate on.

 

[Mikenekro]

Yea I was using it in a copper mech mod.
The battery only lasted about 1 hour or so before it was close to 2.5V. It was running around 15A. I don't have testing equipment but that doesn't seem right. I'm not sure it was soldered, that was just a guess.

EDIT:
The test results I found are here: http://www.candlepowerforums.com/vb/showthread.php?387353-Test-Review-of-Xtar-18650-2100mAh-%28Black%29
I may order another Xtar and test the performance before and after removing the positive in terms of battery life in real world performance.

 

[Ryedan]

Thanks for the information Mike. There goes what I thought was a good source for VTC4s! It will be interesting to hear how yours does without the positive plate.

 

[Sgt.Rock]

I figured they had added the top post but for the life of me I can't understand why they would do that...makes no sense at all.

On the other hand, I have two in my rotation here with a couple VTC5s and 25Rs and really don't see much if any difference in performance or longevity.

Maybe your batt had a bad solder joint?

Love to hear how you make out after removing the top plate. May just have to desolder mine too. The only thing is I do actually prefer the wrap to any other batteries I have here and of course that would be destroyed if I did that dangit.

 

[tFOrRESTee]

I have a pair of Efest 2500s from amazon back in early Oct, a pair of VTC4s and LGHE2 from RTD vapor in late Nov, and a pair of Samsung 25Rs from illumn.com in early Dec. I received my Sigelei 100+ in early Dec as well.

Before I got my Sig100+, I have used each VTC4s only once. Basically they were married. My HE2 were never used until my Sig100+ arrived.

However, my Efests have been used so many times that they were not married.

I marked 1 or 2 on each pair and they took turn to get into my sig100+.

The best performance I observed was SamSung, and Effest and LG came second because all 3 lasted me 2 days. I charge my batteries before they hit below 40% battery life.

Sony, on the other hand, I had to charge them every day.

Granted Sony is 2100 mah and others are 2500 mah, but its battery lives should not be half of the rest.

Sony may be able to handle very low subohm for huge amp draws. But with series box mod, I build large and long wraps to up the resistance and to reduce the amp draw. Therefore, handling huge amp draws is no longer my concern, however, Sony is disappointing in battery life though.

I am waiting for the delivery of my parallel box mod along with my first ever voltage drop meter, and I will buy another pair of vtc4 and a pair of efest 2100 mah (from a different online vendor) to see how they compare, fair and square. Parallel box mods won't up the voltage but double the capacity and split the amp draw, so it will be fun to see the result.

 

[inswva]

I have a pair of Efest 2500s from amazon back in early Oct, a pair of VTC4s and LGHE2 from RTD vapor in late Nov, and a pair of Samsung 25Rs from illumn.com in early Dec. I received my Sigelei 100+ in early Dec as well.

Before I got my Sig100+, I have used each VTC4s only once. Basically they were married. My HE2 were never used until my Sig100+ arrived.

However, my Efests have been used so many times that they were not married.

I marked 1 or 2 on each pair and they took turn to get into my sig100+.

The best performance I observed was SamSung, and Effest and LG came second because all 3 lasted me 2 days. I charge my batteries before they hit below 40% battery life.

Sony, on the other hand, I had to charge them every day.

Granted Sony is 2100 mah and others are 2500 mah, but its battery lives should not be half of the rest.

Sony may be able to handle very low subohm for huge amp draws. But with series box mod, I build large and long wraps to up the resistance and to reduce the amp draw. Therefore, handling huge amp draws is no longer my concern, however, Sony is disappointing in battery life though.

I am waiting for the delivery of my parallel box mod along with my first ever voltage drop meter, and I will buy another pair of vtc4 and a pair of efest 2100 mah (from a different online vendor) to see how they compare, fair and square. Parallel box mods won't up the voltage but double the capacity and split the amp draw, so it will be fun to see the result.

As info, your atomizer resistance doesn't impact amp draw from your batteries with a regulated, variable wattage mod. It only determines the board's output voltage and subsequently, the boards current output. Amp draw from your batteries is the same whether you are using a 1.5ohm build or a .2ohm coil given the same wattage setting. The only determinants of battery current drain (amp draw) are your power setting and the real time voltage of your batteries.

 

[tFOrRESTee]

As info, your atomizer resistance doesn't impact amp draw from your batteries with a regulated, variable wattage mod. It only determines the board's output voltage and subsequently, the boards current output. Amp draw from your batteries is the same whether you are using a 1.5ohm build or a .2ohm coil given the same wattage setting. The only determinants of battery current drain (amp draw) are your power setting and the real time voltage of your batteries.

No way, you are either messing to me, or Ohm's law has been lying to me from its grave. Lol, j/k.

Anyway, I think I am missing something here. I will really appreciate it if you can point out my mistakes.

But here are some of my thoughts.

First of all, what do you mean by real time voltage, what is that?

Now, according to ohm's law, V=I*R, and also we have W=V*R, so we can deduct that W=I*I*R, or I = sqrt(W/R). As you know, V=voltage, R=resistance in ohms, and I = current in Amps.

So if I have 100 watt and my resistance is 0.4, so shouldn't max current output = sqrt(100/.4) ~= 15.8 amp?

Now on my sig100+, when I set it to 100 watts, it shows 6.3 v. 100w/6.3v ~= 15.8 amp as well. So it matches with my resistance's max output. Are you telling me my batteries are outputting less than 6.3? Say it's only outputting 3.5 volts, making it 100/3.5 ~= 28.6 amps? I thought my mod can do bucking so that it can reach higher voltages, am I being lied to?

Also when I set it to 35 watts, it shows 3.7v and the amp draw is 9.4A (=35/3.7). When it's 55 watts, it shows 4.6v and 12 amps, and again, all those numbers match with the resistance, 0.4 ohms.

Please tell me where I got the data wrong, or I am missing some other factors.

Thanks

 

[inswva]

From another thread I posted in:

Coil resistance in a variable wattage, regulated mod like the Sigelei 100 has no bearing on battery current drain. Coil resistance only dictates the voltage ranges you can use.

Notice the amp draw doesn't change in the examples below (.5ohm coil and 1.0ohm coil) but the output voltage does. The second example also shows that a 1ohm build cannot be run at 100W due to the mod's voltage limitations. Additionally, current drain increases as battery voltage decreases. This can be modeled in the calculator by clicking the the down arrow next to the battery voltage and watching the current drain value increase.

Battery drain | Steam Engine | free vaping calculators

Simply put, the basic Ohm's Law calculators aren't applicable for variable wattage devices.

 

[tFOrRESTee]

Ok, I played around that steam-engine spreadsheet a bit and it looks like watts and voltages are dictating the amp draw, not resistance. But how and why? Ohm's law is ohm's law, how couldn't it be applied?

2. "Coil resistance in a variable wattage, regulated mod like the Sigelei 100 has no bearing on battery current drain. Coil resistance only dictates the voltage ranges you can use."
What's the formula? F(v) = R* ( what unknown factors?) When I build 0.3 ohm coils, what voltage is it going to be? how about .4, .5, and so on?

Please clear the puzzled mind, you, Ryeden, any one?

 

[Wolfenstark]

Is it the ohms law applies its just that the mod takes a bit more from your battery due to some loss to get to the settings you put in ?
Depending on the efficiency of your mod how much extra it draws from your battery.

Forget the exact numbers but i put in my setting into the battery drain tab and it gives me whats going to the atomiser and a higher number being taken from the battery.

Or maybe whats being talked about is something entirely different.

 

[Mikey6]

I figured they had added the top post but for the life of me I can't understand why they would do that...makes no sense at all.

On the other hand, I have two in my rotation here with a couple VTC5s and 25Rs and really don't see much if any difference in performance or longevity.

Maybe your batt had a bad solder joint?

Love to hear how you make out after removing the top plate. May just have to desolder mine too. The only thing is I do actually prefer the wrap to any other batteries I have here and of course that would be destroyed if I did that dangit.

The top cap was added to reinforce the post. The additional plate is pretty thick and overall adds about 1mm I think. I have 2 of the xtar vtc4s and wasn't happy with length to start. In my stingray and smpl it pushed the buttons out pretty far and made them look pretty darn ugly imo. Found a thread on some other forum on how to pop off the plate and gave it a shot. Pretty easy to do, didn't even damage the wrapping besides the tiny bit I trimmed off the top to pry to plate off. Glued the insulator back on and respond the wrap and they are good to go.

Thinking the OP may have received some bad cells though as mine performed great out of the mail, more of a cosmetic issue for me.

 

[Stosh]

Ok, I played around that steam-engine spreadsheet a bit and it looks like watts and voltages are dictating the amp draw, not resistance. But how and why? Ohm's law is ohm's law, how couldn't it be applied?

2. "Coil resistance in a variable wattage, regulated mod like the Sigelei 100 has no bearing on battery current drain. Coil resistance only dictates the voltage ranges you can use."
What's the formula? F(v) = R* ( what unknown factors?) When I build 0.3 ohm coils, what voltage is it going to be? how about .4, .5, and so on?

Please clear the puzzled mind, you, Ryeden, any one?

It's very simple, you're mixing up what the input voltage and amp draw from the battery with the output voltage and amperage seen at you atomizer. The electronics between the input and output changes how each if affected.

Go back to the steam engine, set it for a regulated mod using VW, set it for a fixed wattage and vary the resistance all you like. The output voltage and amps will vary but the input side, the battery drain amps will not change. Now leave the wattage AND resistance set, and vary the Battery Voltage, drop it from 4.2 down to 3.7 then 3.3 like the charge is being used up.....the output voltage & amps don't change but the Battery Drain current will increase.

Coil resistance dictates a voltage range only in the sense of what your regulated mod is able to produce. If you use a 100 watt mod set for max wattage and throw a 3Ω atty on it the mod would need to be able to output 17.32 volts, where a 1Ω would only be 10 volts. Depends how high your particular mod can boost the voltage for a particular wattage setting.

 

[Ryedan]

Ok, I played around that steam-engine spreadsheet a bit and it looks like watts and voltages are dictating the amp draw, not resistance. But how and why? Ohm's law is ohm's law, how couldn't it be applied?

2. "Coil resistance in a variable wattage, regulated mod like the Sigelei 100 has no bearing on battery current drain. Coil resistance only dictates the voltage ranges you can use."
What's the formula? F(v) = R* ( what unknown factors?) When I build 0.3 ohm coils, what voltage is it going to be? how about .4, .5, and so on?

Please clear the puzzled mind, you, Ryeden, any one?

Stosh nailed it . In Steam Engine, Battery drain, look at the "Atomizer – what hits your topper" section and that is the volts and amps the mod has to output in order to make the watts set in combination with the resistance in the atty. If the mod can't output either the voltage or the amps, you won't get the watts.

Look at the "Battery drain – what taxes your battery" section and you'll see what's happening to the battery at the same time. If you have two batteries in series, double the battery voltage. If you have two batteries in parallel, you know you have twice the amps available at the battery end.

Keep in mind that battery amp draw will go up as battery voltage goes down (try it in Steam Engine and you'll see what I mean), so I always use 3 volts for one battery and 6 volts for two in series if I'm looking at battery max amp draw, opposite to what we do for mechanical mods.

 

[tFOrRESTee]

I got it. So for variable wattage, when the battery's voltage drops, it has to output more amps to keep up with the wattage required.

So, I have one more question ( sorry for being such a pain in the ..., ). What's the formula for the voltage drop vs amp draw increase in a series box, especially for a vw box.

Also, I guess it's the same for vv box too. To keep it consistent in the desired voltage, it will draw more n more amps as well. So what's the formula?

Thanks guys, I really appreciate it.

 

[Ryedan]

I got it. So for variable wattage, when the battery's voltage drops, it has to output more amps to keep up with the wattage required.

Yup, you've got it

So, I have one more question ( sorry for being such a pain in the ..., ). What's the formula for the voltage drop vs amp draw increase in a series box, especially for a vw box.

I know the basic formulas, but I don't work with them per se. I don't have to because the people who programmed the Steam Engine web site worked out the extended formulas for me

Go to the Steam Engine web site and explore the pages there. I think you'll find all the calculators you need for all the vaping calculations you need to do.

If you have any specific questions, just ask.

In the case of "voltage drop vs amp draw increase in a series box, especially for a vw box", a lot depends on the batteries you're using and the mod. Because of the many variables involved in this scenario, this is not something that a typical calculator can predict.

 

[ronnbert]

Now, according to ohm's law, V=I*R, and also we have W=V*R, so we can deduct that W=I*I*R, or I = sqrt(W/R). As you know, V=voltage, R=resistance in ohms, and I = current in Amps.

Just so you know, your math was slightly off. If V=I*R and W=V*R, then W=I*R*R or I=W/R^2. Not that it has any bearing to the discussion, as others have pointed out resistance has no bearing on amp draw, just thought you should know if you use this formula for a mech, where it is relevant.

 

[tFOrRESTee]

Just so you know, your math was slightly off. If V=I*R and W=V*R, then W=I*R*R or I=W/R^2. Not that it has any bearing to the discussion, as others have pointed out resistance has no bearing on amp draw, just thought you should know if you use this formula for a mech, where it is relevant.

Sorry, that was a typo. W=V*I, not W=V*R. Thanks.