Can't vape at 3.7 volts

[Killjoy1]

Okay, I'm having some trouble with my GGTS, can't get a good vape at 3.7 volts. I never have since I got it, but really only tried once before back when it first arrived. Since that time I've been doing 6 volts to great result, but I ran out of charged 3v batts so when they needed to charge this evening I switched to an LR atty in my UFS and an 18500 batt. The results were pathetic, barely a wisp of vapor and no hit at all, as if the battery were nearly dead.


- tried two different batteries (AW IMR 18500 and Ultrafire 18350, both less than a month old) and got the same results
- batteries are fully charged, checked with multimeter
- all connections are clean on batts and GG
- tried screwing down a little looser and a little tighter on the battery, no change
- tried multiple attys and cartos, all vape fine on other mods (tested them quick on my Darwin @ 3.5 volts output)
- it's not the connector, I tried both the UFS and standard connectors, and both perform the same
- everything was working great with 2 rcr123 batts and a dual coil atty

Like I said, I haven't really used it at 3.7 before, only when I first got it and I thought it wasn't working then because of my n00bishness, which may still be the case

Anybody have any ideas or tips?? There must be something I'm doing wrong or missing . . .

 

[ChaosAffect]

It's probably your springs. Try using it without a battery spring to start off with. If it's hitting better then you need to take some sandpaper to your battery spring. The other possibility is the spring in your button needs to be sanded. That's a bit more complicated.

 

[Skuggi]

hmm apparently I was affected by the same issue, thanks chaos, I just thought I had gotten so used to HV vaping that the 3.7 wasn't cutting it anymore.

 

[fright88]

If you have a multimeter you can always check what the voltage is at the atty connector that might help.

 

[Killjoy1]

Hmm, I didn't even think about the button spring (already polished off the battery spring). I'll have to give that a try

 

[Killjoy1]

If you have a multimeter you can always check what the voltage is at the atty connector that might help.

metered on the connector at 3.96 volts. Gave both springs a polish and this went up to 4.07, but the vape improved significantly (no idea why such a change with virtually the same voltage). Still not spectacular, but it worked and there is a slight boost to 6v performance now, too. I'm about due for a full-on cleaning anyway, so maybe that'll be all I really need now

Thanks again for reminding me about the button spring, Chaos!

 

[Papa Hoyt]

It is all a matter of resistance. You may not have had enough resistance to alter the V much but the current or Amperage was diminished.

Same principle applies in Dual Coil Cartos. Based solely on resistance, there should be no difference in the vape from a 1.5Ω single coil and a 1.6Ω dual coil but there is. The amperage a single 3.7V battery can deliver does not meet the requirements of a "good vape" on 2X3.2Ω coils.

You have now eliminated the resistance and the current is allowed to flow to the coil. I bet with a little cleaning inside the switch you couls tweak just a touch more out of it,

 

[Zogem]

It is all a matter of resistance. You may not have had enough resistance to alter the V much but the current or Amperage was diminished.

Same principle applies in Dual Coil Cartos. Based solely on resistance, there should be no difference in the vape from a 1.5Ω single coil and a 1.6Ω dual coil but there is. The amperage a single 3.7V battery can deliver does not meet the requirements of a "good vape" on 2X3.2Ω coils.

You have now eliminated the resistance and the current is allowed to flow to the coil. I bet with a little cleaning inside the switch you couls tweak just a touch more out of it,

You've hit on a question that I've had for a bit, and haven't understood the answer to.. Perhaps you can explain it..

- I understand Juoles law. V^2 / R = W.. and W translates to BTU.
- Also understand that heat transfer (Watt or BTU) is a function of surface area and contact..

So strictly using Joule I can build a given wattage ( lets say 9 ) with various combinations of v and r, yet the results are not constant; the subjective view is that the results are better as V increases, to a point of diminishing or even negative return.

What part of the equation am I missing?

 

[mendnwngs]

Surface area..

IMO.. Its more coil per ohm gives you the best experience. But, this is neither hard and fast, nor is it any more than my opinion...

 

[Zogem]

Surface area..

IMO.. Its more coil per ohm gives you the best experience. But, this is neither hard and fast, nor is it any more than my opinion...

But I can change the gauge of my resistance and get all the surface area I need.

I'm starting the form the idea that voltage is just a function of existing equipment, and I can build W watts at Y surface area, both constant, and that as long as I meet W/Y, V doesn't matter all that much, as long as V^2/R = W.

Am I missing something?

-Z

 

[mendnwngs]

Smaller wire heats up faster per the same W per mm of coil than larger dia..

Since most are taking 2-6 second drags (of power on) smaller wire is the ticket..

So, short lengths of smaller wire (high resistance) in parallel (DC and TC cartos) to cut down the overall resistance, and suck up the amperage that makes it heat up so wonderfully

gives you very responsive heat.. Small dia. wire, high surface area.. Dunno how to put that into math right now.. Brain is fried.. (Like the pair of 800A SCRs I accidentally fried today at work )

 

[Zogem]

Lol... You "let the smoke out".. that's generally a bad thing..

Love the get the math if you can come up with it.

Disclaimer: I've know mendnwngs for years, and have been pestering him on the concept for a bit, ergo linked him.

 

[Papa Hoyt]

You've hit on a question that I've had for a bit, and haven't understood the answer to.. Perhaps you can explain it..

- I understand Juoles law. V^2 / R = W.. and W translates to BTU.
- Also understand that heat transfer (Watt or BTU) is a function of surface area and contact..

So strictly using Joule I can build a given wattage ( lets say 9 ) with various combinations of v and r, yet the results are not constant; the subjective view is that the results are better as V increases, to a point of diminishing or even negative return.

What part of the equation am I missing?

I hate to derail but I will explain this in a non mathematical way. If this is not clear we can discuss it on another thread or by PM.

We will use cars as most everyone here has driving experience.

Let say you are cruising along and you come up to a hill. The hill is a 7% grade(SINGLE COIL). In your average four cylinder car you will not have to downshift to pull the hill. But if the hill were 15% grade(DUAL COIL) you would not have enough horsepower(VOLTAGE) to pull the hill without downshifting. If you had a V8 engine (5-6 VOLTS) the 15% grade hill would not be as noticeable.

Seeing as a battery cannot downshift, you are stuck in one gear all the time, wide open.

The fact that the resistance is measured @ 1.6Ω is only half the equation, the other parts of the equation is the configuration of the load and voltage capability of the source.

If you were to put two 3.7V ecigs equipped with 3.2Ω cartos in your mouth and draw on both at the same time you would get a cool non eventful vape. Most people that vape have a heat level they enjoy and is an integeral part of the vaping experience. If they performed the same test with a single 1.5Ω they would have heat and vapor. the vapor may be more, it may even be compareable to the 2X3.2Ω test.

Where most people try to figure out the BTU or heat factor and fail is that they forget that this is not a single heat source (1.6Ω) but 2 sources(3.2Ω/2) on the same circuit. The BTU output is directly dependant on the resistance and material of the coil/coils.

The 3.2Ω coil will only heat to a specific temperature range that is dependent on the Voltage. Thus you do not double the heat by adding the second coil, you may create more heat but it is not going to get nearly the amount of heat with a single coil at a lower resistance or higher Voltage.

Dual coils are great for spreading the heat out and not having one hot spot to produce the vapor. This is a great advantage to those that like a warm vape @5V but tend to burn out the filler. The dual hot spots (3.2Ω)produce similar temperatures as the single hot spots(3.2Ω) but they do not have to hit it as long to get the amount of vapor they like.

You are correct in your observation of surface area, the heat is more effectively transferred to the juice at its "wetted surface". Juice close to the coil only heats up as the juice on the coil vaporizes. The way to get the most vapor possible is to maximize the amount of juice that touches the coil at any given time. The way you do this is to either alter the heating element to maximize the amount of juice it comes in contact with or add more material(Coils) to raise the amount of "wetted surface".

I hope you were able to take most of this in and it helps you figure out what you were asking.

 

[Eisofen]

In a nutshell a dc in a Odysseus would be pretty useless since the surface on the wick doesn't expand the same way it has to be, right? I could imagine that the liquid evaporates to fast and the wick isn't capable to transport that much juice in that short time, which results in burnt taste.