Lava tube and Boge

[s]

I'm playing around with my LT... No not like that!

I have LR and S Boges that I am using with it.

They both work fine the LR ones on 3.7v and the S ones on 4.1v.

When I check to see the 'actual' votage that is going to them they are both showing 4.1v. Should this not be different due to the different ressistance? I also put the 'High Voltage' atomisor on to see when that said and that was 4.1v also.

I'm a little confused, I thought that the LT would deliver the max voltage to whatever you put on it even if it was set to 6v as a safeguard.

Am I missing something here?

[dave]

If you mean you are clicking 7 times, that gives the voltage of the battery. The LT boosts the voltage to whatever you have set it. However it will not allow you to draw much more than 2.5 amps, so with a 2 ohm carto even if you set it at more than 5 volts it will automatically reduce the voltage.

[s]

What I couldn't understand was if the Battery is showing 4.1v (7 Presses) how on earth will it produce 6v. Think I'm confusing my self here lol

[leftfield]

I'm not the best person to explain it as I really, really just don't know the electronics side of things so I'll give it as best a shot as I can.

Your battery starts off at 4.2v and will go as low as 3.2v-ish before it cuts out (it's best to charge it when it reaches 3.7v though). That's sort of the "charge" level of the battery, not what the lavatube is putting out.

What the lavatube is putting out you don't really know without some proper kit. For low resistance gear, it'll stop at maybe 3.7 - 4.2v (maybe, I don't know lol) and if you keep upping the voltage to say 5.4v or all the way to 6v it will actually only be putting out 3.7 - 4.2v, it just doesn't tell you. It will keep vaping, but not at the voltage it tells you.

The closer you get to 6v with any resistance atomiser/cartomiser, the more it will dip down in voltage compared to what it's telling you you're using. It does it because your battery is only putting out around 3.7v and the circuitry in the lavatube is "boosting" the voltage up to what you set it to or what it's capped to.

If you use, say a 2.5ohm cartomiser and use it at 5v, it's taking the 3.7v battery, boosting the voltage with its circuitry to 5v and putting that much power out. With that kind of circuit, you get more voltage but less charge time because it's using more current to supply the voltage like it does.

That's as I understand it anyway, probably a little more long winded than it should be. Your 7 clicks just tells you the base voltage of the battery, not what it's actually putting out voltage wise if that makes sense.

[s]

Sussed it. Didn't realise it had an 'inverter' built in, wondered how a 4.2v battery can put out 6v lol

At the end of the day I'm well impressed with it and it works! That's all that matters to me.

Next on the list is a big 'ol tank me thinks...

[FreedomVape]

Jan 27, 2012 17:37:17 GMT leftfield said:

I'm not the best person to explain it .

I think you done a grand job there leftfield, explained better than I could.

[dave]

Good explanation lefty. You are right kerby - the circuitry has the same effect as an inverter.

[2risky]

This may not be any use, but here is some science. My apologies.

In a direct current resistive circuit, which is what we are talking about, there are a couple of useful, simple equations.

The first relates the values of current, voltage and resistance.
Ohms Law

V (voltage, measured in volt) = I (current. measured in amp) x R (measured in ohm)

V= IR.

The second equation uses the same symbols to express the total amount of power used to heat the coil (ie how strong the vape is). Joule's Law

P (Power, watt) = IV
and by substituting Ohm's Law into Joule's

P= IV = IxIxR = (VxV)/R

So it's possible to arrange different values of V,I and R to provide the same 'strength' vape.

If power or (more usually) current is limited, you can easily work out how this affects things, if you know either the target voltage or the resistance.

Again, my apologies.

[Perpetua]

Absolutely no need to apologise 2risky . . . I'm sure the more technical amongst us will appreciate you taking the time to post the science behind what we use.

[2risky]

The more technical will already know it probably. I was trying to tempt the less technical. You never know.

[Perpetua]

Hope springs eternal 2risky . . . personally, I've some sort of technical dyslexia . . . stuff like this goes whoosh over my head.

[Gordy]

HiFi tried explaining on his recent show VTTV

[maccafan]

After trawling the web and reading up on what is fast becoming a science I have read in several places that the sweet spot for vaping is roughly around 7 watts of power calculated by

W=V²/R

So for example on a 3.7 volt battery output the sweet spot can be achieved by using a 2 Ohm resistance coil meaning a current draw of 1.85 amps which would discharge low capacity batteries pretty quickly. My 280 mAh KR808-D1 battery would be deep discharged in about 10 minutes of continuous vaping.

Question is, is this figure of the sweet spot at around 7 watts accurate? Is that what most experienced vapers here have experienced?

[domesticextremist]

I guess different folks will have different preferences, however, the numbers
you give are equivalent to a (partly discharged) Riva with a LR Boge, which is
a pretty sweet spot for me.

[DiscoDes]

2.5 ohm Kanger at 3.7v with Paradise Vapes Cherry Menthol on my Lavatube.