jambandphan03
in flavor country
I updated that message, but thanks for putting links up OF. 
Discontinued ThermoVape Evolution
- Thread starter SameOldTim
- Start date
OF
Well-Known Member
I updated that message, but thanks for putting links up OF.
Yer welcome, bro. I had the info fairly handy, glad to pass the fish to the hungry man this time.
OF
jambandphan03
in flavor country
havealight101
Norski
Haywood
Onward Thru the Fog
Well, I guess this is one area where we will have to disagree.
I don't know what you're basing your "facts" on, but they're not based on science or testing. Look at the discharge curves for the batteries we use. When you tried to translate the formula I used into "words" (I have an EE degree), and then said "there are a couple of things to straighten out" [about what I said in my initial message], I got pissed off. After a little medication, I got over being pissed. I know the tone of this is pretty "in your face" (thanks havalight), and I don't mean any ill will. So take this next sentence with a grain of salt:
The problem isn't that I don't understand amps and volts and watts and watt-hours; the problem is that YOU don't. So let me straighten you out.
You don't have to explain ohms law to me, nor do you have to talk to me like a four year old. "pushing" and "resisting" indeed. In fact, until the last sentence, you just repeated what I said in my original message, just less precisely. Your last sentence, however, is completely wrong, and is the direct opposite of what you said in your previous sentence. Again, as the battery voltage goes down, the current will go up, NOT the other way around.
The problem is that you are totally confusing power, capacity, and current, and you are not taking into consideration what is really happening as the batteries are depleted during discharge. If you look at the 5 amp and 10 amp discharge curves for our batteries, you'll see that for 90%-95% of the discharge, the voltage drops slowly, and for the last 5%-10% of the discharge, it drops precipitously. At that drop off point, you're seeing the battery run out of the capacity to supply the power (not the current, the power) that the load is demanding. In fact, the battery is still supplying 5 amps at that point (or 10 amps if you're looking at the 10 amp discharge curves). That's why the voltage drops so quickly at the end; the current actually remains mostly the same.
Look, this is not really relevant to the thread anymore, and I only mentioned it in the first place due to the discussion about the current requirements for the switch used in a third party power source. I was concerned that someone would buy a third party power handle, and the switch would burn out after a day. I don't want to get into a pissing match about this. (In fact, I think you contribute much more to this forum than most, certainly more than I do).
p.s.
I don't know what you're basing your "facts" on, but they're not based on science or testing. Look at the discharge curves for the batteries we use. When you tried to translate the formula I used into "words" (I have an EE degree), and then said "there are a couple of things to straighten out" [about what I said in my initial message], I got pissed off. After a little medication, I got over being pissed. I know the tone of this is pretty "in your face" (thanks havalight), and I don't mean any ill will. So take this next sentence with a grain of salt:
The problem isn't that I don't understand amps and volts and watts and watt-hours; the problem is that YOU don't. So let me straighten you out.
You don't have to explain ohms law to me, nor do you have to talk to me like a four year old. "pushing" and "resisting" indeed. In fact, until the last sentence, you just repeated what I said in my original message, just less precisely. Your last sentence, however, is completely wrong, and is the direct opposite of what you said in your previous sentence. Again, as the battery voltage goes down, the current will go up, NOT the other way around.
Not only did I never say the core power draw was constant, I made a point of explaining how the power draw changed with temperature. This has nothing to do with the fundamental mistake you're making.
Wrong. Hook up a meter and check yourself if you don't believe the battery manufacturers data, or mine. As the battery discharges, and the voltage decreases, the current will increase, to maintain the 20 watts the core is drawing.
Wrong and wrong. The lower the voltage gets, the more current is used. The reason it takes longer to heat up when the battery is low (it doesn't "seems to take longer", it does take longer) is because the battery is running out of capacity (energy/power), and can't supply 20 watts anymore. The battery is supplying less power at this point. And it's actually not current that sags as the battery dies, it's the voltage. Less voltage, same current, is less power.
The problem is that you are totally confusing power, capacity, and current, and you are not taking into consideration what is really happening as the batteries are depleted during discharge. If you look at the 5 amp and 10 amp discharge curves for our batteries, you'll see that for 90%-95% of the discharge, the voltage drops slowly, and for the last 5%-10% of the discharge, it drops precipitously. At that drop off point, you're seeing the battery run out of the capacity to supply the power (not the current, the power) that the load is demanding. In fact, the battery is still supplying 5 amps at that point (or 10 amps if you're looking at the 10 amp discharge curves). That's why the voltage drops so quickly at the end; the current actually remains mostly the same.
Look, this is not really relevant to the thread anymore, and I only mentioned it in the first place due to the discussion about the current requirements for the switch used in a third party power source. I was concerned that someone would buy a third party power handle, and the switch would burn out after a day. I don't want to get into a pissing match about this. (In fact, I think you contribute much more to this forum than most, certainly more than I do).
p.s.
I really love this typo.
havealight101
Norski
I think a little less "in your face" tone would be beneficial to discuss this constructively. My 2 pennys
Haywood
Onward Thru the Fog
I know, I tried; good thing I didn't post my initial response. I hoped the smiley faces would help a bit. I probably shouldn't have put the part in about talking to me like a four year old... I did just edit the message to explain that I mean no ill will.
Hopefully it won't turn out that I was completely wrong!
Hopefully it won't turn out that I was completely wrong!
JoeKickass
Well-Known Member
Well you don't really expect me to explain something to such an intelligent ee major now do you?
Suffice to say, you contradict yourself several times...
And just fyi: http://lmgtfy.com/?q=formulae
OF
Well-Known Member
Wow. I just stepped out for a bit to supervise the evening and I come back to some fairly heated tech stuff. My kinda chatter, geek speak. I'm there. Well, except I have to go take care of something right now and want to read it all a few times and be sure I follow, but the above seems the core disagreement? It's absolutely wrong. At least that's my take.
More in a bit.....
OF
Haywood
Onward Thru the Fog
Umm, look again how you spelled it. I kind of like your version.
JoeKickass
Well-Known Member
havealight101
Norski
Retracted, please delete
Haywood
Onward Thru the Fog
It means more than one formula
It wasn't the pluralization, which I get. I was amused by the misspelling. Look carefully at how you spelled it again.
havealight101
Norski
Retracted, please delete, sry to clog it up
jambandphan03
in flavor country
not to put too fine a point on it...say I'm the only 
in your bonnet, make a little birdhouse in your soul...
I can feel myself EVOlving...
in your bonnet, make a little birdhouse in your soul...I can feel myself EVOlving...
JoeKickass
Well-Known Member
Ok since you made me laugh since I really did miss that one, this is where I'm coming from:
The only way for the current to increase is for the resistance to decrease. But the resistance stays more or less the same... So when the voltage drops the current has to drop too right? The power drops along with the voltage and the current.
Let the debate begin!
OF
Well-Known Member
OK, I read it a couple times and here's my take: Joe nails it and HW has misinterpreted a few ideas in a kind of interesting way.
When I taught this good stuff more voltage meant more current for the same resistance. And more current and more voltage meant more power. Made stuff get hotter. Blows up a 110 Volt toaster to plug it into 220 every time. There's nothing in the toaster to compensate any more than there's anything in the TV to compensate. If the battery voltage goes up, it will get hotter. It's the way it is.
Most guys understand the flashlight gets colder (dimmer) as the batteries die. Same with our favorite vape.
The language used by Joe to describe 'resisting the push' and the like are not only quite correct but common. He by no means made them up.
Full points to Joe to here.
In a different class of devices, constant power is at least close to possible. It's in fact called "negative Resistance". When the voltage (or current) goes up the other (current or voltage depending) goes down. A few components (like tunnel diodes and some gas tubes) have this characteristic, but we're not there. Your computer is a device that does this to some extent due to it's switching type power supply. It draws less current at 220 volts than at 100. This is because it's designed to do this, 'changing gears' as it were in the process (actually over a smooth curve, but the idea is the same). Such supplies typically 'take' (that is waste) 10 or 20% of the total doing so. That's not what we have either.
However, if we did, some of what HW has to say makes sense to me.
As to the matter of batteries we're a ways off the mark. First off, one of them is never a battery, it's a cell. An individual part, like a monk's cubie in fact. Anyway, two or more cells working together make battery, same as they do with cannons. Since most folks call 'em batteries I usually do the same....sometimes I slip back, sorry if that causes more confusion.
Anyway, as long as the chemicals are present, the cell itself generates the same voltage. Use your DMM on a 'nearly dead' battery some time, near full voltage, right? Put a little load on it and it collapses of course, but the voltage was fine before that. There was enough of the chemicals to drive the meter OK. Electron for electron, atom for atom. Chemicals are making it happen. This 'base voltage' is called the Thevenin voltage for those taking notes. Big V, small th.
Anyway, the missing piece to the puzzle is internal resistance. The insides of the battery (cell, right....) have resistance too. Some metals and other solids, some the chemicals themselves. As the chemicals are used up they are used up. They no longer conduct, the internal resistance rises. This part is sometimes just called 'internal resistance' and is added into the load, wiring, switch, contact and other resistances to find the total for the circuit. However, for our use here, it's the Thevenin Resistance; big R, small th.
Plug Vth and Rth into Ohms law and you get the maximum possible current from the battery at that point. That gets you into Norton stuff, something else that solves this same sort of issues from a different direction.
Anyway, the internal resistance is what causes the offsets in the battery graphs. The 'real' battery (really a cell) voltage is still the same, the 'missing voltage' between the curves is I times R losses inside the battery. This is what makes the battery get hot. The missing Volts times the running Amps makes real Watts happen inside the battery. So as the battery goes flat, self heating ironically goes up for the same current levels. Just what we don't need.
There was a bunch of other stuff, lots to do with spelling and stuff, that got past me. And clearly this saga goes back some. I got no take there, either. But I know Volts and Amps pretty good. I don't know where some of the stuff up above came from but I can assure you none of my students ever said it with a straight face. They heard the stuff I just said (or typed...) above over and over in different terms as necessary until it became second nature. Or they didn't pass. That's how come we were all able to work and play so well together, we spoke the same language....geek speak.
OF
When I taught this good stuff more voltage meant more current for the same resistance. And more current and more voltage meant more power. Made stuff get hotter. Blows up a 110 Volt toaster to plug it into 220 every time. There's nothing in the toaster to compensate any more than there's anything in the TV to compensate. If the battery voltage goes up, it will get hotter. It's the way it is.
Most guys understand the flashlight gets colder (dimmer) as the batteries die. Same with our favorite vape.
The language used by Joe to describe 'resisting the push' and the like are not only quite correct but common. He by no means made them up.
Full points to Joe to here.
In a different class of devices, constant power is at least close to possible. It's in fact called "negative Resistance". When the voltage (or current) goes up the other (current or voltage depending) goes down. A few components (like tunnel diodes and some gas tubes) have this characteristic, but we're not there. Your computer is a device that does this to some extent due to it's switching type power supply. It draws less current at 220 volts than at 100. This is because it's designed to do this, 'changing gears' as it were in the process (actually over a smooth curve, but the idea is the same). Such supplies typically 'take' (that is waste) 10 or 20% of the total doing so. That's not what we have either.
However, if we did, some of what HW has to say makes sense to me.
As to the matter of batteries we're a ways off the mark. First off, one of them is never a battery, it's a cell. An individual part, like a monk's cubie in fact. Anyway, two or more cells working together make battery, same as they do with cannons. Since most folks call 'em batteries I usually do the same....sometimes I slip back, sorry if that causes more confusion.
Anyway, as long as the chemicals are present, the cell itself generates the same voltage. Use your DMM on a 'nearly dead' battery some time, near full voltage, right? Put a little load on it and it collapses of course, but the voltage was fine before that. There was enough of the chemicals to drive the meter OK. Electron for electron, atom for atom. Chemicals are making it happen. This 'base voltage' is called the Thevenin voltage for those taking notes. Big V, small th.
Anyway, the missing piece to the puzzle is internal resistance. The insides of the battery (cell, right....) have resistance too. Some metals and other solids, some the chemicals themselves. As the chemicals are used up they are used up. They no longer conduct, the internal resistance rises. This part is sometimes just called 'internal resistance' and is added into the load, wiring, switch, contact and other resistances to find the total for the circuit. However, for our use here, it's the Thevenin Resistance; big R, small th.
Plug Vth and Rth into Ohms law and you get the maximum possible current from the battery at that point. That gets you into Norton stuff, something else that solves this same sort of issues from a different direction.
Anyway, the internal resistance is what causes the offsets in the battery graphs. The 'real' battery (really a cell) voltage is still the same, the 'missing voltage' between the curves is I times R losses inside the battery. This is what makes the battery get hot. The missing Volts times the running Amps makes real Watts happen inside the battery. So as the battery goes flat, self heating ironically goes up for the same current levels. Just what we don't need.
There was a bunch of other stuff, lots to do with spelling and stuff, that got past me. And clearly this saga goes back some. I got no take there, either. But I know Volts and Amps pretty good. I don't know where some of the stuff up above came from but I can assure you none of my students ever said it with a straight face. They heard the stuff I just said (or typed...) above over and over in different terms as necessary until it became second nature. Or they didn't pass. That's how come we were all able to work and play so well together, we spoke the same language....geek speak.
OF
JoeKickass
Well-Known Member
That debate lasted three minutes... this is why we can't have nice things OF...
JK good job ref!
JK good job ref!
Haywood
Onward Thru the Fog
And you thought it was over. Maybe for tonight.
OF: Your Thevenin comment sure brought back memories. It's been a long time since I studied that; all I could remember from way back then is that to figure out the Thevenin value, I had to put less and less resistance on the cell, until it dropped to half open circuit voltage, at which point the load resistance was equal to the Thevenin value. Never thought I'd have any use for it (or Norton). Refresh my memory; is the Thervenin value (essentially) the internal resistance off the cell?
So what I'm getting here is that my assumption about the core being a constant power device is wrong, rather it's a constant resistance device, and the power it draws corresponds to the voltage applied. (Within reasonable limits, I'm not talking about a 3v core on a 6v battery)
While I sleep and think about this, and think about if the heating core resistance changes to maintain a constant power consumption, or it remains constant to cause the power consumption to go down in concert with the supply voltage, I have a question for you (OF). If you were to hook the T1 heat core up to a power supply that could supply more than ample current, and you fed the core 6 volts, I assume you would see the supply delivering 5 amps, since we believe the heat core draws 30 watts at rated voltage. What would you expect to see if you lowered the voltage to 5 volts? 6 amps? 4.16 amps? to 4 volts? 7.5 amps or 3.33 amps?
Other than to know what current rating the power switch should be, I'm not sure the importance of this all, other than to maybe help me make me look silly. On the other hand, I always like learning. The LiFePO4 cells have a remarkably flat discharge voltage curve, even at 5 amps, with the voltage remaining in the 3.1 to 2.8 volt range for well over 90% of the time. That's about a 10% drop, so if the core load is constant, that means the core is getting 27 watts instead of 30.
And Joe, indeed if the heat core is a constant power device, then my saying its resistance remained mostly constant with differing voltage can't be right. And if the resistance does remain mostly constant, then the core can't be a constant power device. So I am wrong about one of them...
OF: Your Thevenin comment sure brought back memories. It's been a long time since I studied that; all I could remember from way back then is that to figure out the Thevenin value, I had to put less and less resistance on the cell, until it dropped to half open circuit voltage, at which point the load resistance was equal to the Thevenin value. Never thought I'd have any use for it (or Norton). Refresh my memory; is the Thervenin value (essentially) the internal resistance off the cell?
So what I'm getting here is that my assumption about the core being a constant power device is wrong, rather it's a constant resistance device, and the power it draws corresponds to the voltage applied. (Within reasonable limits, I'm not talking about a 3v core on a 6v battery)
While I sleep and think about this, and think about if the heating core resistance changes to maintain a constant power consumption, or it remains constant to cause the power consumption to go down in concert with the supply voltage, I have a question for you (OF). If you were to hook the T1 heat core up to a power supply that could supply more than ample current, and you fed the core 6 volts, I assume you would see the supply delivering 5 amps, since we believe the heat core draws 30 watts at rated voltage. What would you expect to see if you lowered the voltage to 5 volts? 6 amps? 4.16 amps? to 4 volts? 7.5 amps or 3.33 amps?
Other than to know what current rating the power switch should be, I'm not sure the importance of this all, other than to maybe help me make me look silly. On the other hand, I always like learning. The LiFePO4 cells have a remarkably flat discharge voltage curve, even at 5 amps, with the voltage remaining in the 3.1 to 2.8 volt range for well over 90% of the time. That's about a 10% drop, so if the core load is constant, that means the core is getting 27 watts instead of 30.
And Joe, indeed if the heat core is a constant power device, then my saying its resistance remained mostly constant with differing voltage can't be right. And if the resistance does remain mostly constant, then the core can't be a constant power device. So I am wrong about one of them...
OF
Well-Known Member
Guys,
Cool, sounds like we're getting back on the same page. My boss once said to me in a most serious manner, 'don't argue with the electrons, they know what they're doing'..... He was right of course, they won't listen to me so I'd better try to understand them.
I've got a headache that's keeping me from making any sense of this right now, please excuse me until a little later, time to resort to drugs I think....
A few bits are however clear; yes, HW, Rth is the internal resistance for Thevenin based Circuit Analysis. It's also Rn for Norton but you need to short the Thevenin circuit to calculate the Norton current (In) to put in parallel (not series) for that model.
Yes, to a first order the heater is a pure resistor. However (isn't there always?) IIRC the typical alloys all have pretty big positive alphas (positive temperature coefficients) that moderate the rise in output power. That is higher temperatures trigger a rise in resistance that lowers the total rise in current. It still goes up, just not as much. Voltage still goes up as before, but the product makes the 'power output against input voltage' graph flatter.....which is a good thing.
When my head is working better, let me see if I can dig up a few numbers and see exactly what they can tell us?
OF
Cool, sounds like we're getting back on the same page. My boss once said to me in a most serious manner, 'don't argue with the electrons, they know what they're doing'..... He was right of course, they won't listen to me so I'd better try to understand them.
I've got a headache that's keeping me from making any sense of this right now, please excuse me until a little later, time to resort to drugs I think....
A few bits are however clear; yes, HW, Rth is the internal resistance for Thevenin based Circuit Analysis. It's also Rn for Norton but you need to short the Thevenin circuit to calculate the Norton current (In) to put in parallel (not series) for that model.
Yes, to a first order the heater is a pure resistor. However (isn't there always?) IIRC the typical alloys all have pretty big positive alphas (positive temperature coefficients) that moderate the rise in output power. That is higher temperatures trigger a rise in resistance that lowers the total rise in current. It still goes up, just not as much. Voltage still goes up as before, but the product makes the 'power output against input voltage' graph flatter.....which is a good thing.
When my head is working better, let me see if I can dig up a few numbers and see exactly what they can tell us?
OF
Hello everyone, my first post here on FC! I just received a Thermovape Evolution LV today, and been spending some time with it tonight on an existing eCigarette mod, and I wanted to share my thoughts.
I attached the Evo-LV to a Silver Bullet eCigarette mod. This has a 2A switch. Well, here was my own personal math, right or wrong, I'm just sharing my experience here and knowing I may have to deal with the consequences if I didn't figure this right:
- The switch on the Silver Bullet is 2A@48VDC.
- 2 Amps at 48 Volts is 96 watts.
- Performing a conversion to 3.7 Volts at 96 Watts is 25.95 amps, allowing well enough range to get up to 6 amps while performing at low voltage.
Using AW IMR 18650 batteries at full charge (that usually last me 2 days with my eCigarette atomizer, not with this!!), I pack about 3/4 of the chamber up with herbal medicine, press the power button down for a good 30 seconds, and start taking long draws off of it. It produces visible vapor, though a very little amount compared to what you see come out of a Volcano bag or something. It tastes great when you vape off of this. Maybe about 10 puffs and the vapor fades away. I tap the end on the table and all of the vaped medicine comes out. Same color as the vaped weed out of the Volcano, no burning or black.
I tested my battery after a "chamber" full on the LavaTube, and it said the battery was still at 3.8V (a full charge AW IMR is 4.1 for me). After a second "chamber", the battery either is dead or very low, 3.6V. (**** Although I tested my batteries on a LavaTube, it was batteries removed from the Silver Bullet, and the LavaTube will not work with this product ****)
So, I've done about five or six of these trial runs with this thing tonight, and so far the Silver Bullet is holding up just fine. I don't know if it would be the best choice for the Evo-LV or not, or if my Bullet is going to need a replacement someday, but it certainly works now and I'm excited to keep trying it to see what happens.
I attached the Evo-LV to a Silver Bullet eCigarette mod. This has a 2A switch. Well, here was my own personal math, right or wrong, I'm just sharing my experience here and knowing I may have to deal with the consequences if I didn't figure this right:
- The switch on the Silver Bullet is 2A@48VDC.
- 2 Amps at 48 Volts is 96 watts.
- Performing a conversion to 3.7 Volts at 96 Watts is 25.95 amps, allowing well enough range to get up to 6 amps while performing at low voltage.
Using AW IMR 18650 batteries at full charge (that usually last me 2 days with my eCigarette atomizer, not with this!!), I pack about 3/4 of the chamber up with herbal medicine, press the power button down for a good 30 seconds, and start taking long draws off of it. It produces visible vapor, though a very little amount compared to what you see come out of a Volcano bag or something. It tastes great when you vape off of this. Maybe about 10 puffs and the vapor fades away. I tap the end on the table and all of the vaped medicine comes out. Same color as the vaped weed out of the Volcano, no burning or black.
I tested my battery after a "chamber" full on the LavaTube, and it said the battery was still at 3.8V (a full charge AW IMR is 4.1 for me). After a second "chamber", the battery either is dead or very low, 3.6V. (**** Although I tested my batteries on a LavaTube, it was batteries removed from the Silver Bullet, and the LavaTube will not work with this product ****)
So, I've done about five or six of these trial runs with this thing tonight, and so far the Silver Bullet is holding up just fine. I don't know if it would be the best choice for the Evo-LV or not, or if my Bullet is going to need a replacement someday, but it certainly works now and I'm excited to keep trying it to see what happens.
OF
Well-Known Member
Sorry to say it doesn't work that way. You can't get around Ohm's law like some liberals want to do with those nasty old Constitutional rules....no liberal judges to let you skate around the issue and no chance of a new law designed to circumvent the issue. Amps is Amps, you and I don't get to bring Watts into the discussion. The Evolution LV is going to try to draw 5.5 or so Amps and that's all the switch knows.
It will most likely fail before anything else. The likely failure mode will be poor contact causing excessive heating and local 'burn out' of the switch. It could also weld on in the opening sequence. There's just not enough contact in there, nor a stiff enough spring to break the heavier arc involved on openings. I'd plan on the eventual failure or get a unit with a high enough rating. But as you say, for now it's working......
I think a minor adjustment or two to your technique will bring instant and welcome improvements. Please try this instead:
Load a loose load nearly full. With a fresh battery charge key it on and let it run for an honest 15 seconds with no draw. This should have the heat core up to working temperature (glowing nicely). Now start a long, steady, medium speed draw. The kind that takes maybe 20 or even 30 seconds to completely fill your lungs. Taste for the heat and vapor, it will take maybe another 10 or 15 seconds? Puff a bit out to quickly check for vapor (AKA 'test puff'). When you get some vapor, the bowl is now up to (a different) working temperature, time to move to step 3....the hit itself. Quickly blow everything out (I know, there's some goodness there finally, but bear with me?), then go for it. Honk on that sucker if that's your style. The heat you need is already stored in the core, ready to be tapped instantly at this point. Follow your instincts on the hit. Expect great things, this little guy can produce given a chance.
The key is to not be shy with the battery until you've got some experience. Don't afraid to let it run. Your 'bowls per charge' number will come up, but at first accept low numbers for good hits.
Do it in 3 separate steps. If it's warm you can cut the first back a bit. If you get no vapor on step 2 by the end of your inhale, reset. Let it run 10 or 15 seconds again before trying another 'warm the bowl' stage. Then, and only then, take your hit. Clean from the start on a preheated bowl heated by a pre heated core.
BTW, it seems important to keep it loose and well ground and dry as far as fill goes. The unit is designed to be run nearly full, IMO that's the best way to learn it.
Good luck, sounds like you're well down the road.
OF
Dj Bass
Well-Known Member
soooo.... Just discovered e-cig mods... Picked up the Saber Touch 3.7/6.0 with the passthru, the adjustable power supply, and the high power circuitry mod. Heard the switch can stand up to a lot because it is solid state. Seems like a perfect Evo at the desk or on the go solution. Anyone use this? Anything I should know? something said on the site about a wet finger?