SXmini M Class (TEMP CONTROL)

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a tez

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seriously ... ohm's law. you need to know it or you can discuss almost nothing technical about vaping.

I use the calculators. I'm not an electrical engineer. I am learning as I go, hence why I asked this question to learn even more. But, you don't want to help and you would rather say what I should know instead of giving information.

Thanks. No need to reply to someone if they ask for help if you're not going to give it.
 

HolmanGT

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Yes, I know this. But when using these calculators, if I have my mod set to 33J and the max voltage it is putting out at .1ohm is 2.7volts, what would the watts be?

a tez,

You can't use the 0.1 ohm reading to calculate watts. The 0.1 ohm reading is the base resistance and immediately climbs higher when it gets hot. But for all practical purposes the wattage you are applying under the circumstances you describe are 33 watt seconds.
 

dr g

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a tez,

You can't use the 0.1 ohm reading to calculate watts. The 0.1 ohm reading is the base resistance and immediately climbs higher when it gets hot. But for all practical purposes the wattage you are applying under the circumstances you describe are 33 watt seconds.

33 watt-seconds is not a wattage...at least not a wattage in the way needed to answer his question.
 

a tez

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a tez,

You can't use the 0.1 ohm reading to calculate watts. The 0.1 ohm reading is the base resistance and immediately climbs higher when it gets hot. But for all practical purposes the wattage you are applying under the circumstances you describe are 33 watt seconds.

So then is there really no way to calculate this since it climbs and we don't know the end resistance since the device isn't displaying it? And if I take a vape for 3 seconds I would be at 99watts?
 

HolmanGT

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So then is there really no way to calculate this since it climbs and we don't know the end resistance since the device isn't displaying it? And if I take a vape for 3 seconds I would be at 99watts?

No there is no way to know the wattage by calculating the resistance displayed and the voltage applied for the reason you said.

The wattage is the same as the joules value if you factor in the time period contained in the joule measurement which is 1 second. So if 1 joule (a current of 1 amp at 1 volt) is 1 watt second. There for 33 joules is the equivalent of applying 33 watts for a period of 1 second. the following second would be another 33 watts second... or if you held the button for 1 hour your would be applying 33 watt hours.

A joule is just a measurement of Energy per unit time and a Watt is the Power applied with no consideration of time.
 
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HolmanGT

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That example is irrelevant to the topic.

Your unexplained answers to post are so totally nebulous that they mean nothing to anyone that reads them. With out some expansion or clarification of what you are trying to communicate you might as well just whistle into the wind.
 

a tez

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No there is no way to know the wattage by calculating the resistance displayed and the voltage applied for the reason you said.

The wattage is the same as the joules value if you factor in the time period contained in the joule measurement which is 1 second. So if 1 joule (a current of 1 amp at 1 volt) is 1 watt second. There for 33 joules is the equivalent of applying 33 watts for a period of 1 second. the following second would be another 33 watts second... or if you held the button for 1 hour your would be applying 33 watt hours.

A joule is just a measurement of Energy per unit time and a Watt is the Power applied with no consideration of time.

Thanks for the help. I usually take a 3-5 second vape so that helps me better understand.
 

HolmanGT

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So then is there really no way to calculate this since it climbs and we don't know the end resistance since the device isn't displaying it? And if I take a vape for 3 seconds I would be at 99watts?

a tez,

A little expansion on this would be that the unit is holding at 33 joules for as many seconds as you hold the fire button. If you hold the button for 3 seconds you will apply 33 joules for every second held not 99 joules. Therefor you would only apply 33 watts total for a period of 3 seconds. just as you would with any other non TC Mod.

The inability to use the base resistance to calculate the wattage applied IS because the Ni200 resistance increases radically the hotter is gets. With a conventional coil made from Kanthal this is still true but to a much lesser degree. The resistance of Kanthal doesn't change very much with temperature so you can use its base resistance to calculate the wattage without introducing very much error in the calculation.

One great big problem in getting a handle on this is Joules and Watts are to different units of measurement but as I have said before it is practical to just use the joule as a measure of wattage it is just that a joule measurement is cumulative per period of time and a watt is a watt regardless of time. When you apply 1 joule to a circuit you know exactly that you held the button for 1 second. When you apply 1 watt to the same circuit you have absolutely no idea how long the button was held. This is because the time period of 1 second is built into the measurement we call a Joule.
 

a tez

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a tez,

A little expansion on this would be that the unit is holding at 33 joules for as many seconds as you hold the fire button. If you hold the button for 3 seconds you will apply 33 joules for every second held not 99 joules. Therefor you would only apply 33 watts total for a period of 3 seconds. just as you would with any other non TC Mod.

The inability to use the base resistance to calculate the wattage applied IS because the Ni200 resistance increases radically the hotter is gets. With a conventional coil made from Kanthal this is still true but to a much lesser degree. The resistance of Kanthal doesn't change very much with temperature so you can use its base resistance to calculate the wattage without introducing very much error in the calculation.

One great big problem in getting a handle on this is Joules and Watts are to different units of measurement but as I have said before it is practical to just use the joule as a measure of wattage it is just that a joule measurement is cumulative per period of time and a watt is a watt regardless of time. When you apply 1 joule to a circuit you know exactly that you held the button for 1 second. When you apply 1 watt to the same circuit you have absolutely no idea how long the button was held. This is because the time period of 1 second is built into the measurement we call a Joule.

Thanks again. Now I know how to make sense of the difference.
 

dr g

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No there is no way to know the wattage by calculating the resistance displayed and the voltage applied for the reason you said.

What on earth? The most basic electrical equation does exactly that.

The wattage is the same as the joules value if you factor in the time period contained in the joule measurement which is 1 second. So if 1 joule (a current of 1 amp at 1 volt) is 1 watt second. There for 33 joules is the equivalent of applying 33 watts for a period of 1 second. the following second would be another 33 watts second... or if you held the button for 1 hour your would be applying 33 watt hours.

A joule is just a measurement of Energy per unit time and a Watt is the Power applied with no consideration of time.

NO!!!! Joules are not PER UNIT TIME. Joules are a discrete quantity.

And that is exactly why joules cannot express maximum wattage and therefore, amp draw. Joules are not a rate.

33 joules could be 66 watts for .5 seconds. Or 3300 watts for .01 second. Or 1 watt for 33 seconds.
 
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2legsshrt

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I am still lost when it comes to relating Joules, Resistance and Temp to Amperage. With watts I can just plug the numbers in a calculator. I use the AW IMR 18650 that have a maximum continuous amperage of 20. They work fine in a DNA40 and I think they will in the M also but I have no idea how to figure it out. Other then just trying it and see if the protection kicks in. I figure my setting will be about 15-20J and resistance will be about .12. Going to try a temp of maybe 380-400 to start.
 

dr g

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I am still lost when it comes to relating Joules, Resistance and Temp to Amperage. With watts I can just plug the numbers in a calculator. I use the AW IMR 18650 that have a maximum continuous amperage of 20. They work fine in a DNA40 and I think they will in the M also but I have no idea how to figure it out. Other then just trying it and see if the protection kicks in. I figure my setting will be about 15-20J and resistance will be about .12. Going to try a temp of maybe 380-400 to start.

you really can't apply joules to anything without knowing what they are referring to and temp only applies as far as the way the board controls the energy output to achieve that temp.
 

ahgnus

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you really can't apply joules to anything without knowing what they are referring to and temp only applies as far as the way the board controls the energy output to achieve that temp.

I think dr g is trying to incite the community to figure out the secrets of Yihi temperature control algorithm.

Do we have any electrician/programmer to see what's going on in the firmware?

Better yet, does anyone know if there is a toe-to-to 350J vs DNA40 technical discussion thread? I've seen some good discussions about each chip, separately never comparatively, and I am starting to grasp a few things. I want to skip a few lessons and get right into comparison of two chip, juxtaposed in light of versatility/ end-user options. Thi is one huge reason why I chose mini over 40.

I am a damn adult. I dont need a start button. I ruled Windows 3.1!
 
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Technonut

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I think dr g is trying to incite the community to figure out the secrets of Yihi temperature control algorithm.

Do we have any electrician/programmer to see what's going on in the firmware?

Better yet, does anyone know if is there a toe-to-to 350J vs DNA40 technical discussion thread?




As a matter of fact, yes.... Not quite toe-to-toe yet..

http://www.e-cigarette-forum.com/fo...yihi-sx350j-technology-discussion-thread.html

This may be better discussed there...


Just to clarify... Since the M Class has been released, there are folks coming to this thread looking for instructional information / discussion. Most do not want to wade through blocks of postings regarding technical info which does not interest the typical end-user of this device. They simply want to know how to make it work.. :)

I'm officially assigned to both the APV and RBA forums / sub-forums. My job is difficult enough without needing to watch a few select threads like a hawk..;)

I'm not saying it cannot be discussed.. Just clarifying .. The thread title alone is bringing many folks looking for help over this way. Just Google SXmini M Class to see what I mean.. :) As I suggested earlier, perhaps someone could open another thread to discuss the more in-depth technical aspects of the SX350J.. Hopefully it could be discussed in a manner which will not allow it to be closed down like the other..
 
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refuzeandrezizt

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I could be way out of line with my way of doing it, but I always look at the display after a toot and grab the voltage it is telling me it is outputting. I enter that, with the resistance it says, and then I can see the amps and watts in the ohms law calculator. For instance, I will do it now.

Just took a toot at 425 degrees, 20J, with my .09 build. It says the output volts was 2.1. So if I put that voltage and ohm into the calculator, I see that I am pulling 23.33 amps, and 49 watts.

If I up it to 25J, it says it is outputting 2.4 volts. So those numbers give me 26.67 amps, and 64 watts.

So on, and so on. I usually just adjust the joules to taste, but that gives me an idea. Like I said, I could be (and probably am) way off with this thinking. But it works for my brain anyway. As always, YMMV.
 
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