I jumped right i to mechs from eGos but I'm an electrician and ad a good knowledge base to work from. I saw a clone of a REO the other day on here. It looked like a decent device and half the price of an authentic.
OHM law question
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If you have a VW box mod you don't need to worry about the volts the chip inside regulates that for you but what you will soon find out is the lower ohm builds ran @ higher watts ( 75 + ) will diminish your battery quicker as well as becoming a very hot vape and loss of flavor.So to answer your question with two quality battery's ie sony vtc5 @.3 ohm could turn the wattage up to the max level if your chip will run that low of ohms, hope you have plenty of juice and at least four more battery's and a quality charger and the battery's are married together.
You're starting from the wrong place. What are you trying to find out?
The batteries that you are using are 5V, I believe. The only way they could put out 10V is if they are wired in series. Which I suppose they might be, but it seems unlikely.
Anyway...we can assume a max of 5V and we know that you have a .3 ohm coil. Voltage divided by resistance equals current. So the answer there would be 16.7 amps. (5/0.3=16.7)
Power dissipation equals voltage multiplied bu current. 5V x 16.7a=83W.
You're starting from the wrong place. What are you trying to find out?
The batteries that you are using are 5V, I believe. The only way they could put out 10V is if they are wired in series. Which I suppose they might be, but it seems unlikely.
Anyway...we can assume a max of 5V and we know that you have a .3 ohm coil. Voltage divided by resistance equals current. So the answer there would be 16.7 amps. (5/0.3=16.7)
Power dissipation equals voltage multiplied bu current. 5V x 16.7a=83W
At 10V, the current would be 33.3a, so power dissipation would be 333.3W, which would be enough to make any hand held sized battery melt down or explode.
I am a tax guy, but I also have a degree in electrical engineering, which comes in handy on occasion.
The batteries that you are using are 5V, I believe. The only way they could put out 10V is if they are wired in series. Which I suppose they might be, but it seems unlikely.
Anyway...we can assume a max of 5V and we know that you have a .3 ohm coil. Voltage divided by resistance equals current. So the answer there would be 16.7 amps. (5/0.3=16.7)
Power dissipation equals voltage multiplied bu current. 5V x 16.7a=83W.
You're starting from the wrong place. What are you trying to find out?
The batteries that you are using are 5V, I believe. The only way they could put out 10V is if they are wired in series. Which I suppose they might be, but it seems unlikely.
Anyway...we can assume a max of 5V and we know that you have a .3 ohm coil. Voltage divided by resistance equals current. So the answer there would be 16.7 amps. (5/0.3=16.7)
Power dissipation equals voltage multiplied bu current. 5V x 16.7a=83W
At 10V, the current would be 33.3a, so power dissipation would be 333.3W, which would be enough to make any hand held sized battery melt down or explode.
I am a tax guy, but I also have a degree in electrical engineering, which comes in handy on occasion.
The 2 batteries are 4.2v peak, 3.7v nominal together they supply 8.4 ish ~ 7.4 ish volts to the chip, the buck/boost converter boosts the voltage to a maximum output of 10.5v.
Regulated APV's have voltage, wattage & resistance minimums & maximums, all have amp limits.
They will not fire below or above what their built in minimum or maximum limits will allow, no matter what the display shows.
Regulated APV's have voltage, wattage & resistance minimums & maximums, all have amp limits.
They will not fire below or above what their built in minimum or maximum limits will allow, no matter what the display shows.
That all makes sense. I doubt that any available coils could withstand 330W dissipation. Besides, at that point, you would probably create steam and scorch your lungs.
5.0V is a dangerously high voltage for any of the batteries a vaper might use. They shouldn't be charged to anything higher than 4.20V and most calculations for worst-case current draw in an unregulated device use this value.
[edit] There are 4.35V-rated batteries available. If you have the proper charger you could charge them to that voltage safely. Or, you could any good standard Li-Ion charger and enjoy the greatly increased life of your batteries by charging them to 4.20V.
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It just occurred to me that anyone using a mechanical mod needs to be careful about their coil resistance. In the above example, with no regulation, the 0.3 ohm coil would dissipate 83W.
How low does coil resistance go around here, anyway? I'm still new to vaping. 0.3 ohms is getting real close to a short circuit.
Temperature control mods with Ni200 coils fire down as low as .1 or even .06 on some mods, I believe. Standard Kanthal builds don't usually go that low, but I'm sure someone has tried it.
That would depend on how foolish you are, and what kind of mod you're using. On a regulated mod (a good one, of course) you can use any resistance the mod will fire. I have a Snow Wolf that will fire down to .05 Ohms using Ni200 wire (for Temp Control mods). On a mechanical, you are limited to the Ohm's Law amp draw you can safely get from your batteries. I personally don't like to load batteries up much more than 50% or their rating, and with very few exceptions, that's 20 amps. So, a 10 amp load at 4.2 volts is a .42 ohm coil on a single battery mod. If you have a two battery parallel box, with 2 20 amp batteries, you can get 20 amps out of it and still keep a 50% safety margin. 20 amps at full voltage would mean a .21 Ohm coil, and so forth.
Once again, comparing TC resistance levels with a basically Kanthal thread can be very misleading unless the reader is very up on TC. A 0.06 ohm Kanthal build on a single battery Mech would fire at 70 amps and 294 watts!
But, yea, I'm sure someone has tried it. Wait! I posted a pic of the results on the first page here
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hey sorry about the delay, im using a segelei 150w, also im using 2 Samsung 18650's
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