Hello all. I have a LE 80 box mod. It is a dual 18650 box mod. Im told it is wired in series. It has a switch to lock the fire button and show battery voltage. And a dial to adjust the volts. And thats it. No wattage. No menu. No settings. My question is, if i want to calculate the amps being drawed from one of the batteries in the my device, what method do i use? Treat it like a mech mod and go by the resistance and voltage? .5ohms at 4v = 7amps? Or .5ohms at 4v = 32watts. 32/6.4 = 5 amps?
Question about regulated mod
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Its a mech, and a series mod so your available current (amps) is that of one battery. So 4v ÷ 0.5 ohm = 8A.
Just want to make sure I have the right mod ... I believe it's this one.
If it is, it is a variable voltage device which does not do variable wattage. VV was the only game in town for a while before VW became became common.
You set the power by setting the voltage output. The power (watts) is calculated by using the voltage you set combined with your atty resistance using ohm's law. Using the ohm's law calculator on the Steam Engine site if you plug in 0.5 ohms and 4 volts you get a current of 8A and 32 watts. The 8 amps is not what is being pulled from the batteries, it's what is being delivered to the atty through the voltage regulator.
If you want to know the whole picture, go to the Battery Drain page on that site. Click on 'Mode - voltage regulation (VV)', set 'Atomizer resistance' to 0.5 ohms, 'Voltage setting' to 4V and 'Battery voltage' to 8.4V. 8.4V is what you'll get with two fully charged batts (4.2V each) in series. Look in the 'Battery drain' window and you'll see the batts are seeing 4.23A draw each. Change 'Battery voltage' to 6V (each battery at 3V, which is about when the mod should shut you down for low battery voltage) and the battery amp draw is up to 5.93A. Both VV and VW mods draw the most amperage from the batteries when the batts are almost fully discharged. Mechanical mods draw the most amps when the batteries are fully charged.
If you're going to be vaping at high power with this mod you should be using 30A batteries.
If what you have is a version of this mod that is actually mechanical, then you can use the ohm's law calculator using 8.4V as the voltage. OTOH, if your mod is the one I linked to it's VV
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@Ryedan yes thats the one. Its actually not a very old mod i dont think. I've just been calculating it like its a mech mod. This is a confusing one.
Right now i have a 0.7ohm 8 wraps of 24 guage in a lemo 2 and im pushing 5v to it. Using 2 married samsung 25r5's
So im guessing it would be 35w/6.4=5.4amps
Right now i have a 0.7ohm 8 wraps of 24 guage in a lemo 2 and im pushing 5v to it. Using 2 married samsung 25r5's
So im guessing it would be 35w/6.4=5.4amps
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In a series mod, it's 7.14A from both batteries. In a parallel mod, it would draw 3.57A from each.
Where did you get 7 amps from?
Your question is about the two different circuits involved:
1: the battery circuit which is the input to the electronics
2: the output of the electronics going to the coil
Use the output voltage and the coil resistance to calculate your wattage. wattage = outputvoltagesquared/coilresistance
Then use the battery voltage and that wattage you calculated to work out the amp draw on the batteries.
amp draw = wattage/batteryvoltage
All together now: amp draw = outputvoltagesquared/(coilresistance x batteryvoltage)
outputvoltage = 5 volts
coilresistance = 0.7 ohms
batteryvoltage= 6.4 volts
amp draw = 25/(0.7x6.4)= 9.1 amps
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