Has anyone considered using one of these circuits in 3D printing? Just curious.
Originally shared by William L. DeRieux IV
For anyone who is interested I created this simple circuit using an LM317T that can be used to drive 2 to 4 ohm e-cigarette coils such as the commercial offerings from blu-cig, Vuse, or any manufacturer as long as the coil is 2 to 4 ohms.
For example the Vuse Vibe Solo has a 3V battery and uses 2.5 ohm coils.
The current draw is: 3V/2.5ohms = 1.2A with a power draw of 3^2/2.5 = 3.6W.
R2 can be decreased to lower the output voltage and, conversely, R2 can be increased to increase the output.
However, the datasheet for the LM317 states the maximum output current is 1.5A so that must be taken into consideration when setting the value of R2.
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On a side note:
On these vuse cartridges there are three terminals and they are laid out like this:
( x ( + ) x )
The two terminals on outer perimeter are for delivering power to the coil to produce vapor.
The center terminal is the one that enables communication between the cartridge and the battery.
I would bet that if you removed this center terminal that you wold be able use the cartridge indefinitely with a Vuse Solo battery until the cartridge ceases to funcation.
right now I have one of these cartridges jerry rigged to an Eleaf IStick 30W running at 5W and have not had any trouble – although I have only used 2 cartridges so far.
You could connect a battery with a switch to the two outer terminals and do the same.
For the power source I would recommend: 2x AA batteries = 3V and if the cart is 2.5ohms you would be running the cart at ] P=V^2/R = 3V ^ 2 / 2.5 ohms = 3.6 watts ] [ I=V/R = 3V / 2.5 ohms = 1.2 amps].
A single AA cell (at 1.5V) can deliver 2500 mAh or 2.5 Ah if you were to run at 3.6W /1.2 A as above constantly you would get 2.5Ah/1.2A or ~2 hours of usage before you needed to replace or rechange the batteries since the coil won’t be constantly energized for that 2 hours.
2.5 A/hour = <1 mA/second and for 3 seconds it totals to a draw of ~ 2 mA.
If you take a puff for 3 seconds every 10 minutes ( a total of 60min/10puffs or 6 puffs per hour ) you be pulling ~12 mA per hour.
12 mA / hour = 12/1000 Ah = 0.012 A/hour.
2.5Ah / 0.012A/hour = 208.333
That means that a AA battery providing 2.5 Ah would last for at least 200 hours under these conditions – assuming that my math is correct, but the point is that you will at least get a very long runtime.
