Volts/Amps?

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richard-

Full Member
Apr 2, 2009
5
1
Wales -UK
Can someone with a little more knowledge talk about the relationship between volts and amperage as they relate to the requirements of an atomizer?

There are tons of different 3.6v batteries, but a wide swath of mA ratings.

Maybe someone can write a sticky post with the specs of a standard e-cig


Try one of these batteries:
VARTA - 56620201098 - BATTERY, LI-ION, LIP18650
I know Farnell's in the US supply them and they fit into certain maglite shafts and if you cut the maglite shaft from the bulb end to 1 inch larger than the battery you can make a great screwdriver style E-Cig. You will also need a fibre optic chassis connector with screw fitting on one end (also from farnells)
a bit of time with a file (to round the edges of the connector to fit the case) and epoxy resin to glue it all together and fill the gaps. I am getting a drawing made up with the part numbers also so that it can be a home brew job some people might like to try.
 

billyboy689

Senior Member
ECF Veteran
Aug 29, 2008
171
10
australia
if you have a magnum mod that takes 2 x 3.0 16340 bats and you have 2.5 mm spare room left in the diameter of tube you can put an 18650 batt straight in there and make up the extra 5mm of length with rare earth magnets

if your magnum uses 2 x 3.0 17335 type batts then an 18650 should squeeze in there without too much effort at all at the same length

1004914.jpg
 

Vape

Senior Member
ECF Veteran
Mar 13, 2009
88
0
Northern California
Will using a 9v battery kill the atomizer?

I would cation against using a 9V battery.
The coil and is pretty delicate and might go up in a puff of smoke or worse with 9V applied.

Also I saw mention of ma ratings on batteries, if I'm not mistaken you might mean the mah rating.
The mah rating is more of a how long it will last between charges.
In therory a 1800mah battery will last 3 times longer than a 600mah battery between charges.
 

lec666

Full Member
Apr 9, 2009
68
0
somerville,ma
if you have a magnum mod that takes 2 x 3.0 16340 bats and you have 2.5 mm spare room left in the diameter of tube you can put an 18650 batt straight in there and make up the extra 5mm of length with rare earth magnets

if your magnum uses 2 x 3.0 17335 type batts then an 18650 should squeeze in there without too much effort at all at the same length

1004914.jpg

isn't the point of the magnum mod to run the atomizer at 6V?
 

richard-

Full Member
Apr 2, 2009
5
1
Wales -UK
the atomiser runs at 3v-6v maximum without blowing
I have run tests with a calibrated PSU and have replicated the life of an atomiser as follows:

3.5v 650mA 3 second pulse 3 second delay 10 hours = 6000 activations and atomiser OK
4.5v 1A 3 seconds pulse 3 seconds delay 6 hours 11 minutes until atomiser failure = 3710 activations
6v 850mA 3 seconds pulse 3 seconds delay 10 hours = 6000 activations OK

So have decided that a voltage between 3.5-6 and a current between 650-850mA would provide the greatest life span for an atomiser. I did try 9v @ 750mA and the atomiser failed after less than 1 hour (511 activations), i retested just incase it was a duff atomiser and that failed after 90 minutes (897 activations) so i think the voltage must be limited as well as the current. All the info was obtained from new DSE901 atomisers and i cannot comment on other types, also i have found that homebrew mixtures which contain a high level of VG soot over and fail quicker than premixed juices.
 

fadeaway

New Member
Apr 5, 2009
4
0
Skip to the dashed lines if you find this post boring.

Hello people. I don't think I've ever posted anything in this forum yet... can't remember at least. Anyways, I thought I would share a few tidbits since a couple of you seem confused. Great forum by the way. :)

An ampere is in the units of coulombs/seconds so multiplying by a time unit like hour just gives you an energy capacity as some of you have stated in coulombs. So mAh gives you an end unit of coulombs or an amount of charge.

Power = I*V = V*R^2 measured in Joules/second which is a measure of energy dissipation or energy/time.

As for the ideal setup, I am no professional on these vaporizers (actually still waiting for my mail order of my first DSE901b), but it sounds like we're talking about 1A to go through this, right? In that case, you can run whatever voltage battery you want as long as you have roughly 1A or so running through the atomizer. The reason why a 9v battery didn't work is because the current got way too high. You need to put a resistor between the battery and the atomizer at the very least.

If you have a 3.7V battery and 1A current, the atomizer is 3.7 Ohms. I'm sure there are manufacturing variances so let's just say we're talking 2.5-6 Ohms from china.

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I think I would use something between 4.5V and 9V source along with a 0.1-10 Ohm potentiometer and switch in series to dial in the amount of vapor produced. This way you could adjust per hit. Less resistance for stronger and more resistance for weaker.
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In case anyone is wondering, a potentiometer pretty much is a dial. They used them for the knob in old car stereos and such.
 

ISAWHIM

Senior Member
ECF Veteran
Apr 15, 2009
195
1
48
Jacksonville, Florida
www.isawhim.com
Just for an FYI...

The reason higher 3.6 voltage is used, as opposed to 1.5v, is to reduce the amperage running through the device. Lower voltages create higher amps, which heats-up the components between the battery and the high wattage coil. Higher voltage allows for thinner wires between the battery and the coil, since the parts stay cooler, and there is less power lost from less amps flowing through the device. (This is also why power-lines use high voltage over long distances. It is cooler on the wires with less amps flowing through them.)

There is one missing element for any real calculations here. We don't know the wattage of the coil itself. However, we can estimate the wattage by puffs and puff-lengths between charges. (Got a pen? LOL.)

There are three things to determine power consumption.
Volts * Amps = Watts
Watts / Volts = Amps
Watts / Amps = Volts

Uber power pack!
http://www.batteryspace.com/index.asp?PageAction=VIEWPROD&ProdID=2357

Also note...
A 120w device rated for 120v will draw 1A at 120v, but will draw 10A at 12v.
A 12w device rated for 12v will draw 1A at 12v, but will draw 0.1A at 120v.

You know your volts, 3.6v, and you know 1200mAh or 1.2Ah.
(V * A = W) {V * Ah = Wh} Same thing {Hours are assumed}
(3.6v * 1.2A = 4.32W)
At 3.6v, with that 1.2Ah battery, you can power a 4.32W device for an hour.

Reality check...
Most devices require a range of operating voltage. A 3.6v 1200mAh battery, should provide around 3.6v @ 1.2A of power for an hour. However, if the device you are powering is half the wattage (4.32w / 2 = 2.16w), it will operate at 3.6v for 2 hours.

The wattage is the same, but, depending on the voltage, your amps will rise and fall, reverse of the voltage.

Higher voltage will consume less amps, at the same wattage.
Lower voltage will consume more amps, at the same wattage.

A voltage regulator will result in constant voltage that draws more amps from the battery, as the voltage in the battery falls below the output voltage. On the opposite side, the regulator will consume less amps from the battery if the battery voltage is higher than the output voltage.

EG,
A 12v 10Ah battery can handle 120Wh.
If that were regulated to 1.2v output voltage. The 120Wh device would be drawing 100A. Since the battery only has 10Ah, but now draws 1/10 the amps since it is 1/10 the voltage, it can run that 120Wh device for 1 hour.

Battery sees 10Ah load. (Battery stays cool)
Circuit sees 100Ah load. (Circuit is hot, and device is hot.)

Another situation, with a smaller wattage device.

A 12v 10Ah battery can handle 120Wh.
If that were regulated to 1.2v output voltage. A 12Wh device would be drawing 10A. Since the battery has 10Ah, it can only run that 12Wh device for an hour.

Battery sees 10Ah load. (Battery is warm)
Circuit sees 10Ah load. (Circuit is warm, and device is warm.)

Another situation, using higher voltage.

A 12v 10Ah battery can handle 120Wh.
If that were regulated to 120v output voltage. The 120Wh device would be drawing 1A. Since the battery only has 10Ah but now draws 10x more amps in the conversion which is 10x up from 12v, it can only run that 120Wh device for 1 hours.
(12v SLA battery, upped to 120v, will power a 120w light for 1 hour.)

Battery sees 10Ah load. (Battery is hot)
Circuit sees 1Ah load. (Circuit is cool, and device is hot.)

If you add a micro-controller, like this...
http://www.electronicspecifier.com/...eading-Position-in-6-pin-Microcontrollers.asp

You could fully regulate power and control of the device, like the battery that comes with the vaporizers. (Since they use a similar micro-computer controller in the battery.)

That same micro-controller can also be used to charge the battery, or regulate the charge power going into the battery. But you may get the same cruddy results as we get from our battery chargers. It has amperage limitations, so it can not provide the high-amp charge that these batteries like, when they are low on power.

There is another chip that can handle that. (Same chip they use in our cell-phones.) No programming required.
 
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Fadeaway - such a potentiometer would be big to cope with all the wasted heat. However, a potentiometer with power transistor or voltage regulator would allow a variable hit in a larger size mod. Or a multi-way switch with variable numbers of step-down power diodes (about 0.7v step down for each diode with minimal power loss (heat).

The basic idea using a potentiometer (dial/knob) with a power transistor:
4-04.gif

This circuit could be used more of less unchanged. A resistor of about 2.2K on the negative side of the pot (other side from the 470 one) would provide a useful power range.

source (a nice intro to the transistor): Transistors
 
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Vapinginjapan

Senior Member
ECF Veteran
Apr 22, 2009
215
1
39
Would it be possible to deduce the wattage of the coil from the ohms?

I also imagine a multimeter would be able to see the battery draw. I think we need more data on the power consumption characteristics as they scale with voltage.

I've *always* been fascinated with the idea of putting a super capacitor in between the battery and the atomizer. The smaller 150 farad ones are tiny mofos, don't hold much juice (maybe enough for 3-4 puffs), but can shunt out 5+ amps easy and up to 15A peak, and would even out drain on the battery so that instead of one massive power flux when the atty revs up, and then a period of rest, you'd get more or less constant power flow from the battery to the super capacitor, with tons of amps on tap to spare.

Does anyone know how many mAH per 5 second puff we're looking at for a 3.7v new atty?
 

fadeaway

New Member
Apr 5, 2009
4
0
Woah. Glad this sparked a bit of discussion. :D

kinabaloo and kender:
I am definitely too lazy to do multimeter measurements to get this right. This is mostly due to me thinking that I'll be quitting sooner than I would be modding my own equipment. (I HOPE!!!) The only thing I really want is a USB pass-through which I'll end up making with a switch like most everyone else.

I would probably not use BJTs or MOSFETs, but if I did it would be a single NMOS specifically implementing current control in a way that it would not waste current to stray grounding. Since battery life is the main issue, keep it simple stupid (KISS) is what I would try to follow by having the least number of elements while getting the best out of my device. Don't forget the power loss to Rds or Rce if you use transistors.

Vapinginjapan:
I definitely like the cap idea to reduce instantaneous stress on the battery. I don't know enough about battery chemistry and cycle stress to make a good comment on this though. I hope someone chimes in on this one. I'm only used to seeing caps to reduce HF noise on rails.

ISAWHIM:
Current cell phone power regulators are freaking complex. There are too many feedback systems for TX/RX, I/O buffers, system voltage regimes, etc... Use an old phone if you're going to try to take pieces from a phone since most of the older stuff uses a single BJT with a single resistor based feedback circuit for charge control. The other circuit that I think you might be talking about is the actual circuitry in the battery? I'm not certain. (Sorry if this all sounds self-righteous. Used to work in the cell phone chipset industry.)
 
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