How to find out what REGULATED mod is pulling from the battery?

[conanthewarrior]

Hi people.

Now, I went on a bit of a mad one earlier, and used some 24 Gauge to make dual coils at .2 Ohms. Was vaping at 70 watts in my 150, and realised I was using the samsung 25R's, not the sony's.

I am not sure if on the steam engine pages, where you can adjust battery drain to a APV from mechanical mod, you put in the batteries nominal voltage, then do I enter the actual batteries voltage, or what is being displayed on the mod? If I enter, at 70 watts and the voltage displayed, 3.3 volts, I accidentally went over by about 4 watts earlier.

I am just not sure I am using the calculator correctly, and it was pulling this much, or as its regulated it was pulling less than that and I am confusing things?

Thanks people, feel a bit of a donut not knowing how to work out what my reguated mod is pulling, as I am not sure I am doing it correctly. But if I was, I hope the batteries are OK, as it was only about 10 minutes of madness, I was so sure I had the VTC4's in there but alas, I did not.

 

[Zamazam]

You are ok with the Samsungs.

 

[conanthewarrior]

View attachment 27113

You are ok with the Samsungs.

Ahh great, what is that you are using there? It would be nice if I could find out so I can find what my regulated mods are actually pulling from the battery, I also notice you enter the batteries nominal voltage? Is that the guideline to go from when measuring regulated power draw?

EDIT: I also noticed there was a slot for kiloamps and megaamps lol. I wonder the way things are going, a couple of years ago it was ego type batteries and CE4's, I come back to all this technology, with some pushing 200 watts, I wonder if a Kilowatt box will ever be made lol, 1000 watts would be a bit insane ( or 1024 if going along computer terms)

 

[Zamazam]

http://www.onlineconversion.com/ohms_law.htm

You enter 2 of the needed variables, in this case 70 watts and .2 Ohm's resistance equals 18.7 amps and 3.74 volts. Keep in mind that the mod is not 100% efficient in current conversion, I would guess the SIG 150 is probably 95% efficient.

 

[conanthewarrior]

http://www.onlineconversion.com/ohms_law.htm

You enter 2 of the needed variables, in this case 70 watts and .2 Ohm's resistance equals 18.7 amps and 3.74 volts. Keep in mind that the mod is not 100% efficient in current conversion, I would guess the SIG 150 is probably 95% efficient.

Thank you, that will come in handy. I just was confused wether I enter 3.3 volts as displayed on the mod, or the nominal voltage as you have? I guess it is nominal voltage yes?

 

[conanthewarrior]

Ahh I understand now, I enter the watts and resistance and it tells me what is being pulled. Silly me. Thank you for that . A liittle confused why my device displayed it as 3.3 volts though, but as long as I know that is after regulation and I am fairly safe, I feel good.

 

[conanthewarrior]

Ahh, I found out why. I need to rebuild, as it is going between 0.1 and 0.2, when it is 0.1 it displays the lower voltage, and I guess is over driving the batteries, but then jumps to 0.2 and will be at 3.7V at 70 watts vs 3.3 for 0.1

 

[Zamazam]

Tighten up the screws on the posts a bit first and see if that stabilizes the resistance reading.

 

[conanthewarrior]

http://www.onlineconversion.com/ohms_law.htm

You enter 2 of the needed variables, in this case 70 watts and .2 Ohm's resistance equals 18.7 amps and 3.74 volts. Keep in mind that the mod is not 100% efficient in current conversion, I would guess the SIG 150 is probably 95% efficient.

I also meant to ask, where you say the mod is not 100% efficient, would that mean the load is slightly higher, say 5% if it is 95% efficient?

 

[conanthewarrior]

Tighten up the screws on the posts a bit first and see if that stabilizes the resistance reading.

Thank you. That did stabilize it, I usually use 26 and had not tightened down enough for the thicker 24 gauge.

 

[Zamazam]

I also meant to ask, where you say the mod is not 100% efficient, would that mean the load is slightly higher, say 5% if it is 95% efficient?

correct.

 

[conanthewarrior]

correct.

Thank you Zamazam, now I just need to get my head round how much is being pulled under temp control, that will be a bit of a mind killer I think. But tonight you have been more than helpful, and I must thank you for that Zamazam .

 

[conanthewarrior]

I think I may just build at 0.4 and see how I like it. The difference in strain on the battery is immense, and I still think that is quite low (For me). Like in my Lemo V2, I currently have a 0.9 in there, and I can use it at lower wattages on my SIG75TC and still get lovely flavour.

(I mainly use power mode, and see that I can use temperature control as a bonus, as I am not sure how safe it is myself, as I do not understand exactly how it works. When I do understand, I probably will use it all the time, but at the moment seeing the voltages jumping about worries me that it is REALLY over stressing the battery somehow)

And that is the main thing for me, flavour, unless I go crazy like today chasing clouds.

 

[Zamazam]

No need to worry about temp control, the mod is supposed to vary the current to the coil to keep the preset temperature. That's just the processor doing it's thing. The mod will use PWM in temperature control, which is basically pulses of current at a certain frequency to achieve a midpoint with the set watts/joules. Make sense?

 

[NemesisVaper]

Resistance doesn't come into the equation on a regulated device.

The atomiser doesn't touch the cells, it's connected to the board.

Max wattage divided by cutoff voltage is all you need to calculate amp draw on a regulated mod. That's why temp control ultra low ohms doesn't matter. Watts is watts to a regulated mod regardless.

At 100W on a 0.5ohm you draw 15A at 6.4V and 100W at 0.1 with 6.4V you also draw 15A

The board doesn't care and isn't stressed by varying voltage wattage output and amp draw. It's what it's designed to do.

 

[Slurp812]

http://www.onlineconversion.com/ohms_law.htm

You enter 2 of the needed variables, in this case 70 watts and .2 Ohm's resistance equals 18.7 amps and 3.74 volts. Keep in mind that the mod is not 100% efficient in current conversion, I would guess the SIG 150 is probably 95% efficient.

2 batteries don't output 3.7 volts. We have to guess the state of battery charge, and check the discharge curves to estimate the actual voltage of the battery inside the mod while its operating. The OP is actually way better off than 18.7 amps.

Z-mans numbers are good for the atty side of things:

70 watts
0.20 ohms
3.74 volts
18.7 amps

If I were to estimate battery voltage in a dual battery mod doing 70 watts I would use a close to worst case 3.3 volts per cell.

so without losses of the circuit we would have on the battery side:

70 watts
6.6 volts (2 fairly drained cells in series)
10.6 amps
0.622 ohms (the load the batteries see)

Of course, we need to also add the losses of the circuit as well. These numbers are average. These types of circuits draw large current spikes from the batteries in reality. I Geuesstimate we should be under 12 amps (average) at this wattage level with almost dead batteries. So, good to go!