Has Resistance influence Battery Drain? (Regulated Mods)

[greek mule]

The guy in this Video insists resistance has nothing to do with amperage draw from batteries used at regulated devices.Says,only wattage settings does.


Example:Single battery device is set at 100Watt in VW mode
2 identical coils (24awg,3mmID,6wraps,leg Length 5mm)
Current that hit the atty (output side)

1. Kanthal A1 0.572 Ω I=√P/R =13.2 A at 100 W
2. SS316L 0.296 Ω I=√P/R =18.4 A at 100W We see that there is a significant difference at current required to reach the desired 100W Lower resistance for the same wattage=Less Battery life

 

[Zakillah]

You're calculating it the wrong way.

The way to calculate Amp draw on a regulated device is set Watts divided by the Voltage of your battery.
Examples:
100W/4,2 Volts (one fully charged battery)
100W/8,4 Volts (two fully charged batteries in series)
100W/3,2 Volts (one battery, almost empty)

So the Amp draw increases the less Voltage you have left in your cell.

Of course, if the mod runs in parallel you additionally divide the Amp draw by the number of batteries used.

Resistance on regulated devices is (for the most part) completly irrelevant.

 

[greek mule]

You're calculating it the wrong way.

The way to calculate Amp draw on a regulated device is set Watts divided by the Voltage of your battery.
Examples:
100W/4,2 Volts (one fully charged battery)
100W/8,4 Volts (two fully charged batteries in series)
100W/3,2 Volts (one battery, almost empty)

So the Amp draw increases the less Voltage you have left in your cell.

Of course, if the mod runs in parallel you additionally divide the Amp draw by the number of batteries used.

Resistance on regulated devices is (for the most part) completly irrelevant.

This is the required current from the coil (output side).As resistance dicreases,it needs more current.
This reflects that power losses on the input side increase,the lower the resistance.

 

[greek mule]

.

100W/8,4 Volts (two fully charged batteries in series)

That's not correct,at regulated mods it doesn't matter if batteries are connected in series/parallele.
You devide by 4.2 for fresh batteries.

 

[Zakillah]

Well, yeah, 4,2 x 2 = 8,4
Amp draw doesnt change series or parallel. The 8,4 was just for clearer explaination.

 

[Topwater Elvis]

When using a regulated power device amp demand from the cell(s) is highest just before and at the power devices low voltage cutoff.
---> Calculating battery current draw for a regulated mod | E-Cigarette Forum

 

[Punk In Drublic]

With a Regulated device you have 2 sides to the regulated circuit. Call it a coil side and a battery side.

On the coil side you set your prescribed wattage and due to the resistance of the coil, the regulator will assign a set voltage based on that resistance and chosen wattage. If we use 50 watts as our example the voltage and resulting current draw for 2 different coils will be.

0.5 ohm coil.

V = √PxR
V = 5 volts
I = V/R
I = 10 amps

0.1 ohm coil

V = 2.24
I = 22.26 amps

On the battery side of the regulated circuit, the batteries do not see the coil nor the needed current for that coil. The circuit is only requesting power from the cells based on what was prescribed above (50 watts). If the cells were at 4 volts the current draw would be.

I = P/V (50 watts/4volts)
I = 12.5 amps

On the battery side as the voltage drops from the cells, the current increases so the regulated circuit can maintain the prescribed power.

I = 50watts/3.2volts
I = 15.6 amps

This is why when choosing a battery for a regulated device we use power and the cut off voltage to determine the proper CDR. Resistance does not come into the equation. This is done by determining the watts per cell (Wbatt), divide by the cut off voltage of the device and the efficiency of the device (Abatt). 90% efficiency is a safe guesstimate if this value is unknown.

So by using the above 50 watts on a single 18650 device with a cut off voltage of 3.2 volts and efficiency of 90% this would

Wbatt = 50/1 = 50
Abatt = 50/3.2/0.9 = 17.3 amps

With a dual cell device this would be.

Wbatt = 50/2 = 25
Abatt = 25/3.2/0.9 = 8.5 amps per cell.

Way too long of a post for this time of the morning!

 

[Punk In Drublic]

On the battery side of the regulated circuit, the batteries do not see the coil nor the needed current for that coil. The circuit is only requesting power from the cells based on what was prescribed above (50 watts). If the cells were at 4 volts the current draw would be.

I = P/V (50 watts/4volts)
I = 12.5 amps

Just to add – this equation assumes 100% efficiency. Should we use the same 90% efficiency the regulated circuit will take an additional 5.5 watts at 50 watts so the equation will be…

50 watts/90% = 55.5 watts
I = 55.5 watts/4 volts = 13.8 amps

 

[Baditude]

You use Ohm's Law for mechs. Watt's Law for regulated.

 

[zoiDman]

Here is a Cool Video for those into Stuff like this.

 

[greek mule]

I had the "suspicion" that resistance at regulated devices must play a role on amp draw from battery.
Now,i admit that I was wrong.
Deeply sorry for the confusion I made.

 

[ScottP]

Just to add – this equation assumes 100% efficiency. Should we use the same 90% efficiency the regulated circuit will take an additional 5.5 watts at 50 watts so the equation will be…

Honestly I wouldn't sweat the "efficiency". You shouldn't be running your batteries so close to max that this would be a factor anyway.

 

[Punk In Drublic]

Honestly I wouldn't sweat the "efficiency". You shouldn't be running your batteries so close to max that this would be a factor anyway.

I Agree. Efficiency is unknown for many devices so becomes a best guesstimate. But always good to have a little headroom. I just threw it in there for the sake of detail

 

[smacuser]

@Anthony_Vapes

 

[RayofLight62]

The efficiency of run-of-the-mill devices varies dramatically as coil resistance decreases, at parity of power setting.
This lead to lower battery life with the low Ohm coil with SOME mods.
One of my point with reviewers has always been the lack of efficiency tests in almost all device reviews.
Unsurprisingly, devices with DNA and Yihi boards have much longer run times with 0.15 Ohm coils than other brands.

 

[dripster]

Up to a certain point, yes it can matter more than just a little bit although this depends on the hardware's capabilities and how it reacts to the voltage sag of the battery you select. Many regulated mods automatically start to throttle the power down if the battery voltage, due to the battery being discharged, gets too low for the mod to still be capable to deliver the desired power output (i.e. the wattage that you set on the mod) so, if that happens before the low battery voltage cut-off (usually at 3.2 volts) is reached, then it means you won't be able to vape the battery all the way down to the cut-off unless you can live with the reduced power that just keeps getting reduced further until finally the cut-off is reached or until you get fed up and just change out the battery nevertheless so, in this particular sense, the correct answer to your question would be "yes despite some funny people told you no".

 

[Anthony_Vapes]

The guy in this Video insists resistance has nothing to do with amperage draw from batteries used at regulated devices.Says,only wattage settings does.


Example:Single battery device is set at 100Watt in VW mode
2 identical coils (24awg,3mmID,6wraps,leg Length 5mm)
Current that hit the atty (output side)

1. Kanthal A1 0.572 Ω I=√P/R =13.2 A at 100 W
2. SS316L 0.296 Ω I=√P/R =18.4 A at 100W We see that there is a significant difference at current required to reach the desired 100W Lower resistance for the same wattage=Less Battery life

this guy is me, and yes it is correct. you can see the comments in the video as well with mooch confirming it to be true when others didn't want to believe it. the calculations for amp draw from the batteries in regulated mods is (power setting) / (input voltage) / (Chip efficiency) so at 100 watts and assuming it's a dual battery mod and 90% efficient it would be 100/7.4 (i'll use nominal voltage as half charged but whit fully charged batteries the draw is less and with dead batteries the draw is more) / .9 (for 90% efficiency) and the amp draw is 15 amps. as you see there is no resistance value in the formula.

 

[Anthony_Vapes]

I Agree. Efficiency is unknown for many devices so becomes a best guesstimate. But always good to have a little headroom. I just threw it in there for the sake of detail

it's also not a flat number. it's like MPG on a car so fluctuates depending on the usage

 

[Punk In Drublic]

it's also not a flat number. it's like MPG on a car so fluctuates depending on the usage

Yes I agree. There will be optimum ranges where efficiency within a device might be better, which that value is unknown to many. But for the sake of calculating current draw from the cells it is a factor – even if a best estimate. And can explain any discrepancies within calculations such as my (simple) example.

Thx for chiming in.

 

[greek mule]

this guy is me, and yes it is correct. you can see the comments in the video as well with mooch confirming it to be true when others didn't want to believe it. the calculations for amp draw from the batteries in regulated mods is (power setting) / (input voltage) / (Chip efficiency) so at 100 watts and assuming it's a dual battery mod and 90% efficient it would be 100/7.4 (i'll use nominal voltage as half charged but whit fully charged batteries the draw is less and with dead batteries the draw is more) / .9 (for 90% efficiency) and the amp draw is 15 amps. as you see there is no resistance value in the formula.

You have my respect Sir,and of course I know who you are and your condribution to our community!