Question about variable volt real power

[MechaHerc]

hi guys, i have some question about vv mod. in technical view which one is more good for eg: for battery usage (3.7Volt), atomizer coiling, vapor production etc if any..


- low ohm atomizer (1.1ohm) set at 3.5Volt (which have 12.44545W)* VS
- high ohm atomizer (2.8ohm) set at 5.9Volt (which also produce 12.44545W)*


or another example
- low ohm atomizer (1.1ohm) set at 3.06Volt (which have 8.5)* VS
- high ohm atomizer (2.8ohm) set at 4.88Volt (which also produce 8.5W)*




*according ohm law calculator

 

[crxess]

I think you mean LOW ohm

Not sure what you are asking - Battery life or atomizer/flavor performance?

Performance should be about the same as you are controlling identical power output.

 

[MechaHerc]

I think you mean LOW ohm

Not sure what you are asking - Battery life or atomizer/flavor performance?

Performance should be about the same as you are controlling identical power output.

lol yeah, low ohm (why i put slow there? lol) i mean something about battery life.. more watt vs push the volt for 3.7battery

 

[nahoku]

Power is power... as far as the battery is concerned. So, if you're drawing 12 watts from any combination of settings, you're still drawing 12 watts.

VV / VW mods have circuitry. Now whether or not that circuitry has to work harder or not to produce different voltage/wattage outputs is really dependent on how the device was designed.

 

[SoberSnyper]

Power is power... as far as the battery is concerned. So, if you're drawing 12 watts from any combination of settings, you're still drawing 12 watts.

VV / VW mods have circuitry. Now whether or not that circuitry has to work harder or not to produce different voltage/wattage outputs is really dependent on how the device was designed.

This is not an accurate assessment. Running at 12 watts with a 3 ohm coil will not drain your battery as fast as running 12 watts on a 1.5 ohm coil. The lower the resistance of your atty, the faster your battery is being drained. As resistance is decreased, to maintain the same power setting, then current(amps) will increase.

 

[nahoku]

This is not an accurate assessment. Running at 12 watts with a 3 ohm coil will not drain your battery as fast as running 12 watts on a 1.5 ohm coil. The lower the resistance of your atty, the faster your battery is being drained. As resistance is decreased, to maintain the same power setting, then current(amps) will increase.

No, the assessment is spot on. You're missing something here, but I won't tell you what that is. You go think about it, and if you still think I'm wrong, come back and I'll explain it to you.


EDITED... for the benefit of the OP, I decided not to wait for you.

I can understand why you would think that if the load were cut in half (3 ohms to 1.5 ohms), the current in the circuit would increase. This is simple electronics theory, but you cannot only consider load and current when speaking about “power” in a VV mod. Your explanation only applies to non-regulated appliances.

In a VV mod, if you want to maintain 12 watts, then you need to adjust the voltage to the correct level for the load you attached. If you don’t adjust the voltage, then you won’t operate at 12 watts.

Whether you’re running 12 watts with settings of 6 volts and 3 ohms, or 4.24 volts at 1.5 ohms, the battery doesn’t care… all it sees is 12 watts. As long as you continue to run 12 watts, no matter the settings, the run time won’t change.

 

[SoberSnyper]

No, the assessment is spot on. You're missing something here, but I won't tell you what that is. You go think about it, and if you still think I'm wrong, come back and I'll explain it to you.


EDITED... for the benefit of the OP, I decided not to wait for you.

I can understand why you would think that if the load were cut in half (3 ohms to 1.5 ohms), the current in the circuit would increase. This is simple electronics theory, but you cannot only consider load and current when speaking about “power” in a VV mod. Your explanation only applies to non-regulated appliances.

In a VV mod, if you want to maintain 12 watts, then you need to adjust the voltage to the correct level for the load you attached. If you don’t adjust the voltage, then you won’t operate at 12 watts.

Whether you’re running 12 watts with settings of 6 volts and 3 ohms, or 4.24 volts at 1.5 ohms, the battery doesn’t care… all it sees is 12 watts. As long as you continue to run 12 watts, no matter the settings, the run time won’t change.

I don't need an explanation but evidently you do. Using your figures above of 12 watts with a 3 ohm coil vs a 1.5 ohm coil, lets figure what the current draw is on each circuit as your numbers above are correct. Current(amps) is equal to Voltage divided by Resistance(ohms), so with a 3 ohm coil and 6 volts, the current will be 2 amps, and with a 1.5 ohm coil and 4.24 volts, the current will be 2.83 amps, now which setup do you think will drain the battery faster?

 

[Rader2146]

This is not an accurate assessment. Running at 12 watts with a 3 ohm coil will not drain your battery as fast as running 12 watts on a 1.5 ohm coil. The lower the resistance of your atty, the faster your battery is being drained. As resistance is decreased, to maintain the same power setting, then current(amps) will increase.

It was a very accurate assessment. In fact...it's the law.

http://www.e-cigarette-forum.com/forum/blogs/rader2146/3500-calculating-battery-drain-current.html

 

[zapped]

It seems backwards at first but this is definitely true.

When I was using 2.0ohm cartos on my Provari I was switching out 18650s twice a day.At 3.0ohms its only once a day.

Be leery of using lower ohm cartos, atomizers, RBA's and clearomizers in VV and VW devices especially those without built in circuit protection. You can set up a situation where youre discharging the battery at a very unsafe rate that could cause it to fail or explode.

Low Resistance carto's etc were made for 3.7 volt battery type e-cigs .......anything outside of that is asking for trouble.

 

[SoberSnyper]

It was a very accurate assessment. In fact...it's the law.

http://www.e-cigarette-forum.com/forum/blogs/rader2146/3500-calculating-battery-drain-current.html

See my latest post proving it was not an accurate assessment.

 

[Rader2146]

See my latest post proving it was not an accurate assessment.

See (and read) the blog post that explains why it is.

You're only using half of the equation; the output side. You can't ignore the input side of things. Energy is not free, it cannot be created from nothing.

The ONLY reason that a battery would last longer when used at the SAME power level with a different resistance is that the voltage converter is running at a higher efficiency in that particular combination.

 

[Trick]

I can't believe people are arguing that higher resistance will not increase battery discharge time by quoting Ohm's Law, which proves it's true.

In the first example:

- low ohm atomizer (1.1ohm) set at 3.5Volt (which have 12.44545W)*

This will crank out those 12 watts at a rate of 3.36 amps

high ohm atomizer (2.8ohm) set at 5.9Volt (which also produce 12.44545W)*

This will put out essentially identical wattage, but at a rate of 2.1 amps.

Amps are your rate of flow, and what determine how fast your battery drains, not watts. That's why batteries are rated in mAh.

A higher-resistance coil will make your battery last longer on a charge, by reducing the flow of electricity (the amps). In this case, the battery will last quite a bit longer.

It seems a lot of people forget there's a fourth factor involved in Ohm's Law, and it's the one that directly relates to how long a battery charge will last.

 

[SoberSnyper]

See (and read) the blog post that explains why it is.

You're only using half of the equation; the output side. You can't ignore the input side of things. Energy is not free, it cannot be created from nothing.

The ONLY reason that a battery would last longer when used at the SAME power level with a different resistance is that the voltage converter is running at a higher efficiency in that particular combination.

If my APV is putting out 6 volts with a 3 ohm carto, the current is 2 amps, this is fact. If my same APV is putting out 4.24 volts with a 1.5 ohm carto, the current is 2.83 amps, this is also a fact. Both setups put out 12 watts, this is also a fact and is supported by Ohm's Law. Now you please explain to me how the battery discharge is the same with both setups.

 

[Rader2146]

Ohm's law is not the "end all be all" to electricity calculations. A voltage regulator, aka voltage converter, aka DC-DC converter, CONVERTS from one voltage to another voltage. This conversion could be a boost converter that converts low voltage to a higher voltage, or a buck converter than converts high voltage into low voltage. YOU MUST TAKE THE CONVERSION INTO ACCOUNT WHEN CALCULATING BATTERY DRAIN CURRENT. There is a conversion rate, just like dollars to pecos, feet to meters, fahrenheit to celcius. Only difference is that we are converting input to output, and the conversion rate is:

Input Current = (Output Power + Efficiency Losses) / Input Voltage

Here is an older post that breaks it down in a different way than my blog post:

A boost regulator has to transform low voltage into higher voltage. This can only be done by using more current on the input (battery) side. The amount of additional voltage needed is expressed as:

(Volts out - Volts in) / Volts out = Percentage of voltage increase, also know as the switch duty cycle.

Now that we know the duty cycle, we can figure the additional input current required to obtain the desired output voltage.

Amps out / ( 1 - Duty Cycle) = Amps in

Example...

Known factors:
3.7v in
8 watts out
3.0 ohm carto

Ohms Law tells us that we'll need 4.9v and 1.63 amps output to achieve 8 watts.

(4.9-3.7)/4.9 = .24 = 24% increase in voltage = 24% switch duty cycle

1.63 / (1-.24) = 2.16 amps input drawn from the battery.
-----------------------------------------------------------

Now for validation.
As deemed by the Law of Conservation of Energy, power (watts) in must equal power out. (The true statement of the law is power in equals power out + efficiency losses. But for simplicity sake, we'll get to efficiency below.)

Power = Volts * Amps

Input:
3.7 * 2.16 = 8 watts input

Output:
4.9 * 1.93 = 8 watts output
------------------------------------------------------------

But what about efficiency?
Typical efficiency for a boost converter is in the 75-90% range. Efficiency is not constant. It varies with the desired outputs. You can find the efficiency for certain [manufacturer chosen] situations in the regulators data sheet. Using an optimistic value of 90% efficiency we can figure our adjusted input current.

Power out / efficiency = adjusted power in

8 / .9 = 8.89 watts input.

Adjusted power in / Volts in = adjusted amps in

8.89 / 3.7 = 2.4 amps input.
-----------------------------------------------------------
And comparison:
A fixed voltage device @ 3.7v will achieve an 8 watt output using 2.16 amps (Ohms Law)

8 / 3.7 = 2.16 amps input
------------------------------------------------------------
The above calculations explain why I say that boost regulators will get less battery life than a same size fixed volt, and also that you will not achieve better battery life by using a higher resistance coil.

So in reality, and based on the above example: A device using a 3.0Ω coil @ 4.9V and 8 watts output draws MORE current from the battery than a fixed voltage (ie: mechanical) device with a 1.7Ω coil @ 3.7V and 8W output. If you were to do the calculations for two devices with the equal efficiency loss, then the current drain from the battery would be exactly equal.

Bottom line: Efficiency is the only variable that will effect the battery life when the output power remains the same.

Edit: It's come to my attention that I can come across rather brazen. The CAPS are not for yelling, just to emphasize the point.

 

[nahoku]

Wow, I went to dinner and came back and all of this happened! I agree, ohms law is very good for applying to mechs, but can fall short with APV's. It's a different ball game when you're talking PWM, duty cycle and the like. It would make a good study, though!

Peace out!

 

[MechaHerc]

thanks for replying guys.. so in the end which one will save the battery life more? quite still not very understand lol..
if can describe it in layman terms also quite helpful.

 

[SoberSnyper]

Rader2146,

First of all, I don't have a problem with the CAPS, so no worries there. The problem is with your conclusions. Lets forget about power for a second because I agree that output power will always be less than input power, this is a basic physics law.

Lets talk about DC circuits. I have a VV device set at 4.6 volts with a 2 ohm carto. The device is actually delivering 4.6 volts because I can measure it and the 2 ohm carto actually reads 2 ohms. I know what the current is because we can use Ohm's law. 2.3 amps. Using these numbers we can calculate the input power which is 10.58 watts. The actual work being done will be less than 10.58 depending on efficiency, that is true, but that has to do with the efficiency of the coil to produce the heat(watts), this has nothing to do with the circuit. I know what my output voltage is because I can measure it, I know what my resistance is because I can measure it, and from there I can calculate the current(amps), although I don't need to know the current because Power is also equal to Voltage squared divided by resistance(ohms). When I press the fire button I know the input power because I can calculate it, now output power is another discussion.

I also know in practice that this is true, I get longer battery life with a 3 ohm carto than I do with a 2 ohm carto using the same input power.

 

[Trick]

A voltage regulator, aka voltage converter, aka DC-DC converter, CONVERTS from one voltage to another voltage. This conversion could be a boost converter that converts low voltage to a higher voltage, or a buck converter than converts high voltage into low voltage. YOU MUST TAKE THE CONVERSION INTO ACCOUNT WHEN CALCULATING BATTERY DRAIN CURRENT.

If the draw on the battery was somehow kept constant by the circuitry, it's true that higher resistance will cause no difference in battery drain. However, I'm not aware of any PV where that's the case, and it's not like this is a difficult thing to test, and many people have. Lower resistance at the same wattage means more amps, and lower battery life. All theory aside, this is something that's been demonstrated and proven many times over.

 

[andrios]

Interesting stuff!

 

[Rader2146]

Rader2146,

First of all, I don't have a problem with the CAPS, so no worries there. The problem is with your conclusions. Lets forget about power for a second because I agree that output power will always be less than input power, this is a basic physics law.

Lets talk about DC circuits. I have a VV device set at 4.6 volts with a 2 ohm carto. The device is actually delivering 4.6 volts because I can measure it and the 2 ohm carto actually reads 2 ohms. I know what the current is because we can use Ohm's law. 2.3 amps. Using these numbers we can calculate the input power which is 10.58 watts. The actual work being done will be less than 10.58 depending on efficiency, that is true, but that has to do with the efficiency of the coil to produce the heat(watts), this has nothing to do with the circuit. I know what my output voltage is because I can measure it, I know what my resistance is because I can measure it, and from there I can calculate the current(amps), although I don't need to know the current because Power is also equal to Voltage squared divided by resistance(ohms). When I press the fire button I know the input power because I can calculate it, now output power is another discussion.

I also know in practice that this is true, I get longer battery life with a 3 ohm carto than I do with a 2 ohm carto using the same input power.

I think there many be a little difference in our definitions. Input power is from the battery to the voltage converter. Output power is from the converter to the coil. Thermal efficiency of the coil is a completely different conversation, but the efficiency of the converter is very pertinent to this conversation.

If the draw on the battery was somehow kept constant by the circuitry, it's true that higher resistance will cause no difference in battery drain. However, I'm not aware of any PV where that's the case, and it's not like this is a difficult thing to test, and many people have. Lower resistance at the same wattage means more amps, and lower battery life. All theory aside, this is something that's been demonstrated and proven many times over.

Sorry, this isn't thoery. These are the laws of physics. I don't expect anyone to take my word for it. Please, research it yourself if you feel inclined. Just remember that the only way to truely learn is to accept that what you think you know could possibly be wrong. I once thought along the same lines as you, and SoberSnyper, and frankly the majority of this forum. Then I researched, I learned, I tested, and I proved it to myself. The earth was once thought to be flat and the center of the universe.

The only thing that has been demonstrated is that people claim to get better battery life with a certain configuration, and I acknowledge that by saying they are using thier PV in a configuration that runs more efficient. No one has posted any tested, measured, and qualified proof that I have seen. It's all been conjecture. I have proven this to myself by measuring both input and output voltage and current and calculating efficiency loss in certain configurations.