Watts - same same but different?

[fishol]

Hi all! This is my first post on the forum, and I hope some of the vets here can help me out.
I'm using a Provari V2 with a rebuildable atty (Fatty V2 from chris creations) and AW 18490 batts. my question is:
What setup/combo would give a better battery life at a fixed amount of watts?

3.3 volts with a 1.2 ohm coil = 9 watts @ 2.75 amps

vs.

5.2 volts with a 3 ohm coil = 9 watts @ 1.73 amps

Is there a big difference here, considering the gear that I'm using? Or the amp draw would be negligible since the provari has boost circuit thingy?

Cheerios!

 

[gthompson]

Battery life is determined by amps, so the first example will drain your battery faster. Much faster.

 

[justinred]

I noticably get better battery life on my vari when at a higher resistance/voltage combo.

 

[sawlight]

A battery can only store so much power, the faster you use it, the sooner it's gone! In your example, the second figure delivers the same watts, the same heat and same flavor, theoretically at least, but uses half the amperage, thus doubling the life of your battery. This is why VV is so popular.

 

[Rader2146]

Battery life is determined by amps, so the first example will drain your battery faster. Much faster.

A battery can only store so much power, the faster you use it, the sooner it's gone! In your example, the second figure delivers the same watts, the same heat and same flavor, theoretically at least, but uses half the amperage, thus doubling the life of your battery. This is why VV is so popular.

Not true. There is a power conversion that takes place. Output current DOES NOT equal battery drain current. In an ideal (not factoring efficiancy losses) device, battery drain current would be equal in those 2 examples.

 

[gthompson]

Not true. There is a power conversion that takes place. Output current DOES NOT equal battery drain current. In an ideal (not factoring efficiancy losses) device, battery drain current would be equal in those 2 examples.

Haha. Ok.

 

[tnt56]

Just carry a spare battery and don't worry. Sorry but that's what I do. The only time I was left without a PV was coming home from work. My kicked SB died on me and the spares were in the trunk of the car. Kinda hard to reach em when your doing 80mph down the interstate. Lesson learned. Keep a spare handy. So now I got at least 2 PV's within reach.

 

[tnt56]

I'm so sorry. Welcome to ECF. This is a great place. Heck they even allow me to be a member. Hope you enjoy your new journey and I like your choice of PV's.

 

[Rader2146]

Haha. Ok.

I don't see the humor in my post. Care to share the punch line?

 

[sawlight]

Not true. There is a power conversion that takes place. Output current DOES NOT equal battery drain current. In an ideal (not factoring efficiancy losses) device, battery drain current would be equal in those 2 examples.

The power conversion is done with a boost circuit to up the voltage allowing less current drain to achieve the same heat. Why are most large machines wired at 440v? Efficacy, plain and simple, they can achieve the same work load, with less amperage because of the higher voltage.
It's a matter of using the power you have as effiecently as possible, less amperage use, the longer the battery life will be.

 

[gthompson]

I don't see the humor in my post. Care to share the punch line?

I have no desire to argue with someone that knows everything.

 

[Baditude]

I don't see the humor in my post. Care to share the punch line?

Move along little doggy, move along.

 

[Stef64]

Hi all! This is my first post on the forum, and I hope some of the vets here can help me out.
I'm using a Provari V2 with a rebuildable atty (Fatty V2 from chris creations) and AW 18490 batts. my question is:
What setup/combo would give a better battery life at a fixed amount of watts?

3.3 volts with a 1.2 ohm coil = 9 watts @ 2.75 amps

vs.

5.2 volts with a 3 ohm coil = 9 watts @ 1.73 amps

Is there a big difference here, considering the gear that I'm using? Or the amp draw would be negligible since the provari has boost circuit thingy?

Cheerios!

I have not noticed a big difference in battery life on my Provari when using LR or HR attys provided I pull the same amount of power. If you really think about it, it should not make any difference what resistance you are using - the battery still has to provide the same power (W) regardless of where you set the voltage. In both instances, a fully charged battery (4.2V) will provide 2.14A to the boost circuit which will then convert this into 3.3V/2.73A for LR or 5.2V/1.73A for the HR atty. Of course, this is assuming there is no energy loss in the conversion process.

 

[Rader2146]

A wise man once said: "Keep your words sweet, because you'll never know when you'll have to eat them."

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.

-----------------------------------------------------------

Woof, woof.

 

[steved5600]

Welcome to ECF on thing to note that I have found with higher resistance it takes a bit longer for the coils to heat evenly. I have found I need to push the button a bit before taking a drag. Current amps are what determines the rate at what the batter dies. 1500 mah or 1.5 amps will work for 1 hour at a drain of 1.5 amps. Just do what works for you and like tnt56 said carry a spare battery. If I go out for long I carry a spare and a portable 12v plug charger.

 

[sawlight]

A wise man once said: "Keep your words sweet, because you'll never know when you'll have to eat them."



Woof, woof.

But you didn't correct the voltage in said equation. If you up the voltage to 5v and figure in 8.89, you draw 1.778 amps. That calculates in the loss of the boost circuit.
Think for a second, you have a battery rated at 3000mah, that is 3. amps per hour, it will, theoretically pull 3 amps over one hour. Drop it to 1.5 amps, it will pull that load for two hours, cut it again to .75 amps it will run for four hours. This is constant load.
Your quote above figured in loss of conversion, but didn't add the voltage it was converted to, so I fail to see the point of it at all? Is there loss, absolutely, there is loss in switches, wires, connections and conversion boards. 85% is about the going rate for efficiency right now , so figure 15% off the top.
Now lets look at multi-mode led flashlights for giggles. We have a three mode driver board, we loose 15% with the board, have to meet the forward voltage to make it even light up, then vary the amperage we feed it. In the mix of this voltage will change some, within a few volts.
On high, said light will burn 200 lumens at 1.5hrs.
On medium it will go 100 lumens for 5 hours.
On low it will run .5 lumens for 50 hours.
Why? We are lowering the amperage draw on the battery, the low mode needs less energy and uses the battery more efficiently. The same is with a higher resistance atty at higher voltage, you have the loss of the boost circuit, but lower the output amperage.

 

[BikerBob]

The "energy" capacity of a battery is properly quoted in "joules", or "watt-seconds". Multiply by 3600 seconds-per-hour, and get "watt-hours". Multiply by 1000 to get "kilowatt-hours", like is measured by your electrical power utility company. Since there is a relationship between volts, amps, and watts, if you assume a constant (average) battery voltage, the unit "amp-hours" may also be used (see Battery energy storage in various battery types). This is referring to the energy capacity of the battery itself. If all of that energy goes to straight to the coil, there should be no difference, in theory, between your two examples *if* they are run for the same amount of time, since they both consume the same 'watts'. Same amount of energy consumed at the rate of 9 watts.

The wires/circuits between the battery and coil have a bearing here too...they are not really 'zero-resistance' wires, but 'very low resistance', so they will consume some of that 9 watts. Actual amount is proportional to the square of the current, so the points that have lower current will heat up less than those with higher current. First, from the battery to the control circuit will have the same amount of current for each 9-watt coil example, so there would be no difference between them. From the control circuit to the coil will have a different amount, based on the resistance of the coil The example with the lower amount of current would consume less power in the interconnecting wires. Also, there will be lost power within the controller/regulator (can't get more power out than you put in).

Large electrical machinery is designed with this 'lower current' thing in mind. Just consider a 0.1 ohm power cable powering a large 440 volt motor. We can get 440 watts to the motor by supplying either 1 amp at 440 volts, 2 amps at 220 volts, or 4 amps at 110 volt*. The power cable will consume 0.1 watt at 1 amp, 0.4 watt at 2 amps, or 1.6 watt at 4 amps. How much power do you want to waste just heating up the power cable? This is most likely why those who see longer battery life are those using lower currents (higher voltages) through their 510 connector: less power is wasted enroute to the coil.

*As an engineering trade-off, the motor needs 4 times the number of turns of wire (4 times the resistance, hence 4 times the power loss) to run at 1 amp vs 4 amps, to get the same torque.

 

[Thrasher]

LOL dude!!! just buy a couple batts they are 9 bucks, if you can afford a provari im sure you can get a couple batts and a plastic case.

too much thinking when not at work lol.

 

[Rader2146]

But you didn't correct the voltage in said equation. If you up the voltage to 5v and figure in 8.89, you draw 1.778 amps. That calculates in the loss of the boost circuit.

The voltage is regulated based on the desired output voltage. Output is constant, so the losses are calculated on the input side of the equation. Voltage is a set value by the battery charge, so the losses present themselves as increased current, not as increased voltage.

Think for a second, you have a battery rated at 3000mah, that is 3. amps per hour, it will, theoretically pull 3 amps over one hour. Drop it to 1.5 amps, it will pull that load for two hours, cut it again to .75 amps it will run for four hours. This is constant load.
Your quote above figured in loss of conversion, but didn't add the voltage it was converted to, so I fail to see the point of it at all? Is there loss, absolutely, there is loss in switches, wires, connections and conversion boards. 85% is about the going rate for efficiency right now , so figure 15% off the top.

In that post I used 90% as an optimistic value just to prove the point. Without actually doing the efficiency calculations and measurements for your specific devise, it is an arbitrary number. Substitute any number that you see fit. In real world usage, all of the variables that you mentioned would be factored in to the efficiency value.

Now lets look at multi-mode led flashlights for giggles. We have a three mode driver board, we loose 15% with the board, have to meet the forward voltage to make it even light up, then vary the amperage we feed it. In the mix of this voltage will change some, within a few volts.
On high, said light will burn 200 lumens at 1.5hrs.
On medium it will go 100 lumens for 5 hours.
On low it will run .5 lumens for 50 hours.
Why? We are lowering the amperage draw on the battery, the low mode needs less energy and uses the battery more efficiently. The same is with a higher resistance atty at higher voltage, you have the loss of the boost circuit, but lower the output amperage.

Flashlights have a fixed resistance and use pulse width modulation to adjust the average voltage over time. Resistance stays the same, but voltage is changed, which means you are also changing the watts. Not a good analogy in a "battery drain vs fixed wattage" conversation.

 

[CJB]

Thats right. Lower volts on lower ohms will produce less amps prolonging battery life