Question about variable volt real power

[SoberSnyper]

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.

A battery does not provide power, only voltage, and the output from the converter is voltage, not power. Power doesn't enter into the equation until the current passes through the resistor(coil)


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.

If I am using a Vamo, set to 10 watts, will I get more vaping time with a 1.5 ohm coil or with a 3 ohm coil? It is a simple question, the answer should be easy to conclude.

Also, using a mechanical mod with a low res atty to compare to a VV/VW device with a standard or high res atty is an invalid comparison as the VV/VW device is providing a constant supply of voltage and the mechanical mod starts with 4.2 volts of a freshly charged battery and is being continually drained while vaping. As the voltage of the battery decreases, so does the current, as a result, the power of the circuit is continually decreasing as you are vaping.

 

[YKruss]

If I am using a Vamo, set to 10 watts, will I get more vaping time with a 1.5 ohm coil or with a 3 ohm coil? It is a simple question, the answer should be easy to conclude.

42?

Not 42 but the answer is so obvious.

 

[Rader2146]

A battery does not provide power, only voltage, and the output from the converter is voltage, not power. Power doesn't enter into the equation until the current passes through the resistor(coil)

Great googly moogly...is this a joke?

How can you even begin to discuss the subject when you dont know the BASIC FUNDAMENTALS of electricity? I amend my previous statement:

The only way to truly learn is to accept that what you think you know could possibly be IS wrong.

Read up:
Electric power - Wikipedia, the free encyclopedia

 

[Rader2146]

Here, I remembered reading this a while back. If you dont find my explanations of the laws to be worth consideration as true and accurate, or even plausible, statements...maybe the fine folks at Provape Engineering could sway your thinking.

I ran across this message from our engineering team as to why we recommend using high drain batteries.

Yes, this is a technical explaination and my head almost exploded trying to understand it fully. However, I wanted to share it with everyone out there.

Shoot. I'm sure there are many out there that can 'grok' this better than me... and perhaps put it into easier to understand terms?

Why does the Provari need to use HD (High Drain) batteries?

The Provari uses an internal dc/dc converter that continuously monitors the output to the atty load and maintains a constant voltage with
an output current limit of 3.5 amps (each unit is tested to this limit in production, most exceed this limit by .1 to .2 amps). This output voltage
is usually higher than the input battery voltage, depending on what the Provari is adjusted to.

Battery output current is NOT the same as Provari output current.

Here is an example that shows WHY. When the Provari is putting out 4.0 volts at 3.5 amps into an atty load, that calculates to be 14 watts.
That means the battery has to supply 14 watts to the Provari, at a minimum. In reality it is more because nothing is 100% efficient. If the battery voltage
is 3.7 volts it will need to supply 14 watts / 3.7 volts or 3.78 amps minimum. Because nothing is 100% efficient, lets add that in. Most Provari devices
are 90% to 95% efficient. Picking the worst case, 90%, the battery current must now need to be 3.78 amps/ .9 = 4.20 amps. That is a more realistic current draw,

But wait, it could be even higher. WHY? Look at what happens when the battery voltage droops lower than 3.7 volts. At 3.5 volts the battery current goes up to 4.44 amps.

But wait, it can be even higher still when the battery droops to near end of charge, say 3.2 volts. The battery output current now goes to 4.86 amps.

And we have to keep in mind that this is ONLY the average current. Because the Provari converter is a pulsing device, the pulsed battery current can be a factor of 2 higher than the average current. REALLY, the battery needs to supply pulsed currents of over 9 amps. These engineering calculations show why you need a high quality, LOW INTERNAL RESISTANCE, battery to supply what the Provari converter needs.

Where does the internal resistance come into play? According to ohms law, when you are drawing 1 amp out of the battery the battery voltage will droop .1 volts for every .1 ohms of battery internal resistance.

At 4 amps this means the battery voltage is no longer 3.7 volts but 3.3 volts. Next time you look at the battery specification you are buying, see what it says its internal
impedance is. Do the math. Better yet we have done the math AND we have tested what is available. If it won't match up to the demands of the converter in the Provari
we won't sell it.

But wait, why do the cheap batteries work at all? They work when the output is not adjusted to supply high power. They get by running at low power but the full capacity of the battery is NEVER available because of the voltage droop due to the higher internal resistance causes them to shut down early. So if you think you are getting 2500 maH out of a battery with that rating you would be wrong.

Just remember, the Provari, when adjusted to full power out, can supply 6.0 volts at 2.42 amps. That is 14.5 watts. What happens to all these calculations when the output power is now
14.5 watts? This exercise is left for those inclined to want to know.

Provape Engineering

 

[SoberSnyper]

Great googly moogly...is this a joke?

How can you even begin to discuss the subject when you dont know the BASIC FUNDAMENTALS of electricity? I amend my previous statement:

The only way to truly learn is to accept that what you think you know could possibly be IS wrong.

Read up:
Electric power - Wikipedia, the free encyclopedia

Mmm? Then why dear sir are batteries measured in volts and not watts? Maybe because no work is done until a battery is placed across a load and the potential is seen by a circuit, more than voltage is needed to create power, there must be a circuit for power to be created and this relationship we see with Watts law.

 

[Rader2146]

Mmm? Then why dear sir are batteries measured in volts and not watts? Maybe because no work is done until a battery is placed across a load and the potential is seen by a circuit, more than voltage is needed to create power, there must be a circuit for power to be created and this relationship we see with Watts law.

The subject at hand is battery drain current while under a load. Where did you get the idea that there was not a load or a circuit?

P = QV/t = IV

The battery provides the voltage, it also provides the electric current. So if we have voltage from the battery, and current from the battery, if we put them together (IV) you can only conclude that we get POWER from the battery. I know, it's a crazy concept that isn't taught below 7th grade science class.

Oh, BTW......Many batteries are measured in watts, more specifically watt/hours (Wh).......as in how much POWER a battery can provide over time.

I'm done here unless someone else has an earnest question or wants to have a meaningful discussion.

 

[DavidOck]

Batteries are not unlimited power sources, eh? Not like plugging into the wall socket.

Batteries have a time / current discharge curve. The higher the current, the shorter the time. Has to do with the chemistry of the device.

Notes on Batteries provides an interesting overview of the phenomena.

At the same wattage, a higher voltage needs to source less current, so the time goes up.

 

[Slurp812]

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.

Assuming the circuit efficiency is negligible watts in must = watts out. How does one derive 4 volts from a 3.7 volt battery? Easy, more current. Remember, current in the atty/carto is not equal to battery current. As others have stated, power is power. Yes you need to factor in the circuit power as well but in most of these types of circuits, the efficiency is very close regardless of load resistance.

 

[dejo]

If I were to rewire my house to 4160 volts, would I then drop my electric bill by 19 fold? That I would love, but Im afraid there is no free lunch available. Power used is power used

 

[adopogi]

Regarding battery drain...will a single coil and dual coil (in series) with both set at 1 ohm, drain the battery at the same rate or will a dual coil drain the same battery faster?

 

[DavidOck]

Perhaps not 19 fold, but it would be lower. There are system losses that wouldn't be there - fewer transformers, for one thing. Less line loss....