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Boost technology in a regulated mod; who can explain how it works?

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Baditude

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I've had several new vapers ask me how a regulated mod can generate more power than the battery in the mod has to give.

The correct answer is: it doesn't. It will even draw more power from the batteries than it puts out on the firing side.
So how does my single 4.2 volt battery allow me to put out 7 volts to the atomizer with my Provari? :blink:
 

untar

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how does my single 4.2 volt battery allow me to put out 7 volts to the atomizer with my Provari?
Voltage is not power! Voltage times current is power. The voltage is boosted with a magic contraption but that consumes current. In addition, some current will get lost to heating up components.

Let's say you're selecting your provari to fire 7 Volts on a 1.6Ω resistor. That would be 4.375 Amps and the power on the firing side would be 30.625 Watts (7*4.375).

Now let's take a look on the battery side. The mod needs to draw (at least) 30.625 Watts from your single 4.2V battery. 30.625W/4.2V = 7.3A, your mod will draw 7.3A from your battery to be able to deliver what you dialed in for the firing side (that's almost twice the current delivered to your coil).

In reality it will be more since some energy will get lost in the mod and your battery won't stay at 4.2V under load, both means even more power is needed than what I just calculated.

Edit: corrected a typo, not 2.0Ω but 1.6Ω :oops: now it makes more sense ;)
 
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zoiDman

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    (b) When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.

    Boost converter - Wikipedia
     

    Letitia

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    (b) When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.

    Boost converter - Wikipedia
    again:facepalm:
     

    stols001

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    No, it's all perfectly clear:

    en_dJfUQFftLY.png


    Have faith in the boost.... :)

    Anna
     

    zoiDman

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    Voltage is not power! Voltage times current is power. The voltage is boosted with a magic contraption but that consumes current. In addition, some current will get lost to heating up components.

    Let's say you're selecting your provari to fire 7 Volts on a 1.6Ω resistor. That would be 4.375 Amps and the power on the firing side would be 30.625 Watts (7*4.375).

    Now let's take a look on the battery side. The mod needs to draw (at least) 30.625 Watts from your single 4.2V battery. 30.625W/4.2V = 7.3A, your mod will draw 7.3A from your battery to be able to deliver what you dialed in for the firing side (that's almost twice the current delivered to your coil).

    In reality it will be more since some energy will get lost in the mod and your battery won't stay at 4.2V under load, both means even more power is needed than what I just calculated.

    And that All Jives with the Conservation of Energy.

    If Power = Voltage * Current then if I Increase Voltage, Current must Decrease to maintain the Equality.

    Same with the resulting Heat. Since it is a Form of Energy, if Heat is created, that Energy had to come from Somewhere. So something had to Decrease so that everything is the System remains in Equilibrium.

    Else we would have "Free Energy". And that Only happens in the Movies.
     

    Baditude

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    Voltage is not power! Voltage times current is power. The voltage is boosted with a magic contraption but that consumes current. In addition, some current will get lost to heating up components.

    Let's say you're selecting your provari to fire 7 Volts on a 1.6Ω resistor. That would be 4.375 Amps and the power on the firing side would be 30.625 Watts (7*4.375).

    Now let's take a look on the battery side. The mod needs to draw (at least) 30.625 Watts from your single 4.2V battery. 30.625W/4.2V = 7.3A, your mod will draw 7.3A from your battery to be able to deliver what you dialed in for the firing side (that's almost twice the current delivered to your coil).
    Thank you @untar . Your explanation has made the most sense to me so far. We lay people just have to accept that "magic contraption" is going to work its magic. ;)

    I think where many of us get confused (at least me) is that 4.2 volts on the battery wrapper. The really important specification is the amp rating. Mooch kind of alluded to this in one of his recent YT videos. When choosing batteries the primary spec should be the amp rating. The 3.7V or 4.2v merely states what kind of battery it is. The mAh rating provides a very rough estimate of how long the battery will function per charge, but there are so many variables that we can't rely upon that value.

    Voltage is potential energy by definition. Wattage is a calculation. Amperage is electrical current realized (work).

    The only problem is if we use an aftermarket battery with false amp ratings.
     
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    untar

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    Well, instead of "magic contraption" you could call it "deep dark secret of the universe", that's what I do when I don't understand something, it's more sciency-ish :D
    VV/VW Board Schematic...
    I think you're missing a dead possum in the middle of the pentagram. You're lucky, I think there's one in the shinyitis thread :evil:
     
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    Baditude

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    Speaking of magic contraptions, lighted candles, mystical spells, and Provari's. (You only need watch the first couple of minutes of the below video. The original music that was removed from the video was Gregorian chants. For full effect, you can play both videos simultaneously.)



     
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    zoiDman

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    ...

    I think you're missing a dead possum in the middle of the pentagram. You're lucky, I think there's one in the shinyitis thread :evil:

    I actually Omitted it on Purpose. Else these people might Descend on the ECF with Flaming Torches and Pitchforks.

    National Opossum Society
     

    sonicbomb

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    OK if we really want to get deep - all the stuff that 'science' knows about electricity is in a black box. We know what goes in, some magic happens inside the box, and stuff comes out.
    We cannot really understand what electricity is doing because (as yet) quantum phenomena down at the subatomic level are intrinsically unknowable.
    Thank goodness for maths and the scientific method, or we would still be bashing rocks together and probably still be smoking (thanks for nothing Jean Nicot).
     

    Coastal Cowboy

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    If someone could just draw a simple diagram and post that.
    Pretty sure a noob isn't going to understand a single previous post. I don't !.

    Although i do understand a thread like this could be started just for the fun of the outcome.

    I posted a thread about the potential for a "clockwork box mod" a while back. I kept a straight face and enjoyed playing along. But i'm still not entirely sure if readers realised the whole thing was a "wind up".

    Well i thought it was funny !. I still do.
    129090.jpg
    Dude... we are a bunch of tinkerers, hobbyists and ADHD excitables. If you ask us what day it is, we will commence to arguing over the best way to build a calendar.

    And for the love of all that is holy, don't take us fly fishing.
     

    Asbestos4004

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    OK if we really want to get deep - all the stuff that 'science' knows about electricity is in a black box. We know what goes in, some magic happens inside the box, and stuff comes out.
    We cannot really understand what electricity is doing because (as yet) quantum phenomena down at the subatomic level are intrinsically unknowable.
    Thank goodness for maths and the scientific method, or we would still be bashing rocks together and probably still be smoking (thanks for nothing Jean Nicot).
    You took the words right out of my mouth.
     

    Rossum

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    BTW - What Happens if I put Two 18650 Batteries in Series but One Battery is at 4.2 Volts and the Other Battery is at 3.6 Volts?

    giphy.gif


    Now seriously, that's a terrible idea. Under no-load conditions, you'll have 8.0V, but when you start pulling current from them, ugly things can happen. This is why series strings of batteries should always be balance-charged and ideally, the individual cell voltages should be monitored at all times.
     
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    Rossum

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    The correct answer is: it doesn't. It will even draw more power from the batteries than it puts out on the firing side.
    Right. Power Out equals Power In minus losses in the converter

    But Power equals Volts times Amps.

    Losses in a modern DC-DC converter tend to be in the 5-10% range.
     

    zoiDman

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    Now seriously, that's a terrible idea. Under no-load conditions, you'll have 8.0V, but when you start pulling current from them, ugly things can happen. This is why series strings of batteries should always be balance-charged and ideally, the individual cell voltages should be monitored at all times.

    Not Advocating Doing it.

    But the Point of Illustration is that being in Series isn't a so much a Doubling of Voltage. It is a Summation of Voltage.

    Which Ties into Post Number 64.
     
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