So how does my single 4.2 volt battery allow me to put out 7 volts to the atomizer with my Provari?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?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?
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.how does my single 4.2 volt battery allow me to put out 7 volts to the atomizer with my Provari?
again(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
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.
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.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).
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 threadVV/VW Board Schematic...
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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
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.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.
You took the words right out of my mouth.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).
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?
Right. Power Out equals Power In minus losses in the converterThe correct answer is: it doesn't. It will even draw more power from the batteries than it puts out on the firing side.
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.