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

[zoiDman]

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.

BTW - If I have a Single Battery in a Modern VV/VW mod, and it's charge is say 4.0. And I set my mod for an output of 6.0 Volts.

Is the Board somewhere running 4.0 Volts and 2.0 Volts (Not withstanding Efficiency) in series to make a 6.0 Volt Output?

Or is the Board effectively Doubling the Input Voltage to 8.0 Volts, and then Reducing it to 6.0 Volts via Buck?

 

[Rossum]

BTW - If I have a Single Battery in a Modern VV/VW Mod, and it's charge is say 4.0. And I set my Mod for an output of 6.0 Volts.

Is the Boost portion of the Board somewhere running 4.0 Volts and 2.0 Volts (Not withstanding Efficiency) in series to make a 6.0 Volt Output?

Or is the Board effectively Doubling the Input Voltage to 8.0 Volts, and then Reducing it to 6.0 Volts via Buck?

I'm pretty confident modern boards use a single-stage converter. So forget about the latter notion.

However, don't see "series" as the right way to look at a boost converter either.

 

[zoiDman]

I'm pretty confident modern boards use a single-stage converter. So forget about the latter notion.

However, don't see "series" as the right way to look at a boost converter either.

Gotcha.

I'll be Honest with you. I understand about 80% of this.

The Induction or Field Theory isn't the Problem. What I Don't get is How the Range of Voltages can be achieved.

 

[Rossum]

What I Don't get is How the Range of Voltages can be achieved.

By varying the duty cycle on the switch using a feedback loop that monitors the output voltage.

 

[ScottP]

The reason I started this thread is because 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. My stock answer has been either boost converters or magical unicorns, and don't ask me how either one works.

OH You want to know what to tell others, we could have left all the scientific mumbo jumbo out of it. Just tell them there is a chip inside that is able to pull extra current (amps) from the battery and use that to increase the voltage.

If you need more explain with water and a hose. If you think of water moving through a hose, the water is like the current (Amps) and the Pressure is the Voltage and the spout is the battery. The chip would be like adding a vacuum pump to the spout to pull even more water out and force it through the hose. This would increase the pressure in the hose much like the voltage is increased in the circuit.

 

[zoiDman]

By varying the duty cycle on the switch using a feedback loop that monitors the output voltage.

So the Output Voltage is all dependent on the Duty Cycle of the Switch.

OK. That gives me something to Think About.

 

[Rossum]

So the Output Voltage is all dependent on the Duty Cycle of the Switch.

OK. That gives me something to Think About.

Output voltage is controlled by the duty cycle of the switch, but even at a fixed voltage, the duty cycle will vary tremendously based on the load applied at the output.

 

[zoiDman]

Output voltage is controlled by the duty cycle of the switch, but even at a fixed voltage, the duty cycle will vary tremendously based on the load applied at the output.

I think I'm starting to get a Better Grasp on what is Going On.

Watching this Helped...



Thank you for Posting it untar.

 

[Letitia]

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

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

He's not dead yet, work in progress.

 

[Bunnykiller]

No. They use a potentiometer (aka variable resistor) to adjust the voltage going to the light. There is no modulation at all. There is no need to make a complex circuit when all you are doing is reducing voltage (dimming). When the resistance is adjusted (by knob or slider) to 0 Ohms the light is at full brightness. As the resistance is increased the light gets dimmer and dimmer as the voltage it receives get less and less.

and where does the 200W resistor fit in that tiny little box???

 

[Bunnykiller]

Seems like chopped up waveform would cause light bulbs to fail prematurely as well. I mean if you think about it, most of the time a household light fails is when you turn it on. So turning it off and on faster, just seems like a bad plan.

actually, a incandecent bulb turns on and off 120X ( there are 2 peaks in a 60Hz waveform) a second and gets a voltage reversal thru it every 1/60th of a second

 

[Rossum]

actually, a incandecent bulb turns on and off 120X ( there are 2 peaks in a 60Hz waveform) a second and gets a voltage reversal thru it every 1/60th of a second

No, the bulb doesn't turn on and off 120 times a second. The filament's temperature can't change that fast. Yes, there is some brightening and dimming at that rate, which you can see if you use a high speed camera with a sufficient frame rate, but the dips in brightness don't come close to approaching "off" nor does the filament's temperature fall to anywhere near ambient.

 

[Bunnykiller]

And the winner is....



I truly appreciate everyone's input, but being that I have little to no knowledge of electricity. "I flip switch, light comes on. I flip it again, light goes off." That's about the extent that I know.

No one was really able to dumb it down for me. I looked at the wiki page @Rossum suggested. Like @evan le'garde remarked, unless one already has an understanding of electrical science then it is just a big chemistry formula to me. And based upon the varied discussion that followed throughout this thread, maybe there is no simple answer.

The reason I started this thread is because 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. My stock answer has been either boost converters or magical unicorns, and don't ask me how either one works.

terminology... the key word is power... the mod cannot produce more power than the batteries can supply... with the PWM and inductors/caps it can produce more voltage than the batteries can provide thru the board via a transformer effect.
As you increase the output voltage above the batteries voltage, the amps available drop porportionately... so this in effect keeps the power equasion balanced...

lets try this...

you have a mech at 3.7V with a 30A battery ( 3.7X30=111W) not counting resistance
and you place that battery in a regulated and turn up the voltage to 5V

therefore you end up with 111W/5V= 22.2A worth of power, the higher the voltage the less amps available on the output side...

now this is where the volts and amps working in tandem make a coil heat up or not... resistance and wire mass are the key factors here... 2 coils of the same resistance but of different mass can either blow up ( pop) or just barely get warm.

 

[Bunnykiller]

No, the bulb doesn't turn on and off 120 times a second. The filament's temperature can't change that fast. Yes, there is some brightening and dimming at that rate, which you can see if you use a high speed camera with a sufficient frame rate, but the dips in brightness don't come close to approaching "off" nor does the filament's temperature fall to anywhere near ambient.

technically, yes it does turn on and off... when the voltage drops to 0 during the sine wave it is off... it may still be glowing, but there is no voltage flow during that moment.

 

[Rossum]

technically, yes it does turn on and off... when the voltage drops to 0 during the sine wave it is off... it may still be glowing, but there is no voltage flow during that moment.

How long does the voltage & current stay at zero?

In the end, it's not at all the same as turn the light bulb off, not in terms of light output, nor in terms of thermal stress on the filament.

 

[Bunnykiller]

So how does my single 4.2 volt battery allow me to put out 7 volts to the atomizer with my Provari?

via a transformer.
when a source of electricity is pulsed thru a coil of wire it creates an electromagnetic field that can flow thru the core ( ferro metallic) the wire is wrapped around, the other side of the core also was wire wrapped on it to recieve that EM pulse, depending on the number of wraps of both coils, the voltage can be increased or decreased. the 2 coils are called primary and secondary. Primary is usually the one with the electricity supplied to it and the secondary is the one that supplies the changed voltage. For voltage boost, the primary has thicker wire and less turns and the secondary has more turns and thinner wire. The ratio of turns determine the voltage change

for example... a primary has 120 turns of 12Ga wire and the secondary has 1200 turns of 30Ga... this is a 1 to 10 ratio therefore the voltage exiting the transformer will be 10X the input voltage... input 1 volt and you get 10V... input 120V and you get 1200V
this is normally AC voltage... but a pulsed DC will provide the same to a degree... but as I mentioned, the amperage will change with the voltage increase/decrease. This example will change the amperage by a factor of 10 too...
with the 1V at 1A supply will produce 10V @ 1/10th of an amp

 

[Rossum]

via a transformer.

Please look at the model circuits in previous posts or the Wikipedia articles. There is no transformer. A transformer has a minimum of three connections (for an autotransformer, where the primary and secondary share a portion of the same coil), but more typically a transformer has four or more connections and two electrically isolated coils. All we have here is an inductor; a single coil with two connections, and its magnetic core does not conduct power between two coils or even sections of a single coil. Instead, the inductor in a boost converter stores energy while the switch is closed and releases it when the switch is opened.

 

[sonicbomb]

The reason I started this thread is because 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 simple answer is the battery is not providing more power (watts), the regulator is just altering the balance of amps and volts on the atomizer side of the regulator.

Additional answer selection:
1. I don't know, and I don't want to know
2. You don't want to know (refer them to this thread)
3. No one knows, just press the button and vape it.
4. If you are not happy for a chip to calculate your vape for you, use a mechanical device. Cripes, are you people never happy?

 

[Rossum]

Additional answer selection:
1. I don't know, and I don't want to know
2. You don't want to know (refer them to this thread)
3. No one knows, just press the button and vape it.
4. If you are not happy for a chip to calculate your vape for you, use a mechanical device. Cripes, are you people never happy?

1. I do know. I can see the model circuit in my head and understand how it works. In another life, a couple of decades ago, I designed power controls and transformers, but I can't explain it more eloquently than the articles I've linked to.
2. An essential prerequisite is some understanding of the behavior of an inductor in a switched circuit.
3. The only thing that amazes me is how much this stuff has shrunk in size and increased in efficiency since I changed directions in my career over 20 years ago.
4. Despite knowing, I am happiest using a mechanical device.

 

[DaveP]

Voltage across the ecig coil is what produces a current based on the resistive load of the coil.

A variable ecig circuit is changing that voltage as you turn the wattage up and down. Wattage ( power) is a calculation base on voltage and resistance (power = E^2/R (voltage squared/resistance).

Current is calculated using voltage / resistance. Current (Amperage) = E/R

The power value you read on a mod's display is arrived at using the combination of voltage across the coil's resistance. It's just another way to refer to heat in the coil at a given resistance and voltage.

A variable wattage ecig varies the voltage across the coil using a circuit similar to the one in Zoidman's video in an earlier post. The faster the VV circuit pulses the voltage, the more often the capacitor charges and discharges. That happens thousands of times per second. When you adjust wattage on your ecig mod, you are adjusting the frequency (number of times per second) that voltage is applied to the coil. The more often it's pulsed the hotter the coil gets (more on time).