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

Discussion in 'General Vaping Discussion' started by Baditude, Mar 3, 2018.

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  1. stols001

    stols001 Mistress of the Dark Nicotinic Arts Verified Member ECF Veteran

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    May 30, 2017
    Tucson, AZ
    I'm still here.... Hoping for this.... It's becoming clearer....

    [​IMG]

    Who makes up these magic the gathering quotes, I want to know? "Water shifts and confuses but as ice it holds the stillness of truth." I checked, and "Ariel the whisperer" does not appear to be a real person, although she may be eloquent.

    I think I'm getting "You can drop your boost circuit into snow or ice, but don't run it under running water."

    I knew I was getting closer....

    Anna
     
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  2. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

    Supporting member
    Apr 16, 2010
    So-Cal
    This is where I was having the Mental Block.

    Because for Some Reason I had it in my Head that the Frequency of Switch Opened/Closed was Fixed. So I had to come up with Ingenious ways for a VV Output Range to be achieved while still maintaining a Flat DC signal.

    But once Russom said that the Frequency of the Switch could be changed, I went back and looked at the Video that untar posted and it All Made Sense.

    :)
     
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  3. Eskie

    Eskie ECF Guru Verified Member ECF Veteran

    May 6, 2016
    NY
    See? This is why the loss of Heathkit in favor of products assembled from inexpensive mass produced ICs has devastated our technological lead over the Soviet Union and left us vulnerable to the communist threat.

    Oh wait. I don't think they're around anymore. Now it's just hacking credit cards that excites our kids instead of learning how to use a soldering iron.
     
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  4. DaveP

    DaveP PV Master & Musician ECF Veteran

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    May 22, 2010
    Central GA
    Something like this is what you could expect from bargain ecig mod circuits, especially the cheaper ones that use less expensive circuitry. The more you pay, the more technology you can expect (hopefully). The circuit below shows the output waveform with and without the smoothing capacitor.

    This one isn't digital technology. It's just an AC driven circuit using a full wave rectifier circuit that's smoothed and clipped using a filter capacitor. Digital circuits could produce a true square wave right out of the box. The width of the square wave pulse can be controlled through clock speed to raise and lower the effective voltage that the coil sees (the on-off pulse train duty cycle).

    [​IMG]
     
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  5. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    In Calculus terms of, it is Instantaneous.

    BTW - I though I read once that for "120 AC", the Peak Voltage is 170. And that the term "120 Volts" is a Geometric Average of the Area under the Curve.

    Is that True?
     
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  6. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    One Step at a Time Dave.

    I'm still Basking in the Glow of finally Understanding a 1st Semester Electronics problem.

    ;)
     
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  7. DaveP

    DaveP PV Master & Musician ECF Veteran

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    May 22, 2010
    Central GA
    Heath and Knight kits are what helped me solidify my life career ambitions. I was hooked the first time I turned on a kit built from boxed parts. That led to electronics school, computer science, and a life long career.
     
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  8. ScottP

    ScottP Vaping Master Verified Member ECF Veteran

    Apr 9, 2013
    Houston, TX
    I actually still have a Heathkit. Haven't touched it since the early 90's.
     
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  9. Rossum

    Rossum Surly Curmudgeon Verified Member ECF Veteran

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    Dec 14, 2013
    NE FL
    Exactly! In other words, the time for which it's actually OFF is effectively nil.

    Yes. The peak voltage is around 170. The nominal 120V is the RMS value, which coincides with the DC voltage that you would need to use to get the equivalent results in a resistive load like an incandescent bulb.
     
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  10. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    Thanks Rossum.

    Hey BTW, I got thinking More about that Boost Circuit switching On/Off. And realized that not Only does it have to alter the Duty Cycle depending on what Output the user wants. But also due to the Charge State the battery currently has.

    ie: When it is freshly charged at 4.2 Volts verses when it has be discharged down to say 3.4 Volts. And everything in between.
     
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  11. Rossum

    Rossum Surly Curmudgeon Verified Member ECF Veteran

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    Dec 14, 2013
    NE FL
    Or as the voltage sags under load. We tend to ignore the fact that the battery voltages we talk about are no-load voltages that a battery cannot deliver it an actual load. Look at some of Mooch's older pulse tests and you'll see this is quite a substantial factor. Moreover, the amount of sag in a 5 second pulse is not steady either.
     
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  12. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    A Very Good Point.

    It is Easy sometimes to View things from a Static Perspective. And not to see them as a Dynamic Process.
     
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  13. DaveP

    DaveP PV Master & Musician ECF Veteran

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    May 22, 2010
    Central GA
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  14. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    Yeah... That is why I asked Rossum about the Peak Voltage.

    From your Link...

    A periodic sinusoidal voltage is constant and can be defined as V(t) = Vmax*cos(ωt) with a period of T. Then we can calculate the root-mean-square (rms) value of a sinusoidal voltage (V(t)) as:

    rms6.gif



    Integrating through with limits taken from 0 to 360 or “T”, the period gives:

    rms7.gif


    Where: Vm is the peak or maximum value of the waveform. Dividing through further as ω = 2π/T, the complex equation above eventually reduces down too:

    RMS Voltage Equation

    rms8.gif
     
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  15. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    BTW...

    .7071 is the cos(45 degrees)

    It is Also the sin(45 degrees)

    Since 1, 1, 2^1/2 are the Sides of a 45, 45, 90 Triangle.
     
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  16. DaveP

    DaveP PV Master & Musician ECF Veteran

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    May 22, 2010
    Central GA
    In my 1970s electronic school they just told us to use .707 for the constant and gave us a baseline explanation illustrated with graphics as to how and why it works. Rocket scientists take it much further. :)
     
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  17. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

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    Apr 16, 2010
    So-Cal
    Yeah... "Seventy Seventy One" (or ∏/4 when in Radians) comes up a Lot when doing 2D Spatial Math.

    BTW - This is a Good RMS overview which shows the Half-Angle/Double Angle Identity needed and the implied u/du Substitution needed to do that AC RMS Integration.

    https://www.raeng.org.uk/publications/other/8-rms

    Also has a few Interesting Examples of how RMS Value can have some other usages.
     
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  18. Rossum

    Rossum Surly Curmudgeon Verified Member ECF Veteran

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    Dec 14, 2013
    NE FL
    Provided the voltage being measured is a nice, clean sine wave, that's usually true. But cheap meters generally do a poor job if the voltage is anything else, e.g. the signal coming out of a typical lamp dimmer:

    [​IMG]

    If the meter isn't spec'd as being a "True RMS" meter, you won't get useful readings if you try to measure a signal like that.
     
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  19. DaveP

    DaveP PV Master & Musician ECF Veteran

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    May 22, 2010
    Central GA
    If we're designing Voyager 3 the math has to be precise. For most other practical purposes, .707 works every time to approximate the DC equivalent of a sinusoidal waveform..
     
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  20. Rossum

    Rossum Surly Curmudgeon Verified Member ECF Veteran

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    Dec 14, 2013
    NE FL
    Sure, for a clean sinusoidal waveform. But use a 'scope to have a look at the "modified sine wave" that you get on the output of a reasonably priced UPS when it's running on batteries, or the chopped sine wave from a lamp dimmer, or the output of a ferroresonant "constant voltage" transformer. In all of these cases (and many others) you cannot just measure the peak voltage and multiply by 0.707 because doing so will get you terribly inaccurate results.
     
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