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How many steps in a variac

Discussion in 'Electronic Design' started by Gene, Feb 10, 2009.

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  1. Gene

    Gene Guest

    For a lab bench variac that has a dial connected to a brush that
    slides across the winding, does anyone know how many winding/taps
    there typically are? Or another way of putting it, what is the typical
    voltage step between windings? Say there are 100 windings, meaning
    about 1.2 V per winding (ignoring for the moment that it might have
    taps for output voltages > input voltage). Would it really go down to
    1.2 volts (very first tap/winding)? I'm interested in voltage control
    at low voltages.

    Gene
     
  2. Thats simply not true(for the most part). It will take ignore any windings
    after the tap. You are essentially shorting out the windings with the brush
    so how could any of them after the first matter? (they do matter in that
    there is non-zero resistance but 99% of the current will go through the
    first winding.

    It imagine if you think about it for a second you'll realize your error.

    |
    |
    -----
    (think of it as a variable resistor if you like)

    The bush contacts several windings. Any windings after the closes one
    contacted will be ignored(again, for the most part). Simply the past of
    least resistance in action. (the windings are not in parallel.

    similarly you have something like

    |
    |
    +----+----+
    | | |

    The current will ignore R3,R4, and R5 assumine the wire thickness is much
    lower than the resistor/winding resistance.

    i.e., it is identical to

    |
    |
    +----+
    |

    I imagine it might be possible for the windings to share the current do some
    degree because of the contact resistance(specially if lots of current is
    going through it causing it to heat up locally). Obviously the contact(in
    my diagram) forms a voltage divider at the points.

    If I took lessons from the assholes on here I'd bitch you out for being
    wrong... luckily I'm not that much of an asshole and my life doesn't totally
    suck. After all, I could be wrong(something people like Phil never think
    of... even if they did it they would assume they are correct).


    In any case I think if the OP wants to get better resolution he'll need to
    use some other method such as another transformer or add some load.
     
  3. Phil Allison

    Phil Allison Guest

    "Gene" <>


    ** Google Groper Alert !

    ** Connect an AC volt meter to the output and find out for yourself -
    dickhead.

    BTW

    Only an utter moron use a variac to derive a low voltage AC supply.

    Cost it is stupidly dangerous.



    ...... Phil
     
  4. Benj

    Benj Guest

     
  5. Benj

    Benj Guest

    I sure as hell hope YOU aren't giving anyone any lessons! First off,
    you might take a close enough look at a variac to notice that it's a
    TRANSFORMER and not a bunch of resistors. A shorted turn on a
    transformer has a very different effect from shorting out a couple of
    resistors on a voltage divider. The only reason you don't get smoke,
    heat and melted parts because of the bridging brush is because of the
    resistance of the carbon and the quite low voltage generated by any
    given single turn. The BRUSH is what forms the "voltage divider"
    because it is the resistive material. And the device give you
    resolution rather than steps because of the GEOMETRY of the brush to
    copper wires. Think about it asshole.
    A low voltage transformer IS the way to get low voltage output from a
    variac. Not only are you wasting a lot of resolution by only running
    the variac on low settings, but also they are [usually] wound with the
    same gauge wire over the length of the turns. Thus if you are trying
    to get a ratio between the line voltage and the [low voltage] output
    you will be using the same gauge wire for both windings. This make the
    wire gauge for the low voltage section too light and output current is
    thus limited by that deficiency compared to your usual two winding low
    voltage transformer where the low voltage secondary is wound with very
    heavy wire.

    Plus, since a Variac is an autotransformer, there is no isolation from
    the power line which can be a safety factor in some situations. A
    secondary low voltage transformer solves all these problems as others
    have suggested.
     
  6. Greegor

    Greegor Guest

    As a sideline issue, the comment about the wiper arm
    shorting one winding intrigued me.

    That one winding would produce low voltage and
    high current, does that have a history of causing failure?

    Cost is dangerous? Please explain.

    If you've got a variac that outputs the range
    0-line voltage and you output to the primary
    on a 12v transformer you get very fine
    control of the range 0-12 v ac.

    With a 6v transformer instead you get very
    fine control for the range 0-6 v ac.

    How is this so dangerous, aside from cost?
     
  7. Phil Allison

    Phil Allison Guest

    "Greegor"

    As a sideline issue, the comment about the wiper arm
    shorting one winding intrigued me.


    ** There is no hard short.

    See the other replies.


    That one winding would produce low voltage and
    high current,

    ** There is no hard short.

    See the other replies.


    does that have a history of causing failure?


    ** There is no hard short.

    See the other replies.

    Cost is dangerous? Please explain.


    ** Good safety practice is to *never connect* a load to a variac that CANNOT
    handle the FULL output voltage - connecting a load that can only handle a
    tiny fraction of that voltage ( like 1%) is plain NUTS !!!

    PLUS a variac provides NO isolation from the AC supply - so a missing or
    reverse connected neutral conductor means that both connections to the
    variac are at a dangerous ( ie full line ) voltage.


    If you've got a variac that outputs the range
    0-line voltage and you output to the primary
    on a 12v transformer.....

    ** That was never the OP's Q nor relevant to my comment.

    Please learn to read before posting here again and wasting people's time.

    Dick wad.



    ....... Phil
     
  8. Asshole uh? Why can't I say people are wrong? Why can everyone else? Why is
    it ok for Phil, Nobody, Charles, and others like yourself able to reply to a
    post without reading it and say it's wrong and bitch and call people
    assholes(which you seem to be good at) without reading the post correctly?
    Why is there a double standard that allows morons to get away with such
    things?

    Just like what you have done!! I did not say anything incorrect in my post
    yet you try to blast it out of the sky. Also, in your own "blasting" you
    make numerous logical mistakes. One is that you assume you can't replace
    resistance by impedence! This is basic electronics but your allowed to get
    it wrong?

    Cannot a variac be modeled by

    Contact
    +------+--....--+-------+
    | | | |
    | | | |
    R1 R2 ... Rn ... R(n+1)
    | | | |
    V-->---Zi---+--Z1--+--... --+--Zn---+---Zf


    ?? Surely if you thought about it more you would see it is correct and that
    what I have said in my original post is completely correct. (I never said
    the contact had any resistance.

    Note also that I put this little caveat:

    "The current will ignore R3,R4, and R5 assumine the wire thickness is much
    lower than the resistor/winding resistance." (the wire thickness is the
    contact resistance!!! But I guess all those brain cells you have couldn't
    figure that out)

    Which you completely ignore(as expected).

    You people are awefully arrogant. It's quite entertaining. I hope I didn't
    get your panties in a bunch?

    Trust me... there's plenty more where this came from! I'll keep it up as
    long as you people keep up your shit. As long as you guys can read posts
    half-ass and respond by calling people assholes and make worse logical
    mistakes than you assume the other person made I'll call you out on it. (I
    was hoping phil would be the first to respond but I guess he's still trying
    type up his own paper)

    Test: FIND ONE logical mistake I made in the post... not spelling or grammer
    or typo's. If you can do that then I'll submit that you are correct. I've
    already pointed out one of your very serious errors in understand
    electronics.

    Now, as usual we are both right(and that almost always is the case when
    someone is calling someone wrong). Your fallacy is you simply fail to read
    the entire post and when you misunderstand something you have to show your
    ego. (again, find anything I said as being wrong)

    Note I say

    "I imagine it might be possible for the windings to share the current do
    some
    degree because of the contact resistance"

    Which in all your glory you have completely missed and it says exactly what
    you have said. So you have said I am wrong yet I said what you have just
    said... that means your wrong?

    The only thing I'm wrong about is expecting more from a bunch of second
    class humans.

    Of course I'll expect to be hearing from you soon. Completely ignore
    everything I have said and call me wrong and an asshole some more. I guess
    that is what makes your pathetic life interesting?
     
  9. Why do people make aggressive, authoritative, and totally idiotic
    Now that's something I would really like to know!!! I would give my let nut
    for that answer!
     
  10. Phil Allison

    Phil Allison Guest

    "Palinurus"

    ** Is that set the dial to a "fraction of a volt" OR set the AC voltage
    output within such a fraction ??

    ** So the latter ??

    Pointless - as the AC supply does not remain so steady for more than a few
    seconds.



    ...... Phil
     
  11. huh? A turn generates power and dissipates power? As far as I know a winding
    doesn't generate any power.

    The electric field is converted into a magnetic field which is then
    converted back into an electric field at a different intensity.

    http://en.wikipedia.org/wiki/Autotransformer

    So you disagree with wiki? Maybe you should rewrite the page?

    Look at the first figure... do you see that big squiggly line? Do you know
    that is called an inductor? Do you see the taps?

    Do you realize you can draw it as a series of inductors where connected by
    wires andt he taps taken at the wires?

    i.e.

    ----/\/\/\---tap1---/\/\/\/\---tap2---\/\/\/\/\---

    ??? Surely you do get that much?


    Do you realiez that one can conceptualize it as resistors? Maybe this is the
    leap of faith you have? Luckily good basic run of the mill impedence comes
    to the rescue:

    http://en.wikipedia.org/wiki/Electrical_impedance

    OMG!!!!

    "just as impedance extends Ohm's law to cover AC circuits, other results
    from DC circuit analysis such as voltage division, current division,
    Thevenin's theorem, and Norton's theorem, can also be extended to AC
    circuits by replacing resistance with impedance."

    Do you understand that? The autotranformer is just one big voltage divider.
    abstractly at least... it does this a bit different than a resistive network
    but it works out to be the same. After all, we were talking about the
    current instead of the voltage reduction. For resistors the voltage is
    reduced by the drop in potential across a resistive element. For the
    inductors the voltage is converted into a magnetic field but then converted
    back into an electric field. Different principles at work but same abstract
    analysis. (This is because the current through the windings is exactly
    that... regardless of what the inductor is doing to limit the current.


    In fact since it seems to escape you guys:


    V---L1--A--L2---gnd

    What is the voltage at point A? Is it L1*L2/(L1+L2)?? Assuming the inductors
    have the same construction so there inductance per turn is the same then it
    works!!! Again, wiki proves this!


    A = V*L1*L2/(L1 + L2)

    If L1 = L2 then we have

    A = V/2*L!!!!

    Which is exactly which one expects!

    What is that? Voltage divider law? Same principle holds for resistors too,
    and capacitors? right? Get it now? (Note L really is X_L = jwL but in this
    case the magnitudes are identical except for the factor of w!!!!!! Or just
    think of the symbol L as a shorthand for X_L.. nifty huh?)

    I know it's hard but think!!! Or you can just jump on the band wagon... I
    hear Phil "FUCKWIT" Allison is taking applications. (You just have to go
    around calling people Fuckwits and saying they are wrong and prove it by
    changing the hypothesis. Benji is doing a good job of that and I think
    larkin is taking notes.)



    Waiting on Benji to reply and call me an asshole again. This time he
    probably won't even read a tenth of the post.
     
  12. J.A. Legris

    J.A. Legris Guest

    Slaughter murders yet another analysis!

    Here's a puzzle for you:

    Apply your inductive voltage-divider "theory" to an autotransformer
    set anywhere else except at the halfway point - say, at the 1/3 point.
    Then the inductance of the short side is 1/4 of the long side (recall
    that L is proportional to the number of turns SQUARED). Therefore the
    output voltage should be 1/5 not 1/3. Where did you go wrong?
     
  13. Greegor

    Greegor Guest

    On Feb 10, 12:37 am, John Larkin wrote
    Thanks.
     
  14. Rich Grise

    Rich Grise Guest

    When was the last time you set a variac on your bench and actually
    MEASURED the output?

    Thanks,
    Rich
     
  15. Tim Williams

    Tim Williams Guest

    So anyway...

    The number of steps is either the number of turns, N or double, 2N
    (give or take the width of the brush and how the ends are handled).
    Which one depends on the width of the brush; if it spans an integral
    number of turns, it will move off one turn just as it moves onto the
    next, so that the number of contacts is constant and you get N steps.
    If it spans a half-integral number of turns, however, it will move off
    one, then move onto the next, and so on, so that the number of
    contacts alternates between, say, 2 and 3, then you get 2N, double the
    resolution.

    Width of the brush is a continuous variable, so it's worth observing
    what happens at non-half-integral widths: in this case, the phase
    between moving-onto-a-wire and moving-off-a-wire changes as something
    like (Width MOD Spacing) / Spacing.

    Tim
     
  16. Guest

     
  17. Rich Grise

    Rich Grise Guest

    Maybe he means a do-over, like a let serve. ;-)

    Cheers!
    Rich
     
  18. Your right... lets figure it out then, shall we(as if your going to
    following along)?

    Since the point was to find a law relating the voltages we have

    V1 = L1*I1' + M*I2'
    V2 = L2*I2' + M*I1'

    Now I1 = I2 + I3 where I3 is the current into the tap: Z2 = 2nd inductor
    impedence, Z3 = load impedence. The current divder law gives

    I2 = Z3/(Z2 + Z3)*I1 = Z*I1 (Z is unitless)
    I2' = Z*I1' + Z'*I1

    or

    V1 = (L1 + M*Z)*I1' + M*Z'*I1
    V2 = (L2*Z + M)*I1' + L2*Z'*I1

    Canceling the I1's gives

    L2*V1 - M*V2 = [L2*(L1 + M*Z) - M*(L2 + M)]*I1'

    = [L1*L2 + M*Z*L2 - M*L2 - M^2]*I1'

    With me so far?

    We know that M = k*sqrt(L1*L2)

    so

    L2*V1 - M*V2 = [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'

    Now substituting V1 = V - V2 we have

    L2*V - L2*V2 - M*V2 = L2*V - (L2 + M)*V2 = ...

    Solving for (L2 + M)*V2 gives


    (L2 + M)*V2 = L2*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'

    Solving for V2 gives

    V2 = L2/(L2 + M)*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'/(L2 + M)

    V2 = Tau*V - Gamma*I1'


    Note if Gamma is 0 or practically 0 then We have a voltage divider effect.
    I.e., V2 ~= Tau*V. Tau depending on L1 and L2 and there ratio sets tau(it is
    not linear necessarily but we'll investigate that later).

    The main thing is to determine Gamma:

    There are 4 cases

    (Ideal transformer => k = 1, High load => Z = 0)

    For an ideal transformer k = 1 so we have

    Gamma = M*L2/(L2 + M)

    M = sqrt(L1*L2)

    or

    Gamma = 1/(1/L2 + 1/sqrt(L1*L2))

    Note if L2 or L1 is small then Gamma ~= 0. Only for large inductance do we
    get a gamma that is large. (note this doesn't necessarily mean the voltage
    divider effect out of play as we will see shortly)

    Even for 1H inductances we get Gamma = 1/2

    But then

    V2 = Tau*V - I1/2

    But unless Tau*V is compariable to the current being drawn it doesn't have
    much of an effect. Take V = 110V and tau = 0.25 and drawing 3A. Which is
    approximately 5% off the ideal case.

    For your run of the mill transformers obviously L1 and L2 are much much
    smaller and obviously the effect will be insiginificant. (although for very
    low voltages and high current's the effect is much stronger)

    (Ideal transformer => k = 1, Low load => Z = 1)

    Obviously with Z = 1 we at most double the value of I1.


    For the Ideal transformer and practical configurations, while theoretically
    you are correct, for all practical purposes the mutual inductance has no
    effect.



    (Completely Non-ideal transformer => k = 0, High load => Z = 0)

    In this case we have and additional factor of L1*L2 to boost I1. But again,
    for all practical cases it is very small. (M = 0 in this case and the other
    term gets knocked out)


    (Completely Non-ideal transformer => k = 0, Low load => Z = 1)

    Again, this just doubles the effect which is still quite small.


    --------------------

    Conclusion


    While you are theoretically correct about the mutual inductance, it just
    doesn't have any effect on the voltage dividing effect in practical terms(at
    least in all but the most extreme cases).

    So we are left with simply only with V2 = Tau*V

    Tau is the voltage dividing effect. It is set by choosing the ratio of L1
    and L2.

    Tau = L2/(L2 + M)

    If L2 = 0 then Tau = 0 and this is exactly what we expect

    If M = 0 ==> Either no mutual inductance or L1 = 0 then Tau = 1 and this is
    exactly what we expect.


    The curve is nonlinear

    BUT NOTE

    What does L2/(L2 + M) look like??

    How bout the resistive divider law!!!!!!!!!

    Note this is with a load. Without a load the analysis is much easier... This
    is exactly what I said in the beginning.

    Any more bright ideas or will you finally give in?
     
  19. Slaughter murders yet another analysis!

    Here's a puzzle for you:

    Apply your inductive voltage-divider "theory" to an autotransformer
    set anywhere else except at the halfway point - say, at the 1/3 point.
    Then the inductance of the short side is 1/4 of the long side (recall
    that L is proportional to the number of turns SQUARED). Therefore the
    output voltage should be 1/5 not 1/3. Where did you go wrong?

    --

    Actually you went wrong wtih your red herring.

    Where exactly did I talk about the contactor position? This is something you
    created in your mind.

    Just as a voltage divider we get

    V2 = R1/(R1 + R2)*V

    It is not linear.

    You assume way to much to make yourself look intelligent but in the eyes of
    god you are just another imbecile.
     
  20. JosephKK

    JosephKK Guest

    That is the most unbelievably way wrong answer i have seen in a long
    time. Are you trying to out stupid AlwaysWrong?

    You may read, you may ask questions, you may not post answers. STFU.
     
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