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Mathmatical analysis of power transfer on resistive power line

Discussion in 'Electronic Design' started by Mook Johnson, Apr 24, 2007.

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  1. Mook Johnson

    Mook Johnson Guest

    I am trying to do an analysis but have hit a wall.

    Here is the circuit

    VSupply ---------/\/\/\/------------ Vout @ X watts load
    R

    The Vsupply is a power supply whose output voltage follows this equation.
    Vsupply = 500 + Iout * R

    Iout is the supply output current and R is the resistance between the source
    and load.
    R is between 200 and 500 ohms depending on the installation.

    The load is a switching power supply (costant power) that consumes 50 - 400W
    and wants a nominal input voltage of 500V.

    The intention is that as the power consumption goes up the input current
    goes up and the supply voltage goes up at the same time to maintian the
    supply input voltage at ~500V. These loads move slowly so there is plenty of
    time for the source to respond to load changes.

    I'd like to setup some equations to see what happens of the R value used in
    the surface equation is different from the actual resistance between source
    and load. How far off can they before the compensation causes an error of +
    or - 100V at the supply input.

    Seems like a straight forward problem but its got me twisted up somewhat.

    Can any of your guys with sharp math skills help me out.


    thanks
     
  2. Rearrange this to solve for Vout.
    According to the above equation, it doesn't matter what the
    resistance or R is.
    And according to the equation, gets it.
    100V=R*Iout is the limit on R if the maximum voltage
    allowance for R's drop is 100 volts.
    To me, also.
     
  3. Tim Wescott

    Tim Wescott Guest

    Think of it this way: Your Vsupply is just a Thevinin (sp?) voltage
    source, with Rs nominally equal to -R. So your voltage at the load is
    just Vl = Vs - Iout * (Rs + R). When Rs = -R, Vl = Vs - Iout * (R - R)
    = Vs.

    At this point it should be easy to vary Rs and R and try things out at
    maximum Iout.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/

    Do you need to implement control loops in software?
    "Applied Control Theory for Embedded Systems" gives you just what it says.
    See details at http://www.wescottdesign.com/actfes/actfes.html
     
  4. Mook Johnson

    Mook Johnson Guest

    100V=R*Iout is the limit on R if the maximum voltage allowance for R's

    You have to consider that the surface voltage is constantly adjusting as the
    load current goes up so the voltage across R can be greater than 100V.
    These are the numbers I get from Excel by sucessivie guestimation.

    I used a power level of 600 watts instead of 500 and checked R at sever
    spots between 100 and 500 ohms
    Calr R is the number used for calculating the source voltage (Vin) = 500 + I
    * Rcalc. Vo = Vin - actualR * I. I was set to 600W/500V for the nominal
    case, 600W/400V for the low case and 600W/600V for the high case at each
    resistance analyzed. The calc R was manually adjusted to give the Vo for
    that case.

    This tells me what I need to know but I'd like to understand the equations
    behind these guys.



    Vin current calc R actual R Vo Watts
    620 1.2 100 100 500 600
    549.875 1.5 33.25 100 399.875 599.8125
    700 1 200 100 600 600



    Vin current calc R actual R Vo Watts
    740 1.2 200 200 500 600
    699.875 1.5 133.25 200 399.875 599.8125
    800 1 300 200 600 600




    Vin current calc R actual R Vo Watts
    860 1.2 300 300 500 600
    850.025 1.5 233.35 300 400.025 600.0375
    900 1 400 300 600 600



    Vin current calc R actual R Vo Watts
    980 1.2 400 400 500 600
    999.875 1.5 333.25 400 399.875 599.8125
    1000 1 500 400 600 600



    Vin current calc R actual R Vo Watts
    1100 1.2 500 500 500 600
    1149.875 1.5 433.25 500 399.875 599.8125
    1100 1 600 500 600 600
     
  5. I can develop a Vout sum, but have difficulty in
    interpreting it. It looks like there are always
    two stable states? It goes like this........

    Vsupply = 500 + Iout.R1 where R1 is the theoretical
    value of the line resistance. So call R2 the actual
    value of the line resistance.

    Vout = 500 + Iout.R1 - Iout.R2.

    Also P = Vout.Iout, or Iout = P/Vout.

    Vout = 500 + R1.P/Vout - R2.P/Vout.

    Vout^2 = 500.Vout + P.(R1 - R2).

    Vout^2 - 500.Vout - P.(R1 - R2) = 0

    That's a quadratic equation, (a.x^2 + b.x + c) = 0,
    where in this case a= 1, b= -500 and c= -P.(R1 - R2).

    -b (+/-) sqrt(b^2 - 4ac)
    The roots of x = --------------------------
    2a


    500 (+/-) sqrt(500^2 + 4P(R1 - R2))
    Similarly, Vout = ----------------------------------
    2

    In theory, the max and min values should be plugged
    into the expresion to get a range for 4P(R1 - R2).

    But here's where I run into confusion.... it's that
    damned '(+/-)', which says that there are two possible
    answers for Vout for every value of P(R1 - R2).

    Are there two stable states for Vout? I don't know.
     
  6. That's a quadratic equation, (a.x^2 + b.x + c) = 0,

    Since b < 0, -b >0 so avoid the +/- by

    -b + sqrt(b^2 - 4ac)
    First root, x1 = --------------------------
    2a

    2c
    Second root, x2 = -------------------------
    -b + sqrt(b^2 - 4ac)



    Since a > 0 and c < 0, x1 is positive, x2 is negative.
    Now ask yourself, "Which one(s) make sense?" ....

    __________________________________________________________________________

    Now, having gone through all that ... Why is OP not using
    graphical analysis? This is a classic load-line plot.
    You can "what if" the load variations by just plotting
    a variety of load resistance slopes. Use Excel to model
    the source, make a plot, print it out -- then use a
    ruler to draw the various load lines in questions on it.
     
  7. whit3rd

    whit3rd Guest

    Yes, that's exactly right. You need the source to have
    negative resistance.

    Trouble is, all negative resistance schemes are unstable and
    prone to oscillation (in my experience). For a practical device,
    it's OK to identify the suitable range of input voltages (like, 400
    to 550VDC) for the distal device, and make multiple taps
    (or setpoints) for the source (like 500V, 600V, 700V), then
    do some digital magic to select a setpoint. There can be
    hysteresis in such a system, and that stops oscillation.

    In analog terms, latch a current-sense-value, compare it to
    the optimum for the setpoint that is now selected, and if
    it is different from optimum by more than 0.6 setpoint-steps,
    increment (or decrement) to another setpoint. After a few
    seconds have passed, and things have settled down
    repeat the cycle.


    This gives a transfer function with stepwise approximation to
    the negative resistance, but NO points on the transfer function
    curve actually have a local derivative that supports oscillation.
     
  8. Tim Wescott

    Tim Wescott Guest

    -- snip --

    AFAIK just about every mid-budget tape player in the world uses a driver
    with an effective source resistance equal to the negative of the capstan
    motor's drive resistance. Perhaps they're going to brushless motors ala
    CD players, but I wouldn't count on it. As a consequence of the
    negative drive the motor is very stiff, and they don't have to shell out
    for a speed controlled motor.

    But you are correct in being jaundiced -- I wouldn't want to try this
    trick in anything but a well-controlled environment, and oscillation on
    a 500V power line may be just a tad more serious than oscillation on a
    motor drive.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/

    Do you need to implement control loops in software?
    "Applied Control Theory for Embedded Systems" gives you just what it says.
    See details at http://www.wescottdesign.com/actfes/actfes.html
     
  9. Yes, that's exactly right. You need the source to have
    Not necessarily. Depends on the nature and value of the
    negative resistance and value of the load resistance.
    For example, Herbert J. Reich's "Functional Circuits
    and Oscillators" text discusses this in detail. There
    is some, but less material, on it in his "Theory and
    Application of Electron Tubes" which can be had in PDF
    form from Millet's website:
    http://www.pmillett.com/technical_books_online.htm

    Just about any negative feedback loop involving analog
    electronics will appear to have a negative resistance
    when viewed from an appropriately chosen pair of
    terminals. That's why there is such a literature on
    achieving loop stability for negative feedback loops.

    Most analyses will treat the issues in terms of feedback,
    rather than in terms of negative resistance. But that
    is just a matter of convention and what people are used to.
     
  10. whit3rd

    whit3rd Guest

    Yes, that's exactly right. You need the source to have
    negative resistance.

    Trouble is, all negative resistance schemes are unstable and
    prone to oscillation (in my experience). For a practical device,
    it's OK to identify the suitable range of input voltages (like, 400
    to 550VDC) for the distal device, and make multiple taps
    (or setpoints) for the source (like 500V, 600V, 700V), then
    do some digital magic to select a setpoint. There can be
    hysteresis in such a system, and that stops oscillation.

    In analog terms, latch a current-sense-value, compare it to
    the optimum for the setpoint that is now selected, and if
    it is different from optimum by more than 0.6 setpoint-steps,
    increment (or decrement) to another setpoint. After a few
    seconds have passed, and things have settled down
    repeat the cycle.


    This gives a transfer function with stepwise approximation to
    the negative resistance, but NO points on the transfer function
    curve actually have a local derivative that supports oscillation.
     
  11. Ok. Kevin G Rhoads in another post, (Load-line --), has
    stated that only the '+ sqrt..etc' root is reasonable.

    500 + sqrt(500^2 + 4P(R1 - R2))
    So Vout = ---------------------------------
    2

    Making... (2.Vout - 500)^2 = 500^2 + 4P(R1 - R2).

    This expression allows the, (400-600V), allowed Vout range
    to be plugged in, to see the allowed range of P(R1 - R2).

    Vout P(R1 - R2) P(R1 - R2)/400W
    ---- ---------- -----------------

    600 60000 +150 ohms
    500 0 0
    400 -40000 -100

    It is only the maximum power that produces appreciable
    swings in Vout. The OP's original requirement was 400W
    and this produces the example permitted R1-R2 ohm-range.

    Just as a reminder, R1 is the estimated average line
    resistance, used to set up the positive feedback on
    the bench, and R2 is the actual line resistance that
    is subsequently encountered in the field.

    The OP stated a field range of "200 to 500 ohms", so
    set R1= 350 ohms on the bench, and now vary R2.

    500 + sqrt(500^2 + 4.400(350 - R2))
    Vout = --------------------------------------
    2

    R2 (350 - R2) Vout
    -- ---------- ----

    500 -150 300 <-- Vout= Out of spec.
    450 -100 400 <-- Vout= In spec, as predicted.
    400 -50 456 "
    300 +50 537 "
    200 +150 600 "

    So it looks as though a 200-500 ohm line will not quite
    meet the 400-500V range required for Vout, at 400W.
     
  12. whit3rd

    whit3rd Guest

    Sweet; I didn't know that. It has also been used for woofers (linear
    motors, similar reasons), and I've toyed with it for inductive
    pickups (to make a more favorable L/R time constant).

    The application here, though, has a variable-load switching power
    supply connected through enough wire that one expects hundreds
    of ohms. 200 ohms of 20 gage copper is 6km of wire (3km transmission
    line). It isn't a simple matter of compensating a pure-resistor load
    with some negative resistance.
     
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