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How to calculate the peak inductor current?

Discussion in 'Electronic Basics' started by Mike, Nov 8, 2005.

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  1. Mike

    Mike Guest

    Could someone clairify these equations to calculate the peak current
    reached in an inductor when a capacitor is discharged into it
    The equations can be seen at
    http://home.san.rr.com/nessengr/techdata/rlc/rlc.html

    In particular I cannot make out the character that is part of the
    exponent of e in the i(t) equations for all 3 cases of damping. It
    looks like a t, but I'm not sure. Maybe e^((-R/2L)*t) ?
    Also what does the small symbol that looks like a sinewave over the
    omega symbol mean that is in the i(t) equation for the overdamped
    case?

    Thanks
    Mike
     
  2. Yes. t, for time in seconds. These are expressions for some value in
    terms of time.
    It is defined as a function of R, L and C in the previous equation.
    It is the sine component of the decay, even though the real
    exponential dominates the result, because it decays so much in a small
    part of a cycle.
     
  3. The Phantom

    The Phantom Guest

    Looks to me like ~wo = wo * SQRT(-1)
    Wouldn't it be better to call it the "sinh" component?
     
  4. (snip)

    Yes. Thanks.
     
  5. Mike

    Mike Guest

    Thanks, for the clarifications.
    Is there a reason why there is no equation to calculate the peak
    current for the overdamped case?

    Thanks
    Mike
     
  6. I can't see one, except that it is messy.

    I solved for the peak current (with lots of help from Mathcad). I'm
    posting the formula as a graphic on A.B.S.E.

    But it varies little. The extreme values are Vo/R as L approaches
    zero, and about .736 times that for the just barely over damped case.
     
  7. The Phantom

    The Phantom Guest

    The equation for the peak current in the underdamped case is hard to
    read, and I'm not sure it's correct anyway. Can anybody make out the
    expression, and does it work?

    All you have to do is have an expression for the time of the first
    current peak and substitute it in the expression for the current as a
    function of time.

    The correct equations are:

    Underdamped case

    (set your calculator to radians mode, of course)

    Time to first current peak is tp = (1/wo)Arctan(2*L*wo/R)

    Magnitude of peak current = (Vo/L/wo)*Exp(-R*tp/2/L)*Sin(wo*tp)

    Overdamped case

    Time to only current peak is tp = (1/~wo)Arctanh(2*L*~wo/R)

    Magnitude of peak current = (Vo/L/~wo)*Exp(-R*tp/2/L)*Sinh(~wo*tp)

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

    For the critically damped case, his schematic shows a 2 ohm resistor, but
    his text says it's 20 ohms.

    A good test of these expressions is to set the resistor value to 1.99
    ohms (barely underdamped) and compute the peak current. Then set the
    resistor to 2.01 ohms (barely overdamped) and compute the peak current.
    The two peak currents should be nearly the same. I get 3.691096 and
    3.66657 amps for the two cases.
     
  8. I read it as pi.
    I tested the current versus time for the under damped and critical
    cases, with the inductance nearly the same, and got 4 times higher
    current for the critically damped case.

    Could a multiplier of 4 be missing from the underdamped case?
     
  9. The Phantom

    The Phantom Guest

    So does the complete expression look like:

    Ipeak = (Vo/wo/L)*Exp(-R*Pi/(L*wo)) ?

    When I use the values L = 1 uH, C = 1 uF, R = .2 ohms, I get a peak
    current of 5.3448 amps from this expression. But if you look at the
    graphic from the simulation on the web page, the current peak is obviously
    up around 8 amps. I get a peak current of 8.626 amps using the expressions
    given below.
    For this test, you should leave the inductance at 1 uH and vary the
    resistance.
    How did you test? Using the web page expressions? Using a simulator?
    Did you use the L and C from the web page, and if so, what value of R did
    you use for the underdamped case? Did you try the overdamped case too?
    Did you try the values of R that I suggested? And if you did, what peak
    currents did you get for the various cases?

    The expressions for the critically damped case are really simple:

    Time to current peak is tp = 2*L/R

    Magnitude of peak current is Exp(-1)*Vo*SQRT(C/L)

    Using the web page values of L = 1 uH, C = 1 uF and R=2, I get a peak
    current of 3.67879 amps.

    The expression on the web page for the peak current in the critically
    damped case is just the same as above, but with 2/R substituted for
    SQRT(C/L), since the critically damped case is obtained when SQRT(C/L)=2/R.

    The peak currents when R = 1.99 and when R = 2.01 must be very close to
    the 3.67879 amps of the critically damped case.
     
  10. That is what I think I am seeing.
    And Vo = 10 volts, I assume.
    The graphs are so poor that I have been able to make little sense of them.

    I also get a peak value of 8.626 at t=1.478, found by solving for when
    the derivative i(t) = 0 and applying that time to the i(t) formula,
    but 5.345 from the i peak formula. At least one of them is wrong.

    If I solve produce a formula for peak current, based on the i(t)
    formula, I get:

    ipeak=(Vo/(L*w0))*exp((-R/(2*L^2*w0))*atan(2*L^2*w/R))*...

    sin(atan(2*L^2*w)/R)

    and this gives a value of 8.626, which just verifies that I did the
    differentiation and simplification right, since the i(t) it was based
    on looks like it has the same peak value. It doesn't prove that i(t)
    was right to start with.

    (snip)
    Yes. I used values of Vo=1, C=1, R=1 and L=R^2*C/4 and a very
    slightly higher value (1.0001 times) for L, for the under damped case.
    Using Mathcad to evaluate the given expressions for i(t).
    I experimented with the over damped case, earlier, (deriving the i
    peak formula I pasted to A.B.S.E) to solve for i peak) and verified
    that i peak based on the the given over damped i(t) converged to i
    peak given for the critical damped case as critical damping was
    approached from the over damped side.

    So I think the over damped i(t) formula is at least consistent with
    the critical i peak formula. But I haven't yet derived the critical i
    peak formula from the critical damped i(t) formula.
    I've lost track.
    So does this lead you to suspect the formula for i(t) for the under
    damped case is in error?
     
  11. Mike

    Mike Guest

    I found this online simulator
    http://www.oz.net/~coilgun/mark2/rlcsim.htm and it seems to agree
    pretty well with the above equations.
    Mike
     
  12. The Phantom

    The Phantom Guest

    Yes, indeed. You get the same numerical value from his web page formula
    as I do for the peak current in the underdamped case, and I think this
    formula, at least as I am able to read it, is wrong.

    The formulas I posted mostly came from an application note published by
    Sprague concerning their energy storage capacitors.
    ------------------------------------------------------------------------
    I also solved the problem by writing an expression for the current as:

    I(s) = (Vo/L)/(s^2 + s*R/L + 1/L/C)

    and looking up the appropriate Laplace transform. The result is:

    i(t) = (Vo/L/(a-b))*(Exp(a*t)-Exp(b*t))]

    where a and b are the roots of the denominator of the I(s) expression.

    This can be whipped into the form of the expressions for the under and
    overdamped cases from the Sprague app note that I posted yesterday. The
    critically damped case requires taking limits or explicitly making the
    roots equal and then transforming.

    tpeak can be found as: tp = (Ln(a/b) - 2*Pi*j*n)/(b-a), j=SQRT(-1)

    What is neat about that expression is that if you have a calculator or PC
    program that can deal with complex numbers (including taking logs and
    exponentials of complex arguments), you succesively set the variable n to
    0,1,2,... and this expression will give the times for the successive peaks
    (positive and negative) in the underdamped case; and with n=0, it will give
    the peak time for the overdamped case. Then you just substitute those
    times back into the expression for i(t) and you get the peak currents. So
    you can get numerical results without having to manipulate the expression
    for i(t) and possibly making a mistake in the algebra.

    I've done all this and verified that the formulas from the Sprague app
    note that I posted yesterday are correct (or made minor corrections as
    necessary). The numerical results that you have gotten so far are the same
    as mine, and I'm confident the formulas I posted yesterday are correct.
     
  13. I just posted another version of the formula for the peak current in
    the over damped series RLC case on A.B.S.E. Here is a tip of the hat
    to John Woodgate for the suggestions.
     
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