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DC Transient RL Circuit - Current Decay

  1. Oct 3, 2013 #1
    If I have a circuit with

    R = 1Ω
    L = 300μH
    V = 20V
    i0 = 5A

    I know that I can use the equation at the bottom of page 13 to calculate the current rise given any starting current and input voltage:

    i(t) = (V/R)[1-e-t/τ] + i0e-t/τ

    This is fine, and stops increasing at 20A as expected, but then how would I (if I need to) arrange the equation so that it works in a similar way for decay? Say I was starting at a current of 15A, with a driving voltage of 10V?

    I know that if I use:

    i(t) = (V/R)[e-t/τ]

    I will start decaying at 10A down to 0A, but I'm interested in being able to specify a negative voltage to drive the decay faster, with a variable starting current.

    Thanks in advance for any advice!

    Also, how would I work out when it is best to switch between building and decaying current? Would it just be when |V/R| > |i(t)| is less than 0, I build, and when |V/R| < |i(t)| I decay?

    Thanks again
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2
    I don't understand the problem. That equation appears to work regardless of "rise" or "decay", and regardless of what types of inputs you use (provided they are not dynamic).
  4. Oct 3, 2013 #3
    I have been using transient inputs, though every time my input changes, I have resent my t back to 0 and begun recounting, so the dynamic input shouldn't be a problem

    Compared to simulation from ANSYS Maxwell and from test data, this appears to be calculating results which decay much slower, though rise appears relatively similar
  5. Oct 3, 2013 #4
    The equation definitely doesn't have any differences between rise/fall times, it is inherently symmetric. The only thing I can think of is that perhaps you are calculating τ incorrectly? It should be τ = L/R.

    This equation will work fine for multiple step transients provided it has time to fully settle before applying the next one. If you are trying to calculate transients that are something other than step changes, you will want to solve the general differential equation.
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