- #1

LADransfield

- 7

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If I have a circuit with

R = 1Ω

L = 300μH

V = 20V

i

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

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

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!

R = 1Ω

L = 300μH

V = 20V

i

_{0}= 5AI 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}] + i_{0}e^{-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!

**edit**

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 againAlso, 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: