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I'm wondering how you calculate the time taken for the current in a charged inductor to reduce to zero with the only current path being through a capacitor. I'm looking at this from the perspective of an unloaded boost converter where at the end of the switching phase, the inductor is charged up to a certain current with the only current path being through a diode (neglected in my question) and into a capacitor (without an extra load).

Obviously the capacitor voltage will increase as the current through it is positive (the integral of the current will be non zero and positive) but how do I calculate the time for current to hit zero given a peak current in an inductor and the value of both inductance and capacitance, ignoring circuit losses?

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    \$\begingroup\$ Schematic? It worth 1000 words.. \$\endgroup\$
    – Eugene Sh.
    Commented Dec 31, 2015 at 18:17
  • \$\begingroup\$ Highly dependent on the L and C values. I think you're asking the bleed down time of the output of a DC - DC converter, correct? \$\endgroup\$
    – philbrooksjazz
    Commented Dec 31, 2015 at 19:15
  • \$\begingroup\$ An ideal LC will resonate forever without a resistive load. In a real circuit, damping is dependent on effective series resistance of L an C. So your question is impossible to answer. \$\endgroup\$
    – user57037
    Commented Dec 31, 2015 at 19:20
  • \$\begingroup\$ Can you just put a scope on the output and measure it? \$\endgroup\$
    – philbrooksjazz
    Commented Dec 31, 2015 at 19:33

1 Answer 1

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An idealized model during the discharge, when the output diode is on, is a simple LC circuit. The solution to that model for the voltage and current are the sine and cosine resulted from the 2nd order differential equation. so you can use the sine and cosine solution fitted with initial conditions for the calculation.

But in practice, for a boost circuit, you don't see people using the LC model for this calculation. The reason is, the output capacitor is specifically chosen so that the output voltage is near constant (usually less than 1% ripple).

It becomes much simpler with the output voltage (Vout) assumed to be constant: $$ L \frac{dI_L}{dt} = V_L = V_{in} - V_{out} = constant $$ Integrate and fit chosen initial conditions, you get a simple negative ramp: $$ I_L(t) = I_{Lpeak} - \frac{1}{L} (V_{out} - V_{in}) t $$ The typical way of using this relationship for boost circuit design is: $$ I_{Lpeak} = \frac{1}{L} (V_{out} - V_{in}) t_{discharge} $$ \$t_{discharge}\$ is the time it takes for the current \$I_{L}\$ to discharge from \$I_{Lpeak}\$ to 0.

When \$I_L\$ reaches 0, the output diode turns off. \$I_L\$ stays 0 and \$V_L\$ also drops to 0. (Therefore, there is no oscillation.)

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