Electronic – Ideal inductor AC current LTspice

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.)

What I think you're seeing is the start of an exponential where the capacitor is charging. The time constant is 22s and the pulse duration is very small compared with this, so you're only seeing the early stages of the exponential.

In charging a capacitor there is a flow of power in, but the resultant energy is stored and not dissipated.

I don't know why there is no discharge curve; it may be worth decreasing the pulse frequency significantly so that we can see what's happening in the time frame of the time constant. The integration sampling time may also be a factor.