Voltage across inductor is somehow negative despite diodes?

[microdosemishief]

Homework Statement

Trying to make sense of yellow line in simulation graph

Relevant Equations

v = Ldi/dt

I tried to simulate a circuit from MIT Power Electronics Lecture 2, and this is the simulation when it has reached periodic steady state.
I added ground, original circuit didn't have one.

Yellow line - v(vy, vz) - should be voltage across inductor L2.

I know that the negative yellow line corresponds to when current is decreasing, and I know average voltage across L2 is 0 to match ideal inductor characteristics.

But intuitively why is the inductor voltage like this? Is the little bit of current through the diodes Roff significant here? How is the inductor able to have negative voltage with positive current?

 

[Baluncore]

Homework Statement: Trying to make sense of yellow line in simulation graph
Relevant Equations: v = Ldi/dt

How is the inductor able to have negative voltage with positive current?

v = L⋅ di/dt
As the current rises and falls, di/dt changes sign.
What does that do to the voltage across the inductor?

While V2 is positive, current flow increases through D3, L2 and R2.
When V2 goes negative, current through D3 turns off, so the inductor conducts through D2 while the L2 current flow is reducing.

There is no reservoir capacitor in that circuit, instead there is a reservoir inductor to keep the output current flowing while D3 is off.

Ground is needed by SPICE simulations as internal voltages must be relative to something.