Voltage across inductor is somehow negative despite diodes?

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SUMMARY

The discussion analyzes the voltage behavior across inductor L2 in a SPICE simulation of a circuit from MIT Power Electronics Lecture 2. The voltage across L2, represented by the yellow line v(vy, vz), becomes negative during periods when the current through the inductor is decreasing, consistent with the inductor voltage equation v = L⋅di/dt. The simulation includes diodes D2 and D3, where current flow shifts between them depending on the polarity of V2, and the absence of a reservoir capacitor is compensated by the reservoir inductor maintaining output current. Ground reference was added for SPICE simulation requirements, and the negative voltage is explained by the inductor's inherent opposition to changes in current, analogous to mechanical inertia.

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microdosemishief
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Homework Statement
Trying to make sense of yellow line in simulation graph
Relevant Equations
v = Ldi/dt
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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?
 
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microdosemishief said:
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.
 
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An inductor has a property of inertia, analogous to a mass. So when we try to alter the conditions, it will oppose the change. If we try to reduce the current in the inductor, it will create a forward voltage to try and keep the current flowing.
You ask how an inductor can have a negative voltage when the current is "positive", i.e. forward. This will happen if the current is trying to increase, as the inductor tries to oppose the change. If you consider any component, such as a resistor, in a circuit, the voltage drop across it is opposing the current flow. In this way, all the voltages around a circuit add up to zero.
 

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