Consider the circuit shown in Figure (a). A gate pulse \(v_g\) is applied between time instants \(t_0\) and \(t_1\). After \(t_1\), during the MOSFET turn OFF process, it experiences a voltage overshoot.
Based on the \(v_{ds}\) waveforms shown in Figure (b), which one of the following options is correct?
Show Hint
A bigger snubber resistance produces a bigger instantaneous IR spike but decays faster, so rank the resistors directly by the height of each peak in the figure.
Step 1: Understand what happens at MOSFET turn OFF.
While the MOSFET is ON, the inductor \(L\) carries a steady current. The instant the MOSFET turns OFF, that inductor current cannot vanish immediately, so it is forced to keep flowing through the freewheeling diode \(D\) in series with the snubber resistor \(R_f\).
Step 2: Relate the resistor value to the size of the voltage spike.
The moment the current diverts into the \(R_f\text{-}D\) branch, the voltage across \(R_f\) jumps to roughly \(I_L\times R_f\), and this adds on top of \(V_{DC}\) to produce the overshoot seen on \(v_{ds}\). A larger \(R_f\) forces a larger instantaneous voltage drop for the same trapped current, so it produces a taller peak.
Step 3: Relate the resistor value to the decay rate.
After the initial spike, the trapped current decays with a time constant \(L/R_f\). A larger \(R_f\) gives a smaller time constant, so the curve settles back down to \(V_{DC}\) faster; a smaller \(R_f\) decays more slowly.
Step 4: Read the ranking of peaks from the figure.
The curve marked "with \(R_f=R_1\)" has the tallest peak, the curve for \(R_f=R_3\) has the next tallest peak, and the curve for \(R_f=R_2\) has the smallest peak, sitting closest to \(V_{DC}\).
Step 5: Convert the peak-height ranking into a resistance ranking.
Since a bigger \(R_f\) always means a bigger peak, the order of peaks directly gives the order of resistances:
\[
R_1>R_3>R_2
\]
Step 6: Final conclusion.
\[
\boxed{R_1>R_3>R_2}
\]
Hence the correct option is (B).
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