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: Track where the trapped inductor current goes.
Just after $t_1$, the MOSFET opens and stops conducting, but $L$ still wants to push its stored current somewhere. That current is caught by the diode $D$ and routed through $R_f$.
Step 2: Write the spike voltage.
The extra voltage riding on top of $V_{DC}$ right after turn off is approximately $I_L R_f$, so a bigger $R_f$ directly means a bigger spike.
Step 3: Write the decay behavior.
The current in that freewheeling loop dies off with time constant $L/R_f$. A bigger $R_f$ shortens this time constant, so the curve returns to $V_{DC}$ quicker; a smaller $R_f$ lingers longer above $V_{DC}$.
Step 4: Match this to the picture.
The tallest, fastest-decaying curve is labeled $R_f=R_1$, the middle one is $R_f=R_3$, and the shortest, slowest one is $R_f=R_2$.
Step 5: Rank the resistors.
Bigger peak means bigger resistor, so the order from largest to smallest resistance is
\[ R_1>R_3>R_2 \]
\[ \boxed{R_1>R_3>R_2} \]
