GATE 2020 EE Q.20: solution and answer

GATE 2020 · EEQ.20 · 1 mark · NAT

A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode, D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. $L_{p a r}$ is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1, 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.

GATE 2020 EE question 20: Left: plot of diode current (vertical axis) versus time (horizontal axis). — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

ANSWER

The IGBT experiences its highest current stress at point 3, where the diode's reverse recovery current is at its peak.

Official answer: 3Source: Official answer key, GATE 2020 EE (IIT Delhi)

Let $I_{l o a d}$ be the nearly constant current of the inductive load (the load time constant is much longer than the switching interval) and let $i_{D}$ be the diode current, taken positive in the forward (freewheeling) direction. The IGBT, the diode branch and the load branch meet at one node, so by KCL $i_{I GB T} = I_{l o a d} - i_{D} .$
Before the IGBT turns on, the diode freewheels the full load current, so $i_{D} = I_{l o a d}$ and $i_{I GB T} = 0$ . When the IGBT turns on, the voltage across $L_{p a r}$ is about $V_{so u r ce}$ , so the diode current falls at $\frac{d i _{D}}{d t} = - \frac{V _{so u r ce}}{L _{p a r}}$ (point 1 to point 2). The IGBT current rises as $i_{D}$ falls.
At point 2 the diode current is zero, so $i_{I GB T} = I_{l o a d}$ . After point 2 the diode current is negative. Stored charge is being swept out of the diode, and the IGBT supplies this reverse current in addition to the load current: $i_{I GB T} = I_{l o a d} + ∣ i_{D} ∣$ . It keeps increasing.
At point 3 the diode current is at its most negative value, the peak reverse recovery current $I_{r r, ma x}$ . There $i_{I GB T} = I_{l o a d} + I_{r r, ma x},$ which is the largest value of $i_{I GB T}$ over the whole interval.
After point 3 the diode current rises toward zero (point 4), so $∣ i_{D} ∣$ falls and $i_{I GB T}$ falls back to $I_{l o a d}$ . The IGBT current stress is therefore highest at point 3, and the answer is $3$ .

KEY IDEA

At the IGBT turn-on, the IGBT current equals the load current minus the diode current. The IGBT current is largest where the diode current is most negative, which is the peak of the reverse recovery current.

WATCH OUT FOR

Picking point 2, where the diode current crosses zero, or point 1, where the diode current starts to fall. Neither is the peak of the reverse recovery current. The IGBT must also carry the diode's reverse current, so its maximum is at the most negative diode current.

From the GATE 2020 EE paper. Every question of this paper · Download the full paper · All 9 Switches questions, with notes

GATE 2020 EE Q.20: solution and answer

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