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Power Electronics → Power Semiconductor Devices → Switching Characteristics
Last updated 5 September 2026
Question The voltage (
v s v_s v s ) across and the current (
i s i_s i s ) through a semiconductor switch during a turn-ON transition are shown in figure. The energy dissipated during the turn-ON transition, in mJ, is
_______ .
Solution The energy dissipated during the turn-on transition is given by the integral of the instantaneous power:
E = ∫ v s ( t ) i s ( t ) d t E = \int v_s(t) i_s(t) dt E = ∫ v s ( t ) i s ( t ) d t . Based on the provided graphs, the transition can be divided into three distinct intervals:
1. Interval 1 (0 ≤ t ≤ 1 μ s 0 \le t \le 1 \mu s 0 ≤ t ≤ 1 μ s Voltage v s ( t ) = 600 v_s(t) = 600 v s ( t ) = 600 Current i s ( t ) i_s(t) i s ( t ) i s ( t ) = 50 10 − 6 t i_s(t) = \frac{50}{10^{-6}} t i s ( t ) = 1 0 − 6 50 t Energy E 1 = ∫ 0 10 − 6 600 ⋅ ( 50 × 10 6 t ) d t = 30000 × 10 6 ⋅ [ t 2 2 ] 0 10 − 6 = 15 E_1 = \int_0^{10^{-6}} 600 \cdot (50 \times 10^6 t) dt = 30000 \times 10^6 \cdot [\frac{t^2}{2}]_0^{10^{-6}} = 15 E 1 = ∫ 0 1 0 − 6 600 ⋅ ( 50 × 1 0 6 t ) d t = 30000 × 1 0 6 ⋅ [ 2 t 2 ] 0 1 0 − 6 = 15
2.
Interval 2 (1 μ s ≤ t ≤ 2 μ s 1 \mu s \le t \le 2 \mu s 1 μ s ≤ t ≤ 2 μ s
Voltage v s ( t ) = 600 v_s(t) = 600 v s ( t ) = 600 Current i s ( t ) i_s(t) i s ( t ) i s ( t ) = 50 + 50 × 10 6 ( t − 10 − 6 ) i_s(t) = 50 + 50 \times 10^6 (t - 10^{-6}) i s ( t ) = 50 + 50 × 1 0 6 ( t − 1 0 − 6 ) Energy E 2 = ∫ 10 − 6 2 × 10 − 6 600 ⋅ ( 50 + 50 × 10 6 ( t − 10 − 6 ) ) d t = 600 ⋅ [ Average Current ] ⋅ Duration E_2 = \int_{10^{-6}}^{2 \times 10^{-6}} 600 \cdot (50 + 50 \times 10^6 (t - 10^{-6})) dt = 600 \cdot [\text{Average Current}] \cdot \text{Duration} E 2 = ∫ 1 0 − 6 2 × 1 0 − 6 600 ⋅ ( 50 + 50 × 1 0 6 ( t − 1 0 − 6 )) d t = 600 ⋅ [ Average Current ] ⋅ Duration E 2 = 600 ⋅ ( 50 + 100 2 ) ⋅ 10 − 6 = 600 ⋅ 75 ⋅ 10 − 6 = 45 E_2 = 600 \cdot (\frac{50 + 100}{2}) \cdot 10^{-6} = 600 \cdot 75 \cdot 10^{-6} = 45 E 2 = 600 ⋅ ( 2 50 + 100 ) ⋅ 1 0 − 6 = 600 ⋅ 75 ⋅ 1 0 − 6 = 45
3.
Interval 3 (2 μ s ≤ t ≤ 2.5 μ s 2 \mu s \le t \le 2.5 \mu s 2 μ s ≤ t ≤ 2.5 μ s
Voltage v s ( t ) v_s(t) v s ( t ) 0.5 μ s 0.5 \mu s 0.5 μ s Current i s ( t ) = 100 i_s(t) = 100 i s ( t ) = 100 Energy E 3 = ∫ 0 0.5 × 10 − 6 600 ( 1 − t 0.5 × 10 − 6 ) ⋅ 100 d t = 100 ⋅ [ Average Voltage ] ⋅ Duration E_3 = \int_0^{0.5 \times 10^{-6}} 600(1 - \frac{t}{0.5 \times 10^{-6}}) \cdot 100 dt = 100 \cdot [\text{Average Voltage}] \cdot \text{Duration} E 3 = ∫ 0 0.5 × 1 0 − 6 600 ( 1 − 0.5 × 1 0 − 6 t ) ⋅ 100 d t = 100 ⋅ [ Average Voltage ] ⋅ Duration E 3 = 100 ⋅ ( 600 + 0 2 ) ⋅ 0.5 × 10 − 6 = 100 ⋅ 300 ⋅ 0.5 × 10 − 6 = 15 E_3 = 100 \cdot (\frac{600 + 0}{2}) \cdot 0.5 \times 10^{-6} = 100 \cdot 300 \cdot 0.5 \times 10^{-6} = 15 E 3 = 100 ⋅ ( 2 600 + 0 ) ⋅ 0.5 × 1 0 − 6 = 100 ⋅ 300 ⋅ 0.5 × 1 0 − 6 = 15
Total Energy Dissipated: E t o t a l = E 1 + E 2 + E 3 = 15 + 45 + 15 = 75 E_{total} = E_1 + E_2 + E_3 = 15 + 45 + 15 = 75 E t o t a l = E 1 + E 2 + E 3 = 15 + 45 + 15 = 75 mJ.
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Solution The energy dissipated during the turn-on transition is given by the integral of the instantaneous power:
E = ∫ v s ( t ) i s ( t ) d t E = \int v_s(t) i_s(t) dt E = ∫ v s ( t ) i s ( t ) d t . Based on the provided graphs, the transition can be divided into three distinct intervals:
1. Interval 1 (0 ≤ t ≤ 1 μ s 0 \le t \le 1 \mu s 0 ≤ t ≤ 1 μ s Voltage v s ( t ) = 600 v_s(t) = 600 v s ( t ) = 600 Current i s ( t ) i_s(t) i s ( t ) i s ( t ) = 50 10 − 6 t i_s(t) = \frac{50}{10^{-6}} t i s ( t ) = 1 0 − 6 50 t Energy E 1 = ∫ 0 10 − 6 600 ⋅ ( 50 × 10 6 t ) d t = 30000 × 10 6 ⋅ [ t 2 2 ] 0 10 − 6 = 15 E_1 = \int_0^{10^{-6}} 600 \cdot (50 \times 10^6 t) dt = 30000 \times 10^6 \cdot [\frac{t^2}{2}]_0^{10^{-6}} = 15 E 1 = ∫ 0 1 0 − 6 600 ⋅ ( 50 × 1 0 6 t ) d t = 30000 × 1 0 6 ⋅ [ 2 t 2 ] 0 1 0 − 6 = 15
2.
Interval 2 (1 μ s ≤ t ≤ 2 μ s 1 \mu s \le t \le 2 \mu s 1 μ s ≤ t ≤ 2 μ s
Voltage v s ( t ) = 600 v_s(t) = 600 v s ( t ) = 600 Current i s ( t ) i_s(t) i s ( t ) i s ( t ) = 50 + 50 × 10 6 ( t − 10 − 6 ) i_s(t) = 50 + 50 \times 10^6 (t - 10^{-6}) i s ( t ) = 50 + 50 × 1 0 6 ( t − 1 0 − 6 ) Energy E 2 = ∫ 10 − 6 2 × 10 − 6 600 ⋅ ( 50 + 50 × 10 6 ( t − 10 − 6 ) ) d t = 600 ⋅ [ Average Current ] ⋅ Duration E_2 = \int_{10^{-6}}^{2 \times 10^{-6}} 600 \cdot (50 + 50 \times 10^6 (t - 10^{-6})) dt = 600 \cdot [\text{Average Current}] \cdot \text{Duration} E 2 = ∫ 1 0 − 6 2 × 1 0 − 6 600 ⋅ ( 50 + 50 × 1 0 6 ( t − 1 0 − 6 )) d t = 600 ⋅ [ Average Current ] ⋅ Duration E 2 = 600 ⋅ ( 50 + 100 2 ) ⋅ 10 − 6 = 600 ⋅ 75 ⋅ 10 − 6 = 45 E_2 = 600 \cdot (\frac{50 + 100}{2}) \cdot 10^{-6} = 600 \cdot 75 \cdot 10^{-6} = 45 E 2 = 600 ⋅ ( 2 50 + 100 ) ⋅ 1 0 − 6 = 600 ⋅ 75 ⋅ 1 0 − 6 = 45
3.
Interval 3 (2 μ s ≤ t ≤ 2.5 μ s 2 \mu s \le t \le 2.5 \mu s 2 μ s ≤ t ≤ 2.5 μ s
Voltage v s ( t ) v_s(t) v s ( t ) 0.5 μ s 0.5 \mu s 0.5 μ s Current i s ( t ) = 100 i_s(t) = 100 i s ( t ) = 100 Energy E 3 = ∫ 0 0.5 × 10 − 6 600 ( 1 − t 0.5 × 10 − 6 ) ⋅ 100 d t = 100 ⋅ [ Average Voltage ] ⋅ Duration E_3 = \int_0^{0.5 \times 10^{-6}} 600(1 - \frac{t}{0.5 \times 10^{-6}}) \cdot 100 dt = 100 \cdot [\text{Average Voltage}] \cdot \text{Duration} E 3 = ∫ 0 0.5 × 1 0 − 6 600 ( 1 − 0.5 × 1 0 − 6 t ) ⋅ 100 d t = 100 ⋅ [ Average Voltage ] ⋅ Duration E 3 = 100 ⋅ ( 600 + 0 2 ) ⋅ 0.5 × 10 − 6 = 100 ⋅ 300 ⋅ 0.5 × 10 − 6 = 15 E_3 = 100 \cdot (\frac{600 + 0}{2}) \cdot 0.5 \times 10^{-6} = 100 \cdot 300 \cdot 0.5 \times 10^{-6} = 15 E 3 = 100 ⋅ ( 2 600 + 0 ) ⋅ 0.5 × 1 0 − 6 = 100 ⋅ 300 ⋅ 0.5 × 1 0 − 6 = 15
Total Energy Dissipated: E t o t a l = E 1 + E 2 + E 3 = 15 + 45 + 15 = 75 E_{total} = E_1 + E_2 + E_3 = 15 + 45 + 15 = 75 E t o t a l = E 1 + E 2 + E 3 = 15 + 45 + 15 = 75 mJ.
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