ELEC_E8412
Exercise#5
Power Electronics
12.11.2020
Problem 1:
A buck converter has an input of 6 V and an output of 1.5 V. The load resistor
is 3 Ω, the switching frequency is 400 kHz, L = 5 µH, and C =10 µF. Determine
(a) the duty ratio
(b) the average and peak inductor currents
(c) the average source current,
(d) the peak and average diode current.
Solution:
Part (a):
𝐷=
𝑉𝑜
1.5
=
= 0.25
𝑉𝑖𝑛
6
Part (b):
𝐼𝑚𝑎𝑥 + 𝐼𝑚𝑖𝑛 𝑉𝑜 1.5
= =
= 0.5 𝐴
2
𝑅
3
1 1−𝐷
1
1 − 0.25
) = 1.5 ( +
) = 0.78 𝐴
= 𝑉𝑜 ( +
𝑅
2𝐿𝑓
3 2 ∗ 5 ∗ 10−6 ∗ 4 ∗ 105
⟨𝑖𝐿 ⟩ =
𝐼𝑚𝑎𝑥
Part (c):
⟨𝑖𝑖𝑛 ⟩ =?
𝑉𝑖𝑛 . ⟨𝑖𝑖𝑛 ⟩ =
𝑉𝑜2 (𝐷 ∗ 𝑉𝑖𝑛 )2
=
𝑅
𝑅
⟶
⟨𝑖𝑖𝑛 ⟩ =
𝐷2 𝑉𝑖𝑛 (0.25)2 ∗ 6
=
= 0.125 𝐴
𝑅
3
Part (d):
Fig 5-5 b
𝐼𝐷_𝑚𝑎𝑥 = 𝐼𝑚𝑎𝑥 = 0.78 𝐴
⟨𝑖𝐷 ⟩ = ⟨𝑖𝐿 ⟩ ∗ (1 − 𝐷) =
𝑉𝑜
1.5
∗ (1 − 𝐷) =
∗ (1 − 0.25) = 0.375 𝐴
𝑅
3
1
Problem 2:
A buck converter has an input of 50 V and an output of 25 V. The switching frequency is
100 kHz, and the output power to a load resistor is 125 W. Determine
(a) the duty ratio
(b) the value of inductance to limit the peak inductor current to 6.25 A.
(c) the minimum inductor current
Solution:
𝑅=
𝑉𝑜2
(25)2
=
=5𝛺
𝑃𝑜𝑢𝑡
125
Part (a):
𝐷=
𝑉𝑜
25
=
= 0.5
𝑉𝑖𝑛 50
Part (b):
1 1−𝐷
1
1 − 0.5
) = 25 ( +
)
𝐼𝑚𝑎𝑥 = 𝑉𝑜 ( +
𝑅
2𝐿𝑓
5 2 ∗ 𝐿 ∗ 105
𝐼𝑚𝑎𝑥 = 6.25 𝐴
⟶ 𝐿 = 50 µ𝐻
Part (c):
1 1−𝐷
1
1 − 0.5
) = 25 ( −
) = 3.75 𝐴
𝐼𝑚𝑖𝑛 = 𝑉𝑜 ( −
𝑅
2𝐿𝑓
5 2 ∗ 50 ∗ 10−6 ∗ 105
Problem 3:
A boost converter has the following parameters:
Vin = 5 V, Vout = 20 V, and Pout = 40 W
The minimum value of the inductor current must be at least 80% of the average inductor
current. The switching frequency is 85 kHz. Determine the duty ratio and the minimum
inductor value.
2
Solution:
𝐼𝑚𝑖𝑛 > 0.8 ⟨𝑖𝐿 ⟩ ⟶
0.2
⟨𝑖𝐿 ⟩ −
∆𝑖𝐿
> 0.8 ⟨𝑖𝐿 ⟩
2
⟶ 0.2⟨𝑖𝐿 ⟩ >
∆𝑖𝐿
2
𝑉𝑖𝑛
𝑉𝑖𝑛 𝐷𝑇
>
2
(1 − 𝐷) 𝑅
2𝐿
𝐷(1 − 𝐷)2 𝑅
𝐿>
0.4𝑓
D=?
R=?
𝑉𝑜
20
1
=
=
→ 𝐷 = 0.75
𝑉𝑖𝑛
5
1−𝐷
𝑃𝑜𝑢𝑡
𝑉𝑜2 (20)2
= 40 =
=
→ 𝑅 = 10𝛺
𝑅
𝑅
𝐿 >13.79 µH
Problem 4:
A boost converter has parameter Vs =20 V, D =0.6, R =12.5 Ω, L=10 µH, C =40 µF, and
the switching frequency is 200 kHz. Determine
(a) the output voltage
(b) the average, maximum, and minimum inductor currents.
(c) the average current in the diode
Assume ideal components.
Solution:
Part (a):
𝑉𝑜 = 𝑉𝑖𝑛 ∗
1
1
= 20 ∗
= 50 𝑉
1−𝐷
1 − 0.6
Part (b):
⟨𝑖𝐿 ⟩ =
𝐼𝑚𝑎𝑥 + 𝐼𝑚𝑖𝑛
𝑉𝑖𝑛
20
=
=
= 10 𝐴
2
(1 − 𝐷)2 𝑅 (1 − 0.6)2 ∗ 12.5
3
𝐼𝑚𝑎𝑥 =
𝑉𝑖𝑛
𝑉𝑖𝑛
20
20 ∗ 0.6
+
𝐷𝑇
=
+
= 13 𝐴
(1 − 𝐷)2 𝑅 2𝐿
(1 − 0.6)2 ∗ 12.5 2 ∗ 10 ∗ 10−6 ∗ 2 ∗ 105
𝐼𝑚𝑖𝑛 =
𝑉𝑖𝑛
𝑉𝑖𝑛
20
20 ∗ 0.6
−
𝐷𝑇
=
−
=7𝐴
(1 − 𝐷)2 𝑅 2𝐿
(1 − 0.6)2 ∗ 12.5 2 ∗ 10 ∗ 10−6 ∗ 2 ∗ 105
Part (c):
⟨𝑖𝐷 ⟩ = ⟨𝑖𝐿 ⟩ ∗ (1 − 𝐷) = 10 ∗ (1 − 0.6) = 4 𝐴
𝑜𝑟
⟨𝑖𝐷 ⟩ =
𝑉𝑜
50
=
=4𝐴
𝑅 12.5
Problem 5:
A buck-boost converter has the following parameters:
Vin = 24 V, D= 0.65, R = 7.5 Ω, L = 50 μH, C = 200 μF, and switching frequency = 100
KHz. Determine
(a) the output voltage,
(b) the average, maximum, and minimum inductor currents.
Solution:
Part (a):
𝑉𝑜
𝐷
=
, 𝐷 = 0.65, 𝑉𝑖𝑛 = 24 𝑉
𝑉𝑖𝑛 1 − 𝐷
→ 𝑉𝑜 = 44.571 𝑉
Part (b):
⟨𝑖𝐿 ⟩ =
𝑉𝑖𝑛 𝐷
24 ∗ 0.65
=
= 16.98 𝐴
(1 − 𝐷)2 𝑅 (1 − 0.65)2 ∗ 7.5
𝑉𝑖𝑛 𝐷𝑇
24 ∗ 0.65
=
= 3.12 𝐴
𝐿
50 ∗ 10−6 ∗ 105
∆𝑖𝐿
3.12
𝐼𝑚𝑎𝑥 = ⟨𝑖𝐿 ⟩ +
= 16.98 +
= 18.54 𝐴
2
2
∆𝑖𝐿
3.12
𝐼𝑚𝑖𝑛 = ⟨𝑖𝐿 ⟩ −
= 16.98 −
= 15.42 𝐴
2
2
𝛥𝑖𝐿 =
4
Problem 6:
A buck-boost converter has parameters Vs=12 V, D=0.6, R=10 Ω, L =10 µH, C =20 µF,
and a switching frequency of 200 kHz. Determine
(a) the output voltage
(b) the average, maximum, and minimum inductor currents, and
(c) the average value of input current.
Solution:
Part (a):
𝑉𝑜 = 𝑉𝑖𝑛 ∗
𝐷
0.6
= 12 ∗
= 18 𝑉
1−𝐷
1 − 0.6
Part (b):
⟨𝑖𝐿 ⟩ =
𝑉𝑖𝑛 𝐷
12 ∗ 0.6
=
= 4.5 𝐴
2
(1 − 𝐷) 𝑅 (1 − 0.6)2 ∗ 10
𝛥𝑖𝐿 =
𝑉𝑖𝑛 𝐷𝑇
12 ∗ 0.6
=
= 3.6 𝐴
𝐿
10 ∗ 10−6 ∗ 2 ∗ 105
𝐼𝑚𝑎𝑥 = ⟨𝑖𝐿 ⟩ +
∆𝑖𝐿
3.6
= 4.5 +
= 6.3 𝐴
2
2
𝐼𝑚𝑖𝑛 = ⟨𝑖𝐿 ⟩ −
∆𝑖𝐿
3.6
= 4.5 −
= 2.7 𝐴
2
2
Part (c):
⟨𝑖𝑖𝑛 ⟩ =?
𝑉𝑖𝑛 . ⟨𝑖𝑖𝑛 ⟩ =
𝑉𝑜2 (𝐷 ∗ 𝑉𝑖𝑛 )2
=
(1 − 𝐷)2 𝑅
𝑅
⟶ ⟨𝑖𝑖𝑛 ⟩ =
5
𝐷2 ∗ 𝑉𝑖𝑛
(0.6)2 ∗ 12
=
= 2.7 𝐴
(1 − 𝐷)2 𝑅 (1 − 0.6)2 ∗ 10