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In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.
Correct answer is '24'. Can you explain this answer?
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In a DC-DC boost converter, the duty ratio is controlled to regulate t...
Po = 120w, Vs = 24V, Vo = 48V
α = 0.5 [Duty cycle]
Po = VoIo = 120
VSIS = VoIo
At boundary of continuous & discontinuous
α = 0.5 [Duty cycle]
Po = VoIo = 120
VSIS = VoIo
At boundary of continuous & discontinuous
Most Upvoted Answer
In a DC-DC boost converter, the duty ratio is controlled to regulate t...
In a DC-DC boost converter, the value of the boost inductor (in Henrys) can be calculated using the following formula:
L = (V_out * (1 - D)) / (f * ΔI_L)
Where:
L = Inductance value (in Henrys)
V_out = Output voltage (in volts)
D = Duty ratio (unitless)
f = Switching frequency (in Hz)
ΔI_L = Inductor ripple current (in Amperes)
Given:
V_out = 48 V
Duty ratio (D) = ?
f = 50 kHz
P_out = 120 W
To find the duty ratio (D), we can use the formula:
P_out = (V_out * V_in * D) / (1 - D)
Rearranging the formula to solve for D:
D = (P_out / (V_out * V_in + P_out))
Given V_in = 24 V, we can substitute the values:
D = (120 W / (48 V * 24 V + 120 W))
D = 0.2
Now, we can substitute the values of V_out, D, f, and ΔI_L into the formula for L:
L = (48 V * (1 - 0.2)) / (50 kHz * ΔI_L)
Since the converter operates at the boundary between continuous and discontinuous conduction modes, the inductor ripple current (ΔI_L) is equal to half of the average inductor current (ΔI_L = I_L / 2).
The average inductor current (I_L) can be calculated using the formula:
I_L = P_out / V_out
Substituting the values:
I_L = 120 W / 48 V
I_L = 2.5 A
Now we can calculate the inductor ripple current:
ΔI_L = I_L / 2
ΔI_L = 2.5 A / 2
ΔI_L = 1.25 A
Finally, we can substitute the values into the formula for L:
L = (48 V * (1 - 0.2)) / (50 kHz * 1.25 A)
L = 0.768 H (approximately)
Therefore, the value of the boost inductor in this DC-DC boost converter is approximately 0.768 Henrys.
L = (V_out * (1 - D)) / (f * ΔI_L)
Where:
L = Inductance value (in Henrys)
V_out = Output voltage (in volts)
D = Duty ratio (unitless)
f = Switching frequency (in Hz)
ΔI_L = Inductor ripple current (in Amperes)
Given:
V_out = 48 V
Duty ratio (D) = ?
f = 50 kHz
P_out = 120 W
To find the duty ratio (D), we can use the formula:
P_out = (V_out * V_in * D) / (1 - D)
Rearranging the formula to solve for D:
D = (P_out / (V_out * V_in + P_out))
Given V_in = 24 V, we can substitute the values:
D = (120 W / (48 V * 24 V + 120 W))
D = 0.2
Now, we can substitute the values of V_out, D, f, and ΔI_L into the formula for L:
L = (48 V * (1 - 0.2)) / (50 kHz * ΔI_L)
Since the converter operates at the boundary between continuous and discontinuous conduction modes, the inductor ripple current (ΔI_L) is equal to half of the average inductor current (ΔI_L = I_L / 2).
The average inductor current (I_L) can be calculated using the formula:
I_L = P_out / V_out
Substituting the values:
I_L = 120 W / 48 V
I_L = 2.5 A
Now we can calculate the inductor ripple current:
ΔI_L = I_L / 2
ΔI_L = 2.5 A / 2
ΔI_L = 1.25 A
Finally, we can substitute the values into the formula for L:
L = (48 V * (1 - 0.2)) / (50 kHz * 1.25 A)
L = 0.768 H (approximately)
Therefore, the value of the boost inductor in this DC-DC boost converter is approximately 0.768 Henrys.
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In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.Correct answer is '24'. Can you explain this answer?
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In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.Correct answer is '24'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.Correct answer is '24'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.Correct answer is '24'. Can you explain this answer?.
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