A DC-DC Boost converter has Vin = 10 V, L = 100 μH, f = 50 kHz. It delivers 25 W at Vo = 25 V. Verify1.The duty cycle D is 0.6.2.The peak-to-peak inductor current ripple is1.2 A.3.The average inductor current is2.5 A.Which of the statements are correct?
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An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A. Key relation: V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).
Exam focus: For an ideal boost, D=1−Vin/Vo. Common trap: A normal boost converter cannot produce Vo<Vin merely because D<0.5.
Formula / Key Relation
Vo=Vin/(1−D); ΔI_L≈Vin D/(Lfs).
Detailed Explanation
Correct Answer: C - 1, 2 and 3 Quick Concept Explanation An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A. Key relation: V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).Exam focus: For an ideal boost, D=1−Vin/Vo. Common trap: A normal boost converter cannot produce Vo<Vin merely because D<0.5. Statement-wise Verification Statement 1 - Correct.The duty cycle D is 0.6.D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Statement 2 - Correct.The peak-to-peak inductor current ripple is 1.2 A.Ideal average input/inductor current=P/Vin=25/10=2.5 A. Statement 3 - Correct.The average inductor current…
Correct Answer: C - 1, 2 and 3
Quick Concept Explanation
An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A. Key relation: V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).
Exam focus: For an ideal boost, D=1−Vin/Vo. Common trap: A normal boost converter cannot produce Vo<Vin merely because D<0.5.
Statement-wise Verification
Statement 1 - Correct. The duty cycle D is 0.6. D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A.
Statement 2 - Correct. The peak-to-peak inductor current ripple is 1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A.
Statement 3 - Correct. The average inductor current is 2.5 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A.
Core Concept
An ideal boost converter in continuous conduction has an output voltage greater than or equal to input for 0≤D<1. When the switch is off, it must block approximately the output voltage. The output capacitor supplies load current between energy-transfer intervals and reduces output ripple.
Formula / Key Relationship
V_o=V_in/(1−D); ΔI_L≈V_in D/(Lf_s).
Step-by-Step Check
D=1−Vin/Vo=1−10/25=0.6. ΔIL=VinD/(Lf)=10×0.6/(100 μH×50 kHz)=1.2 A. Ideal average input/inductor current=P/Vin=25/10=2.5 A.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 1. Option D is incorrect because it omits correct statement(s) 2.
Exam Shortcut / Approach
For an ideal boost, D=1−Vin/Vo.
Common Mistake
A normal boost converter cannot produce Vo<Vin merely because D<0.5.
Quick Revision
Duty Ratio Inductor Ripple and Input Current is linked with Boost Converter, DC-DC Converters and Power Electronics. For an ideal boost, D=1−Vin/Vo.
Quick Trick
For an ideal boost, D=1−Vin/Vo.
Why Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 1. Option D is incorrect because it omits correct statement(s) 2.
Common Mistake
A normal boost converter cannot produce Vo<Vin merely because D<0.5.
Exam Tip
For an ideal boost, D=1−Vin/Vo.
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