A DC-DC Boost converter has Vin = 10 V, L = 100 μH, f = 50 kHz. It delivers 25 W at Vo = 25 V. Verify
1. The duty cycle D is 0.6.
2. The peak-to-peak inductor current ripple is 1.2 A.
3. The average inductor current is 2.5 A.
Which of the statements are correct?
Answer & step-by-step solution
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.
एक DC-DC बूस्ट कन्वर्टर में Vin = 10 V, L = 100 μH, f = 50 kHz है। यह Vo = 25 V पर 25 W देता है। सत्यापित कीजिए
1. ड्यूटी साइकिल D = 0.6 है।
2. पीक-टू-पीक प्रेरक धारा रिपल 1.2 A है।
3. औसत प्रेरक धारा 2.5 A है।
कौन-से कथन सही हैं?
उत्तर और चरणबद्ध हल
C — 1, 2 और 3। D = 1 − Vin/Vo = 1 − 10/25 = 0.6; ΔIL = Vin·D/(L·f) = 10×0.6/(100 μH × 50 kHz) = 1.2 A; औसत प्रेरक (इनपुट) धारा = P/Vin = 25/10 = 2.5 A।
विस्तृत हल (मूल भाषा में)
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.
એક DC-DC બૂસ્ટ કન્વર્ટરમાં Vin = 10 V, L = 100 μH, f = 50 kHz છે. તે Vo = 25 V પર 25 W આપે છે. ચકાસો
1. ડ્યૂટી સાઇકલ D = 0.6 છે.
2. પીક-ટુ-પીક પ્રેરક પ્રવાહ રિપલ 1.2 A છે.
3. સરેરાશ પ્રેરક પ્રવાહ 2.5 A છે.
કયાં વિધાનો સાચાં છે?
જવાબ અને પગલાવાર ઉકેલ
C — 1, 2 અને 3. D = 1 − Vin/Vo = 1 − 10/25 = 0.6; ΔIL = Vin·D/(L·f) = 10×0.6/(100 μH × 50 kHz) = 1.2 A; સરેરાશ પ્રેરક (ઇનપુટ) પ્રવાહ = P/Vin = 25/10 = 2.5 A.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
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.
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