---
title: "Boost regulator numerical (5 V to 15 V)"
subject: "Advanced Electronics"
chapter: 8
key: "semester_4/advanced_electronics:boost-regulator-5v-15v"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/boost-regulator-5v-15v
site: BCT Engineering
---

# Boost regulator numerical (5 V to 15 V)

Advanced Electronics · Chapter 8: Switched Power Supplies

- Asked in: 2071 Shrawan, 1+2+2+2 marks
- Video: [Mastering Boost Converters : Key Concepts For GATE Power Electronics | Numerical 2](https://www.youtube.com/watch?v=xvV-C3VlTyA) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/boost-regulator-5v-15v
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8

## Question

A boost regulator has an input voltage of Vs = 5V. The average output voltage Va = 15V and average load current Ia = 0.5 A. The switching frequency is 25 KHz. If L = 150 μH and C = 220 μF, determine (i) Duty cycle (ii) The peak current of inductor I2 (iii) Ripple voltage of filter capacitor ΔVc and (iv) Critical value of L and C.

## Short answer

Duty cycle k = 1 − Vs/Va = 0.667. Peak inductor current I2 = Is + ΔI/2 = 1.944 A, output ripple ΔVc = Ia·k/(fC) = 60.6 mV. Critical values: Lc = 133.3 μH, Cc = 0.444 μF.

## Answer

Given: Vs = 5 V, Va = 15 V, Ia = 0.5 A, f = 25 kHz, L = 150 μH, C = 220 μF.

Load resistance R = Va/Ia = 15/0.5 = 30 Ω.

Boost regulator relations (Rashid):

```text
Va  = Vs/(1 − k)
ΔI  = Vs(Va − Vs)/(f·L·Va)
Is  = Ia/(1 − k),   I2 = Is + ΔI/2
ΔVc = Ia·k/(f·C)
Lc  = k(1 − k)R/(2f),   Cc = k/(2fR)
```

#### (i) Duty cycle

```text
k = 1 − Vs/Va = 1 − 5/15
k = 0.6667 (66.67 %)
```

#### (ii) Peak current of inductor I2

```text
ΔI = 5 × (15 − 5)/(25×10³ × 150×10⁻⁶ × 15)
   = 50/56.25 = 0.8889 A
Is = 0.5/(1 − 0.6667) = 1.5 A
I2 = Is + ΔI/2 = 1.5 + 0.4444
I2 = 1.944 A
```

(Valley current I1 = 1.5 − 0.444 = 1.056 A.)

#### (iii) Ripple voltage of filter capacitor

```text
ΔVc = 0.5 × 0.6667/(25×10³ × 220×10⁻⁶)
    = 0.3333/5.5
ΔVc = 60.61 mV
```

#### (iv) Critical values of L and C

```text
Lc = 0.6667 × 0.3333 × 30/(2 × 25×10³)
   = 6.667/50000
Lc = 133.3 μH

Cc = 0.6667/(2 × 25×10³ × 30)
Cc = 0.444 μF
```

Since L = 150 μH > Lc and C = 220 μF > Cc, the inductor current and capacitor voltage are continuous.

**Answer: k = 0.667, I2 = 1.944 A, ΔVc = 60.61 mV, Lc = 133.3 μH, Cc = 0.444 μF**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For Vs = 5 V and Va = 15 V, what is k?
   A. 0.333
   B. 0.667
   C. 0.5
   D. 0.75
   Answer: B. k = 1 − Vs/Va = 1 − 1/3 = 0.667. (Key point: Duty cycle)

2. How is the peak inductor current I2 found?
   A. I2 = Is + ΔI/2 where Is = Ia/(1 − k)
   B. I2 = Ia + ΔI
   C. I2 = Ia·k
   D. I2 = Is − ΔI/2
   Answer: A. The average input current Is = 0.5/0.333 = 1.5 A and I2 = 1.5 + 0.444 = 1.944 A. (Key point: Peak inductor current)

3. What is the ripple voltage ΔVc?
   A. 60.61 mV
   B. 6.06 mV
   C. 0.606 V
   D. 30.3 mV
   Answer: A. ΔVc = Ia·k/(fC) = 0.5 × 0.6667/(25×10^3 × 220×10^-6) = 0.3333/5.5 = 60.61 mV. (Key point: Capacitor ripple)

4. What does comparing L = 150 μH with Lc = 133.3 μH show?
   A. Since L > Lc, the inductor current is continuous
   B. Since L > Lc, current is discontinuous
   C. Lc is not relevant to continuity
   D. The capacitor voltage is discontinuous
   Answer: A. L above the critical value keeps the inductor current continuous; likewise C > Cc keeps the capacitor voltage continuous. (Key point: Critical L)

## Close variants set in other exams

- [Buck-boost regulator numerical (Vs = 10 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Abuck-boost-10v-45-percent) — Advanced Electronics, chapter 8, asked 3×
- [Buck-boost regulator numerical (Vs = 15 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Abuck-boost-15v-40-percent) — Advanced Electronics, chapter 8, asked 3×

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