---
title: "Buck-boost regulator numerical (Vs = 10 V)"
subject: "Advanced Electronics"
chapter: 8
key: "semester_4/advanced_electronics:buck-boost-10v-45-percent"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-boost-10v-45-percent
site: BCT Engineering
---

# Buck-boost regulator numerical (Vs = 10 V)

Advanced Electronics · Chapter 8: Switched Power Supplies

- Asked in: 2074 Asoj, 2+2+3 marks
- Video: [Boost and buck boost converter problems](https://www.youtube.com/watch?v=uVuNUK49ruc) — Eng Samar Abd el fatah
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-boost-10v-45-percent
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8

## Question

The buck-boost regulator has input voltage Vs = 10V, duty cycle k = 45% and switching frequency of 25KHz, the inductance L = 0.1nH, filter capacitance C = 0.2nF and average load current Ia = 1200mA. Determine: i) The average output voltage ii) Peak to peak output ripple voltage iii) Peak to peak output ripple current

## Short answer

Using L=0.1 mH, C=200 μF: average output voltage Va = -Vs·k/(1-k) = -8.18 V; peak-to-peak ripple voltage ΔVc = Ia·k/(f·C) = 108 mV; peak-to-peak ripple current ΔI = Vs·k/(f·L) = 1.8 A.

## Answer

Given: Vs = 10 V, k = 0.45, f = 25 kHz, Ia = 1200 mA = 1.2 A.

Assumption: the values L = 0.1 nH and C = 0.2 nF are not practical (they give a ripple current of 1.8 × 10⁶ A), so they are taken as **L = 0.1 mH** and **C = 0.2 mF (200 μF)**, the usual values for this problem.

Buck-boost relations (Rashid):

```text
Va  = −Vs·k/(1 − k)
ΔVc = Ia·k/(f·C)
ΔI  = Vs·k/(f·L)
```

#### i) Average output voltage

```text
Va = −10 × 0.45/(1 − 0.45) = −4.5/0.55
Va = −8.18 V
```

(The minus sign shows polarity reversal of the output.)

#### ii) Peak-to-peak output ripple voltage

```text
ΔVc = 1.2 × 0.45/(25×10³ × 0.2×10⁻³)
    = 0.54/5
ΔVc = 0.108 V = 108 mV
```

#### iii) Peak-to-peak ripple current

```text
ΔI = 10 × 0.45/(25×10³ × 0.1×10⁻³)
   = 4.5/2.5
ΔI = 1.8 A
```

**Answer: Va = −8.18 V, ΔVc = 108 mV, ΔI = 1.8 A**

## Check yourself

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

1. Why were L = 0.1 nH and C = 0.2 nF reinterpreted?
   A. They give an impractical ripple current of about 1.8 million A, so they are taken as 0.1 mH and 0.2 mF
   B. They make the duty cycle exceed 1
   C. The frequency would become negative
   D. They make the output voltage positive
   Answer: A. The nH and nF values are not practical; the usual intended values are mH and mF. (Key point: Practical values)

2. What is the average output voltage for Vs = 10 V, k = 0.45?
   A. +8.18 V
   B. −8.18 V
   C. −4.5 V
   D. −18.2 V
   Answer: B. Va = −Vs·k/(1 − k) = −4.5/0.55 = −8.18 V, with inverted polarity. (Key point: Output voltage)

3. Which formula gives the output ripple voltage of the buck-boost, and what is its value?
   A. Ia·k/(fC) = 108 mV
   B. Vs·k/(fL) = 1.8 V
   C. Ia/(8fC) = 6 mV
   D. Va/(fC) = 1 V
   Answer: A. ΔVc = 1.2 × 0.45/(25k × 0.2 m) = 0.54/5 = 0.108 V. (Key point: Ripple voltage)

4. What is the inductor ripple current?
   A. 0.108 A
   B. 1.8 A
   C. 4.5 A
   D. 8.18 A
   Answer: B. ΔI = Vs·k/(fL) = 4.5/(25×10^3 × 0.1×10^-3) = 4.5/2.5 = 1.8 A. (Key point: Ripple current)

## Close variants set in other exams

- [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×
- [Boost regulator numerical (5 V to 15 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aboost-regulator-5v-15v) — Advanced Electronics, chapter 8, asked 3×

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