---
title: "Buck regulator numerical (14 V to 6 V)"
subject: "Advanced Electronics"
chapter: 8
key: "semester_4/advanced_electronics:buck-regulator-14v-numerical"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-regulator-14v-numerical
site: BCT Engineering
---

# Buck regulator numerical (14 V to 6 V)

Advanced Electronics · Chapter 8: Switched Power Supplies

- Asked in: 2070 Chaitra, 1+2+2+2 marks
- Video: [Buck Converter | Numerical -1 | Power Electronics | Complete Explanation | Mathspedia |](https://www.youtube.com/watch?v=rqa3i_A7y8U) — Mathspedia
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-regulator-14v-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8

## Question

A buck regulator has an input voltage of Vs = 14V; the required output voltage is Va = 6V at R = 500Ω and peak to peak output voltage ripple is 15mV; the switching frequency is limited to 20KHz. If the peak to peak ripple current of the inductor is limited to 0.7A, determine: (a) the duty cycle (b) Filter inductance L (c) Filter capacitance C and (d) critical values of L and C.

## Short answer

Duty cycle k = Va/Vs = 0.429. Filter inductance L = Va(Vs−Va)/(f·ΔI·Vs) = 244.9 μH, capacitance C = ΔI/(8f·ΔVc) = 291.7 μF. Critical values Lc = (1−k)R/(2f) = 7.14 mH, Cc = (1−k)/(16Lf²) = 0.365 μF.

## Answer

Given: Vs = 14 V, Va = 6 V, R = 500 Ω, ΔVc = 15 mV, f = 20 kHz, ΔI = 0.7 A.

Buck regulator relations (Rashid):

```text
Va  = k·Vs
L   = Va(Vs − Va)/(f·ΔI·Vs)
C   = ΔI/(8·f·ΔVc)
Lc  = (1 − k)R/(2f)
Cc  = (1 − k)/(16·L·f²)
```

#### (a) Duty cycle

```text
k = Va/Vs = 6/14
k = 0.4286 (42.86 %)
```

#### (b) Filter inductance L

```text
L = 6 × (14 − 6)/(20×10³ × 0.7 × 14)
  = 48/196000
L = 244.9 μH
```

#### (c) Filter capacitance C

```text
C = 0.7/(8 × 20×10³ × 15×10⁻³)
  = 0.7/2400
C = 291.7 μF
```

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

```text
Lc = (1 − 0.4286) × 500/(2 × 20×10³)
   = 285.71/40000
Lc = 7.143 mH

Cc = (1 − 0.4286)/(16 × 244.9×10⁻⁶ × (20×10³)²)
   = 0.5714/1567347
Cc = 0.3646 μF
```

Note: L = 244.9 μH is less than Lc = 7.143 mH, so with R = 500 Ω (Ia = 12 mA) the inductor current would actually be discontinuous; C = 291.7 μF is much larger than Cc, so the capacitor voltage is continuous.

**Answer: k = 0.4286, L = 244.9 μH, C = 291.7 μF, Lc = 7.143 mH, Cc = 0.365 μF**

## Check yourself

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

1. For Vs = 14 V and Va = 6 V, what is k?
   A. 0.4286
   B. 0.5714
   C. 0.5
   D. 2.33
   Answer: A. k = 6/14 = 0.4286. (Key point: Duty cycle)

2. Which formula gives L?
   A. Va(Vs − Va)/(f·ΔI·Vs)
   B. ΔI/(8fΔVc)
   C. (1 − k)R/(2f)
   D. Vs·k/(fΔI)
   Answer: A. L = 6 × 8/(20×10^3 × 0.7 × 14) = 48/196000 = 244.9 μH. (Key point: Inductance)

3. What is C for ΔI = 0.7 A, ΔVc = 15 mV, f = 20 kHz?
   A. 29.17 μF
   B. 291.7 μF
   C. 2.917 mF
   D. 145.8 μF
   Answer: B. C = 0.7/(8 × 20×10^3 × 15×10^-3) = 0.7/2400 = 291.7 μF. (Key point: Capacitance)

4. Since L = 244.9 μH is less than Lc = 7.143 mH, what follows?
   A. The inductor current would be discontinuous
   B. The inductor current is continuous
   C. The capacitor voltage is discontinuous
   D. The duty cycle changes
   Answer: A. L below Lc means discontinuous inductor current at the light load (Ia = 12 mA); C far above Cc keeps the capacitor voltage continuous. (Key point: Critical inductance meaning)

## Related in Advanced Electronics

- [Buck regulator numerical (12 V to 6 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Abuck-regulator-12v-numerical) — Advanced Electronics, chapter 8
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