---
title: "Buck regulator numerical (12 V to 6 V)"
subject: "Advanced Electronics"
chapter: 8
key: "semester_4/advanced_electronics:buck-regulator-12v-numerical"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-regulator-12v-numerical
site: BCT Engineering
---

# Buck regulator numerical (12 V to 6 V)

Advanced Electronics · Chapter 8: Switched Power Supplies

- Asked in: 2072 Kartik, 2+2+3 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-12v-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8

## Question

For the Buck regulator if input voltage Vs = 12V, the required average output voltage is 6V at R = 450 ohm and peak to peak output ripple voltage is 20 mV, switching frequency is 20 KHz and peak to peak ripple current of inductor is 0.8A, find: i) Duty cycle ii) Filter capacitance C iii) Filter inductance L. [Figure: buck regulator: switching transistor Q, driven by a control circuit, in series from +Vs to a node with freewheeling diode Dm to the negative rail; inductor L from that node to the output; filter capacitor C and the load in parallel across the output Vo.]

## Short answer

Duty cycle k = Va/Vs = 0.5. Filter capacitance C = ΔI/(8·f·ΔVc) = 250 μF. Filter inductance L = Va(Vs-Va)/(f·ΔI·Vs) = 187.5 μH.

## Answer

Given: Vs = 12 V, Va = 6 V, R = 450 Ω, ΔVc = 20 mV, f = 20 kHz, ΔI = 0.8 A.

Buck regulator relations:

```text
Va  = k·Vs
ΔI  = Va(Vs − Va)/(f·L·Vs)
ΔVc = ΔI/(8·f·C)
```

#### i) Duty cycle

```text
k = Va/Vs = 6/12
k = 0.5 (50 %)
```

#### ii) Filter capacitance C

```text
C = ΔI/(8·f·ΔVc)
  = 0.8/(8 × 20×10³ × 20×10⁻³)
  = 0.8/3200
C = 250 μF
```

#### iii) Filter inductance L

```text
L = Va(Vs − Va)/(f·ΔI·Vs)
  = 6 × (12 − 6)/(20×10³ × 0.8 × 12)
  = 36/192000
L = 187.5 μH
```

Check: average load current Ia = 6/450 = 13.3 mA.

**Answer: k = 0.5, C = 250 μF, L = 187.5 μH**

## Check yourself

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

1. For Vs = 12 V and Va = 6 V, what is the duty cycle?
   A. 0.25
   B. 0.5
   C. 0.75
   D. 2
   Answer: B. k = Va/Vs = 6/12 = 0.5. (Key point: Duty cycle)

2. Which formula gives the filter capacitance?
   A. C = ΔI/(8fΔVc)
   B. C = ΔVc/(8fΔI)
   C. C = ΔI·f/(8ΔVc)
   D. C = 8ΔI/(fΔVc)
   Answer: A. From ΔVc = ΔI/(8fC), C = ΔI/(8fΔVc). (Key point: Capacitor formula)

3. What is C for ΔI = 0.8 A, f = 20 kHz, ΔVc = 20 mV?
   A. 25 μF
   B. 250 μF
   C. 2.5 mF
   D. 250 nF
   Answer: B. C = 0.8/(8 × 20×10^3 × 20×10^-3) = 0.8/3200 = 250 μF. (Key point: Capacitance value)

4. What is the filter inductance L?
   A. 187.5 μH
   B. 375 μH
   C. 18.75 μH
   D. 1.875 mH
   Answer: A. L = Va(Vs − Va)/(f·ΔI·Vs) = 36/(20×10^3 × 0.8 × 12) = 36/192000 = 187.5 μH. (Key point: Inductance value)

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