Skip to main content

Buck regulator numerical (12 V to 6 V)

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.

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.]

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

Buck regulator relations:

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

i) Duty cycle

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

ii) Filter capacitance C

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

iii) Filter inductance L

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

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Advanced Electronics chapter 8 questions →