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Buck regulator numerical (14 V to 6 V)

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.

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.

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

Buck regulator relations (Rashid):

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

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

(b) Filter inductance L

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

(c) Filter capacitance C

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

(d) Critical values of L and C

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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