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Boost regulator numerical (5 V to 15 V)

Short answer

Duty cycle k = 1 − Vs/Va = 0.667. Peak inductor current I2 = Is + ΔI/2 = 1.944 A, output ripple ΔVc = Ia·k/(fC) = 60.6 mV. Critical values: Lc = 133.3 μH, Cc = 0.444 μF.

A boost regulator has an input voltage of Vs = 5V. The average output voltage Va = 15V and average load current Ia = 0.5 A. The switching frequency is 25 KHz. If L = 150 μH and C = 220 μF, determine (i) Duty cycle (ii) The peak current of inductor I2 (iii) Ripple voltage of filter capacitor ΔVc and (iv) Critical value of L and C.

Given: Vs = 5 V, Va = 15 V, Ia = 0.5 A, f = 25 kHz, L = 150 μH, C = 220 μF.

Load resistance R = Va/Ia = 15/0.5 = 30 Ω.

Boost regulator relations (Rashid):

Va  = Vs/(1 − k)
ΔI  = Vs(Va − Vs)/(f·L·Va)
Is  = Ia/(1 − k),   I2 = Is + ΔI/2
ΔVc = Ia·k/(f·C)
Lc  = k(1 − k)R/(2f),   Cc = k/(2fR)

(i) Duty cycle

k = 1 − Vs/Va = 1 − 5/15
k = 0.6667 (66.67 %)

(ii) Peak current of inductor I2

ΔI = 5 × (15 − 5)/(25×10³ × 150×10⁻⁶ × 15)
   = 50/56.25 = 0.8889 A
Is = 0.5/(1 − 0.6667) = 1.5 A
I2 = Is + ΔI/2 = 1.5 + 0.4444
I2 = 1.944 A

(Valley current I1 = 1.5 − 0.444 = 1.056 A.)

(iii) Ripple voltage of filter capacitor

ΔVc = 0.5 × 0.6667/(25×10³ × 220×10⁻⁶)
    = 0.3333/5.5
ΔVc = 60.61 mV

(iv) Critical values of L and C

Lc = 0.6667 × 0.3333 × 30/(2 × 25×10³)
   = 6.667/50000
Lc = 133.3 μH

Cc = 0.6667/(2 × 25×10³ × 30)
Cc = 0.444 μF

Since L = 150 μH > Lc and C = 220 μF > Cc, the inductor current and capacitor voltage are continuous.

Answer: k = 0.667, I2 = 1.944 A, ΔVc = 60.61 mV, Lc = 133.3 μH, Cc = 0.444 μF

✓ Answer checked by Claude Opus 5.5 (high effort)

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