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Buck-boost regulator numerical (Vs = 15 V)

Short answer

Average output voltage Va = −Vs·k/(1−k) = −10 V. Peak-to-peak output ripple ΔVc = Ia·k/(fC) = 125 mV, and inductor ripple current ΔI = Vs·k/(fL) = 2.5 A.

The Buck-Boost regulator has input voltage Vs = 15V, duty cycle K = 40% and switching frequency of 20 KHz. The inductance L = 120 μH, filter capacitance C = 200 μF and average load current Ia = 1.25 A. Determine (i) the average output voltage (ii) peak to peak output ripple voltage and (iii) peak to peak output ripple current.

Given: Vs = 15 V, k = 0.4, f = 20 kHz, L = 120 μH, C = 200 μF, Ia = 1.25 A.

Buck-boost regulator relations (Rashid):

Va  = −Vs·k/(1 − k)
ΔVc = Ia·k/(f·C)
ΔI  = Vs·k/(f·L)

(i) Average output voltage

Va = −15 × 0.4/(1 − 0.4) = −6/0.6
Va = −10 V

The output is 10 V with polarity reversed with respect to the input.

(ii) Peak-to-peak output ripple voltage

ΔVc = 1.25 × 0.4/(20×10³ × 200×10⁻⁶)
    = 0.5/4
ΔVc = 0.125 V = 125 mV

(iii) Peak-to-peak ripple current

ΔI = 15 × 0.4/(20×10³ × 120×10⁻⁶)
   = 6/2.4
ΔI = 2.5 A

Extra check: average input current Is = Ia·k/(1 − k) = 1.25 × 0.4/0.6 = 0.833 A, and peak inductor current I2 = (Is + Ia) + ΔI/2 = 2.083 + 1.25 = 3.33 A.

Answer: Va = −10 V, ΔVc = 125 mV, ΔI = 2.5 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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