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Buck-boost regulator numerical (Vs = 10 V)

Short answer

Using L=0.1 mH, C=200 μF: average output voltage Va = -Vs·k/(1-k) = -8.18 V; peak-to-peak ripple voltage ΔVc = Ia·k/(f·C) = 108 mV; peak-to-peak ripple current ΔI = Vs·k/(f·L) = 1.8 A.

The buck-boost regulator has input voltage Vs = 10V, duty cycle k = 45% and switching frequency of 25KHz, the inductance L = 0.1nH, filter capacitance C = 0.2nF and average load current Ia = 1200mA. Determine: i) The average output voltage ii) Peak to peak output ripple voltage iii) Peak to peak output ripple current

Given: Vs = 10 V, k = 0.45, f = 25 kHz, Ia = 1200 mA = 1.2 A.

Assumption: the values L = 0.1 nH and C = 0.2 nF are not practical (they give a ripple current of 1.8 × 10⁶ A), so they are taken as L = 0.1 mH and C = 0.2 mF (200 μF), the usual values for this problem.

Buck-boost relations (Rashid):

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

i) Average output voltage

Va = −10 × 0.45/(1 − 0.45) = −4.5/0.55
Va = −8.18 V

(The minus sign shows polarity reversal of the output.)

ii) Peak-to-peak output ripple voltage

ΔVc = 1.2 × 0.45/(25×10³ × 0.2×10⁻³)
    = 0.54/5
ΔVc = 0.108 V = 108 mV

iii) Peak-to-peak ripple current

ΔI = 10 × 0.45/(25×10³ × 0.1×10⁻³)
   = 4.5/2.5
ΔI = 1.8 A

Answer: Va = −8.18 V, ΔVc = 108 mV, ΔI = 1.8 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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