Finding the Currents and Voltage in the Buck–Boost Regulator The buck–boost regulator in Figure 5.19a has an input voltage of Vs = 12 V. The duty cycle k = 0.25 and the switching frequency is 25 kHz. The inductance L = 150 µH and filter capacitance C = 220 µF. The average load current Ia = 1.25 A

Question

Finding the Currents and Voltage in the Buck–Boost Regulator

The buck–boost regulator in Figure 5.19a has an input voltage of [latex]V_s = 12 V[/latex]. The duty cycle [latex]k = 0.25[/latex] and the switching frequency is 25 kHz. The inductance [latex]L = 150 µH[/latex] and filter capacitance [latex]C = 220 µF[/latex]. The average load current [latex]I_a = 1.25 A[/latex]. Determine (a) the average output voltage, [latex]V_a[/latex]; (b) the peak-to-peak output voltage ripple, [latex]ΔV_c[/latex] ; (c) the peak-to-peak ripple current of inductor, ΔI; (d) the peak current of the transistor, [latex]I_p[/latex]; and (e) the critical values of L and C.

Accepted Answer

[latex]V_s = 12 V, k = 0.25, I_a = 1.25 A, f = 25 kHz, L = 150 µH, and C = 220 µF[/latex].

a. From Eq. (5.86),

[latex]V_a=-\frac{V_sk}{1 -k}[/latex] (5.86)

[latex]V_a = -12 × 0.25/(1 - 0.25)= -4 V[/latex].

b. From Eq. (5.95),

[latex]\Delta V_c=\frac{I_ak}{fC}[/latex] (5.95)

the peak-to-peak output ripple voltage is

[latex]\Delta V_c =\frac{1.25 imes 0.25}{25,000 imes 220 imes 10^{-6}}=56.8 mV[/latex]

c. From Eq. (5.92),

[latex] \Delta I=\frac{V_sk}{fL}[/latex] (5.92)

the peak-to-peak inductor ripple is

[latex]\Delta I=\frac{12 imes 0.25}{25,000 imes 150 imes 10^{-6}}= 0.8 A[/latex]

d. From Eq. (5.89),

[latex]I_s=\frac{I_ak}{1 -k}[/latex] (5.89)

[latex]I_s = 1.25 × 0.25/(1 - 0.25)= 0.4167 A[/latex]. Because [latex]I_s[/latex] is the average of duration kT, the peak-to-peak current of the transistor,

[latex]I_p=\frac{I_s}{k}+\frac{\Delta I}{2}=\frac{0.4167}{0.25}+\frac{0.8}{2}=2.067 A[/latex]

e. [latex]R= \frac{-V_a}{I_a}=\frac{4}{1.25} =3.2 \Omega [/latex]

From Eq. (5.96), we get [latex]L_c=\frac{(1 -k)R}{2f} =\frac{(1 -0.25) imes 3.2}{2 imes 25 imes 10^3}= 450 \mu H[/latex]

From Eq. (5.97), we get [latex]C_c=\frac{k}{2fR}=\frac{0.25}{2 imes 25 imes 10^3 imes 3.2}=1.56 \mu F [/latex]

Question 5.7: Finding the Currents and Voltage in the Buck–Boost Regulator...

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