Finding the Currents and Voltages in the Cúk Regulator The input voltage of a Cúk converter in Figure 5.20a is Vs = 12 V. The duty cycle is k = 0.25 and the switching frequency is 25 kHz. The filter inductance is L2 = 150 µH and filter capacitance is C2 = 220 µF. The energy transfer

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Finding the Currents and Voltages in the Cúk Regulator

The input voltage of a Cúk converter in Figure 5.20a is [latex]V_s = 12 V[/latex]. The duty cycle is [latex]k = 0.25[/latex] and the switching frequency is 25 kHz. The filter inductance is [latex]L_2 = 150 µH[/latex] and filter capacitance is [latex]C_2 = 220 µF[/latex]. The energy transfer capacitance is [latex]C_1 = 200 µF[/latex] and inductance [latex]L_1 = 180 µH[/latex]. The average load current is [latex]I_a = 1.25 A[/latex]. Determine (a) the average output voltage [latex]V_a[/latex]; (b) the average input current [latex]I_s[/latex]; (c) the peak-to-peak ripple current of inductor [latex]L_1, ΔI_1[/latex]; (d) the peakto- peak ripple voltage of capacitor [latex]C_1, ΔV_{c1}[/latex]; (e) the peak-to-peak ripple current of inductor [latex]L_2, ΔI_2[/latex]; (f) the peak-to-peak ripple voltage of capacitor [latex]C_2, ΔV_{c2}[/latex]; and (g) the peak current of the transistor [latex]I_p[/latex].

Accepted Answer

[latex]V_s = 12 V, k = 0.25, I_a = 1.25 A, f = 25 kHz, L_1 = 180 µ H, C_1 = 200 µF, L_2 = 150 µH[/latex], and [latex]C_2 = 220 µF[/latex].

a. From Eq. (5.108),

[latex]V_a=-\frac{kV_s}{1 -k}[/latex] (5.108)

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

b. From Eq. (5.111),

[latex]I_s=\frac{kI_a}{1 -k}[/latex] (5.111)

[latex]I_s = 1.25 × 0.25/(1 - 0.25) = 0.42 A[/latex].

c. From Eq. (5.114),

[latex]\Delta I_1=\frac{V_sk}{fL_1}[/latex] (5.114)

[latex]\Delta I_1 = 12 × 0.25/(25,000 × 180 × 10^{-6}) = 0.67 A[/latex].

d. From Eq. (5.120),

[latex]\Delta V_{c1}=\frac{I_s(1 -k)}{fC_1}[/latex] (5.120)

[latex]\Delta V_{c1}=0.42 × 11 - 0.252/(25,000 × 200 × 10^{-6}) = 63 mV[/latex].

e. From Eq. (5.117),

[latex]\Delta I_2=-\frac{V_a(1 -k)}{fL_2}=\frac{kV_s}{fL_2}[/latex] (5.117)

[latex]\Delta I_2= 0.25 × 12/(25,000 × 150 × 10^{-6}) = 0.8 A[/latex].

f. From Eq. (5.121),

[latex]\Delta V_{c2}=\frac{1}{C_2}\int_{0}^{T/2}{I_{c2} dt}=\frac{1}{C_2}\int_{0}^{T/2}{\frac{\Delta I_2}{4}dt }=\frac{\Delta I_2}{8fC_2}[/latex] (5.121)

[latex]\Delta V_{c2}=0.8/(8 × 25,000 × 220 × 10^{-6}) = 18.18 mV[/latex].

g. The average voltage across the diode can be found from

[latex]V_{dm}= -kV_{c1}=-V_ak\frac{1}{-k}=V_a[/latex] (5.127)

For a lossless circuit, [latex]I_{L2}V_{dm} = V_aI_a[/latex] and the average value of the current in inductor [latex]L_2[/latex] is

[latex]I_{L2}=\frac{I_aV_a}{V_{dm}} =I_a[/latex] (5.128)

[latex]=1.25 A[/latex]

Therefore, the peak current of transistor is

[latex]I_p=I_s+\frac{\Delta I_1}{2}+I_{L2}+\frac{\Delta I_2}{2} =0.42+\frac{0.67}{2} +1.25+\frac{0.8}{2}=2.405 A[/latex]

Question 5.8: Finding the Currents and Voltages in the Cúk Regulator The i...

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