Finding the Performance Parameters of a Forward Converter The average (or dc) output voltage of the forward–converter circuit in Figure 13.3a is Vo = 24 V at a resistive load of R = 0.8 Ω. The on-state voltage drops of transistors and diodes are Vt = 1.2 V and Vd = 0.7 V, respectively. The duty

Question

Finding the Performance Parameters of a Forward Converter

The average (or dc) output voltage of the forward–converter circuit in Figure 13.3a is [latex]V_o = 24 V[/latex] at a resistive load of [latex]R = 0.8 Ω[/latex]. The on-state voltage drops of transistors and diodes are [latex]V_t = 1.2 V[/latex] and [latex]V_d = 0.7 V[/latex], respectively. The duty cycle is [latex]k = 40\%[/latex] and the switching frequency is [latex]f = 1 kHz[/latex]. The dc supply voltage [latex]V_s = 12 V[/latex]. The turns ratio of the transformer is [latex]a = N_s/N_p = 0.25[/latex]. Determine (a) the average input current [latex]I_s[/latex], (b) the efficiency η, (c) the average transistor current [latex]I_A[/latex], (d) the peak transistor current [latex]I_p[/latex], (e) the rms transistor current [latex]I_R[/latex], (f) the open-circuit transistor voltage [latex]V_{oc}[/latex], (g) the primary magnetizing inductor [latex]L_p[/latex] for maintaining the peak-to-peak ripple current to 5% of the average input [latex]d_c[/latex] current, and (h) the output inductor [latex]L_1[/latex] for maintaining the peak-to-peak ripple current to 4% of its average value. Neglect the losses in the transformer and the ripple content of the output voltage is 3%.

Accepted Answer

[latex]a = N_s/N_p = 0.25[/latex] and [latex]I_o = V_o/R = 24/0.8 = 30 A[/latex].

a. The output power [latex]P_o = V_o I_o = 24 × 30 = 720 W[/latex]. The secondary voltage [latex]V_2= V_o + V_d = 24 + 0.7 = 24.7 V[/latex]. The primary voltage is [latex]V_1 = V_s - V_t = 12 - 1.2 = 10.8 V[/latex]. The turns ratio is [latex]a = V_2/V_1 = 24.7/10.8 = 2.287[/latex]. The input power is [latex]P_i = V_s I_s = V_t kI_s + V_d(1 - k)I_s + V_d I_o + P_o[/latex] which gives

[latex]I_s=\frac{V_dI_o+P_o}{V_s-V_tk-V_d(1-k)}=\frac{0.7 imes 30 + 720}{12 - 1.2 imes 0.4 - 0.7 imes 0.6} =66.76 A[/latex]

b. [latex]P_i = V_s I_s = 12 × 66.756 = 801 W[/latex]. The efficiency [latex]η= 720/801 = 89.9\%[/latex].

c. [latex]I_A = kI_s = 0.4 × 66.76 = 26.7 A[/latex].

d. [latex]ΔIp = 0.05 × Is = 0.05 × 66.76 = 3.353 A[/latex].

e. [latex]I_R=\sqrt{k}[I^2_p+\Delta I_p/3+\Delta I_pI_p]^{1/2}=\sqrt{0.4} imes [66.76^2+3.35/3+3.35 imes 66.76]^{1/2}=44.3 A [/latex]

f. [latex]V_{oc} = V_s + V_2/a = 22.8 V[/latex].

g. [latex] ΔI_{L1} = 0.04 × I_o = 0.04 × 30 = 1.2 A and ΔV_o = 0.03 × V_o = 0.03 × 24 = 0.72 V[/latex].

Using Eq. (13.18),

[latex]I_{L1}(pk)=I_{L1}(0)+\frac{\left(V_s-V_o ight)kT }{L_1}[/latex] (13.18)

[latex]I_{L1}=\frac{V_ok }{f\Delta I_{L1}}=\frac{0.72 imes 0.4}{1 imes 10^3 imes 1.2}=0.24 mH[/latex]

h. Using (13.15),

[latex]I^′_{p(pk)}=I{p(pk)}+\frac{V_skT}{L_p}[/latex] (13.15)

[latex]ΔI_p = a × I_{L1} + (V_s - Vt)kT/L_p[/latex], which gives

[latex]L_p=\frac{(V_s-V_t)k}{f(\Delta I_p-a imes \Delta I_{L1})}=\frac{(12-1.2) imes 0.4}{1 imes 10^3 imes (3.353-2.287 imes 1.2)}=7.28 mH[/latex]

Question 13.2: Finding the Performance Parameters of a Forward Converter Th...

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