Analysis of the Half-Bridge Resonant Inverter

The half-bridge resonant inverter in Figure 7.3 is operated at an output frequency, $f_o = 7 kHz$ . If $C_1 = C_2 = C = 3 µF$ , $L_1 = L_2 = L = 50 µ H$ , $R = 2 Ω$ , and $V_s = 220 V$ , determine (a) the peak supply current $I_{ps}$ , (b) the average transistor current $I_A$ , and (c) the rms transistor current $I_R$ .

Question

Analysis of the Half-Bridge Resonant Inverter

The half-bridge resonant inverter in Figure 7.3 is operated at an output frequency, [latex]f_o = 7 kHz[/latex]. If [latex]C_1 = C_2 = C = 3 µF[/latex], [latex]L_1 = L_2 = L = 50 µ H[/latex], [latex]R = 2 Ω[/latex] , and [latex]V_s = 220 V[/latex], determine (a) the peak supply current [latex]I_{ps}[/latex], (b) the average transistor current [latex]I_A[/latex], and (c) the rms transistor current [latex]I_R[/latex].

Accepted Answer

[latex]V_s = 220 V, C = 3 \mu F, L = 50 \mu H, R = 2 \Omega[/latex], and [latex]f_o = 7 kHz[/latex]. Figure 7.6a shows the equivalent circuit when transistor [latex]Q_1[/latex] is conducting and [latex]Q_2[/latex] is off. The capacitors [latex]C_1[/latex] and [latex]C_2[/latex] would initially be charged to [latex]V_{c1}( = V_s + V_c)[/latex] and [latex]V_c[/latex], respectively, with the polarities as shown, under steady-state conditions. Because [latex]C_1 = C_2[/latex], the load current would be shared equally by [latex]C_1[/latex] and the dc supply, as shown in Figure 7.6b.

Figure 7.6c shows the equivalent circuit when transistor [latex]Q_2[/latex] is conducting and [latex]Q_1[/latex] is off. The capacitors [latex]C_1[/latex] and [latex]C_2[/latex] are initially charged to [latex]V_{c1} [/latex]and [latex]V_S - V_{c1}[/latex], respectively, with the polarities as shown. The load current is shared equally by [latex]C_1[/latex] and [latex]C_2[/latex], as shown in Figure 7.6d, which can be simplified to Figure 7.6e.

Considering the loop formed by [latex]C_2[/latex], dc source, L, and load, the instantaneous load current
can be described (from Figure 7.6b) by

[latex]L\frac{di_o}{dt}+Ri_o+ \frac{1}{2C_2}\int{i_o}dt+\upsilon _{c2}(t=0)- V_s=0[/latex] (7.19)

with initial conditions [latex]i_0(t = 0) = 0[/latex] and [latex]v_{c2}(t = 0) = -V_c[/latex]. For an underdamped condition and [latex]C_1 = C_2 = C[/latex], Eq. (7.5)

[latex]i_1(t)=\frac{V_s+V_c}{\omega _rL}e^{-\alpha t}sin \omega _rt[/latex] (7.5)

is applicable:

[latex]i_o(t)=\frac{V_s+V_c}{\omega _rL}e^{-\alpha t}sin \omega _rt[/latex] (7.20)

where the effective capacitance is [latex]C_e = C_1 + C_2 = 2C[/latex] and

[latex]\omega _r=\left(\frac{1}{2LC_2} -\frac{R^2}{4L^2} ight)^{1/2}[/latex] (7.21)

[latex]=\left(\frac{10^{12}}{2 imes 50 imes 3} -\frac{2^2 imes 10^{12}}{4 imes 50^2} ight)^{1/2}=54,160 rad/s[/latex]

The voltage across capacitor [latex]C_2[/latex] can be expressed as

[latex]\upsilon _{c2}(t)=\frac{1}{2C_2}\int_{0}^{t}{i_0(t)dt-V_c}[/latex]

[latex]= - (V_s + V_c)e^{-\alpha t}1(\alpha sin \omega _rt + \omega _r cos \omega _rt)/\omega _r + V_s[/latex] (7.22)

a. Because the resonant frequency is the same as that of Example 7.1, the results of Example 7.1 are valid, provided that the equivalent capacitance is [latex]C_e = C_1 +C_2 = 6 ΩF[/latex]. From Example 7.1, [latex]V_c = 100.4 V, t_m = 22.47 µs[/latex], and [latex]I_o = 44.1 A[/latex]. From Eq. (7.20), the peak load current is [latex]I_p = 70.82 A[/latex]. The peak supply current, which is half of the peak load current, is [latex]I_p = 70.82/2 = 35.41 A[/latex].

b. The average transistor current [latex]I_A = 17.68 A[/latex].

c. The rms transistor current [latex]I_R=I_o/\sqrt{2}=31.18 A[/latex]

Question 7.2: Analysis of the Half-Bridge Resonant Inverter The half-bridg...

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