Analysis of the Full-Bridge Resonant Inverter with Bidirectional Switches

The full-bridge resonant inverter in Figure 7.9a is operated at a frequency, $f_o = 3.5 kHz$ . If $C = 6 \mu F, L = 50 \mu H, R = 2 \Omega$ , and $V_s = 220 V$ , determine (a) the peak supply current $I_p$ , (b) the average device current $I_A$ , (c) the rms device current $I_R$ , (d) the rms load current $I_o$ , and (e) the average supply current $I_s$ .

Question

Analysis of the Full-Bridge Resonant Inverter with Bidirectional Switches

The full-bridge resonant inverter in Figure 7.9a is operated at a frequency, [latex]f_o = 3.5 kHz[/latex]. If [latex]C = 6 μF, L = 50 μH, R = 2 Ω[/latex], and [latex]V_s = 220 V[/latex], determine (a) the peak supply current [latex]I_p[/latex], (b) the average device current [latex]I_A[/latex], (c) the rms device current [latex]I_R[/latex], (d) the rms load current [latex]I_o[/latex], and (e) the average supply current [latex]I_s[/latex].

Accepted Answer

[latex]V_s=220 V , C = 6 \mu F, L=50 \mu H, R =2 \Omega[/latex],and [latex]f_o=350[/latex] Hz. Eq. (7.21),

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

[latex]\omega _r=54,160 rad/s,f_r=54,160/(2π)=8619.9 Hz ,α=20,000, T_r= 1/8619.9=116 μs, t_1 =116/2=58 μs[/latex], and [latex]T_0=1/3500=285.72 μs[/latex]. The off-period of load current is [latex]t_d = T_0 - T_r = 285.72 - 116 = 169.72 μs[/latex], and the inverter would operate in the nonoverlap mode.

Mode 1. This mode begins when [latex]Q_1[/latex] and [latex]Q_2[/latex] are turned on. A resonant current flows through [latex]Q_1, Q_2[/latex], load, and supply. The equivalent circuit during mode 1 is shown in Figure 7.11a with an initial capacitor voltage indicated. The instantaneous current is described by

[latex]L\frac{di_0}{dt}+Ri_0+ \frac{1}{C}\int{i_0}dt+\upsilon _c(t=0)=V_s[/latex]

with initial conditions [latex]i_0(t = 0) = 0, v_{c1}(t = 0) = -V_c[/latex], and the solution for the current gives

[latex]i_0(t)=\frac{V_s+V_c}{\omega _rL}e^{-\alpha t}\sin \omega _rt[/latex] (7.28)

[latex]v_c(t) = - (Vs + Vc)e^{-αt}( α \sin ω_rt + ω_r \cos ω_rt) + V_s[/latex] (7.29)

Devices [latex]Q_1[/latex] and [latex]Q_2[/latex] are turned off at [latex]t_1 =π / ω_r[/latex], when [latex]i_1(t)[/latex] becomes zero.

[latex]V_{c1} = v_c(t = t_1) = (V_s + V_c)e^{-απ/ω_r} + V_s[/latex] (7.30)

Mode 2. This mode begins when [latex]Q_3[/latex] and [latex]Q_4[/latex] are turned on. A reverse resonant current flows through [latex]Q_3, Q_4[/latex], load, and supply. The equivalent circuit during mode 2 is shown in Figure 7.11b with an initial capacitor voltage indicated. The instantaneous load current is described by

[latex]L\frac{di_0}{dt}+Ri_0+ \frac{1}{C}\int{i_0}dt+\upsilon _c(t=0)=-V_s[/latex]

with initial conditions [latex]i_2(t = 0) = 0[/latex] and [latex]v_c(t = 0) = V_{c1}[/latex], and the solution for the current gives

[latex]i_0(t)=\frac{V_s+V_{c1}}{\omega _rL}e^{-\alpha t}\sin \omega _r t[/latex] (7.31)

[latex]v_c(t) = (V_s + V_{c1})e^{-αt}(α \sin ω_rt + ω_r \cos ω_rt)/ω_r - V_s[/latex] (7.32)

Devices [latex]Q_3[/latex] and [latex]Q_4[/latex] are turned off at [latex]t_1 = π/ω_r[/latex], when [latex]i_0(t)[/latex] becomes zero.

[latex]V_c = -v_c(t = t1) = (V_s + V_{c1})e^{-απ/ω_r }+ V_s[/latex] (7.33)

Solving for[latex] V_c[/latex] and [latex]V_{c1}[/latex] from Eqs. (7.30) and (7.33) gives

[latex]V_c=V_{c1}=V_s\frac{e^z+1}{e^z-1}[/latex] (7.34)

where [latex]z=\alpha \pi /\omega _r[/latex]. For [latex]z = 20,000π/54,160 = 1.1601[/latex], Eq. (7.34) gives [latex]V_c = V_{c1} = 420.9 V[/latex].

a. From Eq. (7.7),

[latex]t_m=\frac{1}{\omega _r} an^{-1}\frac{\omega _r}{\alpha }[/latex] (7.7)

[latex]t_m=\frac{1}{54,160} an^{-1}\frac{54,160}{20,000}=22.47 \mu s[/latex]

From Eq. (7.28), the peak load current [latex]I_p = i_0(t = t_m)= 141.64 A[/latex].

b. A device conducts from a time of t1. The average device current can be found from Eq. (7.28):

[latex]I_A=f_o\int_{0}^{t_1}{i_0(t)dt}=17.68 A[/latex]

c. The rms device current can be found from Eq. (7.28):

[latex]I_R=\left[f_o\int_{0}^{t_1}{i_0^2(t)dt} ight]^{1/2} =44.1 A[/latex]

d. The rms load current is [latex]I_o = 2I_R = 2 × 44.1 = 88.2 A[/latex].

e. [latex]P_o = 88.2^2 × 2 = 15,556 W[/latex] and the average supply current, [latex]I_s = 15,556/220 = 70.71 A[/latex].

Question 7.5: Analysis of the Full-Bridge Resonant Inverter with Bidirecti...

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