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Q. 7.5

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 $I_s$.

Verified Solution

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

$\omega _r=\left(\frac{1}{2LC_2}-\frac{R^2}{4L^2} \right)^{1/2}$

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

Mode 1. This mode begins when $Q_1$ and $Q_2$ are turned on. A resonant current flows through $Q_1, Q_2$, 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

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

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

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

$v_c(t) = – (Vs + Vc)e^{-αt}( α \sin ω_rt + ω_r \cos ω_rt) + V_s$                                                      (7.29)

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

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

Mode 2. This mode begins when $Q_3$ and $Q_4$ are turned on. A reverse resonant current flows through $Q_3, Q_4$, 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

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

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

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

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

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

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

Solving for$V_c$ and $V_{c1}$ from Eqs. (7.30) and (7.33) gives

$V_c=V_{c1}=V_s\frac{e^z+1}{e^z-1}$                    (7.34)

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

a. From Eq. (7.7),

$t_m=\frac{1}{\omega _r}\tan^{-1}\frac{\omega _r}{\alpha }$                                         (7.7)

$t_m=\frac{1}{54,160}\tan^{-1}\frac{54,160}{20,000}=22.47 \mu s$

From Eq. (7.28), the peak load current $I_p = i_0(t = t_m)= 141.64 A$.

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

$I_A=f_o\int_{0}^{t_1}{i_0(t)dt}=17.68 A$

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

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

d. The rms load current is $I_o = 2I_R = 2 × 44.1 = 88.2 A$.

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