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Chapter 7

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.

Step-by-Step

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.