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## Q. 7.4

Analysis of the Half-Bridge Resonant Inverter with Bidirectional Switches

The half-bridge resonant inverter in Figure 7.8a is operated at a frequency of $f_o=3.5 kHz$. If  $C_1=C_2=C=3 \mu F,L_1=L_2=L= 50 \mu H, R = 2 \Omega$, 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

$V_s = 220 V, C_e = C_1 + C_2 = 6 μF, L=50 μH,R=2 Ω$, and $f_o = 3500 Hz$ The analysis of this inverter is similar to that of inverter in Figure 7.3. Instead of two current pulses, there are four pulses in a full cycle of the output voltage with one pulse through each of devices $Q_1, D_1, Q_2$, and $D_2$. Equation (7.20)

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

is applicable. During the positive half-cycle, the current flows through $Q_1$; and during the negative half-cycle the current flows through $D_1$. In a nonoverlap control, there are two resonant cycles during the entire period of output frequency fo. From Eq. (7.21),

$\omega _r=\left(\frac{1}{2LC_2}-\frac{R^2}{4L^2} \right)^{1/2}$                                   (7.21)

$\omega _r=54,160 rad/s$      $f_r=\frac{54,160}{2\pi }= 8619.9 Hz$

$T_r=\frac{1}{8619.9}=116 \mu s$               $t_1=\frac{116}{2} =58 \mu s$

$T_0=\frac{1}{3500}=285.72 \mu s$

$t_d = T_0 – T_r = 285.72 – 116 = 169.72 μs$

Because $t_d$ is greater than zero, the inverter would operate in the nonoverlap mode. From Eq. (7.14),

$V_c=V_s\frac{1+e^{-z}}{e^z-e^{-z}} =V_s\frac{e^{z}+1}{e^{2z}-1}=\frac{V_s}{e^z-1}$                                     (7.14)

$V_c = 100.4 V$ and  $V_{c1} = 220 + 100.4 = 320.4 V$.

a. From Eq. (7.7),

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

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

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

and the peak load current becomes $I_p = i_0(t = t_m) = 70.82 A$.

b. A device conducts from a time of $t_1$. The average device current can be found from

$I_A=f_o\int_{0}^{t_1}{i_0(t)dt}=8.84 A$

c. The rms device current is

$I_R=\left[f_o\int_{0}^{r_1}{i_0^2(t)dt} \right]^{1/2}=22.05 A$

d. The rms load current $I_o = 2I_R = 2 × 22.05 = 44.1 A$.

e. $P_o = 44.12 × 2 = 3889 W$ and the average supply current, $I_s = 3889/220 = 17.68 A$.

Note: With bidirectional switches, the current ratings of the devices are reduced. For the same output power, the average device current is half and the rms current is $1/\sqrt{2}$ of that for an inverter with unidirectional switches. 