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Chapter 7

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

The off-period of load current

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.