Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

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.

Step-by-Step

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.