--- title: "Half-bridge square-wave inverter at 120 Hz" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:square-wave-inverter-120hz" source: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-120hz site: BCT Engineering --- # Half-bridge square-wave inverter at 120 Hz Power Electronics · Chapter 5: Inverter - Asked in: 2069 Chaitra, 8 marks - Video: [Half-Bridge Inverter with MOSFET Switches](https://www.youtube.com/watch?v=jbmI3phNx-0) — Engineering with Prof. Kim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-120hz - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Figure below shows the circuit diagram of single phase square wave inverter with inductive load. The required frequency of output voltage is 120 Hz. Calculate the ON period and OFF period of switches S1 and S2. Derive the time domain equation of load current for first positive half cycle of output voltage. [Figure: half-bridge inverter — two series dc sources +Vs/2 and −Vs/2 with grounded midpoint; transistor switches S1 (upper) and S2 (lower) with antiparallel diodes D1, D2; series R-L load connected between the dc midpoint and the S1–S2 junction, output voltage Vo] ## Short answer At 120 Hz, T=8.333 ms and each switch S1, S2 is ON for 4.167 ms and OFF for 4.167 ms. In the first positive half cycle, i(t)=Vs/(2R)(1-e^-t/tau) - I0 e^-t/tau, with I0=Vs/(2R)(1-e^-T/2tau)/(1+e^-T/2tau); D1 conducts first, then S1. ## Answer In a half-bridge square-wave inverter, S1 and S2 are switched alternately, each for exactly half of the output period, so the load sees $+V_s/2$ and $-V_s/2$ in turn. #### ON and OFF periods of S1 and S2 $$ \begin{aligned} T &= \frac{1}{f} = \frac{1}{120} = 8.333\ \text{ms} \\ T_{ON} &= T_{OFF} = \frac{T}{2} = 4.167\ \text{ms} \end{aligned} $$ - S1: ON for $0 < t < 4.167$ ms, OFF for $4.167 < t < 8.333$ ms. - S2: OFF for $0 < t < 4.167$ ms, ON for $4.167 < t < 8.333$ ms. (In practice a small dead time is left between turning one switch off and the other on, to avoid shoot-through.) ```text S1 gate ____ ____ | | | | ________| |______| |___ S2 gate ______ ___ _____________| |____| 0 T/2 T vo +Vs/2 ____ ____ | | | | -Vs/2 | |______| |__ ``` **Answer: ON period = OFF period = 4.167 ms for each switch.** #### Load current in the first positive half cycle In steady state the current at $t = 0$ is negative, $i(0) = -I_0$ (it flows through D1 until it reverses, then through S1). For $0 \le t \le T/2$, $v_o = +V_s/2$ and the R-L load gives $$ L\frac{di}{dt} + Ri = \frac{V_s}{2} $$ The solution is the steady-state part plus a decaying part, with $\tau = L/R$: $$ i(t) = \frac{V_s}{2R} + A\,e^{-t/\tau} $$ Using $i(0) = -I_0$: $A = -I_0 - \dfrac{V_s}{2R}$, so $$ i(t) = \frac{V_s}{2R}\left(1 - e^{-t/\tau}\right) - I_0\,e^{-t/\tau}, \qquad 0 \le t \le \frac{T}{2} $$ At $t = T/2$ the current reaches $+I_0$ (half-wave symmetry). Putting this in: $$ I_0 = \frac{V_s}{2R}\left(1 - e^{-T/2\tau}\right) - I_0 e^{-T/2\tau} \;\Rightarrow\; I_0 = \frac{V_s}{2R}\cdot\frac{1 - e^{-T/2\tau}}{1 + e^{-T/2\tau}} $$ With $T/2 = 4.167$ ms: $$ i(t) = \frac{V_s}{2R} - \frac{V_s}{2R}\cdot\frac{2}{1 + e^{-0.004167/\tau}}\,e^{-t/\tau}, \qquad 0 \le t \le 4.167\ \text{ms} $$ **Conduction sequence in this half cycle:** from $t = 0$ the current is negative, so D1 conducts and energy returns to the upper source. At $t_1 = \tau \ln\!\left(1 + \dfrac{2R I_0}{V_s}\right)$ the current crosses zero, and S1 then carries the positive current until $T/2$. In the next half cycle D2 and then S2 conduct in the same way with the signs reversed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 120 Hz half-bridge square-wave inverter, what are the ON and OFF periods of S1? A. 8.333 ms ON, 8.333 ms OFF B. 4.167 ms ON, 4.167 ms OFF C. 2.083 ms ON, 6.25 ms OFF D. 4.167 ms ON, 8.333 ms OFF Answer: B. T = 1/120 = 8.333 ms and each switch is ON for T/2 = 4.167 ms and OFF for T/2. (Key point: ON and OFF periods) 2. How are S1 and S2 operated relative to each other? A. Both ON together for half a cycle B. Alternately, each for half the output period (with a small dead time to avoid shoot-through) C. S1 always ON, S2 pulses D. Both ON for a quarter cycle each Answer: B. S1 and S2 are complementary; a dead time prevents both conducting and shorting the source. (Key point: Switch complementarity) 3. For 0 <= t <= T/2 with v_o = +Vs/2 on an R-L load, the current is: A. i = (Vs/2R)(1 - e^(-t/tau)) - I0 e^(-t/tau), tau = L/R B. i = (Vs/2R) e^(-t/tau) C. i = (Vs/R)(1 - e^(-t/tau)) D. i = (Vs/2R) sin(wt) Answer: A. Solving L di/dt + R i = Vs/2 with i(0) = -I0 gives steady state plus a decaying term. (Key point: Load current equation) 4. Why does the diode D1 conduct at the beginning of the positive half cycle? A. The source voltage is negative B. S2 is still ON C. The load current is still negative (i(0) = -I0) and flows back through D1 to the source D. The inductor is fully discharged Answer: C. The inductive current cannot reverse instantly; it flows through D1 until it crosses zero, then S1 carries it. 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