Full-bridge inverter RLC load, THD

The square-wave output of a full-bridge inverter is $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$. Each harmonic current is $V_{nm}/Z_n$, with the series RLC impedance evaluated at $n\omega$.

Given: $R = 10\ \Omega$, $L = 31.5$ mH, $C = 112\ \mu$F, $f = 50$ Hz ($\omega = 314.16$ rad/s), $V_s = 220$ V.

$$\begin{aligned} V_{nm} &= \frac{4\times220}{n\pi} = \frac{280.11}{n}\ \text{V} \\ Z_n &= \sqrt{R^2 + \Big(n\omega L - \frac{1}{n\omega C}\Big)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L - 1/(n\omega C)}{R} \end{aligned}$$

(i) Instantaneous load current (Fourier series)

$$\begin{aligned} i_0(t) ={}& 13.31\sin(314.16t + 61.64^\circ) + 4.14\sin(3\omega t - 63.68^\circ) \\ &+ 1.25\sin(5\omega t - 77.14^\circ) + 0.61\sin(7\omega t - 81.28^\circ) \\ &+ 0.36\sin(9\omega t - 83.36^\circ) + \dots\ \text{A} \end{aligned}$$

At the fundamental the load is capacitive ($X_C > X_L$), so $i_1$ leads; at higher harmonics it is inductive, so they lag.

(ii) RMS of fundamental load current

$$I_1 = \frac{13.306}{\sqrt2} = 9.41\ \text{A}$$

(iii) THD of load current

RMS of each harmonic: $I_3 = 2.93$, $I_5 = 0.88$, $I_7 = 0.43$, $I_9 = 0.25$ A.

$$\begin{aligned} I_0 &= \sqrt{9.41^2 + 2.93^2 + 0.88^2 + 0.43^2 + 0.25^2} = 9.906\ \text{A} \\ THD &= \frac{\sqrt{I_0^2 - I_1^2}}{I_1} = \frac{\sqrt{2.93^2 + 0.88^2 + 0.43^2 + 0.25^2}}{9.41} \\ &= \frac{3.098}{9.41} = 0.329 = 32.9\% \end{aligned}$$

Including all higher harmonics raises it only slightly (about 33.0%).

Answer: $I_{1,rms} = 9.41$ A; $THD_i \approx 32.9\%$ (harmonics up to 9th).