Step-down chopper: fundamental of load current

The output of a step-down chopper with resistive load is a rectangular pulse train (height $V_s$, width $kT$). Its Fourier series gives the dc (average) value and the harmonics; the fundamental is at the chopping frequency.

Given: $V_s = 220$ V, $R = 20\ \Omega$, $k = 0.45$, $f = 1$ kHz.

Average output voltage

$$V_a = kV_s = 0.45\times220 = 99\ \text{V}, \qquad I_a = \frac{99}{20} = 4.95\ \text{A}$$

Fourier series of the output voltage

With the pulse from $t = 0$ to $kT$:

$$v_0(t) = kV_s + \sum_{n=1}^{\infty}\frac{V_s}{n\pi}\Big[\sin(2n\pi k)\cos(n\omega t) + \big(1-\cos(2n\pi k)\big)\sin(n\omega t)\Big]$$

Peak of the $n$th harmonic:

$$V_{nm} = \frac{V_s}{n\pi}\sqrt{\sin^2(2n\pi k) + \big(1-\cos 2n\pi k\big)^2} = \frac{2V_s}{n\pi}\sin(n\pi k)$$

Fundamental component ($n = 1$)

$$\begin{aligned} V_{1m} &= \frac{2V_s}{\pi}\sin(\pi k) = \frac{2\times220}{\pi}\sin(0.45\times180^\circ) \\ &= 140.06\times\sin 81^\circ = 140.06\times0.98769 = 138.33\ \text{V} \\ V_{1,rms} &= \frac{138.33}{\sqrt2} = 97.82\ \text{V} \end{aligned}$$

Fundamental load current (resistive load):

$$\begin{aligned} I_{1m} &= \frac{V_{1m}}{R} = \frac{138.33}{20} = 6.917\ \text{A} \\ I_{1,rms} &= \frac{6.917}{\sqrt2} = 4.891\ \text{A} \end{aligned}$$

Phase: $\tan\phi = \dfrac{\sin 2\pi k}{1-\cos 2\pi k} = \tan(90^\circ - 81^\circ)$, so $\phi = 9^\circ$:

$$i_1(t) = 6.917\sin(2000\pi t + 9^\circ)\ \text{A}$$

Answer: $V_a = 99$ V; fundamental load current $= 6.917$ A peak ($4.891$ A rms) at 1 kHz.