Step-down chopper RLE load continuity

The load current of a step-down chopper feeding an RLE load is continuous when $m = E/V_s$ is less than the critical ratio $\dfrac{e^{\alpha T/T_a}-1}{e^{T/T_a}-1}$ (equivalently, when the computed $I_{min} > 0$).

Given: $V_s = 200$ V, $R = 2\ \Omega$, $L = 10$ mH, $E = 20$ V, $T = 1000\ \mu$s, $T_{on} = 300\ \mu$s.

$$\begin{aligned} \alpha &= \frac{300}{1000} = 0.3, \qquad T_a = \frac{L}{R} = \frac{0.01}{2} = 5\ \text{ms} \\ \frac{T}{T_a} &= \frac{1}{5} = 0.2, \qquad \frac{\alpha T}{T_a} = 0.06 \end{aligned}$$

(i) Continuity of load current

$$\begin{aligned} m &= \frac{E}{V_s} = \frac{20}{200} = 0.1 \\ \frac{e^{0.06}-1}{e^{0.2}-1} &= \frac{1.061837-1}{1.221403-1} = \frac{0.061837}{0.221403} = 0.2793 \end{aligned}$$

Since $m = 0.1 < 0.2793$, the load current is continuous.

Cross-check with the current limits:

$$\begin{aligned} I_{max} &= \frac{V_s}{R}\cdot\frac{1-e^{-0.06}}{1-e^{-0.2}} - \frac{E}{R} = 100\times\frac{0.058235}{0.181269} - 10 = 22.13\ \text{A} \\ I_{min} &= \frac{V_s}{R}\cdot\frac{e^{0.06}-1}{e^{0.2}-1} - \frac{E}{R} = 100\times0.27929 - 10 = 17.93\ \text{A} \end{aligned}$$

$I_{min} = 17.93$ A $> 0$, so the current never falls to zero.

(ii) Average load current

$$\begin{aligned} V_0 &= \alpha V_s = 0.3\times200 = 60\ \text{V} \\ I_0 &= \frac{V_0 - E}{R} = \frac{60-20}{2} = 20\ \text{A} \end{aligned}$$

i0 (A)
 22.13|   /\        /\        /\
      |  /  \      /  \      /  \
 17.93| /    \____/    \____/    \__
      |__________________________________ t
       0  0.3   1   1.3   2   (ms)

Answer: (i) continuous ($m = 0.1 < 0.279$); (ii) average load current $I_0 = 20$ A (current swings between 17.93 A and 22.13 A).