Step-down chopper RLE load, 5 kHz

For a step-down chopper with RLE load in continuous conduction, the load current rises exponentially during $kT$ and decays during $(1-k)T$; the standard steady-state results (Rashid) are used.

Given: $V_s = 220$ V, $R = 10\ \Omega$, $L = 15.5$ mH, $E = 20$ V, $k = 0.5$, $f = 5$ kHz.

$$\begin{aligned} T &= \frac{1}{f} = 200\ \mu\text{s}, \quad kT = 100\ \mu\text{s} \\ \tau &= \frac{L}{R} = \frac{15.5\times10^{-3}}{10} = 1.55\ \text{ms} \\ z &= \frac{TR}{L} = \frac{200\times10^{-6}}{1.55\times10^{-3}} = 0.12903, \quad kz = 0.064516 \end{aligned}$$

Check continuity: $E/V_s = 0.0909$; critical value $\dfrac{e^{kz}-1}{e^{z}-1} = \dfrac{0.066643}{0.137727} = 0.4839$. Since $0.0909 < 0.4839$, current is continuous.

i) Minimum instantaneous load current

$$\begin{aligned} I_1 &= \frac{V_s}{R}\cdot\frac{e^{kz}-1}{e^{z}-1} - \frac{E}{R} = 22\times\frac{1.066643-1}{1.137727-1} - 2 \\ &= 22\times 0.48388 - 2 = 8.645\ \text{A} \end{aligned}$$

ii) Peak instantaneous load current

$$\begin{aligned} I_2 &= \frac{V_s}{R}\cdot\frac{1-e^{-kz}}{1-e^{-z}} - \frac{E}{R} = 22\times\frac{1-0.937521}{1-0.878946} - 2 \\ &= 11.355 - 2 = 9.355\ \text{A} \end{aligned}$$

iii) Maximum peak-to-peak load current ripple

The ripple $\Delta I = I_2 - I_1$ is maximum at $k = 0.5$:

$$\begin{aligned} \Delta I_{max} &= \frac{V_s}{R}\tanh\frac{R}{4fL} = 22\times\tanh\frac{10}{4\times5000\times0.0155} \\ &= 22\times\tanh(0.032258) = 0.709\ \text{A} \end{aligned}$$

Here $k = 0.5$ already, so the actual ripple $9.355 - 8.645 = 0.709$ A equals this maximum. (Approximation for $4fL \gg R$: $V_s/(4fL) = 0.710$ A.)

iv) Average load current

$$I_a = \frac{kV_s - E}{R} = \frac{0.5\times220 - 20}{10} = \frac{110-20}{10} = 9\ \text{A}$$

Answer: $I_{min} = 8.645$ A, $I_{peak} = 9.355$ A, $\Delta I_{max} = 0.709$ A, $I_{avg} = 9$ A.