Power Electronics and Motor Drive Systems

2.11. Single-Phase Half–Wave Diode Rectifier With R–L–E Load

In the circuit of Fig. 2.17(a), when the switch S is closed and the diode D starts conducting, the circuit current is given by:

$\begin{matrix} i_{i} = steady - state component + transient component + dc component \\ = i_{F} + i_{N} + dc component \end{matrix}$

(2.110)

where

$i_{F} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ωt - φ) = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ωt - φ)$

(2.111)

$i_{N} = A e^{- (\frac{ωt}{ωτ})}$

(2.112)

Substituting Eqs. (2.111) and (2.112) into Eq. (2.110) yields:

$i_{i} = i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ωt - φ) + A e^{- (\frac{ωt}{ωτ})} - \frac{E}{R} for α \leq ωt \leq β$

(2.113) ...

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