Single-Phase Diode Rectifier with RL-Load & Freewheeling Diode
The effect of using freewheeling diode are:
Prevents the load voltage from becoming negative
As the energy stored in \(L\) transfers to \(R\) through FD, the system efficiency is improved
The load current is more smooth improving the performance
Single-Phase Diode Rectifier with RLE-Load
Diode will remain off as long as the voltage of the ac source is less than the dc voltage
Diode will conduct at certain \(\alpha\) when \[\begin{aligned} V_m\sin\alpha & = V_{dc} \\ \alpha & = \sin^{-1}\left(\dfrac{V_{dc}}{V_m}\right) \end{aligned}\]
KVL equation: \[V_m\sin(\omega t)=Ri(t)+L\frac{di(t)}{dt}+V_{\mathrm{dc}}\]
Total current response: \[\begin{aligned} i(t) & = i_f(t)+i_n(t) \\ i_f(t) &=\frac{V_m}{Z}\sin(\omega t-\theta)-\frac{V_{\mathrm{dc}}}{R}\\ i_n(t) &=Ae^{-t/\tau} \end{aligned}\]
Complete response: \[\begin{aligned} i(\omega t) &=\begin{cases}\dfrac{V_m}{Z}\sin\left(\omega t-\theta\right)-\dfrac{V_{dc}}{R}+Ae^{-\omega t/\omega\tau}&\text{for}\:\alpha\leq\omega t\leq\beta\\\\0&\text{otherwise}\end{cases}\\ \beta &= \text{Extinction angle} \end{aligned}\]
Solving for \(A\) by substituting \(i(\alpha)=0\) \[A=\left[-\frac{V_m}{Z}\sin(\alpha-\beta)+\frac{V_\mathrm{dc}}{R}\right]e^{\alpha/\omega\tau}\]
The average power absorbed by the resistor \(\Rightarrow I_{rms}^2R\)
\[I_{\mathrm{rms}}=\sqrt{\frac{1}{2\pi}\int\limits_{\alpha}^{\beta}i^2(\omega t)d(\omega t)}\]
The average power absorbed by the dc source \(\Rightarrow~P_{dc} = I_0V_{dc}\)
\[I_o=\frac{1}{2\pi}\int_\alpha^\beta i(\omega t)d(\omega t)\]
The power supplied by the ac source (assuming ideal condition for diode and inductor):
\[\begin{aligned} P_{\mathrm{ac}} & =I_{\mathrm{rms}}^2R+I_oV_{\mathrm{dc}}\\ &=\frac{1}{2\pi}\int\limits_{0}^{2\pi}\nu(\omega t)i(\omega t)d(\omega t)=\frac{1}{2\pi}\int\limits_{\alpha}^{\beta}(V_{m}\sin\omega t)i(\omega t)d(\omega t) \end{aligned}\]