10 best free blogging sites

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aliquam ac vestibulum purus. Donec id tellus consequat, consequat sapien in, fermentum tellus. Pellentesque pulvinar tortor ac magna pellentesque eleifend. Nullam mauris ipsum

Mathematical Analysis

Let e = E0 + sin wt be the input voltage to the rectifier, where E0 be the peak input voltage. When the diode is conducting, let id be the current flowing through the circuit and let the voltage drop across the diode be ed.
Assuming the ohmic resistance of the secondary of the transformer as negligible and applying Kirchhoff's voltage law to this circuit we have,
[ \begin{array}{l} e = e_d + i_d R_L \ = i_d R_f + i_d R_L \end{array} ]
Where Rf is the forward dynamic resistance of the diode.
[ \begin{aligned} \therefore \; e &= i_d (R_f + R_L) \ \text{i.e.,} \quad E_o \sin \omega t &= i_d (R_f + R_L) \ \text{i.e.,} \quad i_d &= i_m \sin \omega t \end{aligned} ]
where ( i_m = \frac{E_o}{R_f + R_L} ) is the peak current.
If the resistance of the diode is negligible compared to RL, then
[{i_m} = \frac{{{E_o}}}{{{R_L}}}]
For a half wave rectifier, we have, (Refer the second graph)
( i_d = \begin{cases} \sin \omega t & \text{for } 0 < \omega t < \pi \ 0 & \text{for } \pi < \omega t < 2\pi \end{cases} )