Question 15.3: A half-wave rectifier circuit has been made using a step-dow......
A half-wave rectifier circuit has been made using a step-down transformer of turn ratio 10:1. The input voltage is ν = 325 sin ωt the diode forward resistance is 25 Ω. A load resistance of 1.2 kW has been connected in the circuit. Assuming a secondary winding resistance of the transformer as 1Ω, calculate the following: (a) Rms value of load current (b) rectification efficiency, and (c) ripple factor.
Input voltage, ν = 325 sin ωt
Input, V_m = 325
Transformer has a turn ratio of 10:1
The output \mathrm{V}_{\mathrm{m}}=\frac{325}{10}=32.5 V
I_m=\frac{V_m}{R_2+R_F+R_L}where R_2 is the secondary winding resistance, R_F is the forward resistance of the diode and R_L is the load resistance.
\begin{aligned} \mathrm{I}_{\mathrm{m}} & =\frac{32.5}{1+25+1200} \\ & =\frac{32.5}{1226} \mathrm{~A} \\ & =26.5 \mathrm{~mA} \end{aligned}Since it is a half-wave rectifier circuit,
\begin{aligned} & \mathrm{I}_{\mathrm{ms}}=\frac{\mathrm{I}_{\mathrm{m}}}{2}=\frac{26.5}{2}=13.25 \mathrm{~mA} \\ & \mathrm{I}_{\mathrm{dc}}=\frac{\mathrm{I}_{\mathrm{m}}}{\pi}=\frac{26.5}{3.14}=8.44 \mathrm{~mA} \end{aligned}output dc power =\mathrm{I}_{\mathrm{dc}}^2 \mathrm{R}_{\mathrm{L}}=\left(8.44 \times 10^{-3}\right)^2 \times 1200
= 85.48 mW
AC input power = \left(\mathrm{I}_{\mathrm{mss}}\right)^2\left[\mathrm{R}_2+\mathrm{R}_{\mathrm{F}}+\mathrm{R}_{\mathrm{L}}\right]
= \left(13.25 \times 10^{-3}\right)^2 \times 1226
= 0.215 W = 215 mW
