Question 15.4: The input to a bridge rectifier is through a step-down trans......
The input to a bridge rectifier is through a step-down transformer of turn ratio 10:1. The supply voltage is 230 V at 50 Hz. The load resistance is 1.2 kΩ secondary winding resistance of the transformer is 4 Ω diode forward resistance is 2 Ω. Calculate the efficiency of the bridge rectifier.
Given V_i (RMS) = 230 V
\mathrm{R}_{\mathrm{F}}=2 \Omega, \mathrm{R}_2=4 \Omega, \mathrm{R}_{\mathrm{L}}=1200 \OmegaThe RMS value of the emf in transformer secondary
\begin{aligned} \mathrm{V}_{\mathrm{s}}(\mathrm{RMS}) & =230\left(\frac{\mathrm{N}_2}{\mathrm{~N}_1}\right) \\ & =230\left(\frac{1}{10}\right) \\ & =23 \mathrm{~V} \end{aligned}Peak secondary voltage, V_m is
\begin{aligned} \mathrm{V}_{\mathrm{m}} & =\sqrt{2} \mathrm{~V}_{\mathrm{s}}(\mathrm{RMS}) \\ & =\sqrt{2} \times 23 \\ & =32.5 \mathrm{~V} \end{aligned}Current through the load will flow from the transformer secondary via two diodes. Therefore, following the current path during the positive half cycle
\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{L}}+2 \mathrm{R}_{\mathrm{F}}+\mathrm{R}_2}=\frac{32.5}{1200+4+4}=26.8 \mathrm{~mA}For a bridge rectifier,
\mathrm{I}_{\mathrm{dc}}=\frac{2 \mathrm{I}_{\mathrm{m}}}{\pi}=\frac{2 \times 26.8}{3.14}=17 \mathrm{~mA}DC power output, \mathrm{P}_{\mathrm{dc}}=\mathrm{I}_{\mathrm{dc}}^2 \mathrm{R}_{\mathrm{L}}=\left(17 \times 10^{-3}\right)^2 \times 1200
= 346.8 mW
AC power input,
\begin{aligned} P_{\mathrm{ac}} & =\left(\mathrm{I}_{\mathrm{ms}}\right)^2\left(\mathrm{R}_{\mathrm{L}}+2 \mathrm{R}_{\mathrm{F}}+\mathrm{R}_{\mathrm{S}}\right) \\ & =\left(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\right)^2\left(\mathrm{R}_{\mathrm{L}}+2 \mathrm{R}_{\mathrm{F}}+\mathrm{R}_{\mathrm{S}}\right) \\ & =\left(\frac{26.8 \times 10^{-3}}{2}\right)^2 \times(1200+2 \times 2+4) \\ & =432 \mathrm{~mW} \end{aligned}Rectifier efficiency, \eta=\frac{P_{d c}}{P_{a c}} \times 100=\frac{346.8 \times 100}{432}=80 \text { per cent }
