Question 15.1: A half-wave diode rectifier has a forward voltage drop, i.e.......
A half-wave diode rectifier has a forward voltage drop, i.e., voltage drop across the diode when conducting is 0.7 V. The load resistance is 600 Ω. The rms value of the ac input is 28.87 V. Calculate I_{dc},\ I_{rms}, PIV, and form factor.
Step-by-Step
\begin{aligned} & \mathrm{V}_{\mathrm{i}}(\mathrm{RMS})=28.27 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{i}}(\max )=\sqrt{2} \mathrm{~V}_{\mathrm{i}}(\mathrm{RMS}) \\ & =1.414 \times 28.27 \\ & \text { i.e., } \mathrm{V}_{\mathrm{m}}=40 \mathrm{~V} \\ & \mathrm{PIV}=\mathrm{V}_{\mathrm{m}}=40 \mathrm{~V} \\ & I_{d c}=\frac{I_m}{\pi} ; \quad I_m=\frac{\mathrm{V}_{\mathrm{m}}-0.7}{\mathrm{R}_{\mathrm{L}}}=\frac{40-0.7}{600}=\frac{39.3}{600} \mathrm{~A} \\ & I_{\mathrm{dc}}=\frac{39.3}{600 \times \pi}=0.0208 \mathrm{~A}=20.8 \mathrm{~mA} \\ & \mathrm{I}_{\mathrm{mms}}=\frac{\mathrm{I}_{\mathrm{m}}}{2} \frac{39.3}{600 \times 2}=0.0327 \mathrm{~A}=32.7 \mathrm{~mA} \\ & \text { Form factor }=\frac{\text { RMS value }}{\text { Average value }}=\frac{I_{\mathrm{ms}}}{\mathrm{I}_{\mathrm{dc}}}=\frac{32.7}{20.8} \\ & =1.57 \\ & \end{aligned}
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