A half-wave rectifier produces a maximum load current (peak value) of 50 mA through a 1200 Ω resistor. Calculate the PIV of the diode. The diode is of silicon material.

Question

Accepted Answer

Assuming a voltage drop of 0.7 V across the silicon diode, the peak value of current, [latex]I_m[/latex] is

[latex]\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}-0.7}{\mathrm{R}_{\mathrm{L}}}=\frac{\mathrm{V}_{\mathrm{m}}-0.7}{1200}[/latex]

[latex]I_m[/latex] is given as 40 mA.

Therefore,

[latex]40 imes 10^{-3}=\frac{\mathrm{V}_{\mathrm{m}}-0.7}{1200}[/latex]

or, [latex]\mathrm{V}_{\mathrm{m}}-0.7=1200 imes 40 imes 10^{-3}=48 \mathrm{~V}[/latex]

or, [latex]\mathrm{V}_{\mathrm{m}}=48+0.7=48.7 \mathrm{~V}[/latex]

PIV = [latex]V_m[/latex] = 48.7 V

Question 15.2: A half-wave rectifier produces a maximum load current (peak ......

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