Electronic – Half Wave Rectifier with Capacitive Filter

circuit analysiscurrentdiodesrectifiervoltage

problem

This is an example problem in my workbook. Please can someone explain me the working of the circuit and how is this formula derived.

I got 1 more solution to the same problem.

ssolution2

Best Answer

I think your workbook is wrong with that formula. They have used the full wave rectifier formula. For HWR, It has to be :

$$V_{dc} = V_m - I_{dc}/2fC$$

Your derivation is correct.

enter image description here From the above waveform,

$$V_{dc} = V_m - V_{rpp}/2$$ from ripple waveform, the amount of charge stored by the capacitor = The charge lost by it in time T seconds. i.e., $$C V_{rpp}= I_{dc}T$$

which gives, $$V_{rpp} = I_{dc}/fC$$ Therefore,

$$V_{dc} = V_m - I_{dc}/2fC$$

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