Question 17.4: A half-wave rectifier connected to a 50 Hz supply generates ......
A half-wave rectifier connected to a 50 Hz supply generates a peak voltage of 10 V across a 10 mF reservoir capacitor. Estimate the peak ripple voltage produced if this arrangement is connected to a load that takes a constant current of 200 mA.
The voltage V across a capacitor is related to its charge q and its capacitance C by the expression
V = \frac{q}{C}Differentiating with respect to time gives
\frac{dV}{dt} = \frac{1}{C} \frac{dq}{dt} = \frac{i}{C}where i is the current into, or out of, the capacitor. In this example, the current is constant at 200 mA and the capacitance is 10 mF, so
\frac{dV}{dt} = \frac{i}{C} = \frac{0.2}{0.01} = 20 V/sTherefore, the output voltage will fall at a rate of 20 volts per second.
During each cycle, the capacitor is discharged for a time almost equal to the period of the input, which is 20 ms. Therefore, during this time, the voltage on the capacitor (the output voltage) will fall by 20 ms × 20 V/s = 0.4 V.
