• In?rectification, to produce a steady?direct current? or voltage from an?alternating current??or voltage, a?smoothing capacitor?is necessary
• Smoothing is defined as:

The reduction in the variation of the output voltage or current

• This works in the following ways:
  • A single capacitor with?capacitance?C is connected in parallel with a load resistor of resistance R
  • The capacitor charges up from the input voltage and maintains the voltage at a high level
  • As it discharges gradually through the resistor when the rectified voltage drops but the voltage then rises again and the capacitor charges up again

A smoothing capacitor connected in parallel with the load resistor. The capacitor charges as the output voltage increases and discharges as it decreases

• The resulting graph of a smoothed output voltage?V_out?and output current against time is a ‘ripple’ shape

A smooth, rectified current graph creates a ‘rippling’ shape against time

• The amount of smoothing is controlled by the capacitance C of the capacitor and the resistance R of the load resistor
  • The less the rippling effect, the smoother the rectified current and voltage output
• The slower the capacitor discharges, the more the smoothing that occurs ie. smaller ripples
• This can be achieved by using:
  • A capacitor with?greater?capacitance?C
  • A resistance with larger resistor?R
• Recall that the product RC is the?time constant?τ of a resistor
• This means that the time constant of the capacitor must be?greater than the time interval?between the adjacent peaks of the output signal

Step 1:???????????

Calculate the time constant with the 60 pF capacitor

τ = RC = (2.6 × 10³) × (60 × 10^-12) = 1.56 × 10^-7?s? = 156 ns

Step 2:???????????

Compare time constant of 60 pF capacitor with interval between adjacent peaks?of the output signal

• • The time interval between adjacent peaks is 80 ms
  • The time constant of 156 ns is too small and the 60 pF capacitor will discharge far too quickly
  • There would be no smoothing of the output voltages
  • Therefore, the 60 pF capacitor is?not suitable

Step 3:???????????

Calculate the time constant with the 800 μF capacitor

τ = RC = (2.6 × 10³) × (800 × 10^-6) = 2.08 s

Step 4:

Compare time constant of 60 pF capacitor with interval between adjacent peaks of the output signal

• • The time constant of 2.08 s is much larger than 80 ms
  • The capacitor will not discharge completely between the positive cycles of the half-wave rectified signal
  • Therefore, the?800?μF capacitor?would be suitable for the smoothing task