CIE A Level Physics (9702) 2019-2021

Revision Notes

26.2.2 Smoothing


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

Worked example

Smoothing_Worked_example_-_Smoothing_Question, downloadable AS & A Level Physics revision notes

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

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