CIE A Level Physics (9702) 2019-2021

Revision Notes

26.2.2 Smoothing

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

Author: Katie

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.
Close

Join Save My Exams

Download all our Revision Notes as PDFs

Try a Free Sample of our revision notes as a printable PDF.

Join Now
Go to Top