PN junctions and diodes

What is the main purpose of placing a capacitor in parallel with the output of a rectifier?

  1. ATo lower the forward voltage of the diode
  2. BTo raise the frequency of the AC
  3. CTo reduce the ripple of the output voltage and make it closer to DC
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Answer: C. To reduce the ripple of the output voltage and make it closer to DC

The output of a rectifier alone is a pulsating voltage. A parallel capacitor charges when the voltage is high and discharges to support the voltage when it falls, reducing the ripple and giving an output close to DC. This is called a smoothing circuit.

  • A. A capacitor does not change the forward voltage of the diode.
  • B. A capacitor does not change the frequency.
  • C. Correct. The capacitor's charging and discharging flattens the ripple.
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More PN junctions and diodes questions

  1. How does the forward voltage of a silicon diode carrying a constant current change as the temperature rises?
  2. For an ideal diode (ideality factor n = 1) at room temperature, roughly how much must the forward voltage increase to raise the forward current tenfold?
  3. Which expression gives the depletion (junction) capacitance C of a reverse-biased PN junction in terms of junction area A, depletion width W and semiconductor permittivity ε?
  4. Why does no current flow externally in a PN junction in thermal equilibrium?
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