Semiconductor Fundamentals Unit 3 – Full-Wave Rectification and Filtering
Exercise 1 – Full-Wave Diode Bridge Rectification
EXERCISE OBJECTIVE
When you have completed this exercise, you will be able to demonstrate full-wave rectification
by using a full-wave bridge rectifier circuit. You will verify your results with an oscilloscope and
a multimeter.
EXERCISE DISCUSSION
- Two types of circuits that utilize diodes for full-wave rectification are illustrated. One
employs a center-tapped full-wave rectifier while the second uses a full-wave bridge rectifier.
The full-wave bridge circuit is used in this exercise. - Diode bridges contain four diodes, designated D1 through D4, configured so that two diodes
conduct during each half-cycle of the input ac signal and produce a pulsating dc output. - The pulsating dc output flows through the load resistance in one direction, independent of
which ac cycle the current is derived. - Two input terminals, usually labeled with a sine wave symbol, and two output terminals,
labeled with positive and negative symbols, are present on the bridge rectifier. - Diodes D1 and D3 are forward biased during the positive half-cycle of the ac input signal.
- Diodes D2 and D4 are forward biased during the negative half-cycle of the ac input signal.
- Each diode pair conducts for one half-cycle of the ac input signal, resulting in full-wave
rectification. - Since there are two dc pulses for one complete cycle of the input ac waveform, the output
pulse frequency of a full-wave rectifier is twice the ac input frequency. - The following relationships apply to full-wave diode bridge rectifiers.
Peak output voltage (Vo(pk)) equals the peak input voltage (Vi(pk)) minus the forward
voltage drop (VF) of the two conducting diodes.
Vo(pk) = Vi(pk) − 2VF
Output rms voltage (Vo(rms) ) equals 0.707 times the peak output voltage.
Vo(rms) = 0.707 x Vo(pk)
Output average (Vo(avg)) voltage equals 0.636 times the peak output voltage.
Vo(avg) = 0.636 x Vo(pk)