Half wave rectifier
OBJECTIVE
• To study the half-wave rectifier.
EQUIPMENT REQUIRED
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Base unit for the IPES system.
Power supply module PSU/EV.
Module holder structure MU/EV.
Individual Control Unit module SIS3/EV.
Experiment module MCM3/EV.
Multimeter.
Oscilloscope.
BASIC THEORY
The rectifier circuit converts the AC voltage furnished by the utilities company into the DC voltage
required to operate electronic equipment. Many common electrical products use voltages provided by a
rectifier. Without the rectifier to convert the AC voltage to the DC voltage required to operate these
electrical units, it would be virtually impossible to have the conveniences that we enjoy today. A
television without a rectifier would require several extremely large batteries. These batteries would
have to be large because of the current that is required. In other words, a television without the rectifier
would be so large that it would occupy an entire room. The rectifier is the heart of the electronic unit.
PRINCIPLE OF CIRCUIT DESIGN
The input block consists of a transformer, normally a power transformer that receives the AC
input signal from some power source. The transformer transfers the electrical energy received to
the rectifier section by electromagnetic induction or mutual inductance. The transformer
performs the transfer of energy without any change in frequency, but it is able to change the
voltage and current from the input source to the voltage and current required by the rectifier
section. The phase relationship of the current in the secondary of the transformer is dependent
upon the phase of the voltage in the primary winding and the direction of the winding in the
secondary. If the secondary windings are wound in the same direction as the primary
windings, the phase between the input signal and the output signal will be the same. If the
secondary windings are wound in the opposite direction of the primary windings, the phase
between the input signal 26 and the output signal will be 180 degrees out of phase. The
schematic drawings of a transformer indicate the phase relationship between the primary and
secondary with the use of dots. The dots on a schematic diagram indicate which windings are in
phase. Figure 2 illustrates this relationship.
The circuit shown in Fig. C4-4 consists of a step-down transformer (240 V to 6.3V) and a half-wave
rectifier. A rectifier is a device that changes an AC (alternating) voltage (average voltage zero) to a DC
voltage (average value non-zero). The transformer secondary output is a sinusoidal signal of nominal
amplitude 8.9V (6.3Vrms) and frequency 50Hz. During the positive half cycles of operation, the diode
conducts so the output signal appears across the load resistor RL. During the negative half cycles the
diode does not conduct so the voltage across RL is zero and a negative voltage drop appears across the
diode. The voltage across RL is therefore a succession of approximate half sinewaves. The circuit here is
known as a half-wave rectifier for reasons which should become clear.
Procedure
1- connect the circuit as shown above.
2. The transformer that will be used is on the bench. Connect the circuit shown in Fig. C4-4 using a value
of RL = 470 and the diode which is about 5 cm long and black - ask your tutor if you’re unsure which it is,
or of its polarity. Using the DC inputs to the oscilloscope, sketch the transformer secondary voltage and
the voltage across RL, labeling axes carefully.
3- Calculate the average DC voltage across RL (try integration). Obtain a voltmeter from the service
counter. Set it for DC and connect it across RL. Compare its reading with the calculated average value.
Why does the DC meter give a steady reading even though the voltage measured is varying with time?
4. Sketch the voltage across the diode vs time (display this voltage on the oscilloscope). Make sure you
get the zero level right! A high oscilloscope sensitivity will be necessary to observe the forward voltage
drop. Comment on the shape of the waveform when the diode is conducting and compare with the
characteristic curve in Fig. C4- 2(c).
1- Construct the circuit of Fig. 2.1 where V is the voltmeter. Note that the resistor limits the
current to a safe value.
2- Switch on the oscilloscope and the sinusoidal supply.
3- With the oscilloscope D.C. coupled adjust the time-base and the Y amplifier
sensitivity
to obtain a steady trace of about 4cm vertical and 5ms/cm horizontal.
4- Measure and record time T and peak voltage Vpk:
5- Sketch the waveform and label it to show the periods when the diode is conducting
and those when it is not. Time T depends upon the frequency of your power supply.
6- Confirm this. Vpk should be very nearly equal to the peak voltage of the alternating
supply.
Questions
1. Why will Vpk not be exactly equal to the peak value of the supply?
2. How much is the difference between the measured and theoretical mean voltage?
(Hint: The mean value of a half-sinusoid can be shown by geometry to be :
𝑉𝑝𝑘
√2
.
But at every half-cycle the voltage is zero. The mean value of the waveform,
𝑉
𝑝𝑘
therefore is: 2√2
. Note the mean voltage indicated by the voltmeter, and compare it
with 0.35Vpk.)
3. The mean voltage you obtain is positive relative to zero. How could you obtain a
negative voltage?
(Hint: Verify your answer by experiment and sketch the waveform.)