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Electronics 1 (EEEN-3421)
Section: 101
Experiment No. 2 : Diode Rectifiers
By:
Faisal Al-Hunsil
200800011
Hussian Al haji
201002375
Hussain Al-marhoon 200600719
To :
Mr. Muhammad Omer
Introduction:
Two of the most important diode application are rectification and regulation. The
primary function of rectification circuits is to establish a DC level from an AC input
signal. Rectifier circuits play an important role in power application and
communication signal detection. Regulation is needed to maintain a constant dc
voltage source and/or load conditions.
Theory :
With four diodes, you can make both halves of the waves positive. This is called a
full-wave rectifier diode bridge. For both positive and negative swings of the input
there is a forward path through the diode bridge. While two of the diodes are forward
biased, the other two are reverse biased and effectively eliminated from the circuit.
Both conduction paths cause current to flow in the same direction through the load
resistor, accomplishing full-wave rectification. Full-wave rectifiers are used in
power supplies to convert AC voltages to DC voltages. A large capacitor in parallel
with the output load resistor reduces the ripple from the rectification process.
Purpose :
In the first part of this experiment we will investigate the half-wave rectification
circuits. In the second part the effect of filter capacitor will be investigated. In the
third part a full wave rectifier circuits will be investigated.
Table 1:
Step
Measurement
1 Measurement diode and resistor
VD= 0.58v
R= 10.36 kΩ
2 contact the diode circuit
Vspp= 10.00v
3 Using oscilloscope
Vspeak= 5.0v
frequency= 49.92 Hz
vopeak= 4.4v
vHWmean= 1.32v
4 Using multimeter
6 Conversion
vrC1= 0.13v
vmeanC1= 4.05V
7 With lowe c2
Vrc2= 0.68v
vfmean=c2= 3.91v
10 Photo oscilloscope
11 Using multimeter
12 Add a capacitor
vFWmean= 2.494v
VrCFW= 0.035V
VmeanCFW= 3.945V
Table 2: Mean, Ripple Measurement Summary
Parameter
Half Wave
Full Wave
Mean v0 without c
1.32v
2.494v
Mean v0 with c=100µF
4.05v
3.945v
Mean v0 with c=10µF
3.91v
None
Ripple, Vr with c=100 µF
0.13v
0.035v
Ripple, Vr with c=10 µF
0.68v
None
Conclusion :
For the half wave voltage, the difference between Vs and Vo
is 0.6 volts which is the voltage required for the diode to
operate. The capacitor increases Vo for half wave and full
wave. The less of value the capacitor has the higher the
voltage will be.