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Laboratory #3
Diode Basics and Applications (I)
I. Objectives
1.
2.
Understand the basic properties of diodes.
Understand the basic properties and operational principles of some
diode-rectifier circuits.
II. Components and Instruments
1.
Components
Transformer × 1, diode (IN4004) × 4, 1 KΩ × 1, 4.7 KΩ × 1, 100
μF × 2, 33μF × 1
2.
Instruments
(1) Function generator
(2) Digital multimeter
(3) Oscilloscope
III. Reading
Concentration will be on Section 3.5 of the Textbook ‘Microelectronic
Circuits Sixth Edition’, a familiarity with the fundamentals of diode being
assumed.
IV. Preparation
1.
Power supply
One of the most important applications of diodes is in the design
of rectifier circuits. A diode rectifier forms an essential building block of the
dc power supplies required to power electronic equipment. A block
diagram of such a power supply is shown in Fig. 3.1. The first block is the
power transformer. It can be used to step the line voltage down to the
value required to yield the particular dc voltage output of supply. And it can
also provide electrical isolation between the electronic equipment and the
power-line circuit. The diode rectifier converts the input sinusoid to a
unipolar output. And the filter makes the output more constant. The
regulator regulates the output against the variation in different condition.
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2.
Fig. 3.1 A block diagram of such a power supply
The relation among Vmax,Vrms,Vav
The relation among maximum voltage (Vmax), peak-to-peak
voltage (Vp-p), Root-Mean-Square voltage (Vrms) and average voltage
(Vav) of a sine wave is shown in Fig. 3.2.
V
0.707 Vmax
0.637 Vmax
Vmax
Vrms
Vav
t
Vp-p
Fig. 3.2 The relation among Vmax, Vp-p, Vrms and Vav of a sinusoid
(1) Maximum voltage (Vmax)
The maximum voltage value of a waveform in one period is
also called peak value.
(2) Peak-to-peak voltage (Vp-p)
The definition of peak-to-peak value is the differential value
between the maximum and minimum value in one period. The
value of Vp-p is equal to 2 times of Vmax.
(3) Root-Mean-Square voltage (Vrms)
Root-mean-square or RMS voltage is the voltage used for
determining the amount of power that a sinusoidal signal is
capable of providing. RMS voltage is defined as follows, where
Vmax is the peak voltage in the AC or sinusoidal signal.
Vrms = Vmax / √2
Note that this equation is only valid for sinusoidal signals.
Using the RMS voltage of a sinusoidal signal, you can determine
the power provided by that signal using the following equation.
P = Vrms2 / R
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(4) Average voltage (Vav)
The relationbetween Vav and Vmax
1 2
2t
2
Vmax sin(
)dt  Vmax

T 2 0
T

T
Vav 
The relationbetween Vav and Vrms
Vav 

3.
2

Vm ax  0.637Vm ax and Vrms 
Vm ax
2
Vrms
V
 1.11 or av  0.9
Vav
Vrms
Half-wave rectifier
The half-wave rectifier utilizes alternate half-cycles of the input of
sinusoid. Fig. 3.3 shows the circuit of a half-wave rectifier. The circuit
consists of the series connection of a diode and a resistor.
During the positive half-cycles of the input sinusoid, the positive
vac will cause current to flow through the diode in its forward direction,
and assume the diode to be ideal. The output voltage will be equal to
the input voltage. On the other hand, during the negative half-cycles
of vac, the diode will not conduct. Thus output voltage will be zero.
Note that while vac alternates in polarity and has a zero average
value, output value is unidirectional and has a finite average voltage
VDC. The relation between vac and VDC is shown in the following
equation.
VDC ,av  0.9 Vrms  2  0.45 Vrms  0.45 vac
Making diode on
Making diode on
+
AC input
vac
IDC
VDC
RL
-
Fig. 3.3 A half-wave rectifier circuit
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4.
Full-wave rectifier
The full-wave rectifier utilizes both halves of the input sinusoid.
To provide a unipolar output, it inverts the negative halves of the sine
wave. Fig. 3.4 shows the circuit of a full-wave rectifier. When the input
line voltage is positive, both of the signals labeled vac1 and vac2 will be
positive. In this case D1 will conduct and D2 will be reverse biased.
During the negative half cycle of ac line voltage, both of the signals
labeled vac1 and vac2 will be negative. The D1 will be cut off while D2
will conduct.
Note that the current through the resistor always flows in the
same direction, and the output voltage will be unipolar. The relation
between vac and VDC is shown in the following equation.
v ac1  v ac 2  v ac , VDC  0.9 v ac
IDC
D1
+
AC input
vac1
VDC
RL
-
vac2
D2
Fig. 3.4 A Full-wave rectifier circuit
5.
Bridge rectifier
An alternative implementation of the full-wave rectifier is shown
in Fig 3.5. The circuit doesn’t require a center-tapped transformer, a
distinct advantage over the full-wave rectifier circuit of Fig. 3.4. The
bridge rectifier requires four diodes as compared to two in the
previous circuit. Observe the Fig 3.6 that there are two diodes in
series in the conduction path, and thus output voltage will be lower
than input voltage by two diode drops.
AC input
D1
D4
IDC
+
vac
D2
D3
RL
-
Fig. 3.5 A bridge rectifier circuit
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Positive cycle
Negative cycle
+
-
D1
D2
-
+
RL
-
D4
D3
+
+
RL
-
Fig. 3.6 The conduction paths in positive and negative cycles
6.
The rectifier with a filter capacitor
The pulsating nature of the output voltage produced by the
rectifier discussed above makes it unsuitable as a dc supply for
electronic circuits. A simple way is to place a capacitor across the
load resistor. To see how the rectifier circuit with a filter capacitor
works, the simple circuit is shown in Fig. 3.7. And the analysis of
output waveform is shown in Fig. 3.8. And the Fig. 3.9 is the output
waveform when output without load resistor. And the value of the
ripple peak-to-peak can be calculated in the following equation.
V p  p  Vmax  Vmax  e
 Vmax (1  (1 

t
RC
 Vmax (1  e

T1
RC
)  Vmax[1  e

T
RC
]
Vmax
V f
T
)) 
 max 0
RC
f in  RC
f in
Charging capacitor
Discharging capacitor
Vin
Vin
Vout
D
+
C
-
Vout
D
RL
+
C
-
RL
Fig. 3.7 A rectifier circuit with a filter capacitor
Vc = Vout
Vc
Discharging
Vmax
Vp-p
Charging
t
t
T1
T
Vin
Fig. 3.8 Analysis of output waveform
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Fig. 3.9 When RL=∞
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7.
Ripple
The definition of ripple factor is shown in the following equation.
V
Vrms of ripple Vripple
Vripple 

and
2 2
Vav of ripple VDC
Assume that the ripple is sinusoid to simplify the calculation. And the
information of the ripple is shown in Fig. 3.10.
Ripple factor 
V
Vmax
ΔV
Vmin
VDC
t
0
Fig. 3.10 Illustration of ripple
8.
Voltage doubler
The half-wave voltage doubler is shown in Fig. 3.11. During the
negative half cycle of ac line voltage vac, the voltage across the
capacitor C1 will be charged to the value about Vmax by the current
through the diode D1. When the positive half cycle of vac, it connects
series with the C1 and supplies the 2 times value of Vmax to the load.
The input vac and output Vout waveforms are shown in Fig. 3.12.
-
C1
D2
Vout
+
vac
+
D1
C2
RL
-
Fig. 3.11 A half-wave voltage doubler
Vout
vac
Vmax
2 Vmax
t
Vmax
VC1
0
Fig. 3.12 The input vac and output Vout waveforms
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V. Explorations
1.
Half-wave rectifiers
(1) Connect the circuit as shown in Fig. 3.13 and use your DMM to
measure the ac voltage of node B (vbc) and the dc voltage of
node D (VDC) which is across the resistance load RL.
(2) Using the oscilloscope to measure the waveforms of node B (Vbc)
and D (Vdc), and observe the waveforms at the same time. Note
the relationship between the waveform of node B (Vbc) and D
(Vdc), and the effect of the diode, both of its orientation and of its
voltage drop.
A
1N4004
B
AC 110 V
60 Hz
D
AC 12 V
RL=1 kΩ
AC 6 V
C
Fig. 3.13
(3) Connect the circuit as shown in Fig. 3.14 and use your DMM to
measure both the dc voltage VDC and the ac voltage vdc which
are across the resistance load RL. What’s the ripple factor value?
And measure the waveform across RL by using the oscilloscope.
(4) Change the capacitor CL to 33μF of the circuit shown in Fig.
3.14. What’s the difference after the alteration? Observe the
differences by using oscilloscope.
(5) What’s the frequency of the ripple?
A
1N4004
AC 110 V
60 Hz
B
D
AC 12 V
+
AC 6 V
RL=1 kΩ
CL=100 µF
C
-
Fig. 3.14
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2.
Full-wave rectifiers
(1) Connect the circuit as shown in Fig. 3.15, and use your DMM to
measure the ac voltage of node A (vab) and C (vcb). And measure
the dc voltage of node D (VDB) which is across the resistance
load RL. VDB/vab=? VDB/vcb=?
(2) Using the oscilloscope to measure the waveform of node D (Vdb).
1N4004
A
D
D1
AC 110 V
60 Hz
B
RL=1 kΩ
AC 12 V
AC 6 V
1N4004
C
D2
Fig. 3.15
(3) Connect the circuit as shown in Fig. 3.16, and use your DMM to
measure both the dc voltage of node D (VDB) and the ac voltage
of node D (vdb) which are across the resistance load RL. What’s
the ripple factor value? VDB/vab=? VDB/vcb=?
(4) And measure the waveform across RL by using the oscilloscope.
(5) What’s the frequency of the ripple?
1N4004
A
D
D1
AC 110 V
60 Hz
B
CL=100 µF
RL=1 kΩ
AC 12 V
1N4004
AC 6 V
C
D2
Fig. 3.16
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3.
Full-wave bridge rectifiers
(1) Connect the circuit as shown in Fig. 3.17, and use your DMM to
measure the ac voltage between node B and C (vbc). And
measure the dc voltage of node D (VD) which is across the
resistance load RL. VD/vbc=?
(2) Using the oscilloscope to measure the waveform of node D (V d).
A
AC 110 V
60 Hz
B
AC 12 V
1N4004
D1
D4
D
AC 6 V
C
D2
D3
RL=1 kΩ
Fig. 3.17
(3) Connect the circuit as shown in Fig. 3.18, and use your DMM to
measure the ac voltage between node B and C (vbc). And
measure the dc voltage of node D (VD) which is across the
resistance load RL. VD/vbc=?
(4) And measure the waveform across RL by using the oscilloscope.
(5) What’s the frequency of the ripple?
A
AC 110 V
60 Hz
B
1N4004
D1
AC 12 V
D4
D
AC 6 V
C
D2
D3
+
RL=1 kΩ
CL=100 µF
Fig. 3.18
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4.
Half-wave doubler [Bonus]
(1) Connect the circuit as shown in Fig. 3.19 with a 6 Vrms sine wave
input at 60 Hz. Using your DMM, measure the ac and dc voltage
at nodes A, B and C.
(2) Using the oscilloscope to measure the waveforms of nodes A, B
and C.
(3) While observing nodes B and C, and measuring the dc output at
node C, connect a 4.7 kΩ load from node C to ground.
(4) Now, raise the frequency by a factor of 10 (to 600Hz) and
observe the results.
(5) Now, with the frequency again at 60Hz, change the input
waveform to square wave, and observe the changes, particularly
in the nature of the output ripple and the output voltage.
A
C1=100 µF
vac
6 Vrms
60Hz
-
D2
C Vout
B
+
1N4004
D1
C2=100 µF
1N4004
+
-
Fig. 3.19
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Laboratory #3 Pre-lab
Class:
Name:
Student ID:
Note: In these simulations, you can replace the voltage vac generated from
transformer by a sine-wave voltage source. Note that the AC 6V means
its root-mean-square value is 6V.
1.
Connect the circuit as shown in Fig. 3.13 and Fig. 3.14 and simulate
the circuits by PSpice. Show the waveform which is across the load
resistor. Please attach the schematic and simulation results.
2.
Connect the circuit as shown in Fig. 3.15 and Fig. 3.16 and simulate
the circuits by PSpice. Show the waveform which is across the load
resistor. Please attach the schematic and simulation results.
3.
Connect the circuit as shown in Fig. 3.17 and Fig. 3.18 and simulate
the circuits by PSpice. Show the waveform which is across the load
resistor. Please attach the schematic and simulation results.
4.
Connect the circuit as shown in Fig. 3.19 and simulate the circuit by
PSpice. Show the waveform of output. Please attach the schematic
and simulation results.
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Laboratory #3 Report
Class:
Name:
Student ID:
 Exploration 1
(1) vbc = _____V, VDC = _____V, VDC/vbc = _____
(2) The effect of the diode: _____________, and the waveform of Vdc:
(3) vdc = _____V, VDC = _____V, kr = vdc / VDC = _____
(4) The difference after alteration is _____, and the waveform of Vdc:
(5) Frequency of ripple = _____Hz
 Exploration 2
(1) vab= _____V, vcb = _____V, VDB = _____V,
VDB/vab = _____, VDB/vcb =____
(2) The waveform of Vdb:
(3) vdb= _____V, VDB = _____V, kr = vdb / VDB = _____
VDB/vab = _____, VDB/vcb =____
(4) The waveform of Vdb:
(5) Frequency of ripple = _____Hz
 Exploration 3
(1) vbc = _____V, VD = _____V, VD/vbc = ____
(2) The waveform of Vd:
(3) vbc = _____V, VD = _____V, VD/vbc = ____
(4) The waveform of Vd:
(5) Frequency of ripple = _____Hz
 Exploration 4 [Bonus]
(1) va = _____V, VA = _____V, vb = _____V, VB = _____V,
vc = _____V, VC = _____V
(2) The waveform of Va, Vb, and Vc:
(3) VC = _____V, and the waveform of Vc:
(4) The waveform of Va, Vb, and Vc:
(5) vc = _____V, VC = _____V, and the waveform of Vc:
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 Problem 1
Please prove the following equation for the sinusoid .
V
2
1. Vav  Vmax
2. Vrms  m ax

2
 Problem 2
What are effects on output average voltage and output ripple with
different load resistance?
 Conclusion
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