Sirindhorn International Institute of Technology
Thammasat University
School of Information, Computer and Communication Technology
___________________________________________________________________________
COURSE
: ECS 304 Basic Electrical Engineering Lab
INSTRUCTOR
: Dr. Prapun Suksompong (prapun@siit.tu.ac.th)
WEB SITE
: http://www.siit.tu.ac.th/prapun/ecs304/
EXPERIMENT
: 06 Diodes and Rectifiers
___________________________________________________________________________
I. OBJECTIVES
1.
To study diodes and their applications to half-wave and full-wave rectifiers.
2.
To study the use of filters and voltage regulators.
3. To study nonlinear properties and characteristics of ideal diode.
II. BASIC INFORMATION
1.
A junction diode shown in Figure 6-1 has unidirectional current characteristics; that is,
it will permit current to flow through in one direction (when forward-biased), but not the
other (reverse-biased). Connection of the negative battery terminal to the N-type and the
positive battery terminal to the P-type silicon results in current flow called forward bias.
Reverse bias is connection of negative battery terminal to the P-type and the positive
battery terminal to the N-type. The turn on or threshold voltage for silicon junction diode
is 0.7 V and for germanium is 0.3 V. Once this potential is applied across the diode, it
will conduct appreciably. Increasing in forward bias voltage causes an increase in the
current through the diode.
N-type material
CATHODE
P-type material
ANODE
Figure 6-1: Circuit symbol for a semiconductor diode
A junction diode can be tested using an ohmmeter. This meter reads the current that the
device allows as determined by the voltage applied from the meter. Through electrical
application of Ohm's law, the current reading is translated into a resistance measurement.
When the ohmmeter leads are connected to the diode so that it is forward biased, a high
current flows, indicating a low resistance. Reversing the ohmmeter leads causes reversebiasing of the diode. This prevents more current from flowing and thus registers a high
resistance reading.
2.
The nonlinear characteristic of the diode is clearly evident in Figure 6-2. When source
voltage vi is positive, iD is positive and the diode is a short circuit (vD = 0), while when vi
is negative, iD is zero and the diode is an open circuit (vD = vi). The diode can be thought
of as a switch controlled by the polarity of the source voltage. The switch is closed for
positive source voltages and open for negative source voltages.
iD
iD
Short circuit
+
vi
vD
vD
-
Open circuit
Figure 6-2: The ideal diode and v-i characteristic.
3.
Rectifiers convert an ac voltage to a dc voltage. Rectifiers are used in both low-power
instrumentation applications and those involving higher powers, such as in dc power
supplies. The output of the half wave rectifier will always be positive or zero depending
on whether the input is positive or negative respectively, as shown in Figure 6-3.
D1
+
+
Vin
RL
_
Vin
Vout
_
Vout
Figure 6-3: Diode half-wave rectifier and rectifier waveforms.
2
For the half-wave rectifier shown in Figure 6-3, the average of the input voltage Vin is
zero. The average value of the output voltage is

1
A
A sin xdx 

2 0

where A is the peak input voltage. In this experiment, we use
A ≈ 12 Vrms  12 2 V.
12 2
So, the average value of the output voltage is
 5.4 V.

These average values can be measured by the DMM in DC mode. In particular,
Suppose a periodic signal x  t  with period T is fed into DMM. In DC mode, the DMM
will measure the average value which is
T
1
x  t dt
T 0
Remark: It is useful to remember that
the average of sin x is 0, the average of sin 2 x is ½, and the average of sin x is 2/.
4.
When the sinusoidal input voltage is positive, the diode is forward biased and therefore
the diode conducts. When the sinusoidal input voltage is negative, the diode is reverse
biased and no current flow through the diode. The current through the resistor in series
with the diode is therefore zero. Hence the voltage across the resistor is zero. In the full
wave rectifier during both positive and negative input voltage the direction of current
flow in the load resistor produces positive output voltage.
5.
Rectifiers output contains considerable voltage variation called ripple. Filters are usually
added to remove the ripple. The simplest filter is a large capacitor connected across the
rectifier's output in parallel with the load resistor.
6.
The polarity of the electrolytic capacitor is almost always indicated by a printed band.
The lead nearest to that band is the cathode lead (-). Additionally, the positive lead is
usually longer.
Positive lead: Longer
Negative lead: Shorter
Polarity band showing
the negative lead
Caution: Most electrolytic capacitors are polarized. Hook them up the wrong way and at
best, you’ll block the signal passing through. At worst (for higher voltage applications)
they’ll explode.
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III. MATERIALS REQUIRED
Power supply:
220-V 50-Hz source
Equipment:
Oscilloscope
Sine-wave generator
Digital multimeter
Resistors:
1 1 k
1 2.7 k 2 5 k
1 5.6 k
1 10 k
Capacitors:
1 100-F 50-V
Diodes:
4 Solid state diodes 1N4001
Miscellaneous:
Power transformer with center tapped secondary, 220/(12-0-12)
IV. PROCEDURE
Remark:
In this experiment, the source will be the output of a transformer. The transformer with label
220/(12-0-12) means that the input voltage is 220V and 12V and -12V output using the center
tap as the ground. From the center tap to either side is 12 volts. Across the entire winding is
24 volts. These values are all RMS values, not peak-peak values.
Warning:
This experiment use high voltage. Damage on any equipment, devices, or any part of your
body is subject to punishment.
4
Part A: Half-Wave and full-wave rectifiers.
1. Connect the circuit of Figure 6-4.
T1
S1
A
D1
1N4001
+
220 V
50 Hz
Bv_in
+
Cv_in
D
S2
D1
1N4001
+
RL
Vout
10 K
_
Figure 6-4: A half-wave and full-wave rectifier circuit.
2. Close S1 and open S2. This means A is connected to the anode of D1 and C is not
connected to the anode of D2.
3. Use the oscilloscope in DC mode, measure the peak-to-peak voltage and waveforms of Vin
and Vout record and draw them in Table 6-1. Indicate where the ground level of the voltage
is.
Note that you can use the scope to display vin and vout simultaneously on channel 1 and 2,
respectively. Because vAB = vBC, you can keep channel 1 of the scope at vAB throughout
part A. Note also that because B is connected to the ground, vC = - vA.
4. With a DMM (in DC mode), measure and record the DC voltage of Vin and Vout in Table
6-1.
5. Open S1 and close S2. Then repeat steps 3 and 4.
6. Close S1 and S2. Then repeat steps 3 and 4.
5
Part B: Effect of a filter capacitor on output of full-wave rectifier.
T1
220 V
50 Hz
D1
1N4001
A
+
D2
1N4001
C1
100 F
50 V
_
+
R1
Vout
2.7K
RL
1K
_
G
Figure 6-5: A full-wave rectifier with capacitor filter.
1. Connect the circuit of Figure 6-5 without the load RL. Note that the capacitor C1 has +/polarity, and its terminals must be connected correctly.
2. With a voltmeter measure Vout, the dc output voltage across R1, and record the result in
Table 6-2. With the oscilloscope in DC mode connected across Rl, observe, measure, and
record the ripple waveform and its peak-to-peak amplitude. in Table 6-2
3. Connect a 1 kΩ load resistor (RL). The voltmeter and oscilloscope are still connected
across RL.
4. Measure Vout and the ripple waveform. Record the results in Table 6-2. Compute the load
current IL in the 1 kΩ resistor and record the results in Table 6-2. Show your
computations.
6
Part C: Operation of a bridge rectifier.
1. Connect the circuit of Figure 6-6. Note that one end of the transformer secondary is open.
T1
C
D1
D4
F
220 V
50 Hz
+
D2
D3
RL
5.6 K
Vout
_
D
open
G
Figure 6-6: A bridge rectifier circuit.
2. Connect the vertical leads of the oscilloscope across RL, the hot lead to F, and the ground
lead to G. Observe and draw the waveform of Vout in Table 6-3.
3. Switch off the power supply.
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TABLE 6-1: Half-wave and full-wave measurements.
Waveform
Vp-p DC (V)
(A.5)
Vin(A to B)
close S1
open S2
(A.5)
Vout
close S1
open S2
(A.5)
Vin(B to C)
close S2
open S1
(A.5)
Vout
close S2
open S1
Vout(full-wave)
close S1
close S2
TA’s Signature: ________________________
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TABLE 6-2.Capacitive Filter.
Load, ohms
None
1K
Vout, V
IL(computed), mA
Ripple Waveform
Vp-p
No load
1 kΩ
TA’s Signature: ________________________
TABLE 6-3. Bridge rectifier output
TA’s Signature: ________________________
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QUESTIONS
1.
After completing Part A of the experiment, determine the meaning of the output and the
obtained graph?
Is there any significant observation? Explain?
2.
D1
10 V
R1
220 AC
10 V
D2
Use the figure above
a) Sketch the outputs in term of voltage across R 1, D1 and D2 on the same time domain
axis. Explain in detail how you arrive at your answer.
b) What is the value of Vpeak-out?
c) What is the value of Vdc?
10
3.
T1
+12 v
C
D1
D4
220 V
50 Hz
0
F
+
D2
D3
-12 v
RL
5.6 K
Vout
_
D
G
From the figure above, sketch the outputs in term of voltage across R L, on the same time
domain axis, explain why?
4.
Why are filters used? How does a capacitive filter affect the output of a full-wave
rectifier. Refer to your experimental results. What is ripple?
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