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Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Experiment 1: Ohm’s Law
Objective:
• To perform an experimental check of Ohm's Law • To practice constructing electric circuits
Apparatus:
• Varied value of resistors
• Power supply
• Connecting wires
• Multimeters
• Rheostat (variable resistance)
Theory:
In this lab, you will construct a simple circuit using a single known resistance, R. Then you will use an ammeter to measure the current, I, through the resistance and a voltmeter to measure the potential difference, V, across the resistance. With this data, you can check the validity of Ohm's Law (V = IR) in the circuit.
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Procedure: 1. Connect the circuit shown below using a fixed resistance R.
2. Set the value of electromotive force into certain voltage (electromotive force). 3. You can get different readings of the current I and voltage V by varying the
battery source. 4. Record the value of the current through the resistance and the voltage across
it. 5. Repeat step 3 to get at least 6 different readings. 6. Record the data in Table-1 7. Repeat the previous steps for another value of the resistance. Record the data
in Table2.
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Experiment 1: Ohm’s Law Name:____________________ ID: _______________________
Instructor:
________________________________________________________
Section:____________________________Date:_________________
DATA SHEET
Table-1 Voltage and current for Resistance R = 1
Current (A) Voltage (V) Resistance R ()
Average Resistance Ravg
Resistance from Slope
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
1. Plot a graph of the voltage ( V ) and the current (I), with V on vertical axis and ( I) on horizontal axis. Draw the best straight line fit of the data. 2. Determine the slope of the straight line which is the resistance R in this case. Slope = _________________
(YOU NEED TO ATTACH A GRAPH PAPER TO THIS DATA SHEET)
Table-2 Voltage and current for Resistance R = 10
Current (A) Voltage (V) Resistance R ()
Average Resistance Ravg
Resistance from Slope
3. Plot a graph of the voltage ( V ) and the current (I), with V on vertical axis and ( I) on horizontal axis. Draw the best straight line fit of the data. 4. Determine the slope of the straight line which is the resistance R in this case. Slope = _________________
1
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Experiment No. 2
The rectifier diode
Objectives:
1. Ability to recognize diodes in various physical forms.
2. Ability to determine the diode polarity and to understand the need for correct
connection.
3. To obtain knowledge of the forward voltage/current characteristic and the
conduction voltage for diodes.
Apparatus:
D.C. power source, ammeter, voltmeter, rheostat, resistors, and rectifier diode.
Introduction
The general from of the current - voltage of a diode is shown in Figure (l). A current
flow in the forward direction is very large compared with that in the reverse direction
and such a device is very useful as a rectifier. The diode is in the forward direction
when an external battery is connected with positive terminal to the (p) region and
negative terminal to the region (n). The reverse current through the diode varies
greatly with temperature and with the semiconductor materiel used.
Fig. 1 V-I characteristics of semiconductor diode
2
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Procedure:
1. Connect the circuit as shown in Figure 2 (a) using silicon diode.
2. Increase the variable DC voltage from zero in steps of (0.2 volts) up to (1 volts),
then in step of (0.5 volt) up to (5 volt), and record the voltage across the resistance.
4. Connect the circuit shown in Figure 2 (b) using Si diode.
5. Increase the variable DC voltage from zero in steps of (0.5 volts) up to (2 volt),
then in steps of (1 volt) up to (10 volts) and for each step record the current.
6. Tabulate your result in a table 1.
(a) Forward bias
Fig. 2 Forward and reverse bias of diode
(b) Reverse bias
3
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Table 1
Forward bias Reverse bias
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
0.2 0.5
0.4 1.0
0.6 1.5
0.7 2.0
0.8 3
1.0 4
1.5 5
2.0 6
2.5 7
3.0 8
3.5 9
4.0 10
4
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Experiment No. 1
The rectifier diode
Name: ـــــــــــــــــــــ Date: ــــــــــــ
Physics 106 Section: ـــــــــــ Instructor’s Name: ــــــــ
----------------------------------------------------------------------------------
Purposes: ………………………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………..
Data and Data Analysis
1. Fill the Table 1, R=…………. Ω
Forward bias Reverse bias
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
0.2 0.5
0.4 1.0
0.6 1.5
0.7 2.0
0.8 3
1.0 4
1.5 5
2.0 6
2.5 7
3.0 8
3.5 9
5
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
4.0 10
2. Plot the diode current Id versus the potential difference across the rectifier diode.
3. From graph find the junction voltage VJ= V
4. Find the slope of curve in each region and calculate the resistance of diode.
Slope= A/V, R=1/slope.
Forward bias:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
Reverse bias (blocking current)
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
1
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
EXPERIMENT 2
HALF-WAVE RECTIFIER
Purposes:
1. To learn a half-wave rectified sinusoidal voltage.
2. To understand the terms mean value and rout mean square for input and output
(rectified) voltage.
3. To understand the effect of reservoir capacitor upon the rectified waveform.
Apparatus:
Electricity & electronics constructor EEC470, basic electricity & electronics kit EEC471-2,
multimeter, oscilloscope.
2
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Fig. 1 Experimental setup
Introduction
The conversion of alternating voltage (changes its direction) to directed voltage. This
conversion is called rectification. To achieve the rectification of alternating voltage you must
to use some circuit. The simplest one is the half-wave rectifier in which a single diode is
used, see Fig. 1.
For a sinusoidal waveform the diode passes a current at first part 1 of waveform, because it
forward-biased. At the same time the diode doesn't pass a current for second part 2, because
it reverse-biased. In this case only the half of the input waveform passes through the diode.
The output voltage (rectified waveform) changes with time which makes it unestablish with
time. To produce a steady output directed voltage and free from variations the load resistor
is connected in parallel with capacitor.
Fig. 2 Half-wave rectification
3
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
To calculate the mean value and root mean square rms of voltage let we must use the
following equations:
=1
𝑇∫ 𝑉(𝑡)𝑑𝑡𝑇
0 (1)
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉2(𝑡)𝑑𝑡𝑇
0 (2)
For sinusoidal waveform we can write (𝑡) = 𝑉𝑃sin(𝜔𝑡) .
=1
𝑇∫ 𝑉𝑃sin(𝜔𝑡)𝑑𝑡
𝑇
0
= −𝑉𝑃𝑇𝜔
(cos(𝜔𝑇) − cos(0)) = −𝑉𝑃2𝜋
(cos(2𝜋) − 𝑐𝑜𝑠0) = 0
The mean value of sinusoidal wave equals zero = 0 (3) .
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉𝑝
2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇
0
= √1
𝑇∫ 𝑉𝑝
2 (1 + sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √1
2𝑇∫ 𝑉𝑝
2𝑑𝑡𝑇
0
+1
𝑇∫ (𝑉𝑝
2sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇∫ 𝑑𝑡
𝑇
0
+𝑉𝑝2
2𝑇∫ sin(2𝜔𝑡) 𝑑𝑡
𝑇
0
= √𝑉𝑝2
2𝑇(𝑇 − 0) +
−1
2𝜔(cos(2𝜔𝑇) − cos(0)) =
For sinusoidal waveform 𝑉𝑟𝑚𝑠 =𝑉𝑃
√2 (4)
For rectified waveform without capacitor we have 𝑉(𝑡) = 𝑉𝑃 sin(𝜔𝑡) 𝑓𝑜𝑟𝑡𝜖[0, 𝑇/2]
𝑉(𝑡) = 0𝑓𝑜𝑟𝑡𝜖[𝑇/2, 𝑇]
4
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
=1
𝑇∫ 𝑉(𝑡)𝑑𝑡 =
1
𝑇(∫ 𝑉𝑃 sin(𝜔𝑡) 𝑑𝑡
𝑇/2
0
+∫ 0𝑑𝑡) =𝑇
𝑇/2
𝑇
0
= −𝑉𝑃𝑇𝜔
cos(𝜔𝑇/2) − cos(0) = −𝑉𝑃2𝜋
cos(𝜋) − cos(0)
For half-wave rectified waveform without capacitor = 𝑉𝑃
𝜋 (5)
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉𝑝
2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇/2
0
= √1
𝑇∫ 𝑉𝑝
2 (1 + sin(2𝜔𝑡)
2)𝑑𝑡
𝑇/2
0
=
𝑉𝑟𝑚𝑠 = √1
2𝑇∫ 𝑉𝑝
2𝑑𝑡𝑇/2
0
+1
𝑇∫ (𝑉𝑝
2sin(2𝜔𝑡)
2)𝑑𝑡
𝑇/2
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇∫ 𝑑𝑡
𝑇/2
0
+𝑉𝑝2
2𝑇∫ sin(2𝜔𝑡) 𝑑𝑡
𝑇/2
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇(𝑇/2 − 0) +
−1
2𝜔(cos(𝜔𝑇) − cos(0))
= √𝑉𝑝2
2𝑇(𝑇/2 − 0) +
−1
2𝜔(cos(2𝜋) − 1)
For half-wave rectified waveform without capacitor 𝑉𝑟𝑚𝑠 =𝑉𝑃
2 (6)
5
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
EXPERIMENT 2
HALF-WAVE RECTIFIER
Name: Date: ــــــــــــ
Electronics laboratory Section: ــ Instructor’s Name:ــ
----------------------------------------------------------------------------------
Procedure:
1. Connect the circuit shown in Fig.1
2. For input signal take the reading of voltmeter Vdc= V (by using dc-voltmeter)
and the reading of Vrms= V (by using ac-voltmeter).
3. Use the oscilloscope to fill the Table 1 and to draw the input signal.
4. For output signal take the reading of voltmeter Vdc= V (by using dc-
voltmeter) and the reading of Vrms= V (by using ac-voltmeter).
5. Use the oscilloscope to fill the Table 2 and to draw the output signal.
6. Connect the capacitor in parallel with load resistor and draw the output signal.
6
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Data and Data Analysis
I) Input signal
1) Take the reading of voltmeter Vdc= V (by using dc-voltmeter) and the reading of
Vrms= V (by using ac-voltmeter).
2) Use the oscilloscope to draw the input signal for that:
a) Fill the Table 1
V (V/cm) t (s/cm)
Y (cm) V (V) V2 (V2) X (cm) t(s)
Average value of V =
Average value of V2 =
b) Calculate the average value for V and compare it with Vdc .
Note that : VAVG=0 =Vdc
7
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………………..…………………
c) Calculate the average value of V2 and find the square root of it. Compare this value
with Vrms= Vp /√𝟐 =√𝑽𝟐 = the reading of ac-voltmeter.
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
d) Draw V versus time and from graph find the area under curve and find:
VAVG=area/(period time) compare this value with result in step b.
e) Draw V2 versus time and from graph find the area under curve and find:
𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞) and then find the square root compare this value with result
in step c.
II) Output signal
1) Take the reading of voltmeter Vdc= V (by using dc-voltmeter) and the reading of
Vrms= V (by using ac-voltmeter).
2) Use the oscilloscope to draw the input signal for that:
a) Fill the Table 2
V (V/cm) t (s/cm)
Y (cm) V (V) V2 (V2) X (cm) t(s)
8
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Average value of V =
Average value of V2 =
b) Calculate the average value for V and compare it with Vdc .
Note that : VAVG=Vp/𝝅 =Vdc
………………………………………………………………………………………………………………………………………………………………………………………………………………
c) Calculate the average value of V2 and find the square root of it. Compare this value
with Vrms= Vp /𝟐 =√𝑽𝟐 = the reading of ac-voltmeter.
d) Draw V versus time. Find from graph the area under curve and find:
VAVG=(area/period time), compare this value with result in step b.
………………………………………………………………………………………………………………………………………………………………………………………………………………
e) Draw V2 versus time and from graph find the area under curve and find:
𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞), compare this value with result in step c.
………………………………………………………………………………………………………………………………………………………………………………………………………………
9
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
III) The effect of a reservoir capacitor
Connect the capacitors (small and big) with output circuit and sketch the output signals.
1
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
EXPERIMENT 3
FULL-WAVE RECTIFIER
Purposes:
1. To learn a full-wave rectified sinusoidal voltage.
2. To understand the terms mean value and rout mean square for input and output
(rectified) voltage.
3. To understand the effect of reservoir capacitor upon the rectified waveform.
Apparatus:
Electricity & electronics constructor EEC470, basic electricity & electronics kit EEC471-2,
multimeter, oscilloscope.
Fig. 1 Experimental setup
Introduction
The conversion of alternating voltage (changes its direction) to directed voltage. This
conversion is called rectification. To achieve the rectification of alternating voltage you must
to use some circuit. The full-wave rectifier consists of four diodes, see Fig. 1.
2
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
For a sinusoidal waveform the diodes D1 and D2 pass a current at first part 1 of waveform,
because they are forward-biased. At the same time the diode D3 and D4 pass a current for
second part 2, because they are forward-biased. In this case the two halves of the input
waveform (Full signal) passes through the bridge of diodes. The output voltage (rectified
waveform) changes with time which makes it unstable with time. To produce a steady output
directed voltage and free from variations the load resistor is connected in parallel with
capacitor.
Fig. 2 Full-wave rectification
To calculate the mean value and root mean square rms of voltage let we must use the
following equations:
=1
𝑇∫ 𝑉(𝑡)𝑑𝑡𝑇
0 (1)
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉2(𝑡)𝑑𝑡𝑇
0 (2)
For sinusoidal waveform we can write (𝑡) = 𝑉𝑃sin(𝜔𝑡) .
3
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
=1
𝑇∫ 𝑉𝑃sin(𝜔𝑡)𝑑𝑡
𝑇
0
= −𝑉𝑃𝑇𝜔
(cos(𝜔𝑇) − cos(0)) = −𝑉𝑃2𝜋
(cos(2𝜋) − 𝑐𝑜𝑠0) = 0
The mean value of sinusoidal wave equals zero = 0 (3) .
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉𝑝
2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇
0
= √1
𝑇∫ 𝑉𝑝
2 (1 + sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √1
2𝑇∫ 𝑉𝑝
2𝑑𝑡𝑇
0
+1
𝑇∫ (𝑉𝑝
2sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇∫ 𝑑𝑡
𝑇
0
+𝑉𝑝2
2𝑇∫ sin(2𝜔𝑡) 𝑑𝑡
𝑇
0
= √𝑉𝑝2
2𝑇(𝑇 − 0) +
−1
2𝜔(cos(2𝜔𝑇) − cos(0)) =
For sinusoidal waveform 𝑉𝑟𝑚𝑠 =𝑉𝑃
√2 (4)
For full-wave rectified waveform without capacitor we have 𝑉(𝑡) = 𝑉𝑃 sin(𝜔𝑡)
=1
𝑇[2∫ 𝑉𝑃 sin(𝜔𝑡) 𝑑𝑡] =
𝑇/2
0
= [−2𝑉𝑃𝑇𝜔
cos (𝜔𝑇
2) − cos(0)] = −
2𝑉𝑃2𝜋
cos(𝜋) − cos(0)
For Full-wave rectified waveform without capacitor = 2𝑉𝑃
𝜋 (5)
4
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
𝑉𝑟𝑚𝑠 = √1
𝑇∫ 𝑉𝑝
2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇
0
= √1
𝑇∫ 𝑉𝑝
2 (1 + sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √1
2𝑇∫ 𝑉𝑝
2𝑑𝑡𝑇
0
+1
𝑇∫ (𝑉𝑝
2sin(2𝜔𝑡)
2)𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇∫ 𝑑𝑡
𝑇
0
+𝑉𝑝2
2𝑇∫ sin(2𝜔𝑡) 𝑑𝑡
𝑇
0
=
𝑉𝑟𝑚𝑠 = √𝑉𝑝2
2𝑇(𝑇 − 0) +
−1
2𝜔(cos(2𝜔𝑇) − cos(0)) = √
𝑉𝑝2
2𝑇(𝑇 − 0) +
−1
2𝜔(cos(4𝜋) − 1)
For Full-wave rectified waveform without capacitor 𝑉𝑟𝑚𝑠 =𝑉𝑃
√2 (6)
5
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
EXPERIMENT 3
FULL-WAVE RECTIFIER
Name: Date: ــــــــــــ
Electronics laboratory Section: ــ Instructor’s Name:ــ
----------------------------------------------------------------------------------
Procedure:
1. Connect the circuit shown in Fig.1
2. For input signal take the reading of voltmeter Vdc= V (by using dc-voltmeter)
and the reading of Vrms= V (by using ac-voltmeter).
3. Use the oscilloscope to fill the Table 1 and to draw the input signal.
4. For output signal take the reading of voltmeter Vdc= V (by using dc-
voltmeter) and the reading of Vrms= V (by using ac-voltmeter).
5. Use the oscilloscope to fill the Table 2 and to draw the output signal.
6. Connect the capacitor in parallel with load resistor and draw the output signal.
6
Kingdom of Saudi Arabia
Ministry of Education
Islamic University in Madinah
Department of Physics
قسم الفيزياء
Data and Data Analysis
I) Input signal
1) Take the reading of voltmeter Vdc= V (by using dc-voltmeter) and the reading of
Vrms= V (by using ac-voltmeter).
2) Use the oscilloscope to draw the input signal for that:
a) Fill the Table 1
V (V/cm) t (s/cm)
Y (cm) V (V) V2 (V2) X (cm) t(s)
Average value of V =
Average value of V2 =
7
Kingdom of Saudi Arabia
Ministry of Education
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b) Calculate the average value for V and compare it with Vdc .
Note that : VAVG=0 =Vdc
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………………..…………………
c) Calculate the average value of V2 and find the square root of it. Compare this value
with 𝑽𝒓𝒎𝒔 =𝑽𝒑/√𝟐 = √𝑽𝟐 = the reading of ac-voltmeter.
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………………..…………………
d) Draw V versus time and from graph find the area under curve and find:
VAVG=area/(period time) compare this value with result in step b.
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………………..…………………
e) Draw V2 versus time and from graph find the area under curve and find:
𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞) and then find the square root compare this value with result
in step c.
………………………………………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………………..…………………
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II) output signal
1) Take the reading of voltmeter Vdc= V (by using dc-voltmeter) and the reading of
Vrms= V (by using ac-voltmeter).
2) Use the oscilloscope to draw the input signal for that:
a) Fill the Table 2
V (V/cm) t (s/cm)
Y (cm) V (V) V2 (V2) X (cm) t(s)
Average value of V =
Average value of V2 =
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b) Calculate the average value for V and compare it with Vdc .
Note that : VAVG=2Vp/𝝅 =Vdc
………………………………………………………………………………………………………………………………………………………………………………………………………………
c) Calculate the average value of V2 and find the square root of it. Compare this value
with Vrms= Vp /𝟐 =√𝑽𝟐 = the reading of ac-voltmeter.
………………………………………………………………………………………………………………………………………………………………………………………………………………
d) Draw V versus time. Find from graph the area under curve and find:
VAVG=(area/period time), compare this value with result in step b.
………………………………………………………………………………………………………………………………………………………………………………………………………………
e) Draw V2 versus time and from graph find the area under curve and find:
𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞), compare this value with result in step c.
………………………………………………………………………………………………………………………………………………………………………………………………………………
III) The effect of a reservoir capacitor
Connect the capacitors (small and big) with output circuit and sketch the output signals.
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Experiment No.4
The zener diode 1 (The Volt-Ampere characteristic curve)
Objectives:
1) Investigate the relationship between current and voltage for a Zener diode in
forward and reverse direction.
2) Find the range of Zener voltage.
3) Find the resistance of Zener diode at Zener voltage.
Apparatus:
D.C. power source, ammeter, voltmeter, rheostat, resistors, zener diode.
Introduction
A Zener diode is a p-n junction device designed to operate in the reverse-breakdown
region. Unlike normal diode reverse breakdown, Zener breakdown occurs at low
reverse voltages. A Zener diode is heavily doped to result in a very thin depletion
region. Therefore, an intense electric field exists within the depletion region near Zener
breakdown voltage (Vz). This intense filed pulls electrons from their valence bands
and create current. If a Zener diode is forward biased, it operates the same as a normal
diode as shown in Fig. 1. At the same time the Zener diodes are specially constructed
to break down at controllable voltages and to do so without damage to the device.
Fig.(4.1)
The resistance of Zener diode is 𝑟𝑧 = 1/𝑠𝑙𝑜𝑝𝑒 of curve at breakdown.
Fig. 1 The Volt-Ampere characteristic of Zener diode.
Breakdown
(Zener region)
Reverse Bias
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Experiment No. 4
The Zener diode
The Volt-Ampere characteristic curve
Name: ـــــــــــــــــــــ Date: ــــــــــــ
Physics 106 Section: ـــــــــــ Instructor’s Name: ــــــــ
----------------------------------------------------------------------------------
Procedure:
1- Set up circuit as shown in Fig. 2 . Connect Z diode in forward direction.
2- Select measurement range of 1 V- and 100 mA-. Make sure polarity on the meter is
correct and that the meters are connected properly.
3- Switch on power supply unit. Increase voltage on the power supply unit to 20 V.
Measure both Vs, Vr. Enter measurements in Table (1).
4- Change the polarity of the Zener diode .
Fig. 2 Setup
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5- Measure both voltage Vs , Vr . Enter measurements in Table 1
6- Calculate the potential difference across the diode by equation Vd=Vs-Vr , and the
current passes through the diode Id=Vd/R= Vd/Rd or (Id= Vr/R= Vr/1000 Ω).
7- Enter your data in Table 1.
Table 1
Forward bias Reverse bias
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
Vs
(V)
Vr
(V)
Vd
(V)
Id
(mA)
Pd
(mW)
0 2
0.30 4
0.50 6
0.55 7
0.60 7.5
0.65 8
0.70 9
0.80 12
1.0 15
2 20
5
10
15
20
8- Graph the diode current as a function of diode voltage in Forward and Reverse bias.
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Data and Data Analysis
- Plot the diode current Id versus the potential difference across the Zener
diode.
9- From graph find:
a) the junction voltage VJ= V
b) the breakdown voltage (the voltage at which a diode connected in reverse
direction loses its blocking ability) Vbreakdown= V.
c) The Zener voltage (the voltage at midpoint of Zener region) Vz= V
10- Find the slope of curve in each region and calculate the resistance of diode.
Slope= A/V, R=1/slope.
Forward bias:
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
Reverse bias (blocking current)
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………
Zener region:
…………………………………………………………………………………………
…………………………………………………………………………………………
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Experiment No.5
The zener diode 2
Regulated Voltage Supply
Objectives:
1) To find the variation of voltage of unregulated voltage supply Vs which can be
tolerated?
2) To find the variation of load current IL which can be tolerated.
Apparatus:
D.C. power source, ammeter, voltmeter, rheostat, resistors, zener diode.
Introduction
The Zener diode has a region in its reverse characteristic of almost constant voltage
regardless of the current trough the diode. This can be used to regulate or stabilize a
voltage source against supply or load variations.
Fig.5.1: Experimental setup
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Fig 5.2 a shows an unregulated voltage source supplying current to a variable load. If
either Vs or RL changes, so the voltage across the load VL will change too. One way
of keeping this voltage more constant is to connect across the load a Zener diode whose
breakdown voltage is the desired constant voltage, Fig 5.2 b.
Fig 5.2: Voltage supply
Figure 5.2 b gives ability to change the voltage of source Vs (0-20 V) and the load
resistance RL . Note that the power of this diode is 0.4 W and VZ=7.5 V that means
the maximum current through the diode must be less than𝐼𝑑 =𝑃
𝑉𝑍=
0.4
7.5= 53 𝑚𝐴.
(a) Unregulated voltage supply
(b) Regulated voltage supply
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Experiment No.5
The zener diode 2
Regulated Voltage Supply
Name: ـــــــــــــــــــــ Date: ــــــــــــ
Physics 106 Section: ـــــــــــ Instructor’s Name: ــــــــ
----------------------------------------------------------------------------------
Procedure:
Set up circuit as shown in Fig. 5.1 (Fig.5.2 b). Select measurement range of 20 V- and
100 mA-. Make sure polarity on the meter is correct and that the meters are connected
properly.
I) Regulation for maximum load current and minimum supply voltage
1- Switch on power supply unit and remove the potentiometer to make IL=0 .
1- Increase slowly the voltage on the power supply unit VS until the diode just
begins to conduct current (1 mA). Record Vs for IL=0 in Table (1).
2- Set the potentiometer to maximum (clockwise) and replace it in the circuit. The
extra current drawn by RL will reduce the diode current bellow 1 mA.
3- Increase the voltage of source VS to 12 V, the diode current will increase above
1 mA. Then adjust RL until the diode current just returns to 1 mA. Record VS
and IL .
4- Repeat step 3 for VS =14, 16, 18, 20 V recording the results in Table 1.
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II) Regulation for minimum load current and maximum supply voltage
1- Reduce the load resistance RL to minimum (anticlockwise) and then set voltage
supply VS to 20 V and increase RL until the diode current Id reads 70 mA.
2- Record the Load current IL against VS=20 V in table 1.
3- Reduce the voltage supply Vs to 19, 18, 17 …. V, each time resetting RL to give
the diode current 70 mA and recording IL . Continue until it is no longer possible
to set the diode current to 70 mA.
Table 1
RL
(Ω)
Vz
(V)
VS
(V)
IL
(mA)
RL
(Ω)
Vz
(V)
VS
(V)
IL
(mA)
∞ 0 All readings for
Vd
just equal to the
zener voltage
Id=1 mA
Maximum IL
And minimum
VS
19 All readings
for
diode current
just equal to
70 mA
Minimum IL
And maximum
VS
12 18
14 17
16 16
18 15
Data and Data Analysis
1) Fill the Table 1
2) Plot the load current IL versus the voltage of voltage supply VS for Id=70 mA.
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(curve 1 - for maximum load current IL and minimum voltage of voltage supply VS).
3) Plot the load current IL versus the voltage of voltage supply VS for Id=1 mA.
(curve 1 - for minimum load current IL and maximum voltage of voltage supply VS).
4) Shadow the area between these two curves. The work area of regulated voltage
supply.
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Experiment No.7
The common emitter
Purposes:
1. To learn the common-emitter output (collector) characteristics.
2. To understand the meaning and the importance of operating point and load
line.
Apparatus:
Electricity & electronics constructor EEC470, basic electricity & electronics kit
EEC471-2, multimeters.
Fig. 1 Experimental setup
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Experiment No.7
The common emitter
Name: ـــــــــــــــــــــ Date: ــــــــــــ
Physics 106 Section: ـــــــــــ Instructor’s Name: ــــــــ
----------------------------------------------------------------------------------
Data and Data Analysis
1. Fill the Table 1
IB
( µA ) 0.4 0.6 1.0 1.8 4.4 6.3
VCE
( V )
0 IC (mA)
10 IC (mA)
45 IC (mA)
2. Plot the collector current IC versus the collector emitter voltage VCE .
3. From table for IB=45 µA find β for this transistor (use a suitable value of Ic for
this purpose).
…………………………………………………………………………………………
…………………………………………………………………………………………
1. From figure for IB = 10 μA , find the slope of linear part of the curve.
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Experiment No. 8:
The Silicon Controller Rectifier
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Experiment No.8
The silicon controlled rectifier
Name: Date:
Physics 106 Section: Instructor’s Name:
----------------------------------------------------------------------------------
Data and Data Analysis
I) Switch on the SCR
a) At first we will apply a voltage across the anode cathode of SCR and increase
this voltage, but this is not enough to switch on the SCR. For this purpose fill
the table 1
Table 1
Vvariable (V) 2 4 6 8 10 12 14
VAK ( V)
IG (mA) 0
IAK (mA)
b) Now for Vvariable=12 V change the gate current and fill the table 2 and note
when the lamp lights.
Table 2
Vvariable (V) 12
VAK ( V)
IG (mA) 1 2 3 4 5 6 8
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IAK (mA)
1. Plot the trigger current IG versus the anode cathode voltage VAK .
2. From graph find the saturation voltage VAK(sat).
…………………………………………………………………………………………
…………………………………………………………………………………………
3. Fill the table 3.
Table 3
Trigger current IGT (mA)
Saturation Voltage VAK(sat) (V)
Holding current IH (mA)
II) Switch off the SCR
a) At first we will apply a voltage across the anode cathode of SCR Vvariable=12 V
and decrease the current IG, but this is not enough to switch off the SCR. For
this purpose fill the table 4
Table 4
Vvariable (V) 12
VAK ( V)
IG (mA) 8 6 5 4 3 2 1
IAK (mA)
b) Now for IG=0 change the voltage Vvariable and fill the table 5 and note when the
lamp switches off.
Table 5
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Vvariable (V) 12 10 7 5 3 2 1
VAK ( V)
IG (mA) 0
IAK (mA)
III) Half-wave rectification for AC-Voltage
a) Apply the AC-Voltage with IG=0 (the SCR is switched off, IAK= 0) and then
connect the oscilloscope across anode and cathode. Draw the signal illustrated by
oscilloscope.
a) Signal for switched off SCR b) Signal for switched on SCR
b) Apply the AC-Voltage with IG=6 mA (the SCR is switched on, IAK= 30 mA)
and then connect the oscilloscope across anode and cathode. Draw the signal
illustrated by oscilloscope.
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Experiment No.6
Transistor familiarization
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----------------------------------------------------------------------------------
Vcc = 20 V and VBB = 10 V
Data and Data Analysis
1. Fill the Table 1
IC
( mA )
VEB
( )
IB
( )
IE=IC+IB
( ) ℎ𝑓𝑒 =
𝐼𝐶
𝐼𝐵 ℎ𝑓𝑏 =
𝐼𝐶
𝐼𝐸
1
5
10
17
23
34
Average value
2. Plot the collector current IC versus the base current IB .
3. Find the slope of the curve.
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…………………………………………………………………………………………
…………………………………………………………………………………………
4. What does the slope mean
…………………………………………………………………………………………
…………………………………………………………………………………………
EXPERIMENT 9
The TRIAC
Purposes:
1. Study the TRIAC device.
2. Understanding of the bi-directional nature of the TRIAC and its applications.
3. To understand the work of the TRIAC in the four operating quadrants.
4. To use the TRIAC as half-wave and full-wave controlled rectifier and to see the
rectified waveforms.
Apparatus:
Electricity & electronics constructor EEC470, basic electricity & electronics kit
EEC471-2, multimeters, oscilloscope.
Introduction
The silicon controlled rectifier can be used to rectify a waveform, but it passes a current
in one direction. The SCR can work as half-wave rectifier and it can be used in a bridge
to achieve controlled full-wave rectification. But a simple and less expensive way of
obtaining bi-directional conduction is to use a TRIAC. A TRIAC is a type of thyristor.
It has a four layers of semiconductor materials. The TRIAC can be triggered into
conduction by gate current for either polarity of the voltage between terminals T1 & T2.
This property makes it possible to work in four modes as shown in Fig.1
Fig. 1 TRIAC triggering modes
These modes are:
in I quadrant, mode I+ (positive gate current, and VT2> VT1 )
mode I- (negative gate current and VT2> VT1)
in III quadrant, mode III+ (positive gate current, and VT2< VT1 )
mode III- (negative gate current and VT2< VT1)
Note: the TRIAC can be triggered in any of four modes. The TRIAC is switched off at
any applied voltage across its two terminals T2 & T1 when the gate current is not big
enough (IG < IGT).
Procedure
I) Four modes of the TRIAC
Connect the circuit shown in Fig. 2 for four modes.
a) Mode I+ (positive gate current, and VT2> VT1 )
1) Connect the two links 1, 2 as shown in Fig.2 to have positive IG and VT2> VT1.
2) Connect the Ammeter with polarity as shown in Fig. 2 to have a positive reading
of gate current.
3) Add to this circuit other ammeter with polarity to have a positive reading of IT2 .
4) Watch the lamp when it will be lighted, to determine IGT and VT2T1(sat).
Fig. 2 Setup for TRIAC modes
5) Fill the table 1
Table 1( mode I+ )
VT2 –VT1 (V)
IG (mA)
IT2 (mA)
Lamp
6) From table 1 find: IGT = mA , and (VT2 –VT1)sat = V , IT2(H)= mA.
7) Decrease the gate current and note that the TRIAC stays switching on.
b) Mode I- (negative gate current, and VT2> VT1 )
1) Connect the two links 1(down), 2(up) in Fig.2 to have negative IG and VT2> VT1.
2) Watch the lamp when it will be lighted, to determine IGT and VT2T1(sat).
3) Fill the table 2
Table 2( mode I- )
VT2 –VT1 (V)
IG (mA)
IT2 (mA)
Lamp
4) From table 1 find: IGT = mA , and (VT2 –VT1)sat = V , IT2(H)= mA.
5) Decrease the gate current and note that the TRIAC stays switching on.
c) Mode III+ (positive gate current, and VT2< VT1 )
1) Connect the two links 1(up), 2(down) in Fig.2 to have positive IG and VT2< VT1.
2) Watch the lamp when it will be lighted, to determine IGT and VT2T1(sat).
3) Fill the table 3
Table 3( mode III+ )
VT2 –VT1 (V)
IG (mA)
IT2 (mA)
Lamp
4) From table 1 find: IGT = mA , and (VT2 –VT1)sat = V , IT2(H)= mA.
5) Decrease the gate current and note that the TRIAC stays switching on.
d) Mode III- (negative gate current, and VT2< VT1 )
1) Connect the two links 1, 2(down) in Fig.2 to have negative IG and VT2< VT1.
2) Watch the lamp when it will be lighted, to determine IGT and VT2T1(sat).
3) Fill the table 4
Table 4( mode III-)
VT2 –VT1 (V)
IG (mA)
IT2 (mA)
Lamp
4) From table 1 find: IGT = mA , and (VT2 –VT1)sat = V , IT2(H)= mA.
5) Decrease the gate current and note that the TRIAC stays switching on.
II) The TRIAC as half-wave and full-wave rectifier.
1) Connect the circuit as shown in Fig.3.
2) Add an ammeter (dc) to measure IG.
3) Connect the oscilloscope to the two ends T1 & T2 of TRIAC.
4) Connect an ammeter (ac) to measure the current IT2.
5) Watch the lamp and note three different stages (no light, small and big brightness).
6) Increase the current IG and fill the table 5 and watch the lamp and oscilloscope
Fig.3 Half-wave and full-wave
rectification V0-20 V
0 V
IT2
IG
Table 5
VT2 –VT1 (V)
(ac)
IG (mA)
(dc)
IT2 (mA)
(ac)
Lamp No
light
No
light
Small
brightness
Small
brightness
Big
brightness
Big
brightness
7) Sketch the shape of signal for three causes of TRIAC:
• switched off,
• switched on with mode I+ (or mode III+) (half-wave rectifier)
• and switched on with two modes I+ & III+ (full-wave rectifier).