experiment 1: ohm’s law

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.






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.






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






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






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






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






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






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


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






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






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






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






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) .


𝑇∫ 𝑉𝑝

2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇

0

= √1

𝑇∫ 𝑉𝑝

2 (1 + sin(2𝜔𝑡)

2)𝑑𝑡

𝑇

0

=


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






=1

𝑇∫ 𝑉(𝑡)𝑑𝑡 =

1

𝑇(∫ 𝑉𝑃 sin(𝜔𝑡) 𝑑𝑡

𝑇/2

0

+∫ 0𝑑𝑡) =𝑇

𝑇/2

𝑇

0


cos(𝜔𝑇/2) − cos(0) = −𝑉𝑃2𝜋

cos(𝜋) − cos(0)

For half-wave rectified waveform without capacitor = 𝑉𝑃

𝜋 (5)


𝑇∫ 𝑉𝑝

2𝑠𝑖𝑛2(𝜔𝑡)𝑑𝑡𝑇/2

0

= √1

𝑇∫ 𝑉𝑝

2 (1 + sin(2𝜔𝑡)

2)𝑑𝑡

𝑇/2

0

=


2𝑇∫ 𝑉𝑝

2𝑑𝑡𝑇/2

0

+1

𝑇∫ (𝑉𝑝

2sin(2𝜔𝑡)

2)𝑑𝑡

𝑇/2

0

=


2𝑇∫ 𝑑𝑡

𝑇/2

0

+𝑉𝑝2


𝑇/2

0

=


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






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







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






………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………..…………………

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




a) Fill the Table 2

V (V/cm) t (s/cm)

Y (cm) V (V) V2 (V2) X (cm) t(s)

8









Note that : VAVG=Vp/𝝅 =Vdc

………………………………………………………………………………………………………………………………………………………………………………………………………………


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.

………………………………………………………………………………………………………………………………………………………………………………………………………………


𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞), compare this value with result in step c.

………………………………………………………………………………………………………………………………………………………………………………………………………………

9






III) The effect of a reservoir capacitor

Connect the capacitors (small and big) with output circuit and sketch the output signals.

1






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.


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






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)


𝑇∫ 𝑉2(𝑡)𝑑𝑡𝑇

0 (2)

For sinusoidal waveform we can write (𝑡) = 𝑉𝑃sin(𝜔𝑡) .

3






=1

𝑇∫ 𝑉𝑃sin(𝜔𝑡)𝑑𝑡

𝑇

0


(cos(𝜔𝑇) − cos(0)) = −𝑉𝑃2𝜋

(cos(2𝜋) − 𝑐𝑜𝑠0) = 0

The mean value of sinusoidal wave equals zero = 0 (3) .


𝑇∫ 𝑉𝑝


0

= √1

𝑇∫ 𝑉𝑝

2 (1 + sin(2𝜔𝑡)

2)𝑑𝑡

𝑇

0

=


2𝑇∫ 𝑉𝑝

2𝑑𝑡𝑇

0

+1

𝑇∫ (𝑉𝑝

2sin(2𝜔𝑡)

2)𝑑𝑡

𝑇

0

=


2𝑇∫ 𝑑𝑡

𝑇

0

+𝑉𝑝2


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







𝑇∫ 𝑉𝑝


0

= √1

𝑇∫ 𝑉𝑝

2 (1 + sin(2𝜔𝑡)

2)𝑑𝑡

𝑇

0

=


2𝑇∫ 𝑉𝑝

2𝑑𝑡𝑇

0

+1

𝑇∫ (𝑉𝑝

2sin(2𝜔𝑡)

2)𝑑𝑡

𝑇

0

=


2𝑇∫ 𝑑𝑡

𝑇

0

+𝑉𝑝2


𝑇

0

=


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






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







I) Input signal




a) Fill the Table 1

V (V/cm) t (s/cm)

Y (cm) V (V) V2 (V2) X (cm) t(s)



7







Note that : VAVG=0 =Vdc

………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………..…………………


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.

………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………..…………………


𝐕𝐫𝐦𝐬𝟐 = 𝐚𝐫𝐞𝐚/(𝐩𝐞𝐫𝐢𝐨𝐝𝐭𝐢𝐦𝐞) and then find the square root compare this value with result

in step c.

………………………………………………………………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………..…………………

8






II) output signal




a) Fill the Table 2

V (V/cm) t (s/cm)

Y (cm) V (V) V2 (V2) X (cm) t(s)



9







Note that : VAVG=2Vp/𝝅 =Vdc

………………………………………………………………………………………………………………………………………………………………………………………………………………


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.

………………………………………………………………………………………………………………………………………………………………………………………………………………


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.

10











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






Experiment No. 4

The Zener diode

The Volt-Ampere characteristic curve



----------------------------------------------------------------------------------

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






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


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.







- 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:

…………………………………………………………………………………………

…………………………………………………………………………………………

1






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

2






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

3






Experiment No.5

The zener diode 2

Regulated Voltage Supply



----------------------------------------------------------------------------------

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.

4






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


1) Fill the Table 1

2) Plot the load current IL versus the voltage of voltage supply VS for Id=70 mA.

5






(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.

1






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.


2






Experiment No.7

The common emitter



----------------------------------------------------------------------------------


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.

1






Experiment No. 8:

The Silicon Controller Rectifier

2






Experiment No.8

The silicon controlled rectifier

Name: Date:

Physics 106 Section: Instructor’s Name:

----------------------------------------------------------------------------------


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

3






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

4






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.

1






Experiment No.6

Transistor familiarization



----------------------------------------------------------------------------------

Vcc = 20 V and VBB = 10 V


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.

2






…………………………………………………………………………………………

…………………………………………………………………………………………

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.


3) Fill the table 2

Table 2( mode I- )

VT2 –VT1 (V)

IG (mA)

IT2 (mA)

Lamp



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.


3) Fill the table 3

Table 3( mode III+ )

VT2 –VT1 (V)

IG (mA)

IT2 (mA)

Lamp



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.


3) Fill the table 4

Table 4( mode III-)

VT2 –VT1 (V)

IG (mA)

IT2 (mA)

Lamp



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).

experiment 1: ohm’s law

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