Basic Electronics - I Lab Manuals

LAB MANUAL

3rd SEMESTER

BASIC ELECTRONICS – I

Lab Engineerer: Subject Teacher

Sir Abdul Kaleem Dr. Manzar Saeed (HOD)

HAJVERY UNIVERSITY LAHORE Basic Electronics – I (LAB)

Section II 1

CONTENTS Exp # List of Experiments

1 Study of Oscilloscope

2 Troubleshooting of diode

3 TO CONSTRUCT A HALF-WAVE RECTIFIER CIRCUIT AND TO CHECK ITS OUTPUT WAVEFORM ON OSCILLOSCOPE

4 TO CONSTRUCT A FULL-WAVE CENTER-TAP RECTIFIER CIRCUIT & TO CHECK AND MEASURE THE INPUT & OUTPUTS WAVE FORMS ON OSCILLOSCOPE

5 INTRODUCTION OF PROTEUS SOFTWARE

6 TO CONSTRUCT A FULL-WAVE BRIDGE RECTIFIER CIRCUIT AND TO CHECK AND MEASURE THE INPUT AND OUTPUTS WAVE FORMS ON OSCILLOSCOPE

7 BIASING TECHNIQUES OF BJT.


Section II 2

EXPERIMENT NO – 01

STUDY OF OSCILLOSCOPE:

Dual Trace Oscilloscope 20MHz (GW INSTEK GOS-620)


Section II 3


Section II 4

Lab Task: 1. Generate the following signals from function generator and verify this on oscilloscope.

Example:


Section II 5


TROUBLESHOOTING OF DIODE

Testing a diode is quite simple, particularly if the multi-meter used has a diode check function. With the diode check function a specific known voltage is applied from the meter across the diode. With the diode check function a good diode will show approximately 0.7 V or 0.3 V when forward biased. When checking in reverse bias the full applied testing voltage will be sent on the display. An ohm-meter can be used to check the forward and reverse resistance of a diode if the ohm-meter has enough voltage to force the diode into conduction. Of course, in forward biased connection low Resistance will be seen and in reverse biased connecting high. OPEN DIODE In the case of an open diode no current flows in either direction which is indicated by the full checking voltage with the diode check function or high resistance using an ohmmeter in both forward and reverse connections. SHORTED DIODE In the case of a shorted diode maximum current flows indicated by 0V with the diode check function or low resistance with an ohm-meter in both forward and reverse connections.


Section II 6


TO CONSTRUCT A HALF-WAVE RECTIFIER CIRCUIT AND TO CHECK ITS OUTPUT WAVEFORM ON OSCILLOSCOPE

THEORY HALF-WAVE RECTIFIER Rectifier is the diode used in converting AC to DC and this process is rectification. The basic way of rectification is half-wave rectifier circuit shown in Figure 1. When the secondary voltage of transformer is positive half period (VAB is +), diode D1 becomes forward bias. Because it represents very low resistance value toward voltage source, so most of the secondary voltage appears both sides of load RL . Silicon and germanium are representative forward biased diode. The step-down range of silicon diode is from 0.5V to 1.0V and that of germanium diode is from 0.2V to 0.6V. Most of stepdown is ignored to make the interpretation of circuit simple. Especially when power supply is very high, forward step-down of diode becomes very small toward output voltage.

Figure 1(b) explains the action of half-wave rectifier. Note that the fact that output becomes ‘0’ when the voltage of transformer (V AB ) is negative (-). It is because diode becomes a backward bias (added anode toward cathode). It is the same as open circuit ideally. Average DC voltage (Vdc ) is the same as 0.318 times ( 0.318 = 1/π ) of maximum value. Most of voltage meter displays average value. So it indicates 0.318 times of maximum voltage toward half-wave rectifier. But effective value must be used to calculate power. The effective voltage toward half-wave rectifier circuit is 0.5 times of maximum value. In case of half-wave;


Section II 7

This 2 way of displaying voltage may cause some confusion. Fortunately, effective value and average value is mostly equal in general DC current. Therefore you may not worry about that. Average current I 0 is the current taken by dividing average voltage of load by load resistance.

Step-down is so small in forward bias. But maximum input voltage appears as the step-down of both sides of diode in backward bias. We call it as Peak Reverse Voltage (PRV). Every diode has maximum allowable PRV rating which must not be exceeded and when the exceeding is happens, the factor extinguished. The voltage of diode VAC in Figure 1(b) follows VAB in backward bias. Therefore diode has very high resistance value. And note that step-down (V AC ) is not ‘0’ but a small positive value. It is a forward step-down of diode and generally it is less than 1V. APPARATUS: 1. Low-voltage AC power supply 2. One 1N4001 diode 3. Resistance 100Ω 4. Oscilloscope The diode need not be an exact model 1N4001. Any of the "1N400X" series of rectifying diodes are suitable for the task. SCHEMATIC DIAGRAM:

PROCEDURE: 1. Connect the diode to the low-voltage AC power supply as shown in a figure. Note that the resistor uses to limit the current. 2. Connect CH1 of oscilloscope to Input and CH2 to Output/Load Resistance of a circuit. 3. Switch on the oscilloscope and the sinusoidal supply. 4. Sketch the input waveform.


Section II 8

INPUT WAVEFORM

5. Measure and record time T , peak voltage Vp and peak to peak voltage Vpp of Input supply

T= _______________ Vp: _______________ Vpp _______________ 6. With the oscilloscope DC. Coupled adjust the time-base and the Y amplifier sensitivity. 7. Sketch the waveform and label it to show the periods when the diode is conducting and those when it is not. Time T depends upon the frequency of your power supply.

OUTPUT WAVEFORM

8. Measure and record time T and peak voltage Vp of output

T= _______________ Vp: _______________ 9. Confirm this Vp should be very nearly equal to the peak voltage of the alternating supply.


Section II 9


TO CONSTRUCT A FULL-WAVE CENTER-TAP RECTIFIER CIRCUIT & TO CHECK AND MEASURE THE INPUT & OUTPUTS WAVE FORMS ON

OSCILLOSCOPE

THEORY: More useful and effective way of converting AC to DC is using both positive and negative range of AC input signal. There are 2 kinds of circuit to use to do this. Figure-1 shows a circuit among the two. Because this method uses all of input wave type as DC output, it is known as full-wave rectification. Center-tap rectifier in Figure-1 uses secondary winding having center-tap. When the polarity of voltage is the same as figure, anode has a positive polarity toward cathode. So D1

becomes a forward bias and conduction status. On the contrary, D2 becomes a backward bias and non-conduction status. Therefore only D1 supplies the current to load.

Figure-1

Because the polarity of secondary voltage of transformer is inverted in the next half cycle of AC, so everything is in opposition to above condition. Therefore, D1 becomes backward bias and D2 becomes forward bias, then D2 supplies current to load. Because each diode is insulation status during only the half cycle (by half cycle in turn), load current which is double the current of half-wave rectifying. Figure -2 shows its output wave type. Note that double the increase of frequency appears in output substantially. It is because cycle of output wave type T is the half of AC input signal. Remember that frequency is the reciprocal of cycle. ( f =1/ T ). Center tap circuit has been the most general full-wave rectifying circuit but, bridge circuit becomes most general owing to the appearance of silicon diode having low price, high reliability and small size.


Section II 10

Figure- 3: I/O waveforms of full wave rectifier

The reason is in the fact that it enables to cut down the size of transformer needed in getting the degree of output as well as center tap. Current flows in turn by dividing secondary side of transformer half and half during each half cycle of main-sub in center tap circuit. During this one cycle of the input sine wave, two positive DC pulses have been developed. ith this Condition, the output frequency has doubled. If the input frequency is 50 hertz, the positive alternation will be present 50 times. After the full-wave rectification, there will be 100 positive pulses at the output. If the DC output signal is measured with a multi-meter, the indication will be the average value of the peak signal. To determine the average value of a full-wave rectified signal, multiply the peak value by 0.636


Section II 11

Example:

VAVG = VP * 0.636 Input peak value = 10 V AC

10 V AC x 0.636 = 6.36 V DC

APPARATUS: 1. Low-voltage AC power supply 2. Two 1N4001 diode 3. Resistance 1KΩ 4. Oscilloscope The diodes need not be exact model 1N4001 units. Any of the "1N400X" series of rectifying diodes are suitable for the task, and they are quite easy to obtain. SCHEMATIC DIAGRAM:

PROCEDURE: 1. Connect the diodes to the low-voltage AC power supply as shown in a figure. 2. Connect CH1 of Oscilloscope to Input and CH2 to Output/Load Resistance of a circuit. 3. Switch on the oscilloscope and the sinusoidal supply. 4. Sketch the input waveform.


Section II 12

INPUT WAVEFORM

5. Measure and record time T , peak voltage Vp and peak to peak voltage Vpp of Input supply

T= _______________ Vp: _______________ Vpp _______________ 6. With the oscilloscope DC. Coupled adjust the time-base and the Y amplifier sensitivity. 7. Sketch the output waveform and label it to show the periods when the diode D1 is conducting and when the diode D2 is conducting those. Time T depends upon the frequency of your power supply.

OUTPUT WAVEFORM

8. Measure and record time T and peak voltage Vp of an output supply.

T= _______________ Vp: _______________


Section II 13


INTRODUCTION OF PROTEUS SOFTWARE


Section II 14


TO CONSTRUCT A FULL-WAVE BRIDGE RECTIFIER CIRCUIT AND TO CHECK AND MEASURE THE INPUT AND OUTPUTS WAVE FORMS ON OSCILLOSCOPE

THEORY: A basic full-wave bridge rectifier is illustrated in Figure 1.

Figure 1 Basic Full-Wave Bridge Rectifier

A full wave bridge rectifier has one advantage over the conventional full-wave rectifier: the amplitude of the output signal. The frequency of the positive pulses will be the same in either rectifier. When the output signal is taken from a bridge rectifier, it is taken across the entire potential of the transformer. Thus, the output signal will be twice the amplitude of a conventional full-wave rectifier. For the first half cycle of a bridge rectifier, refer to Figure 2.

Figure 2 Full Wave Bridge Rectifier (First Half-Wave Cycle Operation)

During the first half cycle of the input signal, a positive potential is felt at Point A and a negative potential is felt at Point B. Under this condition, a positive potential is felt on the


Section II 15

anode of D2 and on the cathode of D1. D2 will be forward-biased, while D1 will be reverse-biased. Also, a negative potential will be placed on the cathode of D3 and the anode of D4. D3

will be forward-biased, while D4 will be reverse biased. With D3 and D2 forward-biased, a path for current flow has been developed. The current will flow from the lower side of the transformer to Point D. D3 is forward-biased, so current will flow through D3 to Point E, from Point E to the bottom of the load resistor, and up to Point F. R3 is forward biased, so current will flow through D2, to Point C, and to Point A. The difference of potential across the secondary of the transformer causes the current to flow. Diodes D 3 and D2 are forward-biased, so very little resistance is offered to the current flow by these components. Also, the resistance of the transformer is very small, so approximately all the applied potential will be developed cross the load resistor. If the potential from Point A to Point B of the transformer is 24 volts, the output developed across the load resistor will be a positive pulse approximately 24 volts in amplitude.

Figure 3. Full-Wave Bridge Rectifier (Second Cycle Operation)

When the next alternation of the input is felt (Figure 3), the potential across the transformer reverses polarity. Now, a negative potential is felt at Point A and a positive potential is felt at Point B. With a negative felt at Point C, D1 will have a negative on the cathode and D2 will have a negative on the anode. A positive at Point D will be felt on the anode of D4 and the cathode of D3. D1 and D4 will be forward-biased and will create a path for current flow. D3 and D2 will be reverse-biased, so no current will flow. The path for current flow is from Point A to Point C, through D1 to Point E, to the lower side of the load resistor, through the load resistor to Point F, through D4 to Point D, and to the lower side of T1. Current flows because of the full potential being present across the entire transformer; therefore, the current through the load resistor will develop the complete voltage potential. The frequency of the output pulses will be twice that of the input pulses because both cycles of the input AC voltage are being used to produce an output. APPARATUS:

1. Low-voltage AC power supply 2. Four 1N4001 diode 3. Resistance 1KΩ 4. Oscilloscope


Section II 16

The diodes need not be exact model 1N4001 units. Any of the "1N400X" series of rectifying diodes are suitable for the task, and they are quite easy to obtain. SCHEMATIC DIAGRAM:

Figure

PROCEDURE: 1. Connect the diodes to the low-voltage AC power supply as shown in a figure. 2. Connect CH1 of Oscilloscope to Input and CH2 to Output/Load Resistance of a circuit. 3. Switch on the oscilloscope and the sinusoidal supply. 4. Sketch the input waveform.

5. Measure and record time T, peak voltage Vp and peak to peak voltage Vpp of Input supply

T= _______________ Vp: _______________ Vpp _______________

6. With the oscilloscope DC. Coupled adjust the time-base and the Y amplifier sensitivity. 7. Sketch the output waveform and label it to show the periods when the diode D1 and D4 are conducting and when the diode D2 and D3 are conducting those. Time T depends upon the frequency of your power supply.


Section II 17

8. Sketch the output waveform during positive Half Cycle.

9. Sketch the output waveform during negative Half Cycle


Section II 18

10. Sketch the output waveform

OUTPUT WAVEFORM

11. Measure and record time T and peak voltage Vp of an output supply.

T= _______________ Vp: _______________

12. Compare Input and output voltages.


Section II 19


Biasing Techniques of BJT.

Components:

Transistor NPN (2N1711)

Resistor (R2 100KΩ)

Resistor (R1 1KΩ)

VCC 10V

Multimeter

Circuit Diagram:

Procedure:

Use a transistor NPN (2N1711), collector resistor R1, base resistor R2 and connecting the

ampere meter in series for IC, IB & IE. Connecting the voltmeter across R1 & R2. Then find

the VCE connecting the voltmeter in terminal of transistor collector to emitter.

Observations & Calculations:

Calculated Values:

Collector Current (IC) Base Current (IB) Emitter Current (IE)


Section II 20

Measured Values:

Collector Current (IC) Base Current (IB) Emitter Current (IE)

Calculated Values:

VCE Β (hfe)

Measured Values:

VCE Β (hfe)

Documents

Basic Electronics - I Lab Manuals