BM 2209 - Lab Manual

Aim To construct a differential amplifier using BJT and to determine the dc collector current of individual transistors and also to calculate the CMRR

APPARATUS REQUIRED

SNo Name Range Quantity1 Transistor BC107 22 Resistor 47kΩ 10kΩ 213 Regulated power supply (0-30)V 14 Function Generator (0-3) MHz 25 CRO 30 MHz 16 Bread Board 1

CIRCUIT DIAGRAM

OBSERVATIONVIN =VO =AC = VO VIN

FORMULA

Common mode Gain (Ac) = VO VIN

Differential mode Gain (Ad) = V0 VIN

Where VIN = V1 ndash V2

Common Mode Rejection Ratio (CMRR) = AdAc

Where Ad is the differential mode gainAc is the common mode gain

THEORY

The differential amplifier is a basic stage of an integrated operational amplifier It is used to amplify the difference between 2 signals It has excellent stability high versatility and immunity to noise In a practical differential amplifier the output depends not only upon the difference of the 2 signals but also depends upon the common mode signal

Transistor Q1 and Q2 have matched characteristics The values of RC1 and RC2 are equal Re1 and Re2 are also equal and this differential amplifier is called emitter coupled differential amplifier The output is taken between the two output terminals

OBSERVATION

VIN = V1 ndash V2

V0 =Ad = V0 VIN

For the differential mode operation the input is taken from two different sources and the common mode operation the applied signals are taken from the same source

Common Mode Rejection Ratio (CMRR) is an important parameter of the differential amplifier CMRR is defined as the ratio of the differential mode gain Ad to the common mode gain Ac

CMRR = Ad Ac

In ideal cases the value of CMRR is very high

PROCEDURE

1 Connections are given as per the circuit diagram2 To determine the common mode gain we set input signal with voltage Vin=2V

and determine Vo at the collector terminals Calculate common mode gain Ac=VoVin

3 To determine the differential mode gain we set input signals with voltages V1 and V2 Compute Vin=V1-V2 and find Vo at the collector terminals Calculate differential mode gain Ad=VoVin

4 Calculate the CMRR=AdAc5 Measure the dc collector current for the individual transistors

RESULT

Thus the Differential amplifier was constructed and dc collector current for the individual transistors is determined The CMRR is calculated as

REVIEW QUESTIONS

1 What is a differential amplifier2 What is common mode and differential mode inputs in a differential amplifier3 Define CMRR4 What is common mode signal

5 Write some applications of Differential amplifier

7 CLASS - A POWER AMPLIFIER

AIM

To construct a Class A power amplifier and observe the waveform and to compute maximum output power and efficiency

APPARATUS REQUIRED

SNo Name Range Quantity1 Transistor CL100 BC558 112 Resistor 47kΩ33Ω220Ω 213 Capacitor 47 microF 24 Signal Generator (0-3)MHz 15 CRO 30MHz 16 Regulated power supply (0-30)V 17 Bread Board 1

CIRCUIT DIAGRAM

Keep the input voltage constant Vin =

Frequency (in Hz) Output Voltage (in volts) Gain= 20 log(VoVin) (in dB)

FORMULA

Maximum power transfer =Pomax=Vo2RL

Effeciencyη = PomaxPc

THEORY The power amplifier is said to be Class A amplifier if the Q point and the input signal are selected such that the output signal is obtained for a full input signal cycle For all values of input signal the transistor remains in the active region and never enters into cut-off or saturation region When an ac signal is applied the collector voltage varies sinusoidally hence the collector current also varies sinusoidallyThe collector current flows for 3600 (full cycle) of the input signal i e the angle of the collector current flow is 3600

PROCEDURE1 Connect the circuit as per the circuit diagram2 Set Vi =50 mv using the signal generator3 Keeping the input voltage constant vary the frequency from 10 Hz to 1M Hz in regular steps and note down the corresponding output voltage4 Plot the graph Gain (dB) vs Frequency(Hz)

RESULTThus the Class A power amplifier was constructed The following parameters were

calculateda) Maximum output power=

` b) Efficiency=

REVIEW QUESTIONS

1 What is meant by Power Amplifier

2 What is the maximum efficiency in class ndash A amplifier

3 What are the disadvantages of Class ndashA amplifier

4 Write some applications of Power amplifier

5 What is the position of Q-point in Class ndashA amplifier

8CLASS B COMPLEMENTARY SYMMETRY POWER AMPLIFIERAIM

To construct a Class B complementary symmetry power amplifier and observe the waveforms with and without cross-over distortion and to compute maximum output power and efficiency

APPARATUS REQUIRED

SNo Name Range Quantity1 Transistor CL100 BC558 112 Resistor 47kΩ15kΩ 213 Capacitor 100microF 24 Diode IN4007 25 Signal Generator (0-3)MHz 16 CRO 30MHz 17 Regulated power supply (0-30)V 18 Bread Board 1

CIRCUIT DIAGRAM

FORMULAInput power Pin=2VccImПOutput power Pout=VmIm2Power Gain or efficiency η=л4(VmVcc) 100

THEORYA power amplifier is said to be Class B amplifier if the Q-point and the input signal are

selected such that the output signal is obtained only for one half cycle for a full input cycle The Q-point is selected on the X-axis Hence the transistor remains in the active region only for the positive half of the input signalThere are two types of Class B power amplifiers Push Pull amplifier and complementary symmetry amplifier In the complementary symmetry amplifier one n-p-n and another p-n-p transistor is used The matched pair of transistor are used in the common collector configuration In the positive half cycle of the input signal the n-p-n transistor is driven into active region and starts conducting and in negative half cycle the p-n-p transistor is driven into conduction However there is a period between the crossing of the half cycles of the input signals for which none of the transistor is active and output is zero

CIRCUIT DIAGRAM

FIG62

OBSERVATIONOUTPUT SIGNALAMPLITUDE TIME PERIOD

CALCULATION

POWER PIN = 2VCC ImлOUTPUT POWER POUT = VmIm2EFFICIENCY η = ( л4)( Vm VCC) x 100

MODEL GRAPH

FIG63

PROCEDURE1 Connections are given as per the circuit diagram without diodes2 Observe the waveforms and note the amplitude and time period of the input signal

and distorted waveforms3 Connections are made with diodes

4 Observe the waveforms and note the amplitude and time period of the input signal and output signal

5 Draw the waveforms for the readings6 Calculate the maximum output power and efficiencyHence the nature of the output signal gets distorted and no longer remains the same as the

input This distortion is called cross-over distortion Due to this distortion each transistor conducts for less than half cycle rather than the complete half cycle To overcome this distortion we add 2 diodes to provide a fixed bias and eliminate cross-over distortion

RESULTThus the Class B complementary symmetry power amplifier was constructed to observe

cross-over distortion and the circuit was modified to avoid the distortion The following parameters were calculated

a)Maximum output power=

b)Efficiency=

9 RC COUPLED AMPLIFIER

Aim To desgn an RC coupled amplifier and to plot its frequency responseComponents and equipments required Regulated Power supply function generator Transistor BC 107 Capacitors and resistorsCircuit diagram

Procedure1 Test all the components using a multimeter Set up the circuit and verify dc biasconditions To check dc bias conditions remove input signal and capacitors inthe circuit2 Connect the capacitors in the circuit Apply a 100 mV peak to peak sinusoidalsignal from the function generator to the circuit input Observe the input andoutput waveforms on the CRO screen simultaneously3 Keep the input voltage constant at 100 mV vary the frequency of the input signalfrom 0 to 1 MHz or highest frequency available in the generator Measure theoutput amplitude corresponding to di_erent frequencies and enter it in tabularcolumn4 Plot the frequency response characteristics on a graph sheet with gain in dB ony-axis and logf on x-axis Mark log fL and log fH corresponding to 3 dB points(If a semi-log graph sheet is used instead of ordinary graph sheet mark f alongx-axis instead of logf)5 Calculate the bandwidth of the ampli_er using the expression BW= fH- fL6 Remove the emitter bypass capacitor CE from the circuit and repeat the steps 3to 5 and observe that the bandwidth increases and gain decreases in the absence

Tabulation

Vin= ___________

Sl no Frequency Output voltage(Vo)

Gain(VoVin) Gain(db)20 log (gain)

QUESTIONS

1 Differentiate between ac and dc load lines Explain their importance in ampli_er analysis2 Why is the center point of the active region chosen for dc biasing3 What happens if extreme portions of the active region are chosen for dc biasing4 Draw the output characteristics of the amplifier and mark the load-line on it Also markthe three regions of operation on the output characteristics5 Which are the different forms of coupling used in multi-stage ampli_ers6 Draw hybrid and hybrid equivalent models of a transistor in the CE con_guration7 Draw the Ebers-Moll model of a BJT8 What are self bias and fixed bias9 Give a few applications of RC-coupled amplifier

10 RC PHASE SHIFT OSCILLATOR

Aim To design and set up an RC phase shift oscillator using BJT and to observe thesinusoidal output waveform

Components and equipments required Transistor dc source capacitors resis-tors potentiometer breadboard and CRO

Theory An oscillator is an electronic circuit for generating an ac signal voltage witha dc supply as the only input requirement The frequency of the generated signal isdecided by the circuit elements An oscillator requires an ampli_er a frequency selec-tive network and a positive feedback from the output to the input The Barkhausencriterion for sustained oscillation is A_ = 1 where A is the gain of the ampli_er and _is the feedback factor The unity gain means signal is in phase (If the signal is 180_out of phase gain will be 10485761) If a common emitter ampli_er is used with a resistive collector load there is a 180_ phase shift between the voltages at the base and the collector Feedback network between the collector and the base must introduce an additional 180_ phase shift at aparticular frequency In the figure shown three sections of phase shift networks are used so that each section introduces approximately 60ס phase shift at resonant frequency By analysisresonant frequency f can be expressed by the equation

The three section RC network o_ers a _ of 129 Hence the gain of the amplifiershould be 29 For this the requirement on the hFE of the transistor is found to be

The phase shift oscillator is particularly useful in the audio frequency range

CIRCUIT DIAGRAM

Procedure The connections are given as per circuit diagram and the ouput sinusoidal waveform is traced using CRO at Vo Viva-voce questions and answers

1 Classify the sinusoidal oscillatorsSinusoidal oscillators can be classi_ed as RC and LC oscillators LC oscillators are usedfor high frequency generation while RC oscillators for audio frequency generation2 Explain Barkhausen criteria for sustained oscillationa) Total loop gain (A_) of the circuit must be exactly unity where A is the gain of theampli_er and _ is the feedback factor b) Total phase shift around the loop must be 360_

3 What are the practical applications of a phase shift oscillatorRC-phase shift oscillator is widely used as audio frequency oscillator

4 What happens when CE is removed WhyWhen CE is removed gain of the ampli_er decreases and oscillation gets damped

5 Why is a minimum hFE value required for the circuit to function as an oscillatorA minimum hFE is required to obtain su_cient gain for the ampli_er part to satisfy theBarkhausen criteria for oscillation

6 How does one RC section generate a phase di_erence of 60_Phase shift introduced by one RC network is tan10485761(RC) Suitable values of R and Cwill provide 60_ phase shift between input and output of one RC network at a particularfrequency

11 HARTLEY OSCILLATOR

Aim - To construct Hartley oscillator using a transistor to find out the frequency ofoscillation and comparing it to that of theoretical frequency

Apparatus - n-p-n transistor Carbon resistors (as shown in circuit) two inductorscapacitors dc power supply CRO and connecting terminals

Formulae - The frequency of oscillation of the oscillator

Theory -The Hartley oscillator is designed for generation of sinusoidaloscillations in the RF range (20 KHz - 30 MHz) It is very popular and used in radioreceivers as a local oscillator

The circuit diagram of Hartley oscillator (parallel or shunt-fed) using BJT isshown in Figure It consists of an R-C coupled amplifier using an n-p-n transistor in CEconfiguration R1 and R2 are two resistors which form a voltage divider bias to thetransistor A resistor RE is connected in the circuit which stabilizes the circuit againsttemperature variations A capacitor CE is connected in parallel with RE acts as a bypasscapacitor and provides a low reactive path to the amplified ac signal The couplingcapacitor CC blocks dc and provides an ac path from the collector to the tank circuit TheL-C feedback network (tank circuit) consists of two inductors L1 and L2 (in series) whichare placed across a common capacitor C and the centre of the two inductors is tapped asshown in fig The feedback network (L1 L2 and C) determines the frequency ofoscillation of the oscillator

When the collector supply voltage Vcc is switched on collector current startsrising and charges the capacitor C When this capacitor is fully charged it dischargesthrough coils L1 and L2 setting up damped harmonic oscillations in the tank circuit Theoscillatory current in the tank circuit produces an ac voltage across L1 which is appliedto the base emitter junction of the transistor and appears in the amplified form in thecollector circuit Feedback of energy from output (collector emitter circuit) to input(base-emitter circuit is) accomplished through auto transformer action The output of theamplifier is applied across the inductor L1 and the voltage across L2 forms the feedbackvoltage The coil L1 is inductively coupled to coil L2 and the combination acts as anauto-transformer This energy supplied to the tank circuit overcomes the losses occurringin it Consequently the oscillations are sustained in the circuit

The energy supplied to the tank circuit is in phase with the generated oscillationsThe phase difference between the voltages across L1 and that across L2 is always 180degbecause the centre of the two is grounded A further phase of 180deg is introduced between

the input and output voltages by the transistor itself Thus the total phase shift becomes3600 (or zero) thereby making the feedback positive or regenerative which is essentialfor oscillations So continuous undamped oscillations are obtained

Procedure -The circuit is connected as shown in figure Connect the CRO across theoutput terminals of the oscillator Switch on the power supply to both the oscillator andCRO Select proper values of C L1 and L2 in the oscillator circuit and get the sine waveform on the screen of CRO The voltage (deflection) sensitivity band switch (Y-plates)and time base band switch (X-plates) are adjusted such that a steady and completepicture of one or two sine waveform is obtained on the screen The horizontal length (l)between two successive peaks is noted When this horizontal length (l) is multiplied bythe time base (m) ie secdiv we get the time-period (T = l x m)The reciprocal of thetime-period(1T) gives the frequency (f) This can be verified with the frequencycalculated theoretically by using the above formula The experiment is repeated bychanging C or L1 or L2 or all The readings are noted in the table given

CIRCUIT DIAGRAM

Result Thus the Hartley oscillator is designed and the theoretical and practical frequency of oscillation are verified

Viva questions

1 Explain the operation of a Hartley oscillators2 What is the difference between LC oscillators and RC oscillators3 Give the applications of Hartley oscillator4 What is the phase shift in Hartley oscillator5 What is the frequency of oscillation

12 Colpitts Oscillator

Aim - To construct Colpittrsquos oscillator using a transistor to find out the frequency ofoscillation and comparing it to that of theoretical frequencyApparatus - n-p-n transistor Carbon resistors (as shown in circuit) inductorcapacitors dc power supply CRO and connecting terminals

Formulae - The frequency of oscillation of the oscillator

Description - The Colpittrsquos oscillator is designed for generation of high frequencysinusoidal oscillations (radio frequencies ranging from 10KHz to 100MHz) They arewidely used in commercial signal generators up to 100MHz Colpitts oscillator is sameas Hartley oscillator except for one difference Instead of using a tapped inductanceColpitts oscillator uses a tapped capacitanceThe circuit diagram of Colpittrsquos oscillator using BJT is shown in Fig It consistsof an R-C coupled amplifier using an n-p-n transistor in CE configuration R1 and R2 aretwo resistors which form a voltage divider bias to the transistor A resistor RE isconnected in the circuit which stabilizes the circuit against temperature variations Acapacitor CE is connected in parallel with RE acts as a bypass capacitor and provides alow reactive path to the amplified ac signal The coupling capacitor CC blocks dc andprovides an ac path from the collector to the tank circuitThe feedback network (tank circuit) consists of two capacitors C1 and C2 (inseries) which placed across a common inductor L The centre of the two capacitors istapped (grounded) The feedback network (C1 C2 and L) determines the frequency ofoscillation of the oscillator The two series capacitors C1 and C2 form the potentialdivider led for providing the feedback voltage The voltage developed across thecapacitor C2 provides regenerative feedback which is essential for sustained oscillations

Theory - When the collector supply voltage Vcc is switched on collector current startsrising and charges the capacitors C1 and C2 When these capacitors are fully charged theydischarge through coil L setting up damped harmonic oscillations in the tank circuit Theoscillatory current in the tank circuit produces an ac voltages across C1 C2 Theoscillations across C2 are applied to base-emitter junction of the transistor and appears inthe amplified form in the collector circuit and overcomes the losses occurring in the tankcircuitThe feedback voltage ( across the capacitor C2) is 180deg out of phase with theoutput voltage ( across the capacitor C1) as the centre of the two capacitors isgrounded A phase shift of 180deg is produced by the feedback network and a further phaseshift of 180deg between the output and input voltage is produced by the CE transistorHence the total phase shift is 360deg or 0deg which is essential for sustained oscillations as

per the Barkhausen criterion So we get continuous undamped oscillationsProcedure -The circuit is connected as shown in figure Connect the CRO across theoutput terminals of the oscillator Switch on the power supply to both the oscillator andCRO Select proper values of L C1 and C2 in the oscillator circuit and get the sine waveform on the screen of CRO The voltage (deflection) sensitivity band switch (Y-plates)and time base band switch (X-plates) are adjusted such that a steady and completepicture of one or two sine waveform is obtained on the screen The horizontal length (l)between two successive peaks is noted When this horizontal length (l) is multiplied bythe time base (m) ie secdiv we get the time-period (T = l x m)The reciprocal of thetime-period(1T) gives the frequency(f) This can be verified with the frequencycalculated theoretically by using the above formula The experiment is repeated bychanging L or C1 or C2 or all The readings are noted in the table given

CIRCUIT DIAGRAM

TABULATION

Result Thus the colpitts oscillator is designed and the theoretical and practical frequency of oscillation is verified

Viva Questions

1 Explain the operation of a Colpitts oscillators2 What is the difference between LC oscillators and RC oscillators3 Give the applications of copitts oscillator4 What is the phase shift in Colpitts oscillator5 What is the frequency of oscillation of Colpitts oscillator

13ZENER DIODE AS VOLTAGE REGULATOR

Aim To determine the line regulation and load regulation characteristics of a zener diode

Apparatus required

Regulated Power supply zener diode resistors milliammeters

Theory

The function of a regulator is to provide a constant output voltage to a load connected in parallel with it in spite of the ripples in the supply voltage or the variation in the load current and the zener diode will continue to regulate the voltage until the diodes current falls below the minimum IZ(min) value in the reverse breakdown region It permits current to flow in the forward direction as normal but will also allow it to flow in the reverse direction when the voltage is above a certain value - the breakdown voltage known as the Zener voltage The Zener diode specially made to have a reverse voltage breakdown at a specific voltage Its characteristics are otherwise very similar to common diodes In breakdown the voltage across the Zener diode is close to constant over a wide range of currents thus making it useful as a shunt voltage regulator

The purpose of a voltage regulator is to maintain a constant voltage across a load regardless of variations in the applied input voltage and variations in the load current The resistor is selected so that when the input voltage is at VIN(min) and the load current is at IL(max) that the current through the Zener diode is at least Iz(min) Then for all other combinations of input voltage and load current the Zener diode conducts the excess current thus maintaining a constant voltage across the load The Zener conducts the least current when the load current is the highest and it conducts the most current when the load current is the lowest

Circuit diagram

Procedure If there is no load resistance shunt regulators can be used to dissipate total power

through the series resistance and the Zener diode Shunt regulators have an inherent current limiting

advantage under load fault conditions because the series resistor limits excess current

A zener diode of break down voltage Vz is reverse connected to an input voltage source V i across a

load resistance RL and a series resistor RS The voltage across the zener will remain steady at its break

down voltage VZ for all the values of zener current IZ as long as the current remains in the break down

region Hence a regulated DC output voltage V0 = VZ is obtained across RL whenever the input voltage

remains within a minimum and maximum voltage

Basically there are two type of regulations

a) Line Regulation

In this type of regulation series resistance and load resistance are fixed only input voltage is

changing Output voltage remains same up to particular low input voltage

Percentage of line regulation can be calculated by =

where V0 is the output voltage and VIN input voltage

b) Load Regulation

In this type of regulation input voltage is fixed and the load resistance is varying Output volt remains

same zener diode maintain the flow of current through the load resistance

Percentage of load regulation =

where is the null load resistor voltage and is the full load resistor voltage

Result Thus a voltage regulator circuit using zener is constructed and line regulation and load regulation are determined

14 MONOSTABLE MULTIVIBRATOR

Aim To study the working principle of monostable multivibrator

Components Required Transistor BC 108 resistors capacitors function generatorDiode 1N4004

Theory The basic collector-coupled Monostable Multivibrator circuit and its associated waveforms are shown above When power is firstly applied the base of transistor TR2 is connected to Vcc via the biasing resistor RT thereby turning the transistor fully-ON and into saturation and at the same time turningTR1 OFF in the process This then represents the circuits Stable State with zero output The current flowing into the saturated base terminal of TR2 will therefore be equal to Ib = (Vcc - 07)RT

If a negative trigger pulse is now applied at the input the fast decaying edge of the pulse will pass straight through capacitor C1 to the base of transistor TR1 via the blocking diode turning it ON The collector of TR1 which was previously at Vcc drops quickly to below zero volts effectively giving capacitorCT a reverse charge of -07v across its plates This action results in transistor TR2 now having a minus base voltage at point X holding the transistor fully OFF This then represents the circuits second state the Unstable State with an output voltage equal to Vcc

Timing capacitor CT begins to discharge this -07v through the timing resistor RT attempting to charge up to the supply voltage Vcc This negative voltage at the base of transistor TR2 begins to decrease gradually at a rate determined by the time constant of the RT CT combination As the base voltage ofTR2 increases back up to Vcc the transistor begins to conduct and doing so turns OFF again transistor TR1 which results in the monostable multivibrator automatically

returning back to its original stable state awaiting a second negative trigger pulse to restart the process once again

Monostable Multivibrators can produce a very short pulse or a much longer rectangular shaped waveform whose leading edge rises in time with the externally applied trigger pulse and whose trailing edge is dependent upon the RC time constant of the feedback components used This RC time constant may be varied with time to produce a series of pulses which have a controlled fixed time delay in relation to the original trigger pulse as shown below

Monostable Multivibrator Waveforms

Result

Thus the waveforms are traced and the working of monostable multivibrator is studied

15 Astable Multivibrator (AMV)

Aim - To construct an astable multivibrator using transistors for getting square waveand to determine the frequency of oscillationApparatus - Two n-p-n transistors two fixed carbon resistors two variable non ndashinductive resistors ( pots) two capacitors dc power supply CRO and connectingterminals

Formula

Circuit Diagram

Theory When power Vcc is applied by closing switch S collector current starts flowing in Q1 and Q2 and the coupling capacitors C1 and C2 start charging up Since the characteristics of no two seemingly similar transistors can be exactly alike one transistorsay Q1 will conduct more rapidly than the other Then the collector current of Ql will rise at a faster rate causing a decrease in its collector voltage The resulting negative signal is applied to the base of Q2 through C2 and drives it towards cut-off Consequently thecollector voltage of Q2 (positive going signal) is fed to the base of transistor Q1 through capacitor C1 As a result of this positive going pulse the collector current of Q1 is further increased The process being cumulative in a short time transistor Q1 is saturated while Q2 is cut-off These actions are so rapid and instantaneous that

C1 does not get a chance to discharge Under this situation whole of Vcc drops across RL1 (since Q1 is saturated or is in ON state) ie Vc1 = 0 and point A is at ground (or zero) potential Also since Q2 is cut-off (OFF state) there is no drop across RL2 and point B is at Vcc Capacitor C2 now begins to discharge through R2 which decreases the reverse bias on base of transistor Q2 Ultimately a forward bias is re-established at Q2 which therefore begins to conduct Consequently Collector of Q2 becomes less positive This negative going voltage signal is applied to the base of transistor Q1 through the capacitor C1 As a result Q1 is pulled out of saturation and is soon driven to cut-off Simultaneously Q2 is driven to saturation Now Vc2 decreases and becomes almost zero volt when Q2 gets saturated Consequently potential of point B decreases from Vcc to almost zero volt The transistor Q1 remains cut-off and Q2 in conduction until capacitor Cl discharges through R1 enough to decrease the reverse bias of Q1 The whole of the cycle is repeated The output of the multivibrator can be taken from the collector of either transistor The output is a square wave with a peak amplitude equal to Vcc Total time of the square wave T = T1 + T2 = 069 (R1C1+R2C2)

Procedure - The two transistors (Q1 and Q2) are connected in CE mode and they are given proper bias with the help of RL1 RL2 and + VCC Collector of each transistor is connected to the base of the other transistor through a condenser The condensers C1 and C2 are connected to the power supply through the variable resistors R1 and R2 The collector of any one of the transistor is connected the Y ndash plates of CRO Switch on the power Vcc and the power supply of CRO Observe the square wave on the screen Adjust the values Rl and R2 and the band switches of X and Y plates of CRO to get at least one complete wave on the screenThen the length of one complete wave ( l ) on screen is measured on horizontal scale this is multiplied with the time base ( t ) The product will give the time period of the wave ( l x t = T )The reciprocal of lsquoTrsquo gives the frequency ( f ) These values are noted in the table This frequency is experimental frequencyNow the Power Vcc is switched off and the resistance values of Rl and R2 are measured using multi-meter The values R1 R2 C1 and C2 are also noted in the table Substituting these values in the above formula we will get the frequency theoretically The theoretical and experimental frequencies are compared They are equalThe experiment is repeated with different values of R1 and R2 (the values of C1 and C2 can also be changed if possible)

Tabulation

Result Thus the astable multivibrator is constructed and the waveforms are traced

16 DARLINGTON AMPLIFIER USING BJT

AIMTo construct a Darlington current amplifier circuit and to plot the frequency response

characteristics

APPARATUS REQUIRED

SNo Name Range Quantity1 Transistor BC 107 12 Resistor 15kΩ10kΩ680Ω6kΩ 11113 Capacitor 01microF 47microF 2 14 Function Generator (0-3)MHz 15 CRO 30MHz 16 Regulated power supply (0-30)V 17 Bread Board 1

CIRCUIT DIAGRAM

MODEL GRAPH

f 1 FIG2 f2 f (Hz)

TAB 41

Keep the input voltage constant Vin =

Frequency (in Hz) Output Voltage (in volts) Gain= 20 log(VoVin) (in dB)

THEORY

In Darlington connection of transistors emitter of the first transistor is directly connected to the base of the second transistor Because of direct coupling dc output current of the first stage is (1+hfe )Ib1If Darlington connection for n transitor is considered then due to direct coupling the dc output current foe last stage is (1+h fe ) n times Ib1 Due to very large amplification factor even two stage Darlington connection has large output current and output stage may have to be a power stage As the power amplifiers are not used in the amplifier circuits it is not possible to use more than two transistors in the Darlington connection In Darlington transistor connection the leakage current of the first transistor is amplified by the second transistor and overall leakage current may be high Which is not desired

PROCEDURE

1 Connect the circuit as per the circuit diagram2 Set Vi =50 mv using the signal generator3 Keeping the input voltage constant vary the frequency from 0 Hz to 1M Hz in regular steps and note down the corresponding output voltage4 Plot the graph Gain (dB) vs Frequency(Hz)5 Calculate the bandwidth from the graphRESULT

Thus the Darlington current amplifier was constructed and the frequency response curve is plotted The Gain Bandwidth Product is found to be =

QUESTIONS

1 What is meant by Darlington pair

2 How many transistors are used to construct a Darlington amplifier circuit

3 What is the advantage of Darlington amplifier circuit

4Write some applications of Darlington amplifier