ec ii lab manual (1)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

IV SEMESTER - ECE ‘B’

EC1256 – ELECTRONIC CIRCUITS – II AND SIMULATION LAB

Cycle – I

1. Current Series feedback amplifier: Frequency response, Input and output impedance calculation

2. Voltage Shunt feedback amplifier: Frequency response, Input and output impedance calculation

3. RC Phase shift oscillator4. Hartley Oscillator5. Colpitts Oscillator 6. CMOS Inverter, NAND and NOR using PSPICE7. Active filter: Butterworth II order LPF using PSPICE8. Differential amplifier using PSPICE

Cycle – II

9. Frequency Response of Single Tuned Amplifier10. Astable multivibrator11. Monostable multivibrator12. Bistable multivibrator13. Wave Shaping Circuits (Integrator, Differentiator, Clippers and Clampers)14. D/A converter using PSPICE15. Astable multivibrator using PSPICE16. Monostable multivibrator using PSPICE

Expt No: 1 CURRENT-SERIES FEEDBACK AMPLIFIER

Aim: To design and test the current-series feedback amplifier and to calculate the

following parameters with and without feedback.1. Mid band gain.2. Bandwidth and cut-off frequencies.3. Input and output impedance.

Components & Equipment required:

Sl. No. Components / EquipmentRange /

SpecificationsQuantity

1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 Connecting wires Accordingly

Circuit diagram:

(i) Without feedback:

2

(ii) With feedback:

Theory:

The current series feedback amplifier is characterized by having shunt sampling and series mixing. In amplifiers, there is a sampling network, which samples the output and gives to the feedback network. The feedback signal is mixed with input signal by either shunt or series mixing technique. Due to shunt sampling the output resistance increases by a factor of ‘D’ and the input resistance is also increased by the same factor due to series mixing. This is basically transconductance amplifier. Its input is voltage which is amplified as current.

Design:

(i) Without feedback:

VCC = 12V; IC = 1mA; fL = 50Hz; S = 2; RL = 4.7K; hfe =

re = 26mV / IC = 26;

hie = hfe re =

VCE= Vcc/2 (transistor Active) =

VE = IERE = Vcc/10

Applying KVL to output loop, we get

VCC = ICRC + VCE + IERE

3

RC =

Since IB is very small when compare with IC,

IC IE

RE = VE / IE =

S = 1+ RB / RE = 2

RB =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

XCi = Zi / 10 = (hie || RB) / 10 =

Ci = 1 / (2f XCi) =

Xco = (RC || RL)/10 =

Co = 1 / (2f XCo) =

XCE = RE/10 =

CE = 1 / (2f XCE) =

(ii) With feedback (Remove the Emitter Capacitor, CE):

Feedback factor, = -RE =

Gm = -hfe / (hie + RE) =

Desensitivity factor, D = 1 + Gm =

Transconductance with feedback, Gmf = Gm / D =

Input impedance with feedback, Zif = Zi D

Output impedance with feedback, Z0f = Z0 D

Procedure:

1. Connect the circuit as per the circuit diagram.2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in

regular steps and note down the corresponding output voltage.3. Plot the graph: Gain (dB) Vs Frequency4. Calculate the bandwidth from the graph.5. Calculate the input and output impedance.6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).

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Tabular column:

(i) Without feedback:Vi =

Sl. NoFrequency

(Hz)Output Voltage

(V0) (volts)Gain = V0/Vi

Gain = 20 log(V0/Vi)(dB)

(ii) With feedback:Vi =

Sl. NoFrequency

(Hz)Output Voltage

(V0) (volts)Gain = V0/Vi

Gain = 20 log(V0/Vi)(dB)

Model graph: (frequency response)

Gain in dB

Without feedback

With feedback

Frequency in Hz

Result:Thus the current series feedback amplifier is designed and constructed and the

following parameters are calculated.

Theoretical PracticalWith feedback Without feedback With feedback Without feedback

Input impedanceOutput impedanceGain (midband)Bandwidth

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Expt. No. 2 VOLTAGE SHUNT FEEDBACK AMPLIFIER

Aim: To design and test the voltage-shunt feedback amplifier and to calculate the

following parameters with and without feedback.1. Mid band gain.2. Bandwidth and cut-off frequencies.3. Input and output impedance.




1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 Connecting wires

Circuit Diagram:

(i) Without Feedback:

6

(ii) With Feedback:

Theory:

In voltage shunt feedback amplifier, the feedback signal voltage is given to the base of the transistor in shunt through the base resistor RB. This shunt connection tends to decrease the input resistance and the voltage feedback tends to decrease the output resistance. In the circuit RB appears directly across the input base terminal and output collector terminal. A part of output is feedback to input through RB and increase in IC decreases IB. Thus negative feedback exists in the circuit. So this circuit is also called voltage feedback bias circuit. This feedback amplifier is known an transresistance amplifier. It amplifies the input current to required voltage levels. The feedback path consists of a resistor and a capacitor.

Design


VCC = 12V; IC = 1mA; AV = 30; Rf = 2.5K; S = 2; hfe = ; =1/ Rf = 0.0004

re = 26mV / IC = 26;

hie = hfe re =


VE = IERE = Vcc/10 =

7

Applying KVL to output loop, we get VCC = ICRC + VCE + IERE

RC =

Since IB is very small when compare with IC, IC IE

RE = VE / IE =

S = 1+ RB / RE

RB =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

(ii) With feedback:

RO = RC || Rf =

Ri = (RB || hie ) Rf =

Rm = -(hfe (RB || Rf) (RC || Rf)) / ((RB || Rf) + hie) =

Desensitivity factor, D = 1 + Rm

Rif = Ri / D =

Rof = Ro / D =

Rmf = Rm / D =

XCi = Rif /10 =

Ci = 1 / (2f XCi) =

Xco = Rof /10 =

Co = 1 / (2f XCo) =

RE’ = RE || ((RB + hie) / (1+hfe))

XCE = RE’/10 =

CE = 1 / (2f XCE) =

XCf = Rf/10

Cf = 1 / (2f XCf) =

Procedure:

1. Connect the circuit as per the circuit diagram.2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in

regular steps and note down the corresponding output voltage.3. Plot the graph: Gain (dB) Vs Frequency4. Calculate the bandwidth from the graph.

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5. Calculate the input and output impedance.6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).

Tabular Column:


Vi = 10 mVFrequency V0

(volts)Gain = V0/Vi Gain (dB) = 20 log(V0/Vi)

(ii) With Feedback:Vi = 10 mV

Frequency V0

(volts)Gain = V0/Vi Gain (dB) = 20 log(V0/Vi)

Model graph: (frequency response)

Gain in dB

Without feedback

With feedback

Frequency in Hz

Result:Thus the current series feedback amplifier is designed and constructed and the

following parameters are calculated.Theoretical Practical

With feedback Without feedback With feedback Without feedbackInput impedanceOutput impedanceGain (midband)Bandwidth

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Expt. No. 3 RC PHASE SHIFT OSCILLATOR

Aim: To design and construct a RC phase shift oscillator for the given frequency (f0).




1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 Connecting wires Accordingly

Circuit Diagram:

10

Theory:

In the RC phase shift oscillator, the required phase shift of 180˚ in the feedback loop from the output to input is obtained by using R and C components, instead of tank circuit. Here a common emitter amplifier is used in forward path followed by three sections of RC phase network in the reverse path with the output of the last section being returned to the input of the amplifier. The phase shift Ф is given by each RC section Ф=tanˉ1 (1/ωrc). In practice R-value is adjusted such that Ф becomes 60˚. If the value of R and C are chosen such that the given frequency for the phase shift of each RC section is 60˚. Therefore at a specific frequency the total phase shift from base to transistor’s around circuit and back to base is exactly 360˚ or 0˚. Thus the Barkhausen criterion for oscillation is satisfied

Design:

VCC = 12V; IC = 1mA; C = 0.01F; fo = ; S = 2; hfe =

re = 26mV / IC = 26;

hie = hfe re =


VE = IERE = Vcc/10



RC =

Since IB is very small when compare with IC,

IC IE

RE = VE / IE =

S = 1+ RB / RE = 2

RB =

VB = VBE + VE =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

Gain formula is given by,

AV = (Av = -29, design given)

Effective load resistance is given by, Rleff = Rc || RL

RL =

11

XCi = {[hie+(1+hfe)RE] || RB}/10 =

Ci = 1 / (2f XCi) =

Xco = Rleff /10 =

Co = 1 / (2f XCo) =

XCE = RE/10 =

CE = 1 / (2f XCE) =

Feedback Network:

f0 = ; C = 0.01f;

fo =

R =

Procedure:

1. Connections are made as per the circuit diagram.2. Switch on the power supply and observe the output on the CRO (sine wave).3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

Vout (Voltage)

Time(ms)

Result:

Thus RC phase shift oscillator is designed and constructed and the output sine wave frequency is calculated as

Theoretical Practical

Frequency

Expt. No.4 HARTELY OSCILLATOR

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Aim: To design and construct the given oscillator for the given frequency (fO).




1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 DIB8 DCB9 Connecting wires Accordingly

Circuit Diagram:

Theory:

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Hartley oscillator is a type of sine wave generator. The oscillator derives its initial output from the noise signals present in the circuit. After considerable time, it gains strength and thereby producing sustained oscillations. Hartley Oscillator have two major parts namely – amplifier part and feedback part. The amplifier part has a typically CE amplifier with voltage divider bias. In the feedback path, there is a LCL network. The feedback network generally provides a fraction of output as feedback.

Design:

VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =

re = 26mV / IC = 26;

hie = hfe re =


VE = IERE = Vcc/10



RC =


RE = VE / IE =

S = 1+ RB / RE = 2

RB =

VB = VBE + VE =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =




RL =

XCi = {[hie+(1+hfe)RE] || RB}/10 =

Ci = 1 / (2f XCi) =

Xco = Rleff /10 =

Co = 1 / (2f XCo) =

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XCE = RE/10 =

CE = 1 / (2f XCE) =

Feedback Network:

f0 = ; L1 = 1mH; L2 = 10mH

A = =

f =

C =

Procedure:

1. Connections are made as per the circuit diagram.2. Switch on the power supply and observe the output on the CRO (sine wave).3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

Vout (Voltage)

Time(ms)

Result:Thus Hartley oscillator is designed and constructed and the output sine wave

frequency is calculated as


Frequency

Expt. No.5 COLPITTS OSCILLATOR

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Aim: To design and construct the given oscillator at the given operating frequency.

Equipments required:



1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 DIB8 DCB9 Connecting wires

Circuit Diagram:

Theory:

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A Colpitts oscillator is the electrical dual of a Hartley oscillator. In the Colpitts circuit, two capacitors and one inductor determine the frequency of oscillation. The oscillator derives its initial output from the noise signals present in the circuit. After considerable time, it gains strength and thereby producing sustained oscillations. It has two major parts namely – amplifier part and feedback part. The amplifier part has a typically CE amplifier with voltage divider bias. In the feedback path, there is a CLC network. The feedback network generally provides a fraction of output as feedback.

Design:

VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =

re = 26mV / IC = 26;

hie = hfe re =


VE = IERE = Vcc/10



RC =


RE = VE / IE =

S = 1+ RB / RE = 2

RB =

VB = VBE + VE =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =




RL =

XCi = {[hie+(1+hfe)RE] || RB}/10 =

Ci = 1 / (2f XCi) =

Xco = Rleff /10 =

Co = 1 / (2f XCo) =

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http://en.wikipedia.org/wiki/Frequency

http://en.wikipedia.org/wiki/Inductor

http://en.wikipedia.org/wiki/Capacitor

http://en.wikipedia.org/wiki/Hartley_oscillator

XCE = RE/10 =

CE = 1 / (2f XCE) =

Feedback Network:

f0 = ; C1 = ; C2 =

A = =

f =

L =

Procedure:1. Rig up the circuit as per the circuit diagrams (both oscillators).2. Switches on the power supply and observe the output on the CRO (sine wave).3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

Vout (Voltage)

Time(ms)

Result:Thus Colpitts oscillator is designed and constructed and the output sine wave

frequency is calculated as


Frequency

Expt. No. 6 CMOS Inverter, NAND and NOR using PSPICE

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Aim:To plot the transient characteristics of output voltage for the given CMOS inverter, NAND and NOR from 0 to 80s in steps of 1s. To calculate the voltage gain, input impedance and output impedance for the input voltage of 5V.

Parameter Table:Parameters PMOS NMOS

L 1 1W 20 5VTO -2 2KP 4.5E-4 2CBD 5p 5pCBS 2p 2pRD 5 5RS 2 2RB 0 0RG 0 0RDS 1Meg 1Meg CGSO 1p 1pCGDO 1p 1pCGBO 1p 1p

Circuit Diagram:

(i) Inverter:

(ii) NAND

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(iii) NOR

Theory:

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(i) InverterCMOS is widely used in digital IC’s because of their high speed, low power

dissipation and it can be operated at high voltages resulting in improved noise immunity. The inverter consists of two MOSFETs. The source of p-channel device is connected to +VDD and that of n-channel device is connected to ground. The gates of two devices are connected as common input.

(ii) NANDIt consists of two p-channel MOSFETs connected in parallel and two n-channel

MOSFETs connected in series. P-channel MOSFET is ON when gate is negative and N-channel MOSFET is ON when gate is positive. Thus when both input is low and when either of input is low, the output is high.

(iii) NOR

It consists of two p-channel MOSFETs connected in series and two n-channel MOSFETs connected in parallel. P-channel MOSFET is ON when gate is negative and N-channel MOSFET is ON when gate is positive. Thus when both inputs are high and when either of input is high, the output is low. When both the inputs are low, the output is high.

Truth Table:

(i) InverterInput Output0 11 0

(ii) NANDV1 V2 Output0 0 10 1 11 0 11 1 0

(iii) NOR

V1 V2 Output0 0 10 1 01 0 01 1 0

Model Graph:

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(i) Inverter

Voltage

Input Waveform5V

time (s) 0 10 20 30 40 50 60 70 80

Output Waveform5V

time (s) 0 10 20 30 40 50 60 70 80

(ii) NAND

Voltage

Input Waveform

time (s) 0 10 20 30 40 50 60 70 80

Output Waveform

time (s) 0 10 20 30 40 50 60 70 80

time (s) 0 10 20 30 40 50 60 70 80

(iii) NOR Voltage

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Input Waveform

time (s) 0 10 20 30 40 50 60 70 80

Output Waveform

0 10 20 30 40 50 60 70 80 time (s)

time (s) 0 10 20 30 40 50 60 70 80

Output:(i) Inverter

Gain = V(2)/Vin = Input Resistance at Vin = Output Resistance at V(2) =

(ii) NANDGain = V(4)/Vin1 = V(4)/Vin2 =

Input Resistance at Vin1 =Input Resistance at Vin2 =

Output Resistance at V(4) =

(iii) NORGain = V(4)/Vin1 = V(4)/Vin2 =

Input Resistance at Vin1 = Input Resistance at Vin2 = Output Resistance at V(4) =

Result:Thus the transient characteristics of output voltage for the given CMOS inverter,

NAND and NOR is plotted and the voltage gain, input impedance and output impedance are calculated. Expt. No.7 SECOND ORDER BUTTERWORTH - LOW PASS FILTER

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Aim:

To design and implement the second order butterworth Low pass filter using PSPICE.

Circuit Diagram:

Theory:

A Low pass filter has a constant gain from 0 to fH. Hence the bandwidth of the filter is fH. The range of frequency from 0 to fH is called pass band. The range of frequencies beyond fH is completely attenuated and it is called as stop band.

Design:

fH = 1000HZ C1= C2 =0.1F RIN=1000

fH = 1 / 2RC

R = 1 / 2CfH

R = R1 = R2 = 1592Gain = 1.586

1.586 = 1 + (RF / RIN)

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RF = 586

Model Graph:

Gain (dB)

3dB

Frequency (HZ) fH

Result:

Thus Low pass filter is designed and implemented using PSPICE.

Expt. No.8 DIFFERENTIAL AMPLIFIER

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Aim:

To implement the differential amplifier using PSPICE.

Circuit Diagram:

Theory:

A differential amplifier amplifies the difference between two voltages V1 and V2. The output of the differential amplifier is dependent on the difference between two signals and the common mode signal since it finds the difference between two inputs it can be used as a subtractor. The output of differential amplifier is

RF

VO = (V2 – V1) R1

Model Graph:

Voltage

26

V1 time

V2

time

V3

time

Calculation:

V1 = 5V V2 = 10V

RF 10KVO = (V2 – V1) = (10 – 5) R1 10K

VO = 5V

Output:

VO = 5V

Result:Thus a differential amplifier is implemented using operational amplifier.

Expt. No.9 FREQUENCY RESPONSE OF SINGLE TUNED AMPLIFIER

Aim:

27

To design and construct a single tuned amplifier and to plot the frequency response.

Equipment Required:



1 Power supply (0-30)V 12 Function generator (0-20M)Hz 13 CRO 14 Transistor BC107 15 Resistors6 Capacitors7 DIB8 DCB9 Connecting wires

Circuit Diagram:

Design:VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =

28

Q = 5; L = 1mH

re = 26mV / IC = 26;

hie = hfe re =


VE = IERE = Vcc/10



RC =


RE = VE / IE =

S = 1+ RB / RE = 2

RB =

VB = VBE + VE =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

RL =

XCi = {[hie+(1+hfe)RE] || RB}/10 =

Ci = 1 / (2f XCi) =

Xco = (RC||RL) /10 =

Co = 1 / (2f XCo) =

XCE = RE/10 =

CE = 1 / (2f XCE) =

Q = RL / L

RL =

f0 =

C = Procedure:

1. Connect the circuit as per the circuit diagram.2. Set Vi = 50 mV (say), using the signal generator.

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3. Keeping the input voltage constant, vary the frequency from 0Hz to3MHz in regular steps and note down the corresponding output voltage.

4. Plot the graph: Gain (dB) Vs Frequency

Tabular Column:

Vi = 50 mVFrequency V0 (volts) Gain (dB) = 20 log(V0/Vi)

Model Graph: (Frequency Response)

Result:Thus single tuned amplifier is designed and constructed for the given

operating frequency and the frequency response is plotted.

Expt. No.10 ASTABLE MULTIVIBRATOR

30

Aim:

To design and construct an astable multivibrator using transistor and to plot the output waveform.

Components / Equipments Required:

Sl. No. Item nameRange /

SpecificationQuantity

1 Transistor BC107 22 Resistors 4.9K, 1.6M 2 each3 Capacitors 0.45nF 24 CRO (0-20M)Hz 15 Power Supply (0-30)V 16 Connecting Wires Accordingly

Circuit Diagram:

Theory:

31

Astable multivibrator is also known as free running multivibrator. It is rectangular wave shaping circuit having non-stable states. This circuit does not need an external trigger to change state. It consists of two similar NPN transistors. They are capacitor coupled. It has 2 quasi-stable states. It switches between the two states without any applications of input trigger pulses. Thus it produces a square wave output without any input trigger. The time period of the output square wave is given by, T = 1.38RC.

Design Procedure:

VCC = 10V; IC = 2mA; VCE (sat) = 0.2V; f = 1KHz; hfe =

VCC - VCE (sat) 12 – 0.2 RC = = = 5.9K IC 0.002

R hfe RC = 315 * 5.9 * 103 = 1.85M

R = 1.5M

T = 1.38RC

C = T / (1.38R) = (1 * 10-3) / (1.38 * 1.5 * 106)= 0.48nF

Procedure:

1. Connections are made as per the circuit diagram.2. Switch on the power supply.3. Note down the output TON, TOFF and output voltage from CRO.4. Plot the output waveform in the graph.

Tabular Column:

Amplitude (in volts)

TON

(ms)TOFF

(ms)Frequency

(in Hz)Vo1

Vo2

Model Graph:

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Vo1 (Volts)

Time (ms)

Vo2 (Volts)

Time (ms)

RESULT:

Thus the astable multivibrator is designed and constructed using transistor and its output waveform is plotted.

Expt. No.11 MONOSTABLE MULTIVIBRATOR

Aim:

33

To design and construct monostable multivibrator using transistor and to plot the output waveform.


Sl. No. Item name Range / Specification Quantity1 Transistor BC107 22 Resistors 4.9K, 1.6M 2 each3 Capacitors 0.45nF 24 CRO (0-20M)Hz 15 Power Supply (0-30)V 16 Connecting Wires Accordingly

Circuit Diagram:

Theory:

Monostable multivibrator has two states which are (i) quasi-stable state and (ii) stable state. When a trigger input is given to the monostable multivibrator, it switches

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between two states. It has resistor coupling with one transistor. The other transistor has capacitive coupling. The capacitor is used to increase the speed of switching. The resistor R2 is used to provide negative voltage to the base so that Q1 is OFF and Q2 is ON. Thus an output square wave is obtained from monostable multivibrator.

Design Procedure:

VCC = 12V; VBB = -2V; IC = 2mA; VCE (sat) = 0.2V; f = 1KHz; hfe =

VCC - VCE (sat) 12 – 0.2 RC = = = 5.9K IC 0.002

IB2(min) = IC2 / hfe = Select IB2 > IB2(min)IB2 =

VCC – VBE (sat) R = = I B2

T = 0.69RCC = T / 0.69R =

-VBBR1 VCE (sat) R2

VB1 = + R1 + R2 R1 + R2

VBBR1 VCE (sat) R2

= (since, V B1 is very less) R1 + R2 R1 + R2

VBBR1 = VCE (sat) R2

R2 =10R1 (since, VBB = 2V and VCE (sat) = 0.2V)Let R1 = 10K, then R2 = 100KChoose C1 = 25pF.

Procedure:

1. Connections are made as per the circuit diagram.2. Switch on the power supply.3. Observe the output at collector terminals.4. Trigger Monostable with pulse and note down the output TON, TOFF and voltage

from CRO.5. Plot the waveform in the graph.

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Tabular Column:

Width (ms)

Input OutputTON

(ms)TOFF

(ms)Voltage (Volts)

TON

(ms)TOFF

(ms)Voltage (Volts)

Model Graph:

Vo1 (Volts)

Time (ms)

Vo2 (Volts)

Time (ms)

Result:

Thus the monostable multivibrator is designed and constructed using transistor and its output waveform is plotted.

Expt. No.12 BISTABLE MULTIVIBRATOR

Aim:

36

To design a bistable multivibrator and to plot its output waveform.


Sl. No. Item name Range / Specification Quantity1 Transistor BC107 22 Resistors 4.9K, 1.6M 2 each3 Capacitors 0.45nF 24 CRO (0-20M)Hz 15 Power Supply (0-30)V 16 Connecting Wires Accordingly

Circuit Diagram:

Theory:

The bistable multivibrator has two stable states. The multivibrator can exist indefinitely in either of the twostable states. It requires an external trigger pulse to change from one stable state to another. The circuit remains in one stable state until an

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external trigger pulse is applied. The bistable multivibrator is used for the performance of many digital operations such as counting and storing of binary information. The multivibrator also finds an applications in generation and pulse type waveform.

Design:

VCC =12V; VBB = -12V; IC = 2mA; VCE (sat) = 0.2V; VBE (sat) = 0.7V

VCC - VCE (sat) 12 – 0.2 RC = = = 5.9K IC 0.002

R2 hfe RC = 315 * 5.9 * 103 = 1.85M

R2 = 1.8M

Let R1 = 10K, C1 = C2 = 50pF

Procedure:

1. Connections are made as per the circuit diagram.2. Set the input trigger using trigger pulse generator.3. Note the output waveform from CRO and plot the graph.

Tabular Column:

Input Voltage (Volts)

Width (ms)

Input OutputTON

(ms)TOFF

(ms)Voltage (Volts)

TON

(ms)TOFF

(ms)Voltage (Volts)

Model Graph:

Vo1 (Volts)

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Time (ms)

Vo2 (Volts)

Time (ms)

Result:

Thus bistable multivibrator has been constructed and its output waveforms are studied.

Expt. No. 13 WAVE SHAPING CIRCUITS(Differentiator, Integrator, Clipper and Clamper)

Aim:

39

To design and implement different wave shaping circuits (Differentiator, Integrator, Clipper and Clamper).




1 Function / Pulse generator (0 – 3M)Hz 12 CRO (0-20M)Hz 13 Resistor 1K / 100K 14 Capacitor 0.1F 15 Connecting wires Accordingly

Circuit Diagram:

(i) Differentiator:

(ii) Integrator:

(iii) Clipper:

(a) Series Positive Clipper:

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(b) Shunt Positive Clipper:

(c) Series Negative Clipper:

(d) Shunt Negative Clipper:

(e) Positive Biased Series Positive Clipper:

41

(f) Positive Biased Shunt Positive Clipper:

(g) Positive Biased Series Negative Clipper:

(h) Positive Biased Shunt Negative Clipper:

42

(i) Negative Biased Series Positive Clipper:

(j) Negative Biased Shunt Positive Clipper:

(k) Negative Biased Series Negative Clipper:

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(l) Negative Biased Shunt Negative Clipper:

(m) Combinational Clipper

(iv) Clamper:

44

(a) Positive Clamper:

(b) Negative Clamper:

Theory:

(i) Differentiator:

The high pass RC network acts as a differentiator whose output voltage depends upon the differential of input voltage. Its output voltage of the differentiator can be expressed as,

dVout = Vin dt

(ii) Integrator:

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The low pass RC network acts as an integrator whose output voltage depends upon the integration of input voltage. Its output voltage of the integrator can be expressed as,

Vout = Vin dt

(iii) Clipper:

This circuit is basically a rectifier circuit, which clips the input waveform according to the required specification. The diode acts as a clipper. There are several clippers like positive clipper, negative clipper, etc. Depending upon the connection of diode it can be classified as series and shunt.

(iv) Clamper:

The clamper circuit is a type of wave shaping circuit in which the DC level of the input signal is altered. The DC voltage is varied accordingly and it is classified as positive clamper or negative clamper accordingly.

Design:

(i) Differentiator:

f = 1KHz = RC = 1msIf C = 0.1FThen R = 10KFor T << , Choose R = 1K and For T >> , Choose R = 100K

(ii) Integrator:

f = 1KHz = RC = 1msIf C = 0.1FThen R = 10KFor T << , Choose R = 1K and For T >> , Choose R = 100K

Procedure:

1. Connect the circuit as per the circuit diagram.2. Set Vin = 5V and f = 1KHz.3. Observe the Output waveform and plot the graph.

Model Graph:

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(i) Differentiator

Vin (Volts)

Time (ms)

Vout (Volts)

Time (ms)

Time (ms)

(ii) Integrator

Vin (Volts)

Time (ms)

Vout (Volts)

Time (ms)

Time (ms)

(iii) Clipper:

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Vin (Volts)

Time (ms)

Vout (Volts)Series Positive Clipper

Time (ms)

Shunt Positive Clipper

Time (ms)

Series Negative Clipper

Time (ms)

Shunt Negative Clipper

Time (ms)

Positive Biased Series Positive Clipper

2V Time (ms)

Positive Biased Shunt Positive Clipper

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2V Time (ms)

Positive Biased Series Negative Clipper

2V Time (ms)

Positive Biased Shunt Negative Clipper

2V Time (ms)

Negative Biased Series Positive Clipper

Time (ms) -2V

Negative Biased Shunt Positive Clipper

Time (ms) -2V

Negative Biased Series Negative Clipper

Time (ms) -2V

Negative Biased Shunt Negative Clipper

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Time (ms) -2V

Combinational Clipper

2V

Time (ms) -2V

(iv) Clamper:

Positive Clamper:

Time (ms)

Negative Clamper: Time (ms)

Result:

Thus different wave shaping circuits are studied and their output waveforms are plotted.

EXPT NO.14 DIGITAL TO ANALOG CONVERTER(R – 2R LADDER TYPE)

Aim:

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To construct a 8 – bit digital to analog converter using R – 2R ladder type.

Circuit Diagram:

Theory:

A DAC accepts an n – bit input word b1, b2, ……, bn in binary and produces an analog signal that is proportional to the input. In this type of DAC, reference voltage is applied to one switch and the other switches are grounded. It is easier to build and number of bits can be expanded by adding more R – 2R sections. The circuit slow down due to stray capacitance.

Observation:

d1 (MSB) d2 d3 (LSB)Output Voltage

VO (Volts)0 0 0 00 0 1 1.250 1 0 2.50 1 1 3.751 0 0 51 0 1 6.251 1 0 7.51 1 1 8.75

Calculation:

Output Voltage, VO = VR (d12-1 + d22-2 + d32-3 )For 100, VO = 5V

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Output:

VO = 5V

Model Graph:

Voltage

5V

time

Result:Thus R – 2R ladder type digital to analog converter is implemented.

EXPT NO.15 ASTABLE MULTIVIBRATOR

Aim:To plot the transient response of voltages at collector terminals of the two

transistors Q1 and Q2. Initial node voltages at collector and base are zero.

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Circuit Diagram:

Theory:

It has two quasi stable states. The transition between the two states occurs automatically due to charging and discharging of the capacitors and not due to any external trigger. Thus none of the transistor is allowed to remain in ON or OFF state.

Design:

VCC = 10V; IC = 2mA; VCE = 0.2V; C = 0.9nF

VCC – VCE (sat) 10 – 0.2RC = = = 4.9K IC 0.002

R hfe RC = 850KT = 1.38 R CT = 1msC = T / (1.38R) = 0.9nF

Model Graph:

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Voltage

V01

Time (ms) 0 0.5 1 1.5

V02

Time (ms) 0 0.5 1 1.5

Result:

Thus astable multivibrator is designed and transient response is plotted.

EXPT NO. 16 MONOSTABLE MULTIVIBRATOR

Aim:To plot the transient response of voltages at collector terminals of Q1 and Q2.

Initial voltages of base and collector of Q1 transistor is zero.

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Circuit Diagram:

Theory:

Monostable multivibrator has two states which are (i) quasi-stable state and (ii) stable state. When a trigger input is given to the monostable multivibrator, it switches between two states. It has resistor coupling with one transistor. The other transistor has capacitive coupling. The capacitor is used to increase the speed of switching. The resistor R2 is used to provide negative voltage to the base so that Q1 is OFF and Q2 is ON. Thus an output square wave is obtained from monostable multivibrator.

Model Graph:

Vo1 (Volts)

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Time (ms)

Vo2 (Volts)

Time (ms)

Result:

Thus monostable multivibrator is designed and transient response is plotted.

13. WEIN BRIDGE OSCILLATOR

AIM: To study and calculate frequency of. Wein Bridge Oscillator.

APPARATUS:

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Transistor (BC 107) 2 NoResistors 10K 4 No

1 K 3 No 2.2 K 2 No 33 K 6.8 K

Capacitors 10 2 No100 0.01 2 No

RPS (0 – 30 V) Potentiometer Bread BoarCRO Connecting wires

CIRCUITDIAGRAM:

THEORY:

The wein bridge oscillator is a standard circuit for generating low

frequencies in the range of 10 Hz to about 1MHz.The method used for getting

+ve feedback in wein bridge oscillator is to use two stages of an RC-coupled

amplifier. Since one stage of the RC-coupled amplifier introduces a phase shift

of 180 deg, two stages will introduces a phase shift of 360 deg. At the

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frequency of oscillations f the +ve feedback network shown in fig makes the

input & output in the phase. The frequency of oscillations is given as

f =1/2π√R1C1R2C2

In addition to the positive feedback

PROCEDURE:

1. Connections are made as per the circuit diagram

2. Feed the output of the oscillator to a C.R.O by making adjustments in the

Potentiometer connected in the +ve feedback loop, try to obtain a stable

sine Wave.

3. Measure the time period of the waveform obtained on CRO. & calculate the

Frequency of oscillations.

4. Repeat the procedure for different values of capacitance.

OBSERVATION:

Given R=10kΩ, C=0.01μF

fT = 1/ 2RC

fP = =

Amplitude,V0 =

MODEL WAVE FORM:

RESULT:

The frequency of the wein bridge oscillator is calculated and is verified

VIVA QUESTIONS:

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1. Give the formula for frequency of oscillations?

2. What is the condition for wien bridge oscillator to generate oscillations?

3. What is the total phase shift provided by the oscillator?

4. What is the function of lead-lag network in Wein bridge oscillator?

5. which type of feedback is used in Wein bridge oscillator

6. What is gain of Wein bridge oscillator?

7. what are the application of Wein bridge oscillator

8. What is the condition for oscillations?

9. What is the difference between damped oscillations undamped

Oscillations?

Wein bridge oscillator is either LC or RC oscillator.

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ec ii lab manual (1)

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