Analog Circuits Ii Lab Manual 1x6r5t

KIT’s COLLEGE OF ENGINEERING Gokul Shirgaon, Kolhapur- 416 234 Ph: 0231- 2638141, Extn: 221 (E&TC Dept.)

A LAB MANUAL ON

ANALOG CIRCUITS-II (As per Shivaji University Syllabus)

PREPARED BY

Prof. Paresh D. Sawant Assistant Professor Email: [email protected] M: 09657895911

Department of Electronics &Telecommunication

Gokul Shirgaon, Kolhapur- 416 234 Ph: 0231- 2638141, Extn: 221 (E&TC Dept.)

CONTENTS EXPT . NO.

NAME OF THE EXPERIMENT

PAGE NO.

01

Design of RC phase shift oscillator using BJT

01

02

Design of Colpitts oscillator using BJT

05

03

Design of Hartley oscillator using BJT

08

04

Design of Astable multivibrator.

12

05

Design of monostable multivibrator using BJT

16

06

Designs of Schmitt trigger using BJT.

19

07

Design and frequency response of two stages RC coupled amplifier.

22

08

Design and frequency response of voltage series amplifier.

26

09

Design and frequency response of current Series amplifier using BJT.

32

10

Design and frequency response of direct coupled amplifier.

37

11

Design of voltage regulator using LM317.

12

Design of voltage regulator using IC723.

Ex.No:01

RC PHASE SHIFT OSCILLATOR

AIM: Design RC phase shift oscillator using BJT, which generates repetitive waveform (Sinusoidal signal) of frequency ____KHz.

OBJECTIVE: To generate a sinusoidal waveform of desired frequency using RC phase shift oscillator. COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM: VCC RC R1 CC

D A

2

Cc

C

B

C

C

C

Q1

1

R R R Manual - 06ESL37EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE 3

____

Analog Circuits-II Laboratory Manual -

R2

1k

RE

CE

THEORY: An oscillator is a circuit that generates the sinusoidal oscillation output, at frequency ranging from audio band up to several GHz. In addition, the frequency may be variable. For lower frequencies, the oscillator usually involves RC component for frequency selection, but for higher frequencies, it rely upon LC resonant circuits.

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Requirement for oscillation: 1. for linear oscillation i.e. oscillation which are relatively free from distortion, the product of factor (k) and (Av) should be greater than one K*Av ≥ 1 This condition is called the Barkhausen criteria of oscillation. 2.

The total phase shift of the circuit should be zero. The output should be given in phase with the input i.e. a positive . A practical circuit of phase shift oscillator using NPN transistor in common emitter configuration is as shown in fig. RL is a load resistor in the collector circuit. Rb1 and Rb2 are bias resistors. The phase shift of signal at input gets reversed when the amplifier amplifies it. The output of the amplifier goes to the network. The network consists of the three identical RC section provides a phase shift of 60 o each. Thus the total of the network provides 3 X 60 = 180 phase shift. The output of the feed back network is the output of this network is in phase with the original input signal to the amplifier. If the condition AvK > =1 is met, oscillation will be sustained.

The frequency at which the RC network provides exactly 180o phase shift is given by: 1 F= 2 RC √ 6+4 (RC /R) This is the frequency of oscillation. Also it can be shown that at this frequency the factor of RC network is: K = 1/29, from the Barkhausen’s criteria we get

Av ≥ 29

This means that the gain Av of the amplifier must be greater than 1/29, only then the oscillation can be started. Advantages of RC phase shift oscillator: 1. It does need transformer and inductor. So less bulky. 2. Circuit is cheap and simple as contains only resistors and capacitors. 3. Waveform obtained from circuit is exceptionally pure and sinusoidal since core saturation effect and harmonic distortion are absent as no transformer is used. 4. This circuit can be used to produce very few frequencies. 5. It provides good frequency stability. Disadvantages of RC phase shift oscillator: 1. It is difficult for circuit to start the oscillation as is generally small. This is because of reactance of R & C. 2. Due to the small , it gives the small output. 3. Circuit needs high voltage battery. It is because of small to provide sufficient output. 4. It cannot be used for higher frequencies. 5. It has high gain. Department of Electronics & Telecommunication – KITCOEK

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PROCEDURE: 1. Make the connections as shown in the circuit diagram. 2. Check the circuit for biasing. 3. Adjust the 1k potentiometer to get sinusoidal waveform at the output. 4. To measure the phase shift Method 1: Connect the channel 1 of the CRO to point D and channel 2 to point A. We will get two sine waves with a phase difference. Measure the difference by converting the time into angle. Method 2: a] Connect channel 1 to point D and channel 2 to point A. Press the XY knob and measure the phase shift. Phase angle фa =  =Sin-1 (a/b)

(approx. =600)

b a

b] Connect channel 2 to point B the graph is as shown

b a

 = Sin-1(a/b);

Phase angle фb =1800-  (approx. = 1200)

C] Connect channel 2 to point C. The transfer function will be almost a straight line and  = 00 and therefore phase angle фc =1800 - 00 = 1800

WAVEFORM:

0

t

f=1/T Department of Electronics & Telecommunication – KITCOEK

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RESULT: Thus a RC phase shift oscillator is designed, constructed and tested. Theoretical frequency fTh

=

Practical frequency fP

=

CONCLUSION:

STUDY QUESTION: 1. 2. 3. 4. 5.

Explain Barkhausen criteria of oscillations. Name the different types of oscillator circuits and mention their range of frequencies. What is an oscillator? List its types? What are RC oscillators? What are its types? What is the phase shift given by each RC section?

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COLPITT’S OSCILLATOR

Ex.No:02 AIM:

Design of Colpitts oscillator for a given Radio frequency of f0 =_____ KHz using BJT.

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors

4.

Inductors

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

Vcc Rc

R1

Cc

VO

CB BC109

Variable 1 K Pot

R2

RE

CE

C2

C1

GND

L

THEORY: An oscillator is a circuit that generates the sinusoidal oscillation output, at frequency ranging from audio band up to several GHz. In addition, the frequency may be variable. For lower frequencies, the

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oscillator usually involves RC component for frequency selection, but for higher frequencies, it rely upon LC resonant circuits. Requirement for oscillation: 1) For linear oscillation i.e. oscillation which are relatively free from distortion, the product of factor (k) and (Av) should be greater than one K*Av ≥ 1 This condition is called the Barkhausen criteria of oscillation. 1) The total phase shift of the circuit should be zero. The output should be given in phase with the input i.e. a positive . A practical circuit of Colpitts oscillator using NPN transistor in common emitter configuration is as shown in fig. R L is a load resistor in the collector circuit. R 1 and R2 are bias resistors. The phase shift of signal at input gets reversed when the amplifier amplifies it. The output of the amplifier goes to the network. The network is designed using two capacitors and an inductor. Voltage across C 2 is 180º out of phase w.r.t. voltage across C 1 and thus 180º phase shift is provided by the network. However, remaining 180º phase shift is provided by CE stage so that total phase shift is in the loop is 0º or 360º. It uses inductor in parallel with series of capacitor and acts as filter to the desired oscillating frequency and block all other frequencies. At resonance

XL = XCT Where XCT is the reactance of the capacitance connected in parallel

with inductance cell. This gives the frequency of oscillation fosc = 1/(2π√ LCT) __________________________1 Where CT = C1 | | C2 The LC network shows the attenuation from amplifier output and is due to potential divider effects of L and C2. Therefore gain of network is K = VF/VO = XC1/ (XL-XC1) ____________ 2 At 180º phase shift the inductive is very small, due to small resistance of coil i.e. Q= ωL/R is large. Therefore From eqn. 2

(XL-XC1) ≈ XC2 _____________________ 3 K = VF/VO = XC2/XC1= C2/C1

Since

XC2 = 1/(2πfC2) and

XC1 =1/(2πfC1)

For the required oscillation the loop gain must be greater than unity. Therefore,

AV > 1/K = C1/C2

PROCEDURE: 1. Rig up the circuit as shown in the circuit diagram. 2. Before connecting the network, check the circuit for biasing conditions i.e. check VCE, and VRE. 3. After connecting the network. Check the output. 4. Check for the sinusoidal waveform at output. Note down the frequency of the output waveform and check for any deviation from the designed value of the frequency. 5. To get a sinusoidal waveform adjust 1K potentiometer. 6. DCB/DIB can be varied to vary the frequency of the output waveform. Department of Electronics & Telecommunication – KITCOEK

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WAVEFORM:

Vo

0 t T  frequency fo = 1/T

RESULT: Thus a Colpitts oscillator is designed, constructed and tested. Theoretical frequency fTh = Practical frequency fP =

CONCLUSION:

STUDY QUESTION: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Explain Barkhausen’s criteria of oscillations. Name the different types of oscillator circuits and mention their range of frequencies. What type of is used in Oscillator circuit? What are the applications of LC oscillators? What type of is used in oscillators? What is the expression for the frequency of oscillations of colpitts oscillator? Is an oscillator DC to AC converter? What is the loop gain and loop phase shift of an oscillator? How does colpitts differ from Hartley? Which pair in circuit forms stabilizing circuit? What is the function of input and output capacitor? What is the condition for sustained oscillations in this oscillator? Output capacitor acts as a?

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HARTLEY OSCILLATOR

Ex.No:04 AIM:

Design of Hartley oscillator for a given Radio frequency of f0 = _____ KHz using BJT.

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors

4.

Inductors

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

THEORY: Oscillators are devices, which generate oscillations. The frequency of oscillations depends on the network. may be of two types namely positive and negative. In positive , the signal is applied in phase with the input signal thus increasing it. In negative , the signal is applied out of phase with the input thus reducing it. The used in oscillators is positive . The oscillators work on the principle of Barkhausen criteria. This states that for sustained oscillations

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i)

Loop gain Av must be equal to 1.

ii) The phase shift around the loop must be 0 deg of 360 deg. Here Av is the gain of the amplifier and  is the attenuation of the network. Consider the network shown in the fig (1) below. Assume an amplifier with input signal Vin. The output signal VO will be 180 deg out of phase with Vin. So to get an in phase output, the network provides 180-deg phase shift. Therefore the output Vf from the network can be made in phase and equal in amplitude to Vin and Vin can be removed. Even then the oscillations continue. Practical oscillations do not need any input signal to start oscillations. They are self-starting due to thermally produced noise in resistors and other components. Only one frequency (fo) of noise satisfies, Barkhausen criteria and the circuit oscillates with that frequency. The magnitude of fo keeps on increasing each time it goes around the loop. The amplification of fo is limited by circuit’s own non-linearities. Therefore to start oscillations Av > 1 and to sustain it, the loop gain Av = 1. Amplifier

Fig 1. Vin

Vo

Av

B

Vf

The network used here consists of L and C. Consider the circuit shown below fig 2. This circuit consists of L and C in parallel. The capacitor stores energy in its electric field whenever there is voltage across it and the inductor stores energy in its magnetic field whenever there is current through it. Initially let us assume that the capacitor has charged to V volts. When S is closed c= 0. When S is closed at t = t0 , capacitor starts charging through the inductor. Thus a voltage gets built up across the inductor due to the change in current through it. If the capacitor was changed with the polarity as shown in the fig 2 the current starts flowing from the positive plate of the capacitor to the negativ4 plate of the capacitor. As shown the voltage across the capacitor reduces during the discharge time v reduces and I increases. At time t1 v will be 0 and I will be maximum as c is fully discharged, the capacitor charges like sinusoidal oscillations. Thus the circuit oscillates with the frequency S

fo = 1/ 2LC

Fig.2

v

L

t = to

C

+ -

i

The Hartley oscillator consists of two inductors and a capacitor. The resonant frequency fo for Hartley oscillator is fo =1/ 2 LeqC ------where Leq = L1 + L2. Department of Electronics & Telecommunication – KITCOEK

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PROCEDURE: 1. Rig up the circuit as shown in the circuit diagram. 2. Before connecting the network, check the circuit for biasing conditions i.e. check V CE, and VRE. 3. After connecting the network. Check the output. 4. Check for the sinusoidal waveform at output. Note down the frequency of the output waveform and check for any deviation from the designed value of the frequency. 5. To get a sinusoidal waveform adjust 1K potentiometer. 6. DCB/DIB can be varied to vary the frequency of the output waveform. WAVEFORM:

Vo

0 t T  frequency fo = 1/T

RESULT: Thus a Hartley oscillator is designed, constructed and tested. Theoretical frequency fTh = Practical frequency fP =

CONCLUSION:

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STUDY QUESTION: 1. Explain Barkhausen’s criteria of oscillations. 2. Name the different types of oscillator circuits and mention their range of frequencies. 3. What are the applications of LC oscillations? 4. What type of is used in oscillators? 5. What the expression for frequency of oscillations? 6. Whether an oscillator is dc to ac converter? 7. What is the loop gain of an oscillator? 8. What is the difference between amplifier and oscillator? 9. What is the condition for oscillations? 10. How many inductors and capacitors are used in Hartley Oscillator? 11. How the oscillations are produced in Hartley oscillator? 10. What is the difference between damped oscillations undamped oscillations?

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Ex.No:04

ASTABLE MULTIVIBRATOR

AIM: Design Astable multivibrator to run at

___ KHz frequency and to observe the response at

base and collector points of the transistors and plot them.

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors

4.

Bread board

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

MODEL WAVEFORMS:

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THEORY: The circuit for Astable Multivibrator is shown in figure above. Since capacitor coupling is used between stages neither transistor can remain permanently in cut off state. The circuit has two quasi-stable states, it makes transistor to change continuously between these two states. Therefore, astable circuit is considered as an oscillator and is used as a generator of “square wave” and since it requires no triggering signal, is itself often a basic source of fast waveforms. As one can see the components in one half of the circuit are considered to be equal to their counterparts in the other half. a)

When the power is applied, one transistor will conduct more than the other due to some circuit imbalance. Assume Q 1 is conducting and Q2 is cut off. Then Vc1 the output of Q1,is approximately zero volt and Vc2 = Vcc.

b) The capacitor C2 will now starter to charge up through Rc and base emitter junction of Q. c)

Mean while C1 had previously charged up to Vcc through Rc of Q 1 when Q2 was conducting. But now when Q1 conducts a path is provided for C1 to discharge through Q1 and R1.The initial pulse of discharge current from C1 through R1 makes the base of Q2 suddenly very negative. (Due to voltage developed across R1) approximately – Vcc, which causes Q2 to go into cut off.

d) The charging current of C2 slowly ceases as the capacitor charges to Vcc. Q 1 is kept in the conducting state by base current provided from Vcc through R2. So we see that initially that Vb 1 is slightly more positive because it has two currents flowing through it. e)

The length of the time for which Q2 is held off is determined by the time constant of discharging of C 1 through Q1. Once the capacitor discharges completely, it starts to charge in the same direction and at the instant when its voltage reaches the forward bias voltage of Q2, Q2 turns on and state to conduct.

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f)

At the same instant, C2 is allowed to discharge through Q2 and R 2, coupling a large negative pulse in the base of Q1 (due to voltage developed across R2 upon initial discharge of C 2). This turns Q1 off quickly and collector voltage Vc1 rises towards Vcc However, this voltage at collector is less than Vcc, at this instant due to voltage drop across Rc occurring while C1 is recharging..This is relatively short time, but it cause a rounding of rising edge in Vc1.

g) Q1 is held in non conducting state for an interval depends upon the R 2 and C2 time constant. After which Q1 begins to conduct, C1 discharges through Q1. It should be noted that the turning ON of the one transistor Q 2 and its falling voltage at the collector permits the discharging of capacitor which drives transistor Q 1 into cut off. The rising voltage at Q 1’s collector ‘feeds back’ to base of Q2 tending to turn it ON more. The process is said to be regenerative or cumulative with quit fast switching results. The length of the time for which Q1 is held off is now decided by the R2C2 time constant. The time of the quasi stable states are: T1 = 0.69 R1C1 For the symmetric square wave:

And

T2 = 0.69 R2 C2

R1 C1 = R2 C2

Therefore total time of square wave: T = T1 + T2

= 1.388 RC

The frequency of oscillation is given by: f = 1/T = 0.72 / RC

PROCEDURE: 1. All the connections are made as per the circuit diagram. 2. Measure different voltages at base and collector points of two transistors w.r.t ground as VC1, VC2, VB1 and VB2. 3. All the waveforms are plotted on the graph sheet, the amplitudes and time periods are noted down. 4. observed frequency and duty cycle of multivibrator with calculated one for symmetrical square wave.

5. Theoretical values of amplitudes and time periods are compared with practical values.

RESULT:

CONCLUSION:

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STUDY QUESTIONS: 1. Define stable state of a transistor? 2. Define semi-stable state of transistor? 3. What are the other names of Astable Multivibrator? 4. Explain the operation of a Astable Multivibrator? 5. How many stable states and semi-stable states present in the Astable Multivibrator? 6. Draw the waveforms of VC1 and VC2 of a Astable Multivibrator? 7. What is the formula for the theoretical value of T in Astable Multivibrator?

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Ex.No:05

MONOSTABLE MULTIVIBRATOR

AIM: Design a monostable multivibrator using transistors to generate a pulse width _____ and to observe the response at base and collector points of the transistors and plot them.

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

MODEL WAVEFORMS:

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THEORY: The circuit diagram of Monostable Multivibrator is as shown in fig. It is also called as collector to base coupled. Monostable Multivibrator as more simply the collector coupled multivibrator. It has a permanent stable state & one quasi stable state. In this configuration a triggered signal is required to include a transition from the stable state to quasistable. The circuit may remain in quasistable state for a time which is very long in comparison with the time of transition between states. Eventually it will return to from the quasistable state to its state, no external signal being required to induce this reverse transition. Since when it is triggered, the circuit returns to its original state by itself after a time ‘T’, it is known as a ‘one shot’ a ‘single cycle’ a ‘single step circuit’ or a’univibrator’. Since it generates rectangular waveforms & hence can be used to gate other circuits, it is also called a ‘gating circuit’. Furthermore, the gate width ‘T’ is given by: T= Rin (2Vcc-VI (sat)-VBE (sat)) /Vcc-Vr = {Rin2+RinVcc-[VCE (sat) +VBE (out)]/2}/ (Vcc-Vr) At room temperature VCE (sat) +VBE (sat) =2vr for either Ce so T=RIn2=0.69(R+R0) C= 0.69RC. Since Ro for a transistor is saturation is small with R. The large Vcc is compared to the function voltage. The better is the approximation. The internal T is not particularly stable against temperature variation. The base to emitter voltage VBE (sat) decreases as temperature increases at the rate of about 2mv/c, where as VCE (sat) has a temp coefficient which is at opposite sign and substantially smaller the gate width‘t’ decreases as the temperature increases. The value of VCC, smaller is this effect. We consider how the effect of the reverse saturation current on the time ‘T’ of stable state during the internal ,when Q2 is cut-off a nominal current I CBO flows out of base of transistor of Q2.Then the voltage at the base of Q2 with C disconnected would not be VCC but ICBO*R . The initial voltage VC in the stable state is not VCC but is VCC-I CBO*RC, where now ICBO is the collector current of Q1 when it is off. Consequently the drop in voltage VCC when multi trigger is obtained. It has one stable and one quasi stable state. It is also called as ‘one shot’ or ‘delay circuit’ or ‘uni vibrator’. An external trigger signal is required to be applied to an appropriate in the circuit to cause transition from stable state to quasi stable state and after a time determine by the state circuit components. It returns to its original stable state an its called as Monostable Multivibrator. Monostable Multivibrator has the application in pulse circuitry. It may be used to establish a fixed time interval, one beginning and end of which are marked by an abrupt discontinuity in a voltage waveform.

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The pulse width of Monostable Multivibrator is given by T = 0.693 RC. Thus by changing the values of RC the pulse width of Monostable Multivibrator can be adjusted.

PROCEDURE: 1.

Connect the circuit on the breadboard.

2.

Give the power supply.

3.

Give the trigger pulse.

4.

Observe the wave form on CRO.

5.

Plot the graph.

CALCULATION & OBSERVATION: Calculations: Tgate pulse = 0.693RC= _____________=__________

Observations: Tgate pulse = _____________=__________

CONCLUSION:

STUDY QUESTIONS: 1. 2. 3. 4. 5.

Define stable state of a transistor? What is the significance of the term Monostable? Explain the Stable, Quasi stable, unstable. What is the use of commutating capacitors? While giving the trigger for the transistor, Transistor to be switched from ON to OFF or from OFF to ON state? Justify your answer. 6. What are the other names of Monostable Multivibrator? 7. How many stable and semi stable states present in the Monostable Multivibrator? 8. Explain the operation of Monostable Multivibrator? 9. What is the theoretical value of T ? 10. What is the name of base capacitor and what is the purpose of base capacitor?

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Ex.No:06

SCHMITT TRIGGER

AIM: To study and Design Schmitt trigger circuit.

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

MODEL WAVEFORMS:

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THEORY: Schmitt trigger is generally bistable circuit. It is also called as emitter – coupled binary. The circuit diagram shows Schmitt trigger using transistor Q1 & Q2. This circuit is an amplitude comparator to mark the movement at which an arbitrary wave form attains a particular reference level. The existence of any two states result from the fact that the Forward Biasing is incorporated into the circuit &from the further fact loop gain of the circuit is greater than unity. If transistor ‘Q2’ is conducting, this will cause a volume drop across R which will elevate the emitter at Q1 consequently. If Vin is small enough in volume, Q1 will be in cut off. As Vin rises, the circuit will not respond until Q1 reaches it’s cut in point. With Q1 conducting, the circuit will amplifying & since the gain ∆Vo /∆Vin is +ve the output will rise in response to the rise of Vin. As Vin continues to rise RB drop also increases. Therefore voltage of Vin will be achieved at which Q2 is turned OFF .At this point Vo = Vrms & output again is no longer responds to the input . A plot of Vo against V is drawn from readings. The voltage at which Q1 reaches cut in marked V = V1 hysteresis. In many instance the hysteresis of the Schmitt Trigger circuit is not a matter of concern should be the case if we had a periodic signal of amplitude large in comparison with the hysteresis range V T and our interest is using the circuit as one way comparing another application. Thus even when used as one way comparator if the signal were smaller than +ve voltage then the comparator having responded by the transition in one direction would never reset itself to hysteresis, may be eliminated by adjusting the loop gain of the circuit to be unity.

CIRCUIT OPERATION: To study the circuit operation of Schmitt trigger consider the following cases. Case 1: Vs = 0 In this Q1 is OFF & Q2 is ON. This is the first stable state. As no input signal is there, because of the voltage division across Vcc due to Rc1 and Rc2 and Rc we get Q1 in OFF state & Q2 is in ON state. Case 2: Upper transition point (UTP) Apply the input signal & increase the magnitude. The magnitude of input signal to make first transition because of which Q1 will be in ON state & Q2 will be in OFF state is called as UTP. Case 3: Lower transition point (LTP) Let us consider applied input signal amplitude we are decreasing then after particular amplitude value, there is again transition happens which cause Q1 to be in OFF state & Q2 to be in ON state. The magnitude of input signal at which second transition takes is called as LTP.

PROCEDURE: 1.

Connect the circuit on the breadboard.

2.

Give the power supply and apply sinusoidal input voltage.

3.

Observe the input and output waveform on CRO.

4.

Also observe hysteresis loop on CRO and find UTP & LTP

5.

Finally plot the graph.

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CONCLUSION:

.

STUDY QUESTIONS: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Explain the Hysteresis loop of Schmitt trigger. Explain the working of Schmitt trigger. Define upper trigger potential? Define lower trigger potential? Define hysteresis? What are the other names of Schmitt trigger? For any type of i/p what is the o/p of a Schmitt trigger? Explain the operation of a Schmitt trigger? What is the theoretical value of UTP? What is the theoretical value of LTP?

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Ex.No:07

MULTISTAGE AMPLIFIER WITHOUT

AIM: To Design and study frequency response of two stages RC coupled amplifier and determine the effect of cascading on gain and bandwidth.

OBJECTIVE: a) To study and design two stages RC coupled amplifier without .

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

THEORY: This is most popular type of coupling as it provides excellent audio fidelity. A coupling capacitor is used to connect output of first stage to input of second stage. Resistances R1, R2,RE form biasing and stabilization network. Emitter by capacitor offers low reactance paths to signal coupling Capacitor transmits ac signal, blocks DC. Cascade stages amplify signal and overall gain is increased total gain is less than product of gains of individual

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stages. Thus for more gain coupling is done and overall gain of two stages equals to A=A1*A2 where A1=voltage gain of first stage and A2=voltage gain of second stage. When ac signal is applied to the base of the transistor, its amplified output appears across the collector resistor Rc. It is given to the second stage for further amplification and signal appears with more strength. Frequency response curve is obtained by plotting a graph between frequency and gain in db .The gain is constant in mid frequency range and gain decreases on both sides of the mid frequency range. The gain decreases in the low frequency range due to coupling capacitor Cc and at high frequencies due to junction capacitance Cbe. PROCEDURE: 1. Apply input by using function generator to the circuit. 2. Observe the output waveform on CRO. 3. Measure the voltage at a. Output of first stage b. Output of second stage. 4. From the readings calculate voltage gain of first stage, second stage and overall gain of two stages. Disconnect second stage and then measure output voltage of first stage calculate voltage gain. 5. Compare it with voltage gain obtained when second stage was connected. 6. Note down various values of gain for different frequencies. 7. A graph is plotted between frequency and voltage gain. MODEL GRAPH: INPUT WAVE FORM:

FIRST STAGE OUTPUT:

SECOND STAGE OUTPUT:

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FREQUENCY RESPONSE:

OBSERVATION TABLE: Sr. No

Freq . (Hz)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

10 20 30 50 70 90 100 200 300 500 700 900 1K 2K 3K 5K

Vo (V)

Gain |Av|= Vo/Vin

Vinpp= _________V Gain in dB |Av| dB= 20log10(Vo/Vin)

Sr. No

Freq. (Hz)

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

7K 9K 10K 20K 30K 50K 70K 90K 100K 200K 300K 500K 700K 900K 1M

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

RESULT AND CONCLUSION:

.

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STUDY QUESTIONS: 1. What is the necessity of cascading? 2. What is 3dB bandwidth? 3. Why RC coupling is preferred in audio range? 4. Which type of coupling is preferred and why? 5. Explain various types of Capacitors? 6. What is loading effect? 7. Why it is known as RC coupling? 8. What is the purpose of emitter by capacitor? 9. Which type of biasing is used in RC coupled amplifier?

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Ex.No:08

MULTISTAGE AMPLIFIER WITH

AIM: To Design and study frequency response of two stages RC coupled amplifier with voltage series and determine the effect of cascading on gain and bandwidth. OBJECTIVE: b) To study and design two stage RC coupled amplifier with voltage series .

COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

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MODEL GRAPH: INPUT WAVE FORM:

FIRST STAGE OUTPUT:

SECOND STAGE OUTPUT:

FREQUENCY RESPONSE:

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Analog Circuits-II Laboratory Manual -

THEORY: In voltage amplifier without , the output voltage Vout = Av * Vin, where the amplification Av (also known as open loop gain) may in general be a function of both frequency and voltage. Open loop gain Av is given by: Av = Vout/Vin

….1

Suppose we have the loop so that a fraction βVout of the output is added to the input. Β is known as the factor and is determined by the network that is connected around the amplifier. The network may be modified using reactive elements like capacitors or inductors to give frequency dependent closed loop gain as in equalization tone control circuits or construct oscillators. The input to the amplifier is now, Vin’, where Vin’ = Vin + βVout

…2

The closed loop gain Avf is given by, Avf = Vout/ Vin’

…3

Substituting for Vin’ from 2 Avf = Vout/ Vin + βVout

…4

Rearranging and dividing both sides by Vin 1+ βVout / Vin = Vout / Vin * Avf

…5

Since Av = Vout/ Vin Then

1+ βAv = Av/ Avf

…6 and

Avf = Av/ 1+ βAv …7

If Av>>1, then Avf ~ 1/ β and the effective amplification (or closed loop gain) Avf is set by the characteristics of constant β, thus making linearising and stabilizing the amplification characteristics straightforward. If there are conditions where βAv = -1, the amplifier has infinite amplification and amplifier starts oscillating. Now amplifier works as an oscillator and the system is unstable. Advantages of negative : 1) Improves stability of gain

2)Increases input impedance

3) Decreases output impedance

4) Reduces distortion & internally generated noise.

5) Increases the bandwidth. A frequency response curve is graphical representation of the relationship between the amplifier gain and operating frequency. The circuit power gain remains relatively constant across the midband range of frequencies. As operating frequency decreases from midband area of the curve, a point is reached where the power gain begins to drop off. The frequency at which the power gain equals 50% of its midband value is called ‘Lower cut off frequency(fcl)’. As operating frequencies increases from midband area of the curve, a point is reached where the power gain begins to drop off again. The frequency at which the power gain equals 50% of its midband value is called ‘Upper cut off frequency(fcu)’. Note the bandwidth of the circuit is found as the difference between the cut off frequencies. So the bandwidth is: Department of Electronics & Telecommunication – KITCOEK

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Analog Circuits-II Laboratory Manual -

BW= (fcu-fcl) NEED OF CASCADING: A= [-hfe x RL1Ri] It becomes very difficult to get very high gain using single stage amplifier. Thus it is necessary to go for the cascading. TYPES OF COUPLING: While coupling the output of first stage as input to the other different types of coupling are used. Direct coupling DC coupling tuned inductive coupling, transformer coupling and optical coupling are few important types. The amplifier used for practical purpose is RC coupling method. Due to cascading effective bandwidth of the amplifiers gets reduced because gain has been boosted. To get proper boosting in gain it is necessary to match the impedance i.e. output impedance of first stage with input impedance of the second. For practical circuit as shown in figure voltage series is used improve stability. FREQUENCY RESPONSE: 1) (<50Hz): The reactance of coupling capacitor Cc is quite high and hence very small range of signal will from one stage to next stage. 2) At high frequency(>20KHz): The reactance of Cc is very small and it behaves as a short circuit. This increases the loading effect of next stage and causes to reduce the voltage gain. Moreover, at high frequency capacitive reactance of base-emitter junction is low which increases the base current. Two reason the gain drop off at high frequency.

PROCEDURE: 1. Use the components as per design and build the circuit on breadboard. 2. Apply Vcc to the circuit. 3. Check the DC voltage across collector and emitter. Confirm that transistor is operated in active region. 4. Apply the input voltage of amplitude 50mV and frequency 1KHz. 5. Observe output on CRO. 6. Observe output with and without . 7. Calculate gain as Av = Vout / Vin 8. To plot the frequency response, vary the frequency in the regular intervals from 20 Hz to 20 KHz. Measure the output voltage and calculate the voltage gain. 9. Do the voltage gain conversion in the form of decibels as dB= 20 log(Vout/Vin) 10. Calculate Bandwidth from frequency response.

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OBSERVATION TABLE: Sr. No

Freq . (Hz)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

10 20 30 50 70 90 100 200 300 500 700 900 1K 2K 3K 5K

Vo (V)

Gain |Av|= Vo/Vin

Vinpp= _________V Gain in dB |Av| dB= 20log10(Vo/Vin)

Sr. No

Freq. (Hz)

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

7K 9K 10K 20K 30K 50K 70K 90K 100K 200K 300K 500K 700K 900K 1M

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

RESULTS AND CONCLUSION:

.

STUDY QUESTIONS: 1. What is meant by ? 2. What are the types of amplifiers? Explain? 3. Draw the circuit for voltage series ? 4. What are the differences between positive and negative ? 5. What is the effect of negative on gain of an amplifier? 6. What is the formula for voltage gain with negative ? 7. What are the other names for positive and negative circuits? 8. What is the formula for input resistance of a voltage series ? 9. What is the formula for output resistance of a voltage series ?

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Analog Circuits-II Laboratory Manual -

Ex.No:09

CURRENT SERIES AMPLIFIER

AIM: To Design and study frequency response current series amplifier using BJT and determine the effect of on gain and bandwidth. OBJECTIVE: a) To measure the voltage gain of current - series amplifier. b) To measure the bandwidth, with and without . COMPONENTS REQUIRED:

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Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

MODEL GRAPH:

INTPUT WAVEFORM

OUTPUT WAVEFORM

FREQUENCY RESPONSE

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THEORY: When any increase in the output signal results into the input in such a way as to cause the decrease in the output signal, the amplifier is said to have negative . The advantages of providing negative are that the transfer gain of the amplifier with can be stabilized against variations in the hybrid parameters of the transistor or the parameters of the other active devices used in the circuit. The most advantage of the negative is that by proper use of this, there is significant improvement in the frequency response and in the linearity of the operation of the amplifier. This disadvantage of the negative is that the voltage gain is decreased. In Current-Series , the input impedance and the output impedance are increased. Noise and distortions are reduced considerably. PROCEDURE: 1) Connections are made as per circuit diagram. 2) Keep the input voltage constant at 30mV peak-peak and vary the frequency from 20Hz to 1MHz and note down the output voltage and calculate the gain by using the expression Av = 20log (V0 / Vi) dB. 3) Remove the emitter by capacitor (CE) and repeat STEP 2.And observe the effect of on the gain of the amplifier. 4) For plotting the frequency response the input voltage is kept constant at 30mV peak to peak and the frequency is varied from 100Hz to 1MHz. 5) Note down the value of output voltage for each frequency. All the readings are tabulated and the voltage gain in dB is calculated by using expression Av = 20log (V0 / Vi ) dB. 6) The Bandwidth of the amplifier is calculated from the graph using the expression Bandwidth B.W = f2 – f1. Where f1 is lower cut off frequency of CE amplifier and f2 is upper cut off frequency of CE amplifier. 7) The gain-bandwidth product of the amplifier is calculated by using the expression Gain-Bandwidth Product = 3-dB midband gain X Bandwidth. Department of Electronics & Telecommunication – KITCOEK

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Analog Circuits-II Laboratory Manual -

OBSERVATION TABLE (without ):

Vinpp= _________V

Sr. No

Freq . (Hz)

Sr. No

Freq. (Hz)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

10 20 30 50 70 90 100 200 300 500 700 900 1K 2K 3K 5K

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

7K 9K 10K 20K 30K 50K 70K 90K 100K 200K 300K 500K 700K 900K 1M

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

OBSERVATION TABLE (with ): Sr. No

Freq . (Hz)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

10 20 30 50 70 90 100 200 300 500 700 900 1K 2K 3K 5K

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

Vinpp= _________V Sr. No

Freq. (Hz)

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

7K 9K 10K 20K 30K 50K 70K 90K 100K 200K 300K 500K 700K 900K 1M

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

RESULTS AND CONCLUSION:

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Analog Circuits-II Laboratory Manual -

STUDY QUESTIONS: 1) What is the effect of Current-Series amplifier on the input impedance of the amplifier? 2) What is the effect of negative on the Bandwidth of an amplifier? 3) State the reason for the usage of negative in an amplifier? 4) What are the fundamental assumptions that are made in studying amplifiers? 5) What are the advantages of providing negative amplifier? 6) What are the ideal characteristics of a voltage amplifier? 7) Draw the circuit for the current series ? 8) What is the other name for current series amplifier? 9) What is the formula for input resistance of a current series ? 10) What is the formula for output resistance of a current series

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Analog Circuits-II Laboratory Manual -

Ex.No:10

DIRECT COUPLED AMPLIFIER

AIM: To Design and study frequency response of direct coupled amplifier. OBJECTIVE: a) To measure the voltage gain of direct coupled amplifier. b) To measure the bandwidth of direct coupled amplifier. COMPONENTS REQUIRED: Sl. No.

Components Details

1.

Transistor

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, CRO with Probe

CIRCUIT DIAGRAM:

MODEL GRAPH

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Analog Circuits-II Laboratory Manual -

THEORY: As shown in the below diagram, the direct coupled amplifier (DC) is consist of two transistors Q1 and Q2 , a voltage divider base bias resistor network (R1, R2) which is provided on the transistor Q1 base two collector resistors Rc1and Rc2, the transistor Q2 is self biased, we also use tow emitter by resistors RE1and RE2. The direct-coupled amplifier is operated without the using of frequency sensitive component like capacitor, inductor and Transformer etc. The direct coupled amplifier amplifies the A.C signal with frequency as low a fraction of Hertz (Hz). First of all when we applied a +ve half cycle at the I/P of Q1 transistor, which is already biased through the divider bias network. The +ve half cycle forwarded bias the transistor Q1 which start the conduction and give an inverted and amplified O/P at the collector. As we know that, VCE= Vcc – IcRc. This amplified -ve signed is provided to the base of Q2 transistor, which is self-bias (because they are connected in cascade condition). The base of Q2 transistor is a reversed and did not conduct, the O/P of transistor Q2 is amplified signal (inverting to I/P of Q2) when the Q2 did not conduct and the voltage drop across collector emitter will be zero, therefore the VCC is equal to IcRc. The O/P equal to the voltage drop across the collector resistors.

Application of Direct coupled Amplifier. 1. Pulse amplifier 2. Differential Amplifier 3. Regulator circuits of electronic power supply 4. Computer circuitry 5. Jet engine control 6. Electronic instruments

PROCEDURE: 1. Construct the circuit as shown in the diagram. 2. Set the input value in the mill volts range. 3. Vary the frequency and note the output Vo in the CRO. 4. By taking the frequency in X-axis and gain in Y-axis draw the graph. 5. Note the lower and upper cutoff frequency by drawing the 3db line. 6. From these two frequencies calculate the band width of the DC amplifier.

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Analog Circuits-II Laboratory Manual -

OBSERVATION TABLE : Sr. No

Freq . (Hz)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

10 20 30 50 70 90 100 200 300 500 700 900 1K 2K 3K 5K

Vo (V)

Gain |Av|= Vo/Vin

Vinpp= _________V Gain in dB |Av| dB= 20log10(Vo/Vin)

Sr. No

Freq. (Hz)

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

7K 9K 10K 20K 30K 50K 70K 90K 100K 200K 300K 500K 700K 900K 1M

Vo (V)

Gain |Av|= Vo/Vin

Gain in dB |Av| dB= 20log10(Vo/Vin)

RESULTS AND CONCLUSION:

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Analog Circuits-II Laboratory Manual -

Ex.No:11

VOLTAGE REGULATOR USING LM723

AIM: To Design and construct adjustable voltage regulator using IC LM723. OBJECTIVE: a) To design low voltage regulator using LM723. b) To design high voltage regulator using LM723. COMPONENTS REQUIRED: Sl. No.

Components Details

1.

IC

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, connecting wires etc.

CIRCUIT DIAGRAM: a) To design low voltage low current (LVLC) regulator using LM723.

Positive Low voltage regulator circuit diagram and pin connection using LM723

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b) To design high voltage low current (HVLC) regulator using LM723.

THEORY: A voltage regulator is a circuit that supplies a constant voltage regardless of changes in load current and input voltage variations. Using IC 723, we can design both low voltage and high voltage regulators with adjustable voltages. For a low voltage regulator, the output VO can be varied in the range of voltages Vo < Vref, where as for high voltage regulator, it is VO > Vref. The voltage Vref is generally about 7.5V. Although voltage regulators can be designed using Op-amps, it is quicker and easier to use IC voltage Regulators. IC 723 is a general purpose regulator and is a 14-pin IC with internal short circuit current limiting, thermal shutdown, current/voltage boosting etc. Furthermore it is an adjustable voltage regulator which can be varied over both positive and negative voltage ranges. By simply varying the connections made externally, we can operate the IC in the required mode of operation. Typical performance parameters are line and load regulations which determine the precise characteristics of a regulator. The pin configuration is shown in in the diagram. PROCEDURE: a) Line Regulation: For Vo = 5V and Vo = 13 V 1. Construct the circuit as shown in the diagram. 2. Obtain R1 and R2 for Vo = 5V / 13 V 3. By varying Vin from 3 to 12V, measure the output voltage Vo for LVLC regulator. 4. By varying Vin from 10 to 20V, measure the output voltage Vo for LVLC regulator. 4. Draw the graph between Vin and Vo as shown in model graph (a) 5. Calculate the line regulation. b) Load Regulation: For Vo = 5V and Vo =13 1. Set Vi such that Vo= 5 V / 13V 2. By varying RL, measure IL and Vo. 3. Plot the graph between IL and Vo as shown in model graph(b) 4. Calculate the load regulation.

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Analog Circuits-II Laboratory Manual -

MODEL GRAPHS: a) Line Regulation

b) Load Regulation

OBSERVATION TABLE: a) Low voltage low current voltage regulator. Line regulation : RL = Vin

Vo

IL

Load regulation : Vo = RL

IL

Vo

b) High voltage low current voltage regulator Line regulation : RL = Vin

Vo

IL

Load regulation : Vo = RL

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IL

Vo

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RESULTS AND CONCLUSION:

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Analog Circuits-II Laboratory Manual -

Ex.No:12

VOLTAGE REGULATOR USING LM317

AIM: To Design and construct adjustable voltage regulator using IC LM317. OBJECTIVE: a) To design low voltage regulator using LM317. COMPONENTS REQUIRED: Sl. No.

Components Details

1.

IC

2.

Capacitors

3.

Resistors Bread board

4.

Specification

Qty

DC Supply, connecting wires etc.

CIRCUIT DIAGRAM:

THEORY: LM317 adjustable positive voltage regulator. The connection diagram is shown above. The resistors R1 and R2 determine the output voltage Vout. The resistor R2 is adjusted to get the output voltage range between 1.2 volts to 57 volts. The output voltage that is required can be calculated using the equation: Vout = Vref (1+R2/R1) + Iadj R2 In this circuit, the value of Vref is the reference voltage between the adjustment terminals and the output taken as 1.25 Volt. The value of Iadj will be very small and will also have a constant value. Thus the above equation can be rewritten as Vout = 1.25 (1+R2/R1). Department of Electronics & Telecommunication – KITCOEK

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In the above equation, due to the small value of Iadj, the drop due to R2 is neglected. The load regulation is 0.1 percent while the line regulation is 0.01 percent per volt. This means that the output voltage varies only 0.01 percent for each volt of input voltage. The ripple rejection is 80 db, equivalent to 10,000. The LM 337 series of adjustable voltage regulators is a complement to the LM 317 series devices. The negative adjustable voltage regulators are available in the same voltage and current options as the LM 317 devices. PROCEDURE: a) Line Regulation: For Vo = 6V 1. Construct the circuit as shown in the diagram. 2. Obtain R1 and R2 for Vo = 6V. 3. By varying Vin from 5 to 15V, measure the output voltage Vo for LVLC regulator. 4. Draw the graph between Vin and Vo as shown in model graph (a) 5. Calculate the line regulation. b) Load Regulation: For Vo = 6V 1. Set Vi such that Vo= 6 V. 2. By varying RL, measure IL and Vo. 3. Plot the graph between IL and Vo as shown in model graph (b) 4. Calculate the load regulation. MODEL GRAPHS: a) Line Regulation

b) Load Regulation

OBSERVATION TABLE:

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RESULTS AND CONCLUSION:

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