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Experiment No. 1
Objective: To study the effect of frequency on BJT Amplifier circuits.
Aim:To study the frequency response of CE Amplifier.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 CRO 2 channel/30 MHZ 1
3 Power supply 0-30 V DC/1A 1
4 BJT BC147B/BC547B
5 Resistors 33K,6.8K ,1.5 K,1 K 1
6 Capacitors 100 f, 22 f, 22 f -
7 Function generator 3 MHz -
Theory:
Frequency response of the RC coupled amplifier is as shown below
The frequency response of the R.C. coupled amplifier is divided in three regions
i. Low frequency regionii. Mid frequency region
iii. High Frequency region
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Low Frequency Region:
In this region gain of amplifier is less. This is due to effect of and .Since and are coming in serial with signal and output respectively, highreactance offered by these capacitors at low frequencies reduces the gain. Sincereactance of capacitor is high, it reduces the net input voltage to the amplifierand hence the output voltage decreases. Similarly because of high reactance
offered by net output voltage decreases. Therefore the gain of amplifierdecreases at low frequencies. Again at low frequency reactance offered by isalso high and because of which is not effectively bypassed, hence there a.c.negative feedback through reducing the input voltage and hence gain of theamplifier. Hence and , determines low frequency response of amplifier.Mid frequency Region:
In this mid frequency region the gain of the amplifier remains constant and is
maximum. In this region the reactance offered by the coupling capacitors and
bypass capacitor is negligible. Thereby increasing the net input voltage and hence
output voltage. In this region the amplifier circuit can be considered purely
resistive. Here small signal low frequency equivalent circuit is used for transistor
analysis.
High frequency Region:
In the high frequency region, gain of amplifier decreases this is because of internal
capacitance of the transistor and external wiring (stray, parasitic) capacitance. The
reduction of the high frequency gain is due to.
A.Effect of internal capacitance and
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Since the base emitter junction is forward biased and collector emitter junction is
reverse biased, these are known as (diffusion capacitance) and (transitionCapacitance) respectively. Because of low reactance at high frequency capacitor
provides low reactance path to input current and hence effective input currentto the base of amplifier reduces and hence causes gain to reduce. Since there isphase shift of between input and output of CE amplifier, and since thereexists a feedback path between the collector and base through . The feedbackcurrent is out of phase with respect to input current and hence base current to the
transistor decreases. This decreases the input voltage and hence output voltage
decreases. So the gain of amplifier decreases at high frequencies.
Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Supply input voltage from function generator, Vin= 50 mVpp.
3. Supply 12 V DC from dc regulated power supply.
4. Measure the output voltage at one frequency.
5. Repeat the observation for frequency ranges from 30 Hz to 3MHz.
6. Plot the frequency response on semi-log graph.
7. Calculate fL, fHand bandwidth of CE amplifier.
Result:
Sr. No. Parameters Theoretical Value (Hz) Practical Value (Hz)
1. Lower cut-offfrequency
2. Upper cut-offfrequency
3. Bandwidth
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Conclusion:
Hence we have studied the frequency response of CE Amplifiers. By plotting the
gain versus frequency plot on semi-log graph we have observed that gain is low at
low frequency due to blocking, coupling and bypass capacitors and the gain is lowat higher frequencies due to miller, junction and wiring capacitors but the gain is
high or maximum at mid frequency range. The value of lower cut-off frequency,
upper cut-off frequency and bandwidth of CE Amplifier were calculated from
graph. Theoretical and practical values are not same due the following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. Draw and explain the frequency response of RC coupled CE Amplifier.2. Draw the equivalent circuit of BJT amplifier at high frequency. Derive the
expression for upper cutoff frequency.
3. Draw equivalent circuit of BJT amplifier at high frequency for the givenamplifier and hence draw low frequency response.
4. Show the effect of low frequency and high frequency on coupling andbypass capacitors.
5. Describe the general frequency response of an amplifier and define the lowfrequency, mid-band, and high-frequency ranges.
6. Describe the general characteristics of the equivalent circuits that apply tothe low-frequency, mid-band, and high-frequency ranges.
7. Describe the short circuit current gain versus frequency response of a BJTand define the cutoff frequency.
8. Show the effect of source and load resistance on frequency response of RCcoupled CE Amplifier.9. Describe the Miller effect and the Miller capacitance.10.What effect does the Miller capacitance have on the amplifier bandwidth?
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Experiment No. 2
Objective: To study the effect of frequency on FET Amplifier circuits.
Aim:To study the frequency response of CS Amplifier.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 CRO 2 channel/30 MHZ 1
3 Power supply 0-30 V DC/1A 1
4 FET BFW11 1
5 Resistors 1M,1.5 K,1 K 1
6 Capacitors 47 f, 1f,1 f 1
7 Function generator 3 MHz 1
Theory:
Frequency response of the RC coupled amplifier is as shown below
The frequency response of the R.C. coupled amplifier is divided in three regions
iv. Low frequency regionv. Mid frequency region
vi. High Frequency region
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Low Frequency Region:
In this region gain of amplifier is less. This is due to effect of and .Since and are coming in serial with signal and output respectively, highreactance offered by these capacitors at low frequencies reduces the gain. Sincereactance of capacitor is high, it reduces the net input voltage to the amplifierand hence the output voltage decreases. Similarly because of high reactance
offered by net output voltage decreases. Therefore the gain of amplifierdecreases at low frequencies. Again at low frequency reactance offered by isalso high and because of which is not effectively bypassed, hence there a.c.negative feedback through reducing the input voltage and hence gain of theamplifier. Hence and , determines low frequency response of amplifier.Mid frequency Region:
In this mid frequency region the gain of the amplifier remains constant and is
maximum. In this region the reactance offered by the coupling capacitors and
bypass capacitor is negligible. Thereby increasing the net input voltage and hence
output voltage. In this region the amplifier circuit can be considered purely
resistive. Here small signal low frequency equivalent circuit is used for transistor
analysis.
High frequency Region:
In the high frequency region, gain of amplifier decreases this is because of internal
capacitance of the transistor and external wiring (stray, parasitic) capacitance. The
reduction of the high frequency gain is due to.
B.Effect of internal capacitance and
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Since the base emitter junction is forward biased and collector emitter junction is
reverse biased, these are known as (diffusion capacitance) and (transitionCapacitance) respectively. Because of low reactance at high frequency capacitor
provides low reactance path to input current and hence effective input currentto the base of amplifier reduces and hence causes gain to reduce. Since there is
phase shift of between input and output of CS amplifier, and since thereexists a feedback path between the collector and base through . The feedbackcurrent is out of phase with respect to input current and hence base current to the
transistor decreases. This decreases the input voltage and hence output voltage
decreases. So the gain of amplifier decreases at high frequencies.
Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Supply input voltage from function generator, Vin= 50 mVpp.
3. Supply 12 V DC from dc regulated power supply.
4. Measure the output voltage at one frequency.
5. Repeat the observation for frequency ranges from 30 Hz to 3MHz.
6. Plot the frequency response on semi-log graph.
7. Calculate fL, fHand bandwidth of CS amplifier.
Result:
Sr. No. Parameters Theoretical Value (Hz) Practical Value (Hz)
1. Lower cut-offfrequency
2. Upper cut-off
frequency
3. Bandwidth
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Conclusion:
Hence we have studied the frequency response of CS Amplifiers. By plotting the
gain versus frequency plot on semi-log graph we have observed that gain is low at
low frequency due to blocking, coupling and bypass capacitors and the gain is lowat higher frequencies due to miller, junction and wiring capacitors but the gain is
high or maximum at mid frequency range. The value of lower cut-off frequency,
upper cut-off frequency and bandwidth of CS Amplifier were calculated from
graph. Theoretical and practical values are not same due the following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. Draw and explain the frequency response of RC coupled CS Amplifier.2. Draw the equivalent circuit of FET amplifier at high frequency. Derive the
expression for upper cutoff frequency.3. Draw equivalent circuit of FET amplifier at high frequency for the given
amplifier and hence draw low frequency response.
4. Show the effect of low frequency and high frequency on coupling andbypass capacitors.
5. Describe the general frequency response of an amplifier and define the lowfrequency, mid-band, and high-frequency ranges.
6. Describe the general characteristics of the equivalent circuits that apply tothe low-frequency, mid-band, and high-frequency ranges.
7. Show the effect of source and load resistance on frequency response of RCcoupled CS Amplifier.
8. Describe the short circuit current gain versus frequency response of a FETand define the cutoff frequency.9. Describe the Miller effect and the Miller capacitance.
10.What effect does the Miller capacitance have on the amplifier bandwidth?
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Experiment No. 3
Objective: To study the performance and analysis of two stage BJT Amplifiers
Aim:To study the a.c analysis of CE-CE amplifier circuit.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 CRO 2 channel/30 MHZ 1
3 Power supply 0-30 V DC/1A 1
4 BJT BC147B/BC547B 2
5 Resistors 56K,33K,6.8K ,1K,3.3K,
10 K, 4.7K
1
6 Capacitors 47 f, 1 f, 10 f -
7 Function generator 3 MHz -
Theory:
Multistage Frequency Effects:
For a second transistor stage connected directly to the output of a first stage there
will be a significant change in the overall frequency response. In the high
frequency region the output capacitance must now include the wiringcapacitance, parasitic capacitance and miller capacitance of thefollowing stage. Further there will be additional low frequency cutoff levels due to
the second stage that will further reduce the overall gain of the system in the
region.
For each additional stage the upper cutoff frequency will be determined primarily
by that stage having the lowest cutoff frequency. The low frequency cutoff isprimarily determined by that stage having the highest low frequency cutoff
frequency. If non-identical stages with different lower and higher cutoff
frequencies are cascaded then highest among the lower cutoff frequencies will be
effective and lowest among higher cutoff frequencies will be effective and hence
overall band width decreases. It is shown in following figure
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Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Supply input voltage from function generator, Vin= 50 mVpp,10 KHz.
3. Supply 12 V DC from dc regulated power supply.
4. Measure the output voltage and calculate its voltage gain.
5. Calculate the input and output impedance.
6. Compare the obtained result with theoretical values.
Result:
Sr. No. Parameters Theoretical Value Practical Value
1. Voltage Gain (AV)
2. Input Impedance(ZI)
3. Output Impedance(Zo)
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Conclusion:
Hence we have studied the performance and analysis of two stage BJT Amplifier
circuit. The gain of CE-CE Amplifier is high, input impedance is fairly high and
output impedance is low. By comparing practically observed values withtheoretical values we have concluded that the theoretical and practical values are
not same due the following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. Explain the concept of multistage amplifier circuits.2. What is the need of multistage amplifiers.3. Explain the various types of coupling methods used in multistage amplifiers.4. Draw and explain frequency response of RC coupled CE-CE amplifier.5. What are the advantages of CE-CE amplifier circuits6. Derive the expression for voltage gain, current gain, input and output
impedance of CE-CE amplifier.
7. List the application of CE-CE amplifier.
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effective and lowest among higher cutoff frequencies will be effective and hence
overall band width decreases. It is shown in following figure
Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Supply input voltage from function generator, Vin= 50 mVpp,10 KHz.
3. Supply 12 V DC from dc regulated power supply.
4. Measure the output voltage and calculate its voltage gain.
5. Calculate the input and output impedance.
6. Compare the obtained result with theoretical values.
Result:
Sr. No. Parameters Theoretical Value Practical Value1. Voltage Gain
(AV)
2. Input Impedance
(Zi)
3. Output Impedance(Zo)
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Conclusion:
Hence we have studied the performance and analysis of two stage FET-BJT
Amplifier circuit. The gain of CS-CE Amplifier is moderately high, input
impedance is very high (in mega-ohms range) and output impedance is low. Bycomparing practically observed values with theoretical values we have concluded
that the theoretical and practical values are not same due the following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. Explain the concept of multistage amplifier circuits.2. What is the need of multistage amplifiers.3. Explain the various types of coupling methods used in multistage amplifiers.4. Draw and explain frequency response of RC coupled CS-CE amplifier.5. What are the advantages of CS-CE amplifier circuits6. Derive the expression for voltage gain, current gain, input and output
impedance of CS-CE amplifier.
7. Explain why the input impedance of CS-CE amplifier is very high.8. List the application of CS-CE amplifier.
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Experiment No. 5
Observation: To study the performance and analysis of multistage buffer
amplifier circuits or multistage common collector amplifier.
Aim: To study ac analysis of Darlington pair amplifier circuit and plot its
frequency response.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 CRO 2 channel/30 MHZ 13 Power supply 0-30 V DC/1A 1
4 BJT BC147B/BC547B 2
5 Resistors 100 K ,100 K,5.6K 1
6 Capacitors 1f -
7 Function generator 3 MHz -
Theory:
In electronics, the Darlington pair is a compound structure consisting of two
bipolar junction transistors (either integrated or separated devices) connected in
such a way that the current amplified by the first transistor is amplified further by
the second one with much higher current gain.
When using an emitter follower in a circuit, the level of current gain, and the input
impedances of the circuit is limited by the current gain that can be achieved using a
single transistor. The gain of the Darlington transistor pair is that gain of the two
individual transistors multiplied together.
Current Gain = hfe1* hfe2
The basic Darlington transistor circuit is formed by taking the emitter of the input
transistor and connecting it such that its emitter drives the base of the second and
then connecting both collectors together. This circuit can be used as any single
transistor would be in a variety of circuits, but particularly as an emitter follower.
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Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Supply input voltage from function generator, Vin= 50 mVpp,10 KHz.
3. Supply 12 V DC from dc regulated power supply.
4. Measure the output voltage and calculate its voltage gain.
5. Calculate the input and output impedance.
6. Compare the obtained result with theoretical values.
7. Repeat the observation for frequency ranges from 30 Hz to 3MHz.
8. Plot the frequency response on semi-log graph.
9. Calculate fL, fHand bandwidth of CE amplifier.
Result:
Sr. No. Parameters Theoretical Value Practical Value1. Voltage Gain (AV)
2. Input Impedance
(ZI)
3. Output Impedance
(Zo)
4. Lower cut-off
frequency (fL)
5. Upper cut-off
frequency (fH)6. Bandwidth (B.W)
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Conclusion:
Hence we have studied the performance and analysis of Darlington pair amplifier.
The voltage gain of CC-CC amplifier is close to unity, current gain is very high,
input impedance is high and output impedance is low. By plotting the gain versusfrequency plot on semi-log graph we have observed that gain is low at low
frequency due to blocking, coupling and bypass capacitors and the gain is low at
higher frequencies due to miller, junction and wiring capacitors but the gain is high
or maximum at mid frequency range. The value of lower cut-off frequency, upper
cut-off frequency and bandwidth of CC-CC amplifier were calculated from graph.
By comparing practically observed values with theoretical values we have
concluded that the theoretical and practical values are not same due the following
reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. What is Darlington pair? What is the need of Darlington pair amplifier?2. Draw the equivalent circuit of Darlington pair amplifier at high frequency.
Derive the expression for upper cutoff frequency.3. Draw equivalent circuit of Darlington pair amplifier at high frequency for
the given amplifier and hence draw low frequency response.4. Discuss Darlington pair. What are its primary features? Obtain expressions
for, Av, Ai and Ri.5. Explain the practical cascode amplifier and derive the expression for Av, Ri
and Ro. List the applications of cascode amplifier.
6. List the application of Darlington pair amplifier.7. Explain the advantages and disadvantages of Darlington pair amplifier.
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Experiment No. 6
Objective: To study the performance of Power amplifier circuits.
Aim:To study class B and class AB amplifier.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 CRO 2 channel/30 MHZ 1
3 Power supply 0-30 V DC/1A 1
4 BJT NPN-CL100, PNP-CK100 1
5 Resistors 22 K,10 K 26 Diode 1N4007 2
7 Function generator 3 MHz 1
Theory:
The final stage in any analog system is known as output stage. This output stage
usually consists of a power amplifier or a driver to deliver a high power output
signal to the load without loss of gain. Since the power amplifiers or the deriversdeliver a relatively high power signal they are known as large-signal amplifiers.
Linearity and power-conversion efficiency of the power amplifier are the main
requirements.
a) Harmonic distortion
The total harmonic distortion (THD) imposed by the amplifier on the output signal
is used to evaluate the power amplifiers after the design. The total harmonic
distortion (THD) is expressed in a percentage (measured in rms values) of theharmonic components with respect to the fundamental of the output signal.
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b) Power efficiency
Power-conversion efficiency is a measure of the ability of an active device to
convert the dc power of the supply into ac (signal) power delivered to the load and
is given by:
According to energy conservation law
input power = output power + power losses,
the most efficient power amplifier is the one that is capable of converting the
power of the dc power source (PS) into the ac power without losses. The power loss
is the portion of power that dissipates or consumes by the active devices (PD)
during the conversion process.
The average power dissipated in the class B stage is given by
LSD PPP
wherePLis power dissipation in the load LR . The above equation satisfies the law
of energy conservation. Likewise small-signal amplifier classifications, power
amplifiers are classified into classes according to the initial location of the dc
operating point Q. The four classes are; class A, class B, class AB and class C.
Class B Ampl if ier
Class B Amplifier has major disadvantages of very audible distortion with small
signal. This distortion can be so hard that it is objectionable even with large Signal.
This distortion is called as cross over distortion, because it occurs at point when
output stage crosses sourcing and sinking current. Class B amplifier consists of
driven transistor connected from output to negative power supply signal drive one
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transistor ON, while other OFF So in class B amplifier, no power is wasted going
from positive to negative power supply.
Class AB Ampli fi er
Class AB Amplifiers are almost same to Class B amplifier in that they are two
driver transistors. However Class AB Amplifier drifts from Class B amplifier in
that they have small current flowing from positive Supply to negative supply, even
when no input signal is provided that idle current slightly increases.
Procedure: -
1.Make connections for Class B amplifier as shown in the figure.2.Apply Vin= +5V3.Observe the cross over distortion On CRO.4.Make Connection for Class AB Amplifier as Shown in the figure.5.Apply Vin= +5V6.Observe the Output On the CRO Screen.7.Compare the obtained output with expected output.8.Observe the differences and comment on them in conclusion part.
Result:
Sr. No. Parameters Theoretical Value Practical Value
1. Class B
2. Class AB
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Conclusion
Class AB amplifier duo is used to overcome cross over distortion, which occurs in
Class B Amplifier.Small difference in practical and theoretical values is observed
which is due to the following reasons. In the calculation, there is
i. Resistance tolerance,ii. Capacitive tolerance,
iii. Power supply variation and calculation erroriv. Human error andv. Faulty instruments are responsible for the difference.
vi. Transistor is assumed to be of ideal nature.
Questionnaire
1. Explain working and analyze class B power amplifier2. Write a short note on design of Heat Sink.3. What is a heat sink? Why is it required for power amplifiers?4. Show the relationship between thermal and electrical analogy with a neat
sketch.5. Explain harmonic distortion and crossover distortion in power amplifier.
How are they overcome?
6. Explain why a voltage amplifier cannot be used as a good power amplifier.7. Differentiate between small signal and large signal amplifier8. Write a short note on Class AB push-pull power amplifier9. Explain working of transformer less push-pull amplifier.10.Describe the safe operating area for a transistor.11.Define and describe the power de-rating curve for a transistor.12.Describe the operation of an ideal class-B output stage.13.Explain the concept of crossover distortion.14.What is meant by harmonic distortion?15.Sketch a class-AB complementary BJT pushpull output stage using a VBE
multiplier circuit.
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Experiment No. 7
Objective: To study the performance of high efficiency power amplifier circuit.
Aim:To study class C Amplifier circuit.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 Power supply 0-30 V DC/1A 1
3 BJT NPN CL-100 1
4 Resistors
5 Capacitors 100 f, 0.1f
6 Inductors -
7 Function generator 3 MHz 1
8 CRO 2 channel/30 MHZ 1
Theory:
Class-C amplifiers conduct less than 50% of the input signal and the distortion at
the output is high, but high efficiencies (up to 90%) are possible. The usualapplication for class-C amplifiers is in RFtransmitters operating at a single fixed
carrier frequency, where the distortion is controlled by a tuned load on the
amplifier. The input signal is used to switch the active device causing pulses of
current to flow through atuned circuit forming part of the load.
The class-C amplifier has two modes of operation: tuned and un-tuned. The
diagram shows a waveform from a simple class-C circuit without the tuned load.
This is called un-tuned operation, and the analysis of the waveforms shows the
massive distortion that appears in the signal. When the proper load (e.g., aninductive-capacitive filter plus a load resistor) is used, two things happen. The first
is that the output's bias level is clamped with the average output voltage equal to
the supply voltage. This is why tuned operation is sometimes called a clamper.
This allows the waveform to be restored to its proper shape despite the amplifier
having only a one-polarity supply. This is directly related to the second
http://en.wikipedia.org/wiki/Transmitterhttp://en.wikipedia.org/wiki/LC_circuithttp://en.wikipedia.org/wiki/LC_circuithttp://en.wikipedia.org/wiki/Transmitter -
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phenomenon: the waveform on the center frequency becomes less distorted. The
residual distortion is dependent upon thebandwidth of the tuned load, with the
center frequency seeing very little distortion, but greater attenuation the farther
from the tuned frequency that the signal gets.
The tuned circuit resonates at one frequency, the fixed carrier frequency, and so the
unwanted frequencies are suppressed, and the wanted full signal (sine wave) is
extracted by the tuned load. The signal bandwidth of the amplifier is limited by
theQ-factor of the tuned circuit but this is not a serious limitation. Any residual
harmonics can be removed using a further filter.
In practical class-C amplifiers a tuned load is invariably used. In one common
arrangement the resistor shown in the circuit above is replaced with a parallel-
tuned circuit consisting of an inductor and capacitor in parallel, whose componentsare chosen to resonate at the frequency of the input signal. Power can be coupled to
a load by transformer action with a secondary coil wound on the inductor. The
average voltage at the drain is then equal to the supply voltage, and the signal
voltage appearing across the tuned circuit varies from near zero to near twice the
supply voltage during the RF cycle. The input circuit is biased so that the active
element (e.g. transistor) conducts for only a fraction of the RF cycle, usually one
third (120 degrees) or less.
The active element conducts only while the drain voltage is passing through its
minimum. By this means, power dissipation in the active device is minimized, and
efficiency increased. Ideally, the active element would pass only an instantaneous
current pulse while the voltage across it is zero: it then dissipates no power and
100% efficiency is achieved. However practical devices have a limit to the peak
current they can pass, and the pulse must therefore be widened, to around 120
degrees, to obtain a reasonable amount of power, and the efficiency is then 60-
70%.
http://en.wikipedia.org/wiki/Bandwidth_(signal_processing)http://en.wikipedia.org/wiki/Q-factorhttp://en.wikipedia.org/wiki/Q-factorhttp://en.wikipedia.org/wiki/Bandwidth_(signal_processing) -
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Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Apply DC power to the amplifier. Measure and record the values VB, VE, VC.
3. Calculate and record the resonant frequency of the LC circuit (this is the centre
or mid frequency of the amplifier).
4. Adjust the function generator to a 1.5Vpp amplitude and an operating frequency
equal to the center frequency calculated in above step.
5. Connect the function generator to the input of the amplifier.
6. Connect the CRO to the output, and change the function generator frequency
until a maximum output is obtained. Measure this frequency.
7. Without changing the function generator frequency, adjust the output amplitude
of the generator to the largest value that wont produce an output signal at the
collector. Measure and record this peak to peak minimum signal level Vin (with no
input signal).
8. Increase the function generator level while observing the amplifier output.
Adjust the function generator signal level to the minimum value that just produces
a complete sinusoidal output signal. Measure and record the peak to peak input andoutput signal levels and record these values Vin (with output signal) and Vout.
Result:
The following results were observed:
VB = ____________ VE = ____________ VC = ____________ VIN(With no output signal ) = ____________ VIN (With output signal) = ____________ VOUT(With output signal) = ____________
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Conclusion:
Hence we have studied the operation and performance of Class C power amplifiercircuit. Small difference in practical and theoretical values is observed which is
due to the following reasons. In the calculation, there is
i. Resistance tolerance,ii. Capacitive tolerance,
iii. Power supply variation and calculation erroriv. Human error andv. Faulty instruments are responsible for the difference.
vi. Transistor is assumed to be of ideal nature.
Questionnaire
1. Explain working and analyze class C power amplifier2. Write a short note on design of Heat Sink.3. Write short notes on Class C amplifier4. Explain why a voltage amplifier cannot be used as a good power amplifier.5. Differentiate between small signal and large signal amplifier6. What are the advantages of Class C power amplifier?7. What are the applications of Class C amplifier?8. Why the efficiency of Class C amplifier is high?9. Explain the working and region of operation of Class C amplifier.
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Experiment No. 8
Objective: To study the OP-Amp based voltage amplifiers.
Aim:To study the OP-AMP based inverting and non inverting voltage amplifiers.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 Power supply 0-30 V DC/1A 1
3 OP-AMP IC-741 1
4 Resistors 15K, 10K, 1K,2.2 K 1
5 Function generator 3 MHz 1
6 CRO 2 channel/30 MHZ 1
Theory:
i. Inverting AmplifierInverting amplifier is one of the most basic amplifier circuits. One advantage
of this inverting amplifier is that its voltage gain equals the ratio of the
feedback resistance to the input resistance. The closed-loop voltage gain ofoperational amplifier is defined as
We know that the operational amplifier has very large open loop gain A, for
the large open loop gain the two inputs V1 and V2 must be nearly equal. In
circuit V2 is connected with ground potential. So V1 must be nearly zero.
Here the inverting terminal input V1 is said to be virtual ground, it means
that the inverting input V1 is essentially be zero volts, but it does not provide
a current path to ground.
i.e., V1=0
From figure, we can write the branch currents
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assume current in the op-amp is zero. Thus, the current i1must flow through
resistor R2,i.e., i1=i2
From this, we can say that the inverting amplifier closed loop gain is not
dependant on transistor parameters. It is the function of the ratio of feedback
resistor and input resistor. And the minus sign implies the phase reversal of
inverting amplifier. In inverting amplifier if the input voltage is +ve the
output must be ve, because op-amp input V1 is virtually ground.
And also note that if the output terminal V0 is open-circuited, the current i2must flow back into op-amp. Here the output does not depend on the current
i2 and load resistance (feedback resistance R2). It always depends on the
input resistor R1 because op-amp input V1 is virtually ground, V2=0 and
i1=i2=V1/R1.
ii. Non-Inverting AmplifierNon-inverting amplifier is another basic op-amp circuit. The main advantage
of this circuit is stable voltage gain, high input impedance and low outputimpedance.
Here the input signal V1 is directly applied to the +ve terminal and the other
terminal connected to ground through resistor R1. As we discussed
previously in inverting amplifier, in this circuit the negative feedback
connection forces the terminal voltage V1 and V2 to be equal. Such a
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condition is referred as virtually short, it means the voltage difference
between V1 and V2 is nearly zero. Here by using this virtual short concept,
the ideal non inverting amplifier is analyzed as follows. Assume, there is no
current flow into the input terminals. Since V1=V2 i.e., V1=Vi and the
branch current i1 is given by
Branch current i2 is given by
By the virtual short, we can write i1=i2
{ }
* +
we can say that the gain of non-inverting amplifier is always greater than unity and
the output is in phase with input. And also note that the input impedance is very
large and the input current is zero because the input signal V1 is connected directly
to the non-inverting terminal.
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Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Apply DC voltage of 15V.
3. Connect the input sine wave of 1 KHz, 200 mVpp from function generator.
4. Observe the output waveform and verify that output is in phase with the input.
5. Compute the gain of non inverting amplifier and compare it with theoretical
reading.
6. Select the new value of RF and measured the gain again and record the
observations.
7. Repeat the same procedure from step-1 to step-6 also for inverting amplifier
circuit.
Result:
Sr.
No.
Amplifier Value of Feedback
resistor (RF)
Voltage gain
Theoretical
Value
Voltage Gain
Practical
Value1. Non Inverting
Amplifier
10 K
15 K
2. Inverting
Amplifier
10 K
15 K
Conclusion:
Hence we have studied the op-amp based inverting and non-inverting voltage
amplifier circuits and measure its voltage gain. It was observed that the gain of
amplifier depends on feedback resistor (RF). The output of inverting amplifier is
out of phase by 180 while output of non-inverting amplifier is in phase.
Theoretical and practical values are not same due the following reasons:
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i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. List the ideal characteristics of Op-amp.2. Explain the working of op amp inverting amplifier with derivation of voltage
gain.
3. Explain the working of op amp non inverting amplifier with derivation ofvoltage gain.
4. Design the op-amp circuit which can give the output as VO= 2V1+4V2-V35. Compare active filter and passive filter.6. Find the output voltage for a difference amplifier7. Draw the block diagram of a typical Op-amp and explain the function of
each block.8. List the application of Op-amp9. Explain the virtual ground concept10.Explain the role of feedback resistance in inverting and non inverting
amplifier11.Derive the expression for voltage gain of inverting amplifier12.Derive the expression for voltage gain of non inverting amplifier
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Experiment No. 9
Objective: To study and analyze the application of OP-Amp.
Aim:To study the OP-AMP based practical integrator and differentiator circuits.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 Power supply 0-30 V DC/1A 1
3 OP-AMP IC-741 1
4 Resistors 10 K, 1K 1
5 Capacitors 0.01f, 0.47f,0.47f 1
6 Function generator 3 MHz 1
7 CRO 2 channel/30 MHZ 1
Theory:
i. DifferentiatorDifferentiator is a circuit that performs a mathematical calculus operation
called differentiation and that produces an output voltage proportional to theinstantaneous rate of change of the input voltage. In the differentiator circuit
resistor and capacitor is used along with the op-amp to obtained
differentiating amplifier. The output of differentiator is given by:
Thus, the output voltage of differentiator is equal to RC time instantaneous
rate of change of input voltage with respect to time.
ii. IntegratorAn integrator is a circuit that performs a mathematical operation called
integration and is producing a sine or cosine or ramp of output voltage which
is a linearly decreasing or increasing voltage. This is also called as miller
integrator. An integrator circuit includes an op-amp, the feedback
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component resistor is replaced by capacitor. The output of integrator is given
by:
Where Vin(0) is the integration constant, and RC is called a time constant of
integrator.
Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Apply DC voltage of 15V.
3. Connect the input square wave of 1 KHz, 2 Vpp from function generator.
4. Connect CRO in dual trace mode in order to observe both input and output
waveform.
5. Switch the function generator to triangular waveform mode of 2 Vpp at 1 KHz.
Observe the output waveform on CRO.
6. Repeat the same procedure from step-1 to step-5 for the integrator circuit andobserve the input and output waveform on CRO screen.
Result:
Sr.
No.
Circuit Type Input Waveform
Type
Output Waveform
Type
Peak Voltage
& Phase Shift
1. Differentiator Square wave
Triangular wave
2. Integrator Square wave
Triangular wave
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Conclusion:
Hence we have studied the op-amp based practical differentiator and integrator
circuit. The output of differentiator and integrator is observed for the input
triangular and square waveform. It was observed that the output of differentiatorand integrator is out of phase by 90. Theoretical and practical values are not same
due the following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
Questionnaire
1. List the ideal characteristics of Op-amp.2. Draw the circuit of basic integrator using op-amp. Find the expression for
the output voltage.3. Explain the disadvantages of basic integrator circuit.4. Draw the circuit of basic differentiator using op-amp. Find the expression
for the output voltage.
5. Explain the disadvantages of basic differentiator circuit.6. Give the difference in frequency response of integrator and Low-pass filter.7. Draw and explain the circuit of practical integrator8. Draw and explain the circuit of practical differentiator9. What are the practical applications of integrator and differentiator circuits?
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Experiment No. 10
Objective: To study the performance of regulated power supply.
Aim:To study series and shunt voltage regulators.
Apparatus Required:
Sr. No Instruments Specification Quantity
1 Breadboard - 1
2 Power supply 0-30 V DC/1A 1
3 Zener Diode 6 V 1
4 Power BJT 1
5 Resistors 100 ,220 16 Multi-meter 1
Theory:
i. Series RegulatorSeries regulator circuit is also known as emitter-follower regulator because the
voltage at the emitter follows the base voltage. In this set up the transistor behaves
like a variable resistor whose resistance is determined by the base current. It iscalled as pass transistor because total current to be regulated passes through it. The
expression for output voltage is given by:
Vo = VinVs
Vs is the voltage drop across control element. The sampling network produces
feedback voltage which is proportional to the output voltage, depending upon the
values of control element is adjusted in order to maintain output voltage.
ii. Shunt RegulatorThe zener regulator circuit is modified to obtain the transistor shunt, ransistor act
as a shunt control element. The output voltage is given by:
Vo = Vz + VBE
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If the output voltage decreases due to any reason then (Vz + VBE) will also
decrease. But VZ is constant so more current will flow through load and will
increase. If output voltage increases, the exactly opposite action will take place to
regulate output voltage.
Procedure:
1. Mount the circuit as per circuit diagram on breadboard.
2. Input is connected to dc source that is varied form 7V to 15V.
3. Input is applied across 100 ohm resistor.
4. Measure the output voltage across RLby using multi-meter.
5. Repeat the same procedure for series regulator circuit.
6. Plot the graph and calculate % regulation and line regulation.
Result:
Sr. No. Parameters Theoretical Value
(Sv)
Practical Value
(Sv)
1. Series Regulator2. Shunt Regulator
Conclusion:
Hence we have studied the series and shunt regulator, it is possible to provide
temperature compensation without additional circuitry. Better regulation is
obtained by zener regulation this is because due to presence of transistor, the
variation in Iz is small. Theoretical and practical values are not same due the
following reasons:
i. Instrumental Error.ii. Human Error
iii. Resistance tolerance.iv. Aging of components.
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Questionnaire
1. What are voltage regulators?2. What is the application of regulator?3. Draw the circuit diagram of series regulator. Derive the expression of output
impedance and line regulation.4. Draw the circuit diagram of shunt regulator. Derive the expression of output
impedance and line regulation.
5. What is load regulation and line regulation?6. Describe the operation of BJT series regulator and derive the expression for
line regulation and load regulation.
7. Compare series and shunt regulator circuits using BJT.8. What are the practical applications of Voltage regulators?