15. CURRENT-SERIES FEEDBACK AMPLIFIER

15. CURRENT-SERIES FEEDBACK AMPLIFIER

AIM: To measure the voltage gain of current - series feed back amplifier.

APPARATUS: Transistor BC 107

                          Breadboard

                          Regulated Power Supply (0-30V,1A)

                          Function Generator

                          CRO(30 Mhz,dualtrace)

                          Resistors 33kΩ,3.3kΩ,330Ω,1.5kΩ,2.2k Ω,4.7k Ω, 1 k Ω.

                         Capacitors   10µF                                           - 2Nos

                                             100µF                                        

CIRCUIT DIAGRAM:

        

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 feedback.

The advantages of providing negative feedback are that the transfer gain of the amplifier with feedback can be stablised against varations in the hybrid parameteresof the transistor or the parameters of the other active devices used in the circuit. The most advantage of the negative feedback is that by propere use of this, there is significant improvement in the frequency respponse and in the linearity of the operation of the amplifier.This disadvantage of the negative feedback is that the voltage gain is decreased.

                  In Current-Series Feedback, the input impedance  and the output impedance are increased.Noise and distortionsare reduced cosiderably.

PROCEDURE:

1.    Connections are made as per circuit diagram.

2.    Keep the input voltage constant at 20mV peak-peak and 1kHz frequency.For different values of load resistance, note down the output voltage and calculate the gain by using the expression

                                   A_v = 20log(V₀ / V_i ) dB

3.    Remove the emitter bypass capacitor and repeat STEP 2.And observe the effect of feedback on the gain of the amplifier.

4.    For plotting the frquency the input voltage is kept constant at 20mV peak-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                                      A_v = 20log (V₀ / V_i ) dB

6.    A graph is drawn by takung frquency on X-axis and gain on Y-axis on semi log graph sheet

7.    The Bandwidth of the amplifier is calculated from the graph using the expression Bandwidth B.W = f₂ – f₁.

Where f₁ is lower cutt off frequency of CE amplifier

            f ₂ is upper cutt off frequency of CE  amplifier

8.    The gain-bandwidth product of the amplifier is calculated by using the expression

            Gain-Bandwidth Product = 3-dB midband gain X Bandwidth.

OBSERVATIONS:

Voltage Gain: V_i = 20 mV

S.NO	Output Voltage (Vo) with feedback	Output Voltage (Vo) without  feedback	Gain(dB) with feedback	Gain(dB) without feedback

Frquency Response: 

S.NO	Frequency (Hz)	Output Voltage (Vo­)	Gain A = Vo­/Vi	Gain in dB 20log(Vo­/Vi)

MODEL WAVEFORM:

Frequency response

PRECAUTIONS:

1.    While taking the observations for the frequency response , the input voltage must be maintained constant at 20mV.

2.    The frequency should be slowly increased in steps.

3.    The three terminals of the transistor should be carefully identified.

4.    All the connections should be correct.

RESULT:

                The effect of negative feedback (Current-Series Feedback ) on the amplifier is observed. The voltage gain and frquency response of the amplifier are obtained.Also gain-bandwidth product of the amplifier is calculated.

VIVA QUESTIONS

1.    What is the effect of Current-Series Feedback amplifier on the input inmpedance of the amplifier?

2.    What is the effect of negative feedback on the Bandwidth of an amplifier?

3.    State the reason for the usage of negative feedback in an amplifier?

4.    What are the fundamental assumptions that are made in studying feedback amplifiers?

5.    What are the advantages of providing negative feedback amplifier?

6.    What are the ideal characteristics of a voltage amplifier?\

7.    Draw the circuit for the current series feedback?

8.    What is the other name for current series feedback amplifier?

9.    What is the formula for input resistance of a current series feedback?

10. What is the formula for output resistance of a current series feedback?

BE Lab TRANSISTOR CE AMPLIFIER

9. TRANSISTOR CEAMPLIFIER

AIM:     1. ToMeasure the voltage gain of a CE amplifier

           2. To draw the frequency responsecurve of the CE amplifier

APPARATUS:

                          TransistorBC-107                                          

                           Regulated power Supply(0-30V, 1A)           

                           FunctionGenerator                                        

                           CRO                                                               

                           Resistors    [33KΩ, 3.3KΩ, 330Ω, 1.5KΩ

                                            1KΩ, 2.2KΩ, 4.7KΩ]                    

                          Capacitors- 10µF                 -2No

                                              100µF                                

                           Bread Board

                           Connecting Wires

THEORY:

               The CE amplifier provides highgain &wide frequency response. The emitter lead is common to both input& output circuits and is grounded. The emitter-base circuit is forwardbiased. The collector current is controlled by the base current rather thanemitter current. The input signal is applied to base terminal of the transistorand amplifier output is taken across collector terminal. A very small change inbase current produces a much larger change in collector current. When +VEhalf-cycle is fed to the input circuit, it opposes the forward bias of thecircuit which causes the collector current to decrease, it decreases thevoltage more –VE. Thus when input cycle varies through a -VE half-cycle,increases the forward bias of the circuit, which causes the collectorcurrent  to increases thus the outputsignal is common emitter amplifier is in out of phase with the input signal.

CIRCUIT DIAGRAM:

PROCEDURE:

1.   Connectthe circuit as shown in circuit diagram

2.   Applythe input of 20mV peak-to-peak and 1 KHz frequency using     Function Generator

3.   Measurethe Output Voltage Vo (p-p) for various load resistors

4.   Tabulatethe readings in the tabular form.

5.   Thevoltage gain can be calculated by using the expression                               A_v=(V₀/V_i)

6.   Forplotting the frequency response the input voltage is kept Constant at  20mV peak-to-peak and the frequency is variedfrom 100Hz to 1MHz Using function generator

7.   Notedown the value of output voltage for each frequency.

8.   Allthe readings are tabulated and voltage gain in dB is calculated by  Using The expression A_v=20 log₁₀(V₀/V_i)

9.   Agraph is drawn by taking frequency on x-axis and gain in dB on y-axis

          On  Semi-log graph.

The band widthof the amplifier is calculated from the graph

           Using the expression, 

                          Bandwidth, BW=f₂-f₁

                         Where f₁ lowercut-off frequency of CE amplifier, and 

                         Where f₂ upper cut-offfrequency of CE amplifier  

The bandwidthproduct of the amplifier is calculated using the

            Expression           

          Gain Bandwidth product=3-dBmidbandgain X Bandwidth  

OBSERVATIONS:

Input voltageVi=20mV

LOAD RESISTANCE(KΩ)	OUTPUT VOLTAGE (V0)	GAIN     AV=(V0/Vi)	GAIN IN dB Av=20log10 (V0/Vi)

FREQUENCY RESPONSE:       Vi=20mv

FREQUENCY(Hz)	OUTPUT VOLTAGE (V0)	GAIN IN dB Av=20 log10 (V0/Vi)

MODELWAVE FORMS:

INPUT WAVE FORM:

OUTPUTWAVE FORM

FREQUENCY RESPONSE

   

RESULT: The voltage gain andfrequency response of the CE   amplifierare obtained. Also gain bandwidth product of the amplifier is calculated.

VIVA QUESTIONS:

1.   Whatis phase difference between input and output waveforms of CE amplifier?

2.   Whattype of biasing is used in the given circuit?

3.   Ifthe given transistor is replaced by a p-n-p, can we get output or not?

4.   Whatis effect of emitter-bypass capacitor on frequency response?

5.   Whatis the effect of coupling capacitor?

6.   Whatis region of the transistor so that it is operated as an amplifier?

7.   Howdoes transistor acts as an amplifier?

8.   Drawthe h-parameter model of CE amplifier?

9.   Whattype of transistor configuration is used in intermediate stages of a multistageamplifier?

10.Whatis Early effect?

CSE & IT 3rd SEMESTER

[EURCS 305/EURIT 305]

B.Tech. DEGREE EXAMINATION

CSE & IT

III SEMESTER

PROBABILITY & STATISTICS

(Effective from the admitted batch 2007–08 onwards)

Time: 3 Hours                                                           Max.Marks: 60

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Instructions:  Each Unit carries 12 marks.

                       Answer all units choosing one question from each unit.

                       All parts of the unit must be answered in one place only.

                       Figures in the right hand margin indicate marks allotted.

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UNIT-I

1.   a)   What are mutually exclusive and independent events?  State

            Bayes theorem on probability.                                                 6

      b)   A problem in statistics is given to three students A, B and C,

            whose chances of solving it are respectively. 

            What is the probability that the problem will be solved?          6

OR

2.   a)   Explain Poisson distribution.  Give its applications.                 6

      b)   A manufacturer of cotter pins knows that 5% of his product is

            defective.  If he sells cotter pins in boxes of 100 and guarantees

            that not more than 10 pins will be defective, what is the

            probability that a box will fail to meet the guaranteed quality?                        6

UNIT-II

3.   a)   Explain an exponential distribution.  Obtain its mean and

            variance.                                                                                  6

      b)   What are the important characteristics and applications of

            normal distribution?                                                                 6

OR

4.   a)   What is the principle of least squares?  How do you fit a power

            curve  to the given data?                                             6

      b)   Fit a power curve is of the form  to the following data:                     6

X:	1	2	3	4	5
Y:	8	15	30	60	125

UNIT-III

5.   a)   From the following data, compute the coefficient of correlation

            between X and Y.                                                                    6

No. of Items		X Series	Y Series
	:	15	15
Arithmetic Mean	:	25	18
Sum of squares of deviations from mean	:	136	138
Sum of products of deviations of X and Y from their means	:	122

      b)   Fit a linear regression equation of Y on X to the following data:                     6

X:	5	8	7	6	4
Y:	3	4	5	2	1

OR

6.   a)   Explain the terms (i) Population (ii) Sample (iii) Parameter and

            (iv) Statistic.                                                                             6

      b)   Describe the method of Maximum likelihood estimation.       6

UNIT-IV

7.   a)   What are the steps involved in test of significance?                   6

      b)   A machine puts out 16 imperfect articles in a sample of 500.

            After machine is overhauled, it puts out 3 imperfect articles in

            a batch of 100.  Has the machine improved?                              6

OR

8.   a)   A random sample of 400 students is found to have a mean

            height of 171.38 cms.  Can it be reasonably regarded as a

            sample from a lalrge population with mean height 171.17 cms.

            and standard deviation 3.30 cms. (Test at 5% level of

            significance).                                                                           6

      b)   A random sample of 1200 households from one town gives the

            mean income as Rs.500 per month with a standard deviation of

            Rs.70 and a random sample of 1000 households from another

            town gives the maximum income as Rs.600 per month, with a

            standard deviation of Rs.90.  Test whether the mean income of

            households from two towns differ significantly or not?

            (Test at 5% level of significance).                                           6

UNIT-V

9.   a)   Two independent samples of 8 and 7 items respectively had the

            following values:                                                                    

Sample I	:	9	11	13	11	15	9	12	14
Sample II	:	10	12	10	14	9	8	10

            Is the difference between the means of samples significant?               6

      b)   A random sample of 11 pairs of observations gives a correlation

            coefficient 0.52.  Is the correlation coefficient significant? 

            (Test at 5% level of significance).                                           6

OR

10. a)   Describe the c² – test of goodness of fit.                                 6

      b)   In an experiment of immunization of cattle from Tuberculosis,

            the following results were obtained:                                        6

		Affected	Unaffected
Inoculated	:	12	28
Not-inoculated	:	13	7

             Examine the effect of vaccine in controlling the incidence of

             the disease.

                                                                                                               [3,7/IIIS/109]