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    EXPERIMENT 1: SMALL SIGNAL AMPLIFIERS

    Part A: Single Stage Transistor Amplifier Circuits

    Introduction

    A small signal amplifier (also called preamplifier) is an amplifier whose signal level is very small

    compared to DC potentials. It is always found at the input stage of the amplifying systems. Most

    preamplifiers are class A type. The transistors used in small signal amplifiers carry a relatively low power

    and usually dont require heat sinks for cooling purposes because they dont dissipate much heat during

    operation. They are called small signal transistors. Examples of small signal transistors are BC 107, BC

    558 and BC 547.

    Preamplifiers are further divided into three types on the basis of what parameter they amplify. These

    are:

    i. Voltage amplifierii. Current amplifieriii. Charge sensitive amplifier

    Our discussion will concentrate on the first two preamplifiers, voltage amplifier and current amplifier.

    Before explaining voltage and current amplifiers, let us observe the main configurations of transistor

    amplifier circuits. There are three fundamental linear transistor amplifier circuit configurations that one

    can study. These are Common Base (CB), Common Collector (CC) and Common Emitter (CE). Each

    configuration has its own merits and demerits in circuit applications as far as voltage gain, current gain

    and power gain is concerned. The frequency response, input and output impedance of each

    configuration differ. Figure 1.1 (a) through (c) show the three basic configurations of transistor amplifiercircuits. Note that the common terminal is held to ground directly or through a capacitor. Dont worry if

    you cannot comprehend whether the given configuration is a voltage and current amplifier at this stage.

    It will become clearer later.

    Q1C1

    C2

    C3

    Figure 1.1 (a): Common Emitter configuration

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    Q1

    R1

    C2

    C3

    C1

    Figure 1.1 (b): Common Base configuration

    Q1C1

    C2

    C3

    Figure 1.1 (c): Common Collector configuration

    Voltage Amplifier

    A voltage amplifier is characterized by having high input impedance. Ideally it does not draw any current

    at its input. The main parameter of interest here is voltage, so the system should not draw current

    because if it does so it will reduce the signal voltage level and hence gain. The system does not qualify to

    be a voltage amplifier if it draws significant amount of current at its input. The system designer should

    ensure that the signal voltage levels are preserved in case the input signal has a very small voltage. For

    example, a signal from a microphone is too small and thus needs to be preserved during amplification.

    The voltage amplifier can be prepared for DC or RC coupling, as we shall see in the section of multistage

    amplifiers. Single stage amplifying circuits prepared for RC coupling use a lot of components and

    therefore they are not widely used in the fabrication of ICs because of bulkiness and noise that can be

    introduced by components during operation. On the other hand, DC coupling are very useful in the

    fabrication of ICs because they utilize very few components and hence they are less prone to noise due

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    to components. However, temperature stability is very critical in DC coupled amplifiers than RC coupled

    amplifiers.

    Single stage voltage amplifier prepared for RC coupling

    Figure 1.2 shows a single stage CE voltage amplifier prepared for RC coupling

    Figure 1.2: Single stage CE voltage amplifier prepared for RC coupling

    Lets design a CE amplifier with a voltage gain of -100, an f3dB point of 100Hz, and a quiescent current

    IQ=1mA, where hFE=100 and Vcc=20V. The following are the basic steps:

    Step 1: Choose RC to centre Vout (or VC ) to VCC to allow for maximum symmetrical swings in the

    output. In this example, this means VCshould be set to 10V. Using Ohms law, we can find RC as follows:

    (1.1)

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    Step 2: Next we select RE to set VE=1V for temperature stability. Using Ohms law, and taking IQ=IE=1mA,

    we get RE=VE/RE=1V/1mA=1K.

    Step 3: Now, choose R1 and R2 to set the voltage divider to establish the quiescent base voltage of

    VB=VE+0.6V, or 1.6V. To find the proper ratio between R1 and R2, use the voltage divider rule:

    (1.2)

    This means that R1=11.5R2. The parallel resistance should be less than or equal to 0.1Rin(base),ac.

    Therefore:

    (1.3)

    After plugging R1=11.5R2 into this equation and using Rin(base),ac=hFERE, you find that R2=10K, which in

    turn means R1=115K (lets say, 110K is close enough for R1).

    Step 4: Next, choose R3 for the desired gain, where:

    (1.4)

    The double line means to take RE and (rtr+R3) in parallel). To find rtr, use

    rtr=0.0026V/IE=0.026V/IC=0.026V/1mA=26. Now, you can simplify the gain expression by assuming that

    REdisappears when the signals are applied. This means that the gain is simplified to:

    (1.5)

    Solving this equation for R3, you get R3=74

    Step 5: Next, choose C1 for filtering purposes such that:

    (1.6)

    Here, Rin is the combined parallel resistance of the voltage-divider resistors R1 and R2, and Rin(base), ac

    looking in from the left into the voltage divider:

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    (1.7)

    Solving this equation, you get Rin=5K. This means:

    (1.8)

    Step 6: To choose C2, treat C2 and (rtr+R3) as a high pass filter (again, treat RE as being negligible during ac

    conditions). C2 is given by:

    (1.9)

    Figure 1.3 shows a CE transistor configuration with the calculated values above.

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    Figure 1.3: A practical example of a single stage CE voltage amplifier

    However, you should note that we always escape from using transistor parameters, such as h FE and hie,

    because they vary widely. For example, it is impractical to find two identical transistors with the same

    hFE and hie. Temperature is one of the determinant factors which cause variation in transistor

    parameters. It is practical and a good practice for a system designer you should use the externalcomponents, such as resistors, to determine the gain of an amplifier.

    Single stage voltage amplifier prepared for DC coupling

    Figure 1.4 shows a circuit diagram of a single stage CE voltage amplifier prepared for DC coupling. The

    emitter resistor is not coupled in order to increase the input impedance to qualify the circuit a voltage

    amplifier.

    Q1C1

    Figure 1.4: Single stage CE voltage amplifier prepared for DC coupling

    Current Amplifier

    On the other hand, the current amplifier is characterized by having very small input impedance. The

    parameter of interest in this case is current, and therefore we should allow as much current as possible

    to flow into the system for amplification purposes. Figure 1.5 shows a small signal current amplifier. In

    this circuit (figure 1.5) the base bias current flows through the resistor RB1 which is straight connected

    RE

    RCRB

    +VCC

    Vout

    Vin

    IB IC

    IE

    Vin

    VCC

    0.5VCC

    0

    0V

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    from the supply voltage. This circuit, however, has temperature stability problems. There is no self

    mechanism of stabilizing the circuit when, for example, the collector current increases as a result of

    temperature rise. The modified version of this circuit is shown in figure 1.6, where the negative

    feedback resistor RfB1 is introduced between the base and the collector terminal of the transistor. This

    circuit stabilizes the operating conditions for the stage and compensate for variations in transistor

    parameters.

    Q1C1

    C2

    Figure 1.5: Single stage current amplifier

    In figure 1.6, the base bias current is derived from the collector voltage which is dependent on collector

    current which, in turn, depends on base current. A negative feedback loop introduces a certain degree

    of self-regulation. If the collector current increases for whatever reasons the collector voltage will fall

    and the base current will be reduced. The reduction in base current will produce a corresponding

    decrease in collector current to offset the original change. Conversely, if the collector current falls for

    whatever reasons the collector voltage will increase and the base current will rise. This leads to increase

    in collector current to offset the original change.

    The negative feedback of the circuit shown in figure 1.6 involves an a.c. signal component as well as a

    d.c. bias current. This results in the reduction of the signal gain. To solve this problem a bypass capacitor

    CB is required as shown in figure 1.7. The value of the bypass capacitor CB is chosen so that it exhibits

    very low reactance at the lowest frequency of operation compared to the series resistor network.

    RB1 RC1

    Vin

    Vo

    IB1 IC1RB1=

    1BI

    inVCCV

    +VCC

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    Q1

    C2

    C1

    Figure 1.6: Current amplifier with negative feedback network

    Q1

    C2

    C1

    Figure 1.7: Modified version of figure 1.6

    VC

    VinVo

    RfB1

    RC1

    IC1

    IfB1

    RfB1=

    1fBI

    inVCV

    +VCC

    RfB1 RfB1

    CBVin

    Vo

    IfB1

    +VCC

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    EXPERIMENT 1.1: COMMON EMITTER AMPLIFIER WITH FIXED BIAS

    Apparatus and Components

    Digital Multimeter (DMM), Protoboard, Power Supply, Function Generator, Oscilloscope, Transistors,

    Capacitors, and Resistors

    Theory of the Experiment

    The DC and AC analysis of a simple CE amplifier stage, shown in figure 1.11, is to be studied. To stabilize

    the Q-point the fixed bias circuit is modified by attaching an external resistor to the emitter, as

    illustrated in figure 1.12. This resistor introduces a negative feedback which regulates the circuit against

    temperature variations.

    Q1BC107

    C1

    C2

    Figure 1.11: Fixed Bias Common Emitter Amplifier

    RB RC

    Vin

    Vout

    +VCC

    0V

    IB IC

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    Q1BC107

    C1

    C2

    B

    E

    C

    Use the datasheet of the NPN transistor BC 107A (or its equivalent) to recognize the important transistor

    parameters which will assist you in finding the unknown values of resistors and capacitors. Choose

    experimental VCEQ such that the Q-point is symmetrically positioned within the active region of the

    output characteristics. Then calculate RB and RC and mount the circuit.

    Refer to figure 1.12, from Kirchoffs voltage law,

    BEEECCRB VRIVV (2.0)

    From Ohms law, the base current is:

    B

    RBB

    R

    VI (2.1)

    The way feedback controls the bias point is as follows. If V BE is he held constant and temperature

    increases, emitter current increases. However, the larger IE increases the emitter voltage VE=IERE, which

    in turn reduces the voltage VRB across the base resistor. A lower base-resistor voltage drop reduces the

    base current, which results in less collector current, because I C= IB. Collector current and emitter

    current are related by IC= IE, with 1, so increase in emitter current in emitter current with

    temperature is opposed, and operating point is kept stable.

    Vin Vout

    IB IC RCRB

    IERE

    +VCC

    0V

    VCC

    0.5VCC

    0

    Vin

    Vout

    Figure 1.12: Self Bias Common Emitter Amplifier with Emitter resistor

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    Similarly, if the transistor is replaced by another, there may be a change in I C (corresponding to a change

    in -value, for example). By similar process as above, the change is negated and operating point is kept

    stable.

    From the given circuit in figure 1.12,

    EB

    BECCB

    RR

    VVI

    1 (2.2)

    Merits of the circuit presented in figure 1.12

    The circuit has a tendency to stabilize the operating point against changes in temperature and -value.

    Demerits of the circuit presented in figure 1.12

    In this circuit, to keep IC independent of the following condition must be met:

    E

    BECC

    EB

    BECC

    BCR

    VV

    RR

    VVII

    1

    (2.3)

    This is approximately the case if BE RR 1

    As -value is fixed for a given transistor, this relation can be satisfied either by keeping R E verylarge, or making RB very low.

    If RE is of large value, high VCC is necessary. This increases cost as well precautions necessarywhile handling.

    If RB is low, a separate low voltage supply should be used in the base circuit. Using two suppliesof different voltages is impractical.

    In addition to the above, RE causes a.c. feedback which reduces the voltage gain of the amplifier.Usage

    The feedback also increases the input impedance of the amplifier when seen from the base, which can

    be advantageous. Due to the above advantages, this type of biasing circuit is used only with careful

    considerations of the trade-offs involved.

    Design

    Use the datasheet of the NPN transistor BC 107A and figure 1.12 above to recognize the important

    transistor parameters which will assist you in finding the unknown values of resistors and capacitors.

    Choose experimental VCEQsuch that the Q-point is symmetrically positioned within the active region of

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    the output characteristics. Note that, to provide maximum symmetrical swing of the output voltage

    without clipping or bottoming of the output amplified signal, the collector voltage should be half the

    supply voltage, VCC, as illustrated in figure 1.12.

    By applying KVL to outside side,

    CECCCC

    CECCCC

    VRIV

    VRIV

    0 (2.4)

    Assuming equal drops across RC and VCE ,

    CCCECC

    RC RIVV

    V 2

    (2.5)

    By applying KVL to the input side,

    B

    BECC

    B

    C

    B

    BEBBCC

    I

    VVR

    II

    VRIV

    0

    (2.6)

    The input capacitor can be calculated as follows:

    12

    1

    ChF

    ie (2.7)

    where F is the lowest signal frequency and h ie is the input resistance observed in the emitter region.

    Thus,

    ieFhC

    2

    11 (2.8)

    The Gain of the amplifier can be calculated as follows:

    S

    O

    V

    VGain (2.9)

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    Where VO is the amplitude of the output voltage of the amplifier and V S is the amplitude of output

    voltage of the Function Generator.

    S

    O

    V

    VdBinGain log20)( (3.0)

    Determine the values of RB, RC , RE, C1 and C2 and mount the circuit

    Frequency Response Model Graph

    The graph of Gain (in dB) against Frequency (in Hz) is shown in figure 1.4

    Figure 1.4: Frequency Response characteristics of the amplifier

    (a) DC MeasurementsCheck the voltage drop VCEQ. If it is outside 10% accuracy as compared to theoretical choice, repeat

    the design taking into consideration the factors contributing to your errors. Proceed with the

    experiment if you have fallen within the limits of the allowed errors.

    (i) Measure and record the DC voltages VB and VC(ii) Using the measured values, calculate V

    BEQ, V

    CEQ, I

    CQ, I

    BQand h

    FEand present them in tabular

    form to compare with the theoretical design values. Comment on their differences if any.

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    (b) AC MeasurementsPart I: Observing waveforms distortions and phase angle relationship

    Apply a small amplitude 3KHz sinusoidal signal, Vin, at the input of the amplifier. Use the two channels of

    the CRO to observe the input and output signals simultaneously. Increase the amplitude of the input

    signal until some clipping or bottoming (or both) of the output signal, Vout , occurs

    i. Copy both the input and output waveforms and label their amplitudesii. Which part of the waveform is distorted first?iii. Explain, with illustrations, the causes of the distortions observed in (b)(i) aboveiv. What is the phase relationship between Vin and Vout

    Part II: Observing the Frequency Response of the amplifier

    (i)The frequency response curve is plotted on a semi-log scale(ii)The mid frequency voltage gain is divided by 2 and these points are marked in the

    Frequency Response curve.

    (iii)The high frequency point is called the upper 3dB point(iv)The lower frequency point is called the lower 3dB point(v) The difference between the upper 3dB point and the lower 3dB point in the frequency scale

    gives the bandwidth of the amplifier.

    (vi)From the plotted graph the bandwidth is obtained from the formula; Bandwidth=fH-fL(vii)Apply large amplitude sinusoidal signal to the amplifier input. You may optionally start with a

    50mV sine wave signal. Ensure that it is not distorted. Keeping the input voltage constant, vary

    the frequency from 1Hz to 1MHz in regular steps of 10Hz. Ensure that no significant distortion of

    the output signal is observed; otherwise reduce the amplitude of the input signal to reduce the

    noticeable distortion of the output signal.

    (viii)Note down the corresponding output voltage(ix)Plot the graph of (a)Voltage Gain in dB Versus Frequency (b) Voltage Gain in dB Versus

    Frequency in logarithmic scale

    (x)Calculate the bandwidth from the response graph in (a)

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    (xi)Deduce the voltage gain of the amplifier at 3KHz(xii)What causes the drop of the amplifier gain at lower and higher frequencies

    (c) The Input ResistanceThe input resistance of the amplifier can be measured using the circuit shown in figure 1.13. From

    the circuit, one can see that the input resistance is given by:

    V

    inS

    in

    in RVV

    VR

    (3.1)

    (i) Derive equation 1.1 from the simplified equivalent circuit(ii) Use the variable resistance or discrete resistances to obtain the suitable values of R V for

    different values of Vin

    (iii)Calculate the average value of R iin(iv)Calculate the theoretical value of Rin using the h-parameters of the transistor given in the

    Datasheet in appendix A

    (d) The Output ResistanceThe output resistance RO of the amplifier can be measured using the same principle adopted in the

    determination of Rin. Set the circuit shown in figure 1.14 on the Protoboard. First set R V at infinity

    and measure the output voltage Vout=VO. Then vary RV to obtain three different values of Vout for

    different values of RV.

    RV

    VSVin

    AMPLIFIER

    Rin

    Figure 1.13: Input Resistance measurement

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    (i) Find a suitable mathematical relationship of RO, RV, and Vo from the simplified equivalentmodel.

    (ii) Using the mathematical relationship derived in (d) (i) above, plot the graph(iii) Deduce from the graph plotted in (d) (ii) above the average value of R O.(iv) Use the h-parameters given in the datasheet to calculate RO

    VO VO RV

    RO

    Figure 1.14: Output Resistance measurement

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    EXPERIMENT 1.2: SELF BIAS (VOLTAGE DIVIDER BIAS) SINGLE STAGE CE AMPLIFIER

    A combination of

    EXPERIMENT 1.3: COMMON COLLECTOR AMPLIFIER

    Apparatus and Components

    Transistor (BC 107A or its equivalent), Regulated Variable Power Supply, Audio Frequency Oscillator

    (Function Generator), Resistors, Capacitors and CRO

    Theory of the Experiment

    An amplifier configuration commonly used at outputs of many multistage amplifiers is a Common

    Collector configuration. One may be inquisitive, that how is it possible to ground a common collector

    terminal which is responsible to draw current from the supply voltage? It has no meaning, and in fact itis impractical to ground a collector leg which is responsible to deliver current to the transistor. Because

    of this, Common Collector is often configured in emitter follower mode, as illustrated in figure 1.3.1. The

    term Emitter Follower implies that the output signal is an exactly copy of the input signal.

    Q1

    R1

    R2

    C1

    Vin Vo

    Figure 1.3.1: Common Collector Amplifier

    Emitter Follower configuration is not used as an amplifying circuit because it voltage gain is less than 1

    (AV )1 . However, it has a lot of functions to serve. Its main usefulness is in the translation of a higher

    output resistance of an amplifier to a lower output resistance. Another function of an Emitter Follower

    is to shift the level of a signal (level shifter)

    +VCC

    0V

    RE

    CE

    RL

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    Since voltage amplification is done in the transistor amplifier circuit, we assume equal drops across

    Collector-Emitter terminals (VCE) and Emitter. Use the datasheet of BC 107A (or its equivalent) to choose

    the quiescent collector current, ICQ, and the supply voltage. Note that the supply voltage is chosen such

    that the transistor is not damaged. The parameter VCEO can be useful when choosing appropriate supply

    voltage. Use the information obtained from the datasheet to calculate the values of unknown resistors

    and capacitors.

    Drop across RE is assumed to be half the supply voltage VCC. The drop across VCE is VCC-VRE. We know that

    ICQ=IE. Now,

    E

    RE

    EI

    VR . The voltage across R2, VR2=VBE+VRE=0.6+VRE.

    From potential divider rule:

    CCR VRR

    RV

    12

    22

    Hence, the values of R2 and R1 can be chosen.

    Mount and test the circuit, with all calculated values of capacitors and resistors, on a Protoboard.

    Measure the DC voltages VB and VE with reference to ground in the absence of the signal source.

    In this experiment, the frequency response, voltage gain, input and output resistances of the Emitter

    Follower amplifier are measured and compared with those of CE configuration.

    (a) Frequency Response MeasurementsProcedures

    (i) Set the Audio Frequency Oscillator (AFO) to sine wave type of peak to peak amplitude of 5mV.(ii) Keeping the input voltage constant, vary the frequency from 0 Hz to 1 MHz in regular steps and

    note down the corresponding output voltage

    (iii)Plot the graph of gain against frequency.(iv)Calculate the bandwidth from the graph.

    The model graph of the frequency response is shown in figure 1.3.2

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    Figure 1.3.2: Frequency Response Model Graph

    (b) A.C. MeasurementsIntroduce a large but non-distorted 3 KHz sinusoidal signal Vi , at the input of the amplifier

    (i) Measure the gains of the amplifier when the load RL is connected to the circuit.(ii) Repeat (b) (i) above but with RL not connected to the circuit.(iii)What is the relationship between Vi and VO ?

    (c) Input Resistance(i) Use the procedures mentioned in Experiment 1.1 (c) to measure the input resistance, R i , of the

    amplifier.

    (ii) How does Ri of the configuration compare with that of the common emitter amplifier tested inexperiment 1.1 ?

    (d) Output Resistance(i) Use the procedures mentioned in experiment 1.1 (d) to measure the output resistance R O of the

    amplifier.

    (ii) How does this resistance value compare with that of the common emitter amplifier tested inExperiment 1.1?

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    (iii) If you were to drive a 4 speaker, what configuration will you choose between the EmitterFollower and Common Emitter amplifier configurations? Explain.

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    EXPERIMENT 2: MULTISTAGE AMPLIFIERS

    Two-Stage Voltage Amplifier

    The two-stage voltage amplifier can be formed by combining a single stage voltage amplifier and a single

    stage current amplifier. Figure 3 shows a direct coupled two-stage voltage amplifier. The base bias

    current of transistor Q1 is provided by the feedback resistor R B1 connected between the base terminal of

    Q1 and the junction of the resistors RE2 and RE2. The capacitor C3 bypasses any a.c. component to

    ground and hence making the second stage have low input impedance (current amplifier

    characteristics).

    The direct coupled amplifier, like that shown in figure 3, has distinct advantages over RC coupled

    amplifier because it uses fewer components and thus noise due to components is intensively reduced.

    Because of this, direct coupled amplifiers are usually preferred in the fabrication of ICs. The RC coupled

    amplifiers are not very practical because of a large number of components involved which impose noise

    problems to the system.

    Q1 Q2

    C2

    C3

    C1

    RC1 RC2

    RE1RE2

    RE2

    RB1

    RB2

    Vin

    Vout

    +VCC

    0V

    VE1

    VBE1

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    Figure 5: Two-Stage voltage amplifier

    From figure 3;

    Vin=VE1+VBE1, but VBE1=0.6

    =VE1+0.6 VE1

    =ou t

    BE

    E xVRR

    R

    21

    1

    Thus,

    1

    21E

    BV

    in

    ou t

    R

    RA

    V

    V

    Or

    1

    21E

    BV

    R

    RA

    Example

    Consider the voltage preamplifier of class A shown in figure 4. The necessary voltages, currents and

    capacitances are as indicated. The hFE for transistor Q2 is 100. Assume the voltage gain of 100, calculate

    the values of the resistors RC1, RC2, RE1, RE2, RE2, RB1, RB2 and the current ICQ1.

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    Q1 Q2

    C2

    C3

    C1

    Figure 6: Class A voltage preamplifier

    Solution

    RC2=mA

    V

    1

    5=5K

    IB2= mAmAh

    I

    FE

    CQ 01.010012

    ICQ1=10ICQ1=0.1mA

    AV= 1001

    2 E

    B

    R

    R

    RB2=100RE1 (Choose RE1=1K, RB2=100K)

    25V

    ICQ2=1mAICQ1RC1 RC2

    RE1RE2

    RE2

    RB1

    RB2

    Vin

    Vout

    VCC=+10V

    0V

    VE1

    VBE1

    IfB1

    IfB2

    1.65V

    1.3V

    100F

    1F

    2V

    1F

    5V

    IB2

    VC1

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    RC1=

    1

    1

    CQ

    CCC

    I

    VV

    =mA

    VV1.0

    210

    =80K (Take 82K standard resistor)

    Two-Stage current amplifier

    The two-stage current amplifier is constructed by cascading a single stage current amplifier followed by

    a single stage voltage amplifier, as shown in figure 5. In this circuit, the base bias current of the second

    stage transistor Q2 is obtained from the collector of the first stage transistor Q1. The first stage

    transistor Q1obtains its base bias current through a negative feedback network provided by series

    resistors RfB1 and RfB1. The capacitor Cfboosts the gain of the first stage transistor Q1.

    The circuit shown in figure 5, however, has problems in terms of stability. There is no proper

    communication between the two stages in the light of stability. Although the second stage gets

    information of the first stage through its base bias current, but the first stage knows nothing about the

    first stage. For example, if temperature rise causes increase in collector current of the second stage

    transistor, neither the first stage nor the second stage has a self-regulatory mechanism to suppress the

    effect of change to maintain stability. To provide stability the circuit in figure 5 should be modified as

    shown in figure 6.

    In figure 6, the emitter resistor of the second stage is splitted into two resistors, R E2 and RE2, and the bas

    bias of the first stage is derived from the junction of the resistors R E2 and RE2 of the second stage. The

    second stage emitter resistor RE2 is coupled while RE2 is not coupled because it is required in the current

    amplification process. The capacitors C1 and C2 act as high pass filters to bypass high frequency signals

    in order to overcome Miller effect. Recall that Miller effect is caused by the presence of a stray capacitor

    between the base and collector. The reactance of this capacitor decreases as signal frequency increases,

    hence distorting signal amplification process.

    The current gain can be found by observing that;

    Iin+IfB1=IB1, but IB1

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    Hence

    1

    1

    E

    fB

    i

    in

    o

    R

    RA

    I

    IGainCurrent

    Or

    1

    1

    E

    fB

    i

    R

    RA

    For temperature stability conditions and to ensure that the output voltage attains its maximum

    symmetrical swing to avoid clipping, VE=1V and RE=RE.

    Q1 Q2

    C2

    1uF

    C1

    Figure 7: Two stage current amplifier

    RB1 RC1

    RE1

    RfB1RfB1

    CfVin

    Vo

    IfB1

    IC1

    IC2

    IB2

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    Q1

    C1

    Q2

    C2 C3

    C4

    Figure 8: Modified version of figure 5

    Practical Examples of Preamplifiers

    Preamplifiers have a wide range of applications. But why should we have preamplifiers in amplifying

    circuits? Preamplifiers are used to boost weak signals from an output device. For example, the signals

    from a microphone (MIC) are very weak in terms of voltage and current and their signal levels are

    comparable to noise. Therefore, pre-amplification is necessary to make the signals stronger than noise.

    Because we cannot simultaneously amplify voltage and current then we have to amplify each entity

    separately. That is, we amplify voltage using a voltage amplifier followed by current amplification using a

    current amplifier. A typical MIC can deliver 10mV signal level. Figure 7 shows a pre-amplifier used to

    amplify the signals from a MIC.

    0V

    RE2

    RE2IfB1

    RfB1

    IB2

    IE2

    IO

    IC1IC2

    +VCC

    RC1 RC2

    IB1Iin

    IE1

    VinVo

    PRE-AMPLIFIER

    VOLTAGE

    AMPLIFIER

    CURRENT

    AMPLIFIER10mV 100mV

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    Figure 9: Application of a Preamplifier in a MIC

    Aim of the experiment: The aim of this experiment is to design a two-stage small signal voltage

    amplifier and the gain, input characteristics, output characteristics, bandwidth and frequency response.

    Materials and Apparatus: Oscilloscope, Function Generator, Digital Multimeter, Power Supply,

    Breadboard, connecting wires, wire cutter, wire stripper/shaper, Resistors, Capacitors and BC 107.

    Circuit Diagram

    Figure 8 shows a two stage voltage preamplifier.

    Q1 Q2

    C2

    C3

    C1

    Figure 10: Two stage voltage preamplifier

    FUNCTION

    GENERATOR

    ICQ2ICQ1RC1 RC2

    RE1RE2

    RE2

    RB1

    RB2

    VinVout

    +VCC

    0V

    VE1

    IfB1

    IfB2

    VE2

    IB2

    VC1

    BC 107BC 107

    IE1

    VB1

    VBE1

    VS=50mV

    (0-1)MHz(0-1)MHz

    RS

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    Design Procedures

    Use the datasheet of BC 107 NPN transistor shown in Appendix A to design a two-stage voltage amplifier

    shown in figure 8, by calculating the values of unknown resistors and capacitors. The following

    guidelines may help you during the designing process. Note that the transistors in both stages (first and

    second stage) are identical.

    a) Choose the supply voltage VCC . Use the data sheet to confirm that the chosen supply voltage issafe and cannot damage the transistor. This is observed by looking at the maximum collector-

    emitter voltage with the base open (VCEO) that a transistor can handle without being destroyed.

    b) Find the collector voltage (VC) such that the output voltage attains its maximum symmetricalswing without clipping.

    c) Use the datasheet to recognize the maximum collector current that a transistor can handle. Usethis value to determine the quiescent collector current (ICQ2) for the second stage transistor.

    d) Use the results in (a) and (b) above to determine the second stage collector resistor (R C2)e) Use the voltage gain formula, VA -

    1

    2

    E

    B

    R

    R, to select the values of RB2 and RE1. Assume the

    overall voltage gain is 100.

    f)

    Calculate the values of the resistors, RE2 and RE2, of the emitter of the second stage transistor.For stability conditions set VE2=1.3V and RE2=RE2. The simplest approach to find RE2 and RE2 is to

    treat them in such a way that they form a potential divider network. Hence, RE2=RE2=0.5VE2.

    g) Use the current gain formula,2

    1

    E

    B

    iR

    RA , and the result obtained in part (f) above to find the

    value of RB1. Theoretically, the current gain should not exceed 50; otherwise the system will

    oscillate and thus become unstable. Choose a current gain of 30.

    h) Assume the minimum hFE of BC 107 transistor (consult a datasheet), find the base current, IB2,flowing into the second stage transistor by using the formula

    2

    12

    FE

    CQB

    h

    II , where hFE2 is the

    minimum current gain of the second stage transistor.

    i) Determine the collector current, ICQ1, of the first stage transistor. This current should be 10 timesthe base current of the second stage transistor. That is, ICQ1=10IB2.

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    j) Determine the collector voltage, VC2, of the first stage transistor. Use this value together withthe result obtained in (i) above to find the value of the collector resistor, R C1, of the first stage

    transistor.

    k) Calculate RE1. Note that VE1=1V for stability conditions and RE1=1

    1

    E

    E

    I

    V

    1

    1

    CQ

    E

    I

    V

    l) Calculate the values of the capacitors C1, C2 and C3 such that they exhibit very low impedanceat the lowest frequency of operation. Assume the lowest frequency, fL, is 20Hz

    m) Set the voltage source, VS =50mV (assume) using a signal generatorn) Keeping the input voltage constant, vary the frequency from 0Hz to 1MHz in regular steps of 10

    and note down the corresponding output voltage.

    o) Plot the frequency response: Gain (dB) versus Frequency (Hz)p) Find the input and output impedancesq) Calculate the bandwidth from the graphr) Note down the phase angle, bandwidth, input and output impedances

    General Procedures for calculation

    1. Input impedancea) Connect a Decade Resistance Box (DRB) between input voltage source and the base of the

    first stage transistor (series connection)

    b) Connect the Voltmeter (0-100mV)across the biasing resistor RB1c) Vary the value of the DRB such that the ac Voltmeter reads the voltage half the input signald) Note down the resistance RB, which is the input impedancee) Compare and contrast DRB obtained in (d) above with that of Common Emitter

    configuration

    2. Output impedancea) Measure the output voltage when the amplifier is operating in the mid-band frequency with

    the load resistance connected (Vload)

    b) Measure the output voltage when the amplifier is operating in the mid-band frequencywithout the load connected (V no-load)

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    c) Substitute these values in the formula Zo= %100xV

    VV

    load

    loadnoload to get output resistance

    d) Compare and contrast Zo obtained in (c) above with that of Common Emitter configuration

    3. Bandwidtha) Plot the frequency responseb) Identify the maximum gain regionc) Draw the horizontal line bi -3dBd) The -3dB line intersects the frequency response plot at two pointse) The lower intersection point of -3dB line with the frequency response plot gives the lower

    cut-off frequency

    f)

    The upper intersection point of -3dB line with the frequency response plot gives the uppercut-off frequency

    g) The difference between the upper cut-off frequency and the lower cut-off frequency iscalled the bandwidth. Thus bandwidth = fH-fL

    h) Compare and contrast the bandwidth obtained in (g) above with that of Common Emitterconfiguration

    Model Graph (Frequency Response)

    Supplementary Questions

    1. Briefly explain how temperature stabilization is achieved in this circuit2. What role is played by the C1 and C2 and C3 in the system?3. Describe the AC and DC characteristics of the amplifier.

    fL fH

    |A|max (dB)

    Gain (dB)

    Bandwidth

    Frequency (Hz)

    -3dB

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    4. Can this amplifier be used to amplify very small signals from a microphone to a magnitude thatis sufficient to drive a loud speaker? Explain. (Assume a typical microphone delivers 10mV signal

    level)

    5. Is this a small signal or large signal amplifier? Explain