Experiment 1_ Final

7/31/2019 Experiment 1_ Final

1/31

1

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


2/31

2

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


3/31

3

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)


4/31

4

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:


5/31

5

(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.


6/31

6

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


7/31

7

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


8/31

8

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


9/31

9

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


10/31

10

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


11/31

11

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


12/31

12

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)


13/31

13

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.


14/31

14

(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)


15/31

15

(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


16/31

16

(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


17/31

17

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


18/31

18

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


19/31

19

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?


20/31

20

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


21/31

21

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


22/31

22

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.


23/31

23

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


24/31

24

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


25/31

25

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


26/31

26

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


27/31

27

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


28/31

28

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.


29/31

29

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)


30/31

30

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


31/31

31

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

Documents

Experiment 1_ Final