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RF ELECTRONIC PREPARATORY II

AMPLIFIER AND FILTER LABORATORY EXERCISEProject report 17.11.2009

Viktor Mihaylov 219566


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1.IntroductionThe following paper examines the design of an simple common-emitter amplifier and

two filter for the purpose of an laboratory work. Principles and main characteristics are

observed in the first part of the study. The second one is concerned with the actual design.

Matlab is used for the interpretation of the filters.

2.Theoretical Background2.1Amplifier. Transistor.

A basic component found in everydays electronics is the active, three terminal device

known as the transistor. The one employed in our studies is an NPN BJT Bipolar

Junction Transistor. It is a three layer semiconductor sandwich with an emitterand a

collectorin the ends and a base in the middle. Bipolar transistors work as current-

controlled current regulators. They restrict the amount of current passed according to a

smaller, controlling current. The main current that is controlled comes from emitter to

collector(for NPN only).

Figure1.BJTstructure

For the purpose of our investigation the operation mode of the amplifier that we areusing is active mode. We use the simple circuit model when the transistors both signal

source and load share its emitter lead as a common connection point this is called the

common-emitter configuration. It is commonly used for voltage and current amplification

because it exhibits good properties when it is used so. Another specific characteristic is

the that the output signal is inverted with a 180 degree angle (180 out of phase )

inverting amplifier circuit.

All types of signal amplifiers operate using AC signal inputs which alternate between

above and below zero value so some way of presetting the amplifier circuit to operate

between these two maximum or peak values is required. This is achieved through biasing.

It establishes the correct operating point of the amplifier thus leading to reduced

distortions to the input signal. In the active region of a common-emitter amplifier the


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collector-base and the emitter-base are oppositely biased the first is opposite and the

second forward biased. This point is called quiescent point and its best possible position is

as close to the centre position of the load line as possible. On Figure 2 a model of the

common emitter amplifier circuit is shown.

UEUB

UC

Figure2 Common emittercircuit

For the large signal model we can make the following assumptions. The quiescent

Base voltage (Vb) is determined by the potential divider network (R1 and R2) and the

power supply voltage Vcc :

1

,The collector current can be found using :

,though can be assumedand

Where IE is the emitter current and is the current gain. In BJTs it is usually more

than 100 but for high power applications it can be less.


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In large signal analysis the capacitance are bypassed (CE) or act as blocking (C2 and

C1).

As set in [2] those equations can be derived and used:

,where Uc would be 1/3 by presumptions

For R2 we assume that the current is 10 times bigger than IB. For R1 we have , where UB is the voltage across the .For the small signal analysis a simplified model is used [2]:

Figure3 Smallsignalmodel

For the voltage gain AV we have the equation. , where .and the input impedance is 1

he selection of the capacitive elements depends on the lower frequency response.

Val

2

T

ues for every capacitor are calculated by excluding the the other open circuits.1

cut off frequency for output capacitance and then the capacitance=>


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2 1

1

2

cut off frequency for input capacitance and then the capacitance=>

2 1

12cut off frequency for bypass capacitance and then the capacitance=>

2 1 where the 1 1

ed to be or

The lower half power frequency could be assum and it will be 1/3 of our lowest frequency(20Hz) => = 6,66

3.Design and cahe starting parameters are:

the formula 20 then AV = 2.510mA base current

nd VEQ = 5V

oned equation the following results are derived:

lculations.

TAv = 8dB and when calculated using

= 100 - the minimum for 1

Vcc = 15V

Rload = 3k

Bf (20Hz 20KHz)

VCQ = 10V a

IC = 10mA

Using the above menti

= , 1 50,01 500

100 1

5,7

0,001 5,7


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155,70,001 9,3

and for RE we calculate it from AV:

1 2,6 11

169from this we can derive

331Th values for the capacitors are found when the :

12100030006,66

5,97 2

1 12.3,14.2927.6.6 8,2

1

2

1

6,28.11300.6,66 200,11

RC The maximum expected amplitude that is expected to be amplified without

distortion cannot be more than the supplied voltage which is VCC=15V and since it is a

common emitter amplifier and the output is derived from the UE = 5V then it is 2

e definition of the small sign mplifier is presented insec

atio

which in

The effect of the load impedance on the voltage gain can be defined in section 2 in the

form

t matched with the

uld be lesser.

The expected cut off frequencies are determined by the capacitances of the circuit. in

section 2 the formula for those are presented. For higher frequencies there should be

additional computation due to parasitic capacitances and impurities.

Th al input impedance of the ation 2. It is the voltage ratio between input voltage signal and current(AC).

Components that determine it are R1,R2, REF and the thermal voltage. 1 2,927.The definition of the small signal output impedance of the amplifier is the voltage r

between input voltage signal and current(AC). It is determined by RC. this case is RC= RC||RL=428

ula for AV. Deducting from the formula the they are proportional with the increase

of RL the gain will be increased. For every value of the load that is no

transistors the output power wo


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Because C1 and C2 are blocking capacitors higher values should be used for the

purpose of lowering the leakage currents.

The desired specifications needed should be measured Gain, Rin, Rout- etc.

4.hey are

idering the source determines

ers. The capacitor acts as open circuit for low

frequencies and as short circuit for high frequencies. When in parallel to the source it

shu

he cutoff f

2

Filters. Theory and design.

First order filters are a simple combination of a resistor and a capacitor. T

link d in terms of a voltage divider and their relation cons

whether they are high or low pass filt

e

nts the higher frequencies to the ground making the circuit a low-pass filter. When

placed otherwise, in series, it blocks the lower frequencies making a high-pass filter. On

figure 4 and 5 low-pass and high-pass filter are represented.

T

Figure 4 First order Low-pass filter

Figure 5 First order High-pass filter

requency(3dB frequency or half power frequency) for the Low-pass and

the High-pass is given by the same equation:

1


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Bellow the cutoff frequency signal is pass without attenuation adn above it with

decreased amplitude converging to zero. This statement is valid for the low-pass, and

opposite of this for the high-pass.

to be assumed it should not be too high because the

resi ity will drop, so is C=470pF thus leading to the R value of:

For the design of the low-pass we have the following parameters:

6 11000 7000Since a capacitor value have

stiv

48,4 which is rounded to the nearest R = 56k.Since we have a tolerance of 10% then the minimum and maximum values would be

423pF and Ch = 517pF.Cl =

For the design of the high-pass we have the following parameters:

F thus leading to the R value of:

6 1100 700The value for C=100n 2,27 which is rounded to the nearest R = 2.7k.

m values would be

Cl = 90nF and Ch = 110nF.

by Matlab simulations,. The

transmition functions H are u

Since we have a tolerance of 10% then the minimum and maximu

The amplitude and phase responses are determined

sed and simple commands as tf(), bode() etc.

For low-pass: and for the high-pass: . The followinggraphics include superimposed responses the one calculated and three others for the

practical values and capacitive tolerance.

Figure 6. Low-pass amplitude response.


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Figure 7 Low-pass frequency response

Figure 8 High-pass amplitude response


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For the estimation of the amplitude responses of the signal chains the following

deductions are made.

source =>high pass => low pass

Basically a two amplitude responses band pass response. Both with cutoff

frequencies that respond to their characteristics.

source=>low pass=high pass This is analogically derived as the previous one. Both

are band pass. The transmission equations of the low pass and the high pass would be

summed.

source=>low pass=> amplifier=> high pass This would be again a band pass filter

but with amplification,. Lower cutoff frequency determined by the low-pass filter and upper

by the high-pass filter.

Figure 9 High-pass frequency response

5.References.

[1] Electrical and electronic principles and technology, John Bird

[2] RF-electronics preparatory II, Jari Kangas

[3] http://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.html

[4]Practical Analog And Digital Filter Design, Les Thede, 2004
http://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.htmlhttp://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.htmlhttp://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.html


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6.Matlab CodeFor estimation of the mixed curves of the high - pass filter:

R1=2270

R12=2700C1=100*(10^(-9))C2=90*(10^(-9))C3=110*(10^(-9))

RC1=R1*C1RC2=R12*C1RC3=R12*C2RC4=R12*C3

num1=[RC1 0]den1=[RC1 1]sys1=tf(num1, den1)




bode(sys1, sys2, sys3, sys4)

and fo the analogical low-pass filter:

R1=48400R12=56000C1=470*(10^(-12))C2=423*(10^(-12))C3=517*(10^(-12))

RC1=R1*C1

RC2=R12*C1RC3=R12*C2RC4=R12*C3

num1=1/RC1den1=[1 1/RC1]sys1=tf(num1, den1)




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bode(sys1, sys2, sys3, sys4)

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