RF ELECTRONIC PREPARATORY II

AMPLIFIER AND FILTER LABORATORY EXERCISE

Project report 17.11.2009















Viktor Mihaylov 219566


1. Introduction
The 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 Background

2.1 Amplifier. 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 emitter and a
collector in 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 are
using 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
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.

U
E

U
B

U
C


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 :
I
B
=
v
cc
R
2
R
1
+R
2

I
C
= I
L
[1 -
1
[
,
The collector current can be found using :

C
= I
L
I
C
I
B
,though I can be assumed
and [ =

Where I
E
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.
In large signal analysis the capacitance are bypassed (C
E
) or act as blocking (C
2
and
C
1
).
As set in [2] those equations can be derived and used:

R
C
=
0
c
I
C
,
where Uc would be 1/3 by presumptions
R
L
=
u
L
I
L

1
=
cc B
I
2
For R
2
we assume that the current is 10 times bigger than I
B
.
v -0
For R
1
we have R , where U
B
is the voltage across the R
2
=
0
B
I
2
.

For the small signal analysis a simplified model is used [2]:

















Figure3 Smallsignalmodel

For the voltage gain A
V
we have the equation.
v
A
v
=
cut
v
in
=
L C

n
+(1+[)R
EF
[R R
, where r
n
=
T
I
BQ
v
.
and the input impedance is R

n
= R
b
[|r
n
+([ +1)R
LP
]
he selection of the capacitive elements depends on the lower frequency response.
Val

o
=
2n(R
C
+R
L
)C
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=>
C
2
=
2n(R
C
+R
L
)
o
1

n
=
1
2nR
n
C
1

cut off frequency for input capacitance and then the capacitance=>
1
2nR
n

n
C =
1

bp
=
1
2nR
L
i
C
L

cut off frequency for bypass capacitance and then the capacitance=>
2nR
L

bp
C
L
=
1 +[
i

where the R
L
i
= | 1 +[)R
LB
[|(1 +[)R
LP
+r
n
]
ed to be
L
=
bp n
+
o
or
]
L
(

+ The lower half power frequency could be assum

o
=
n
=
bp
=
3
and it will be 1/3 of our lowest frequency(20Hz) =>
3
]
L
=
20
3
= 6,66
3. Design and ca
he starting parameters are:
the formula A
v
|JB] = 2ulog(A
v
) then A
V
= 2.51
0mA base current
nd V
EQ
= 5V

oned equation the following results are derived:
R
C
=
0
c
I
C

lculations.
T
A
v
= 8dB and when calculated using
= 100 - the minimum for 1
Vcc = 15V
Rload = 3k
Bf (20Hz 20KHz)
V
CQ
= 10V a
I
C
= 10mA
Using the above menti
=
10
0,01
= 1k0
I
L
R
L
=
u
L
=
S
u,u1
= Suu0
[ =
I
C
I
B

=> I
B
=
I
C
[
= 1uupA => I
2
= 1mA
R
2
=
u
B
I
2
=
LB
+u
L
I
2
u
=
S,7
u,uu1
= S,7k0
R
1
=
I
cc
-u
B
I
2
=
1S -S,7
u,uu1
= 9,Sk0
and for R
E
we calculate it from A
V
:
A
v
=
[R
L
R
C
)R
LP
r
n
+(1 +[
wbcrc r
n
=
I
1
I
B
= 2,60
tbcn R
LP=
1
1 +[
_
[R
C
[R
L
A
v
-r
n
_ = 1690
from this we can derive
LB
= R
L
-R
LP
= SS10

Th values for the capacitors are found when the :
C
2
=
1
2n(1uuu +Suuu)6,66
R
= S,97pF
C
1
=
2nR
n

n
1
=
1
2.S,14.2927.6.6
= 8,2pF

C
L
=
1
2nR
L
i

bp
=
1 +[
6,28.11Suu.6,66
= 2uu,11pF
R
C
The maximum expected amplitude that is expected to be amplified without
distortion cannot be more than the supplied voltage which is V
CC
=15V and since it is a
common emitter amplifier and the output is derived from the U
E
= 5V then it is <5V. And
since A =
v
cut

v
in
=
5
v
in
=> I =
5
n
A
v
in our case A
V
is 2,51 =>I = 2

e definition of the small sign mplifier is presented in
sec
atio
C
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.
n
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 a
tion 2. It is the voltage ratio between input voltage signal and current(AC).
Components that determine it are R1,R2, R
EF
and the thermal voltage. R
n
= R
b
[|r
n
+
([ +1)R
LP
] = 2,927k0.
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 R
C
. R
out
= R
this case is R
C
= R
C
||R
L
=428
ula for A
V
. Deducting from the formula the they are proportional with the increase
of R
L
the gain will be increased. For every value of the load that is no
transistors the output power wo
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
C
2nRC
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

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:
]
C
C2n
For the design of the low-pass we have the following parameters:

C
= (6 +1)1uuu = 7uuuEz
Since a capacitor value have
stiv
R =
1
= 48,4k0 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 C
h
= 517pF.

C
l
=
For the design of the high-pass we have the following parameters:
F thus leading to the R value of:
=

C
= (6 +1)1uu = 7uuEz
The value for C=100n
1
R
]
C
C2n
= 2,27k0 which is rounded to the nearest R = 2.7k.
m values would be
C
l
= 90nF and C
h
= 110nF.
by Matlab simulations,. The
transmition functions H are u
1
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: E(s) =
1+sCR
and for the high-pass: E(s) =
1+sCR
sCR
. The following
graphics include superimposed responses the one calculated and three others for the
practical values and capacitive tolerance.
Figure 6. Low-pass amplitude response.
Figure 7 Low-pass frequency response

Figure 8 High-pass amplitude response



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





6. Matlab Code

For estimation of the mixed curves of the high - pass filter:
R1=2270
R12=2700
C1=100*(10^(-9))
C2=90*(10^(-9))
C3=110*(10^(-9))

RC1=R1*C1
RC2=R12*C1
RC3=R12*C2
RC4=R12*C3

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

num2=[RC2 0]
den2=[RC2 1]
sys2=tf(num2, den2)

num3=[RC3 0]
den3=[RC3 1]
sys3=tf(num3, den3)

num4=[RC4 0]
den4=[RC4 1]
sys4=tf(num4, den4)

bode(sys1, sys2, sys3, sys4)

and fo the analogical low-pass filter:

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

RC1=R1*C1
RC2=R12*C1
RC3=R12*C2
RC4=R12*C3

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

num2=1/RC2
den2=[1 1/RC2]
sys2=tf(num2, den2)

num3=1/RC3
den3=[1 1/RC3]
sys3=tf(num3, den3)

num4=1/RC4
den4=[1 1/RC4]
sys4=tf(num4, den4)

bode(sys1, sys2, sys3, sys4)