transistorckk.com.tr/ders/analog/00 transistor 02.pdfbjt transistors • a transistor is a device...
TRANSCRIPT
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Transistor
Dr. Cahit Karakuş
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BJT Transistors
• A transistor is a device which acts like a controlled valve. The current flow permitted can be controlled.
• The bipolar junction transistor (BJT) is a three-terminal electronic valve - the output (collector) terminal current-voltage characteristics are controlled by the current injected into the input port (base). The BJT is a semiconductor device constructed from two pn junctions. There are two types of BJT: pnp and npn. Figures 1 and 2 show the circuit symbols and common current and voltage polarities during normal (active) operation.
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DC Analiz
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Vcc
VBB
Rc
Ic
C
E
BRB
VBE
VCE
IE
IB
CECCCC VIRV *
BEBBBB VIRV *
BC II *
B
BEBBB
R
VVI
C
CCSATC
R
VI
.0;0 OlurVVSaturasyonIIdayaVV CESATccCE
.0;0 OlurAIIKesmedeiseAI CBB
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RC
VCE
VBE
Ic
IE
IB C
E
B
RE
RB
Vcc
EECECCCC IRVIRV **
EEBEBBCC IRVIRV **
BC II *
BBBBCE IIIIII *)1(*
CC
CE II
II *)1
(
-
RC
VCE
VBE
Ic
IE
IB C
E
B
RE
RB
VccEECEBCCCC IRVIIRV *)(*
EEBEBBBCCCC IRVIRIIRV **)(*
BC II *
BBBBCE IIIIII *)1(*
CC
CE II
II *)1
(
-
Vcc
RC
VCE
VBE
Ic
IE
IB C
E
B
RE
R1
R2
Vcc
VBB
Rc
Ic
C
E
BRB
VBE
VCEIB
IE
RE
21
2*RR
RVV CCBB
21
21 *
RR
RRRB
EECECCCC IRVIRV **
EEBEBBBB IRVIRV **
BC II *
BBBBCE IIIIII *)1(*
CC
CE II
II *)1
(
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Sem I 0809/rosdiyana
Introduction to BJT Small Signal Analysis
CHAPTER 5
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15
Introduction
•To begin analyze of small-signal AC response of BJT
amplifier the knowledge of modeling the transistor is
important.
•The input signal will determine whether it’s a small signal
(AC) or large signal (DC) analysis.
•The goal when modeling small-signal behavior is to make of
a transistor that work for small-signal enough to “keep
things linear” (i.e.: not distort too much) [3]
•There are two models commonly used in the small signal
analysis:
a) re model
b) hybrid equivalent model
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16
Disadvantage
Re model
• Fails to account the output impedance level of device and feedback effect from output to input
Hybrid equivalent model
• Limited to specified operating condition in order to obtain accurate result
Introduction
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17
• The transistor can be employed as an amplifying device. That is, the output sinusoidal signal is greater than the input signal or the ac input power is greater than ac input power.
• How the ac power output can be greater than the input ac power?
Amplification in the AC domain
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18
Amplification in the AC domain
Conservation; output
power of a system
cannot be large than its
input and the efficiency
cannot be greater than 1
The input dc plays the
important role for the
amplification to
contribute its level to the
ac domain where the
conversion will become
as η=Po(ac)/Pi(dc)
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19
• The superposition theorem is applicable for the analysis and design of the dc & ac components of a BJT network, permitting the separation of the analysis of the dc & ac responses of the system.
• In other words, one can make a complete dc analysis of a system before considering the ac response.
• Once the dc analysis is complete, the ac response can be determined using a completely ac analysis.
Amplification in the AC domain
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20
BJT Transistor Model • Use equivalent circuit
• Schematic symbol for the device can be replaced by this equivalent circuits.
• Basic methods of circuit analysis is applied.
• DC levels were important to determine the Q-point
• Once determined, the DC level can be ignored in the AC analysis of the network.
• Coupling capacitors & bypass capacitor were chosen to have a very small reactance at the frequency of applications.
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21
The AC equivalent of a network is
obtained by:
1. Setting all DC sources to zero & replacing them by a short-circuit equivalent.
2. Replacing all capacitors by a short-circuit equivalent.
3. Removing all elements bypassed by short-circuit equivalent.
4. Redrawing the network.
BJT Transistor Model
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22
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23
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24
Example
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25
Example
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26
Example
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27
The re transistor model
• Common Base PNP Configuration
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28
Common Base PNP Configuration
• Transistor is replaced by a single diode between E & B, and control current source between B & C
• Collector current Ic is controlled by the level of emitter current Ie.
• For the ac response the diode can be replaced by its equivalent ac resistance.
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29
Common Base PNP Configuration
• The ac resistance of a diode can be determined by the equation;
Where ID is the dc current through the diode at the Q-point.
EI
mVre
26
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30
• Input impedance is relatively small and output impedance quite high.
• range from a few Ω to max 50 Ω
• Typical values are in the M Ω
CBi reZ
CBZo
Common Base PNP Configuration
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31
The common-base characteristics
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32
Voltage Gain
re
RA
re
R
rI
RI
V
VA
rI
ZI
ZIV
RI
RI
RIV
LV
L
ee
Le
i
OV
ee
ie
iii
Le
LC
Loo
:gain voltage
: ageinput volt
)(
: tageoutput vol
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33
Current Gain
1
i
e
e
e
C
i
oi
A
I
I
I
I
I
IA
• The fact that the polarity of the Vo as determined by the current IC is the same as defined by figure below.
• It reveals that Vo and Vi are in phase for the common-base configuration.
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34
Common Base PNP Configuration
Approximate model for a common-base npn transistor configuration
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35
Example 1: For a common-base configuration in figure
below with IE=4mA, =0.98 and AC signal of 2mV is
applied between the base and emitter terminal:
a) Determine the Zi b) Calculate Av if RL=0.56k
c) Find Zo and Ai
e
b b
c
ec I αI
IcI
e
common-base re equivalent cct
re
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36
Solution:
5.6m4
m26
I
26mr Za)
Eei
43.845.6
)k56.0(98.0
r
RA b)
e
Lv
98.0I
IA
Ω Zc)
i
oi
o
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37
Example 2: For a common-base configuration in previous
example with Ie=0.5mA, =0.98 and AC signal of 10mV is
applied, determine:
a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib
20m5.0
m10
I
V Za)
:Solution
e
ii
88mV5
(1.2k)0.98(0.5m)
RIRIV b) LeLco
8.58m10
m588
V
VA c)
i
ov
98.0A d) i
A10
)98.01(m5.0
)1(m5.0
I-I
I-II e)
ee
ceb
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Common Emitter NPN Configuration
• Base and emitter are input terminal
• Collector and emitter are output terminals
38
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Common Emitter NPN Configuration
• Substitute re equivalent circuit
• Current through diode
39
bc II
bbe
bbbce
III
IIIII
)1(
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40
• Input impedance
ei
ei
b
ebi
eb
eei
b
be
i
ii
rZ
rZ
I
rIZ
rI
rIV
I
V
I
VZ
; 1an greater thusually
)1(
)1( that so
)1(
:ageinput volt
:impedanceinput
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41
The output graph
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42
bI
c
e
bIi=I
b
re model for the C-E transistor configuration
re
ro
e
0AbI
c
e
bI
i=I
b
re
ro
e
Vs=0V
= 0A
oZ
impedance)high cct,(open ΩZ
the thusignored is r if
rZ
o
o
oo
Output impedance Zo
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43
e
LV
eb
Lb
i
oV
eb
iii
Lb
Lco
Loo
r
RA
rI
RI
V
VA
rI
ZIV
RI
RIV
RIV
that so
:ageinput volt
: tageoutput vol
i
b
b
b
C
i
oi
A
I
I
I
I
I
IA
Voltage Gain Current Gain
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re model for common-emitter
44
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45
Example 3: Given =120 and IE(dc)=3.2mA for a common-
emitter configuration with ro= , determine:
a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load
975)125.8(120rZ
125.8m2.3
m26
I
26mr a)
ei
Ee
:Solution
15.246125.8
k2
r
Rb)A
e
Lv
120I
IA c)
i
oi
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46
Example 4: Using the npn common-emitter configuration,
determine the following if =80, IE(dc)=2 mA and ro=40 k
a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k
k04.1)13(80rZ
13m2
m26
I
26mr a)
ei
Ee
:Solution
bI
cbIi=I
b
re model for the C-E transistor configuration
re
ro
e
RL
Io
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47
67.77
)80(k2.1k40
k40
Rr
r
I
Rr
)I(r
A
Rr
)I(rI
I
I
I
IiAb)
(cont)Solution
Lo
o
b
Lo
bo
i
Lo
boL
b
L
i
o
6.8913
k40k2.1
r
rRvAc)
e
oL
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48
Common Collector Configuration
• For the CC configuration, the model defined for the common-emitter configuration is normally applied rather than defining a model for the common-collector configuration.