# Chapter 9 Common Emitter Amplifiers (PowerPoint)

Shared by:
Categories
-
Stats
views:
138
posted:
4/23/2011
language:
English
pages:
55
Document Sample

```							            Chapter 9
Common-Emitter Amplifiers
Pictures are redrawn (with some modifications) from
Introductory Electronic Devices and Circuits
By
Robert T. Paynter

1
Objectives
• Describe AV, AI, AP associated with the three
amplifier configurations.
• Describe the input/output voltage and current
phase relationships.
• Calculate the ac emitter resistance of a transistor.
• Discuss two roles of the capacitors in the circuits.
• Derive the ac equivalent circuit for a given
amplifier.
• Explain how the voltage gain drift due to
temperature occurs.
• Discuss the relationship between the load
resistance and voltage gain of a CE amplifier.
2
Objectives (Cont.)
• Calculate Zin(base) and Zin for a CE amplifier
• Discuss the effects of swamping on the ac
characteristics of a CE amplifier
• List and describe the four ac h-parameters

3
Fig 9.1 Common-emitter input-
output phase relationship.
VB
+VCC
IB   15 A
10 A
5 A              RC
RB

Q1

VC   8V
6V
Ai = 100             4V
IC   1.5 mA
1 mA
500 A   4
AC Emitter Resistance

25mV
re 
IE

where r’e = ac emitter resistance
IE = the dc emitter current, found as VE / RE for
example.

5
Fig 9.2 Example 9.1.
R2              2.2kΩ
VB  VCC            10V          1.8V
+10 V                R1  R2          12.2kΩ

VE VB  0.7V
IE                  1.1mA
RC                        RE    1kΩ
R1
10k        4k
25mV 25mV
re                 22.7Ω
RL               IE   1.1mA
hFE = 300   15k
R2
2.2k       RE
1k
6
Fig 9.3 Graphical determination of
ac emitter resistance.
IE

IE                 Q2

VBE
re 
Q1                    I E

IE
VBE
VBE VBE                   7
Fig 9.4 The determination of ac beta.
IC

I C ic
 ac      
I B ib

Q
hFE = dc beta
IC
hfe = ac beta

IB
IB
8
AC Model of A BJT
c

ic = acib
C         C
B         B         b
ib
E         E            r'e
npn       pnp
e

9
Roles of Capacitors in Amplifiers

1. A coupling capacitor passes an ac
signal from one amplifier to another,
while providing dc isolation between the
two.
2. A bypass capacitor is used to “short
circuit” an ac signal to ground while not
affecting the dc operation of the circuit.

1   The higher the freq.,
XC 
2 fC the lower the capacitor impedance.
10
Fig 9.5 Coupling capacitors in a
multistage amplifer.
VCC

CC3
CC1
CC2

11
Fig 9.6a AC coupling.

GND

CC3
CC1
CC2

12
Fig 9.6b DC isolation.

VCC

CC3
CC1
CC2

13
Fig 9.7 Capacitive vs. direct
coupling.
VCC             VCC

CC

14
Fig 9.8-9 Bypass capacitors.
VCC               GND                  VCC

CB                     CB                     CB

For AC analysis        For DC analysis
15
Fig 9.10 Typical common-emitter
amplifier signals.
VCC      1.8V
5.6V
5.6V
1.8V
0V
0V

1.1VDC   1.1VDC
16
Fig 9.11 Deriving the CE ac-
equivalent circuit.

R1   RC                  R1     RC   R1         RC

RS = 0 
VCC   (ideally)

R2   RE                  R2     RE   R2         RE

(a)                      (b)         (c)

17
Fig 9.12a Example 9.2.
VCC

RC
R1           CC2

CC1
Q1               RL

R2
RE      CB

18
Fig 9.12b Example 9.2.

GND
RC
R1

Q1          RL

R2
RE

19
Fig 9.12c-d Example 9.2.

Q1    RC      RL
(c)
R1   R2

Q1   RC||RL
(d)
R1||R2

20
Voltage Gain of CE
Q1           RC||RL

R1||R2

vout      vout  ic  RC RL 
ic = ib    RC||RL         ic rC where rC  RC RL
vin      ib
Q1                    vin  ie re
r'e
vout   ir      r
R1||R2                              Av           c C  C
vin     ie re re
21
Fig 9.13 Example 9.4. (1)
+20 V          Transform the base circuit to its
Thevenin equivalent.
R2           20kΩ
Vth  VCC            20V
R1       RC                                    R1  R2       170kΩ
150k       12k                          2.353V
Rth  R1 R2  20kΩ 150kΩ
RL              17.65kΩ
hFE = 200      50k
R2
RE                       Vth  VBE          2.353V  0.7V
20k                      IB                      
2.2k                Rth  (hFE  1) RE 17.65kΩ  201 2.2kΩ
 3.595μA

22
Fig 9.13 Example 9.4. (2)
+20 V          I C  hFE I B  200  3.595μA  718.9μA
I E   hFE  1 I B  201 3.595μA  722.5μA

VCE  VCC  I C RC  I E RE
R1       RC
 20V  718.9μA 12kΩ  722.5μA  2.2kΩ
150k       12k
 9.784V      active 
RL         25mV    25mV
re                 34.6Ω
hFE = 200   50k         IE   722.5μA
rC  RC RL  12kΩ 50kΩ  9.677kΩ
R2
20k        RE                         rC     9.677kΩ
2.2k             Av                   279.7
re     34.6Ω

23
Fig 9.14 Example 9.5. (1)
+10 V
Transform the base circuit to its
Thevenin equivalent.
R2           4.7kΩ
RC                     Vth  VCC            10V
R1                                           R1  R2       22.7kΩ
1.5k
18k                                   2.070V
Rth  R1 R2  4.7kΩ 18kΩ
RL
hFE = 30                    3.727kΩ
hfe = 200    5k
R2
4.7k       RE                     Vth  VBE         2.070V  0.7V
IB                      
1.2k              Rth  (hFE  1) RE 3.727kΩ  311.2kΩ
 33.49μA

24
Fig 9.14 Example 9.5. (2)
+10 V          I C  hFE I B  30  33.49μA  1.005mA
I E   hFE  1 I B  31 33.49μA  1.038mA

VCE  VCC  I C RC  I E RE
RC              10V  1.005mA 1.5kΩ  1.038mA  1.2kΩ
R1        1.5k
18k                        7.247V      active 

hFE = 30       RL re  25mV  25mV  24.08Ω
5k        IE  1.038mA
hfe = 200              rC  RC RL  1.5kΩ 5kΩ  1.154kΩ
R2
4.7k       RE                 Av  
rC

1.154kΩ
 47.91
1.2k                       re     24.08Ω

25
CE Current Gain
iout
Ai 
iin

Ai is always less than hfe due to two factors:
1.The input ac current is divided between
the transistor and the biasing network.
2.The output collector current is divided
between the collector resistor and the

26
Power Gain
Ap  Ai Av

Example 9.7 The amplifier shown in Fig. 9.5 has
values of Av = 45.3 and Ai = 20. Determine the
power gain (Ap) of the amplifier and the output
power when Pin = 80 W.

Ap  AAv  2  4 . 0
i     0 53 9 6

Pot  A P  9 6 8 μ  7 . 8 W
u     pi
n   0 0 W 24 m

27

RC      RL                         RC     RL
3k     12k                       3k    6k

R1||R2   r'e=25                   R1||R2   r'e=25

rC  RC RL  2.4kΩ                 rC  RC RL  2kΩ
rC                                rC
Av    96                      Av    80
re                               re

The lower the load resistance is, the lower the voltage gain.
28
Example 9.8
The load in Fig. 9.16 is open. Calculate the
open-load voltage gain of the circuit.

rC  RC  3kΩ

rC   3kΩ
Av          120 (max. gain)
re  25Ω

29
Calculating Amp.
Input Impedance
VCC
Z in  R1 R2 Z in(base)
R1

DC: RIN(base)   hFE  1 RE
 hFE RE
R2
Zin                 Zin(base)   AC: Z in(base)   h fe  1 re
 h fe re

Biasing circuit
30
Fig 9.17 Example 9.9.
+10 V

RC
R1                    1.5k
18k

hFE = 30       RL          Z in(base)  h fe re
5k
hfe = 200
 200  24Ω  4.8kΩ
R2
Zin             Zin(base)      RE
4.7k
1.2k                     Zin  R1 R2 Zin(base)
 18kΩ 4.7kΩ 4.8kΩ
Q1
RC       RL          2.1kΩ
1.5k    5k
R1       R2
Zin                     Zin(base)
18k    4.7k
r'e = 24
hfe = 200                                                 31
Calculating the Value of Ai
iout
ic
iin        ib
Q1        RC v            RL
out
vin
R1    R2
Zin               Zin(base)

i                    i                                vin  iin Zin  ib Zin(base)
Ai  out           h fe  c
iin                 ib                                     ib Zin(base)       ib  h fe  1 re
 iin                    
vout  ic  RL RC   ic rC  iout RL                             Zin                  Zin
ic rC                                                        Z in rC
 iout                                                Ai   h fe
RL                                                       Z in(base) RL
32
Example 9.10
Calculate the value of Ai for the circuit shown in Fig.
9.17.

RC          RL
Q1
1.5k       5k
R1        R2
Zin                       Zin(base)
18k     4.7k
r'e = 24
hfe = 200

Z in rC
Z in(base)  h fe re  200  24Ω  4.8kΩ     Ai  h fe
Z in(base) RL
Z in  R1 R2 Z in(base)  2.1kΩ
2.1kΩ 1.15kΩ
 200 
rC  RC RL  1.15kΩ                                            4.8kΩ  5kΩ
 20.2                     33
Multistage Amp.
Gain Calculations
AvT  Av1 Av 2 Av 3     Procedure:

AiT  Ai1 Ai 2 Ai 3     1. Do dc analysis
2. Find r’e for each stage
ApT  AvT AiT
3. Find rC for each stage
4. Using r’e and rC to find Av
for each stage

Input impedance of next stage is the load of current stage.
(Zin of next stage is RL of current stage)
34
Fig 9.18 Example 9.11. (1)
+15V
R3                           R7
R1       5k              R5          5k
CC1    22k                      15k

Q1   CC2                     Q2 CC3             RL
10k
R2       R4               R6         R8
CB1                         CB2 hFE = 150
3.3k    1k              2.5k      1k
hfe = 200

Determine Av of the 1st          The input impedance of the 2nd stage:
stage. Assume that r’e
for the 1st stage is 19.8     Zin(base)   h fe  1 re  20117.4  3.497kΩ
and r’e for the 2nd stage is
found to be 17.4 . For        Z in  R5 R6 Z in ( base )  1.329kΩ
the 2nd stage, hfe is 200.
35
Fig 9.18 Example 9.11. (2)
+15V
R3                         R7
R1        5k              R5        5k
CC1    22k                       22k

Q1   CC2                  Q2 CC3         RL
10k
R2        R4               R6      R8
CB1                     CB2 hFE = 150
3.3k     1k              3.3k   1k
hfe = 200

rC  R3 Zin  5kΩ 1.33kΩ=1.05kΩ

Finally, Av for the 1st stage is found as

rC     1.05kΩ
Av                  53.03
re    19.8Ω
36
Example 9.12. (1)
Determine the value of AvT for the amplifier in Figure
9.18.
rC for the 2nd stage can be found as

rC  R7 RL  3.33kΩ

Av for the 2nd stage is found as
rC   3.33kΩ
Av             191.38
re  17.4Ω

AvT  Av1 Av 2   53.03 191.38  10.15 103

37
Fig 9.19 The swamped CE amplifier
and its ac equivalent ckt.
+VCC              Swamped amplifier is an
amplifier that uses a partially
bypassed emitter resistance to
RC           increase ac emitter resistance.
R1                 Also referred to as a gain-
C1                      stabilized amplifier.
Q1 C2    RL

R2         rE                               rC
Q1
R1//R2
RE           CB                   rE

38
Av of Swamped Amp.
rC
Q1
vout    ic rC
Av       
R1//R2
rE                           vin ie  re  rE 

rC
vout      Av  
re  rE
ic = acib    rC
vin                  Q1
R1//R2    ib    r'e

rE

39
Fig 9.20 Example 9.13. (1)
+10V
R2          4.7kΩ
Vth  VCC            10V         2.070V
R1  R2       22.7kΩ
RC           Rth  R1 R2  18kΩ 4.7kΩ  3.727kΩ
R1        1.5k
18k                                               Vth  VBE
IB 
hFE = 200     RL                 Rth   hFE  1 rE  RE 
hfe = 150     10k
2.070V  0.7V
R2        rE                          
4.7k     300                          3.727kΩ   2011210Ω 
 5.550μA
RE
910
CB               I C  hFE I B  200  5.550μA
 1.110mA
I E   hFE  1 I B  201 5.550μA
 1.116mA                          40
Fig 9.20 Example 9.13. (2)
+10V       VCE  VCC  I C RC  I E  rE  RE 
 10V  1.110mA 1.5kΩ  1.116mA  1210Ω 

RC            6.985V (active)
R1        1.5k
18k                                    25mV     25mV
re                   22.41Ω
hFE = 200      RL                IE   1.116mA
hfe = 150      10k
R2        rE                    rC  RC RL  1.304kΩ
4.7k     300
rC         1.304kΩ
RE                    Av                              4.046
910
CB              re  rE    22.41Ω  300Ω

41
Example 9.14
Determine the change in gain for the amplifier in Example
9.13 when r’e doubles in value.

rC         1.304kΩ
Av                              3.782
re  rE    44.82Ω  300Ω

Av  4.046  3.782  0.2639

Swamping improves the gain stability of a CE
amplifier when rE >> r’e.

42
The Effect of Swamping on Zin
c

b
Z in(base)   h fe  1 re
r'e
Zin(base)
 h fe re                             e           c

b
Zin(base)   h fe  1  re  rE 
r'e
Zin(base)
 h fe  re  rE                              e
rE
43
Fig 9.22 Gain stabilization.

rE

RE                           RE

Av                        -rC / r’e               -rC / (r’e+rE)

Zin(base)                  hfer’e                  hfe(r’e+rE)

Advantage      Higher values of Av than     Relatively stable Av.
the swamped amplifier.       Much smaller distortion.
Disadvantage   Relatively unstable values   Lower Av than the
of Av.                       standard amplifier.
44
The Hybrid Equivalent Model

I1                      I2
1                                           2
Two-Port
V1          System                V2

1'                                          2'

Hybrid model is derived from two-port system.

45
Six Circuit-Parameter Models
for Two-Port Systems
Independent   Dependent    Circuit Parameters
Variables    Variables

I1, I2       V1, V2     Impedance Z

V1, V2        I1, I2    Admittance Y

V1, I2       I1, V2     Inverse Hybrid g

I1, V2       V1, I2     Hybrid h

V2, I2       V1, I1     Transmission T

Inverse Transmission
V1, I1       V2, I2     T’
46
Equations for Hybrid Model
V1  h11I1  h12V2
I 2  h21I1  h22V2

Let V1 = Vi, I1 = Ii, V2 = Vo, and I2 = Io.
Then
Vi  h11Ii  h12Vo
I o  h21Ii  h22Vo

47
Equivalent Circuit for
Hybrid Model

Ii                                           Io
1                                                                2
hi
Vi           hrVo               hfIi      ho           Vo

1'                                                               2'

Vi  h11I i  h12Vo  hi I i  hrVo
I o  h21I i  h22Vo  h f I i  hoVo
48
h-Parameters
Vi                    Vi
h11                  h12 
Ii   Vo  0           Vo    Ii  0
Io                     Io
h21                   h22 
Ii   Vo  0            Vo   Ii  0

h11 = hi = Input Resistance
h12 = hr = Reverse Transfer Voltage Ratio
h21 = hf = Forward Transfer Current Ratio
h22 = ho = Output Admittance
49
h-Parameters for CE Amp.
•   hie = the base input impedance
•   hfe = the base-to-collector current gain
•   hoe = the output admittance
•   hre = the reverse voltage feedback ratio

vbe  hieib  hre vce
ic  h feib  hoe vce

50
Hybrid Model for
CE Configuration
ib                                     ic
b                                               c
ic                 hie
ib                 c                               hfeib   hoe
b                                        hrevce

ie                vbe                                   vce
ie
e
e
vin                                  ic
hie      (output shorted)           hoe      (input open)
ib                                   vce                  May be
neglected.
ic                                  vbe
h fe     (output shorted)           hre      (input open)
ib                                  vce
51
h-parameters of 2N3904

52
Hybrid Model without hre and hoe
b                                          c
ib                      ic
hie              hfeib               h fe   ac

hie   hfe  1 re  hfe re  Z in(base)
ie

c                                     c
ic                                            ic                   h fe rC
acib                                         acib       Av  
ac+1)r'e                              hie
b                                          b
ib                                     ib
r'e                                                                    Zin rC 
ie                                            ie           Ai  h fe         
 hie RL 
e                                             e
53
Determining h-Parameter Values

Use geometric means if given max. and min.
values.

hie  hie(min)  hie(max)

h fe  h fe(min)  h fe(max)

See examples 9.18 and 9.19.

54
Summary
• AC concepts
• Roles of capacitors in amplifiers
• Common-emitter ac equivalent
circuit
• Amplifier gain
• Gain and impedance calculations
• Swamped amplifiers
• h-parameters

55

```
Other docs by gjjur4356
Chapter 82011455721