Chapter 10 Other BJT Amplifiers

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```					          Chapter 10
Other BJT Amplifiers
Common-Collector & Common-Base Amp.
Pictures are redrawn (with some modifications) from
Introductory Electronic Devices and Circuits
By
Robert T. Paynter

1
Objectives
• Perform a complete dc analysis of an emitter
follower.
• Describe the use of the emitter follower as a
current amplifier and as a buffer.
• Perform a complete dc analysis of a common-
base amplifier.
• Describe the use of the common-base amplifier
as a voltage amplifier and a buffer.
• List the overall gain and impedance
characteristics of all commonly used BJT
amplifiers.

2
Fig 10.1 The emitter follower (common-
collector amplifier).
VCC 1. Convert voltage divider circuit to its
DC                        corresponding Thevenin equivalent circuit:
Analysis:                                    R2
Vth  VCC
R1  R2
R1
1V      CC1                      Rth  R1 R2
2. Find IB and IE.
CC2                0.95V
Vth  0.7V
IB 
Rth   hFE  1 RE
I E   hFE  1 I B

3. Check VCEQ whether the transistor is active or not.
VCEQ  VCC  I E RE
3
Fig 10.2 Example 10.1.
10V            Step 1:
R2               20kΩ
Vth  VCC            10V               5V
R1  R2         20kΩ  20kΩ
Rth  R1 R2  20kΩ 20kΩ  10kΩ
R1
CC1   20k                         Step 2:
hFE = 150
Vth  0.7V
IB 
CC2                     Rth   hFE  1 RE
5V  0.7V
R2     RE                                              5.621μA
RL            10kΩ  151 5kΩ
20k   5k                  I E   hFE  1 I B  151 5.621μA
 848.8μA

Step 3:
VCEQ  VCC  I E RE  10V  848.8μA  5kΩ  5.756V
4
Fig 10.3 Example 10.2.
Derive and draw the dc load line for the circuit
shown in Figure 10.2.

IC(mA)                                                              VCC 10V
I C  sat             2mA
RE 5kΩ

3                      VCC                         VCE (off )  VCC  10V
I C (sat )   
2                      RE
1                               VCE (off )  VCC
VCE(V)
2    4       6   8 10

5
Fig 10.5 Emitter follower and
its ac equivalents.
12V

ib          ic = hfeib
R1
CC1   25k
vin          R1||R2          r'e
CC2
vout
R2     RE      RL
rE  RE RL
33k   2k     5k

vout    r
Av          E
vin re  rE
1      rE   re
6
Ai of CC Amp.
From CE amp.,

Z in rC
Ai   h fe                     rC  RC     RL 
Z in(base) RL

So for CC amp.,

Z in rE
Ai  h fc                      rE  RE    RL 
Z in(base) RL
where

h fc  h fe  1 (from Appendix C)
Z in(base)  h fc (re  rE )  h fe (re  rE )
Z in  R1 R2 Z in(base)                       7
Example 10.5.
Determine the input impedance of the amplifier
shown in Figure 10.5a. Assume that the value of
hfe for the transistor is listed on the spec sheet as
220.
R2             33kΩ
VB  VCC            12V              6.828V
R1  R2       25kΩ  33kΩ
VE  VB  0.7V  6.828V  0.7V  6.128V
VE 6.128V
IE            3.064mA
RE   2kΩ

re 
25mV

25mV
 8.16Ω        Z in(base)  h fc (re  rE )   h fe  1 (re  rE )
IE   3.064mA
 2211.437kΩ  317.5kΩ
rE  RE RL  2kΩ 5kΩ  1.429kΩ
Z in  R1 R2 Z in(base)
rE    1.429kΩ
Av                    0.9943                  25kΩ 33kΩ 317.5kΩ
re  rE 1.437kΩ
 13.61kΩ                                       8
Zout of CC Amp.
RS        ib
+8V                                        b              c

R1||R2      hie        ic = hfeib   h fc  h fe  1  h fe
R1                                                                           hrc  1  hre  1
e
RE       Zout     hic  hie
RS                                                                                   hoc  hoe
R2                              RS        ib
RE                                   b                    e
vS                        Zout
R1||R2       hic        ie = hfcib       RE
Zout

hrcvec
      R                      Rin 

Z out  RE     re  in      RE    re           
      h fc                  h fe  1              c
                                     
where          
Rin  RS R1 R2
9
Fig 10.6 Example 10.6.
Determine the output impedance (Zout) of the
amplifier shown in Figure 10.6.
+8V                      hie    hic 4kΩ
re                     20Ω
h fe  1 h fc 200

R1                     
Rin  R1 R2 RS
hic = 4k
3k       hfc = 200        3kΩ 4.7kΩ 600Ω  451.9Ω
hFC = 180
RS                                               Rin 

Z out    RE  re       
600   R2                                        h fc 
4.7k     RE                                   
390                                   451.9Ω 
 390Ω  20Ω              
           200 
 21.06Ω                       10
Applications for CC Amp.
• Current amplifier (Av  1)
• Buffer – a circuit used to compensate
for an impedance mismatch between

11
Fig 10.12 Using a buffer to overcome an
impedance mismatch.
Zout
10k

ZL
12V
1k

Zout              Av = 0.98
10k

Zin         Zout          ZL
12V
180k       20           1k

Fig 10.14 Darlington emitter-
follower. (1)
VCC
DC Analysis:
IC1+IC2            VB1  2VBE
IC1                       IE2   
R1                                           RE
IB1                                   I E1         I
Q1     IC2        I B1          , IB2  E 2
hFC1         hFC 2
R2
Q2         Rin 2  hFC 2  re2  RE   hFC 2 RE
IE1=IB2               Rin1  hFC1 (re1  Rin 2 )
IE2                     hFC1 Rin 2  hFC1hFC 2 RE
RE         RL

13
Fig 10.14 Darlington emitter-
follower. (2)
VCC
AC Analysis:
IC1+IC2
IC1                     Zin(base)  hic1  h fc1  hic 2  h fc 2rE 
R1
IB1                          Z in  R1 R2 Z in(base)
Q1    IC2
        re1   Rin / h fc1  

R2                                   Zout  RE        re2                         
R’in                                                     h fc 2         
Q2                                                        
IE1=IB2
Z in rE

Rin  R1 R2 RS           IE2              Ai  h fc1h fc 2
Z in(base) RL
RE         RL
rE  RE RL

14
Fig 10.16 Typical common-
base amplifier.
+VCC

RC          vin                                 vout
RE                RC
R1
vin                        vout                                +VCC
RE
CB     R2

-VEE

Emitter bias                  Voltage-divider bias

15
Fig 10.17 The common-base ac
equivalent circuit.
iin    ie                                      iout
vout

vin                RE         r'e         ic = hfeib   RC           RL
Zin                                                Zout
ib
rC  RC RL

Av 
vout ic rC rC
                    iout    vout / RL     rC  RE re
Ai                     
vin   ie re re          iin vin /  RE re re    RL
Zin  re RE  re                    RC RL 1 RE re 1                   RC RE
                               
 RC  RL  re  RE  re RL  RC  RL  RE  re
1
Z out  RC         RC               RC
hob                
RC  RL                                                 16
Fig 10.18 Example 10.10.
Determine the gain and impedance values for the
circuit shown in Figure 10.18.
+10V
Zin  re  75.58Ω
RS
RC              Zout  RC  10kΩ
10k
75
rC  RC RL  10kΩ 5.1kΩ
RL
100V                RE
5.1k        3.377kΩ
13k
r    3.377kΩ
Av  C            44.69
-5V                                  re 75.58Ω
0.7V   VEE        0.7V  5V
IE                                       330.8μA
RE                 13kΩ
25mV    25mV
re                   75.58Ω
IE   330.8μA                                                            17
Common-Base Applications

• Voltage amplification without current
gain
• High-frequency applications (more
common)

18
BJT Amplifiers: A Summary

CE              CC*         Darlington CC      CB

Av    Midrange            <1                <1         Midrange

Ai    Midrange         Midrange       Extremely high     <1

Ap       High          Midrange             High       Midrange

Zin   Midrange           High         Extremely high     Low
Zou   Midrange            Low               Low          High
t

*Another name for CC is emitter follower.

19
Summary
• The emitter-follower (common-
collector amplifier.)
• Emitter-follower ac analysis.
• Emitter-follower applications.
• Darlington emitter-follower.
• The common-base amplifier and their
applications.

20

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