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AC Analysis of
BJT and MOSFET
Inverting Amplifiers
Assoc Prof Chang Chip Hong email: echchang@ntu.edu.sg
EE2002 Analog Electronics
EE2002 Analog Electronics
Lesson Objectives
AC Analysis of BJT and MOSFET Inverting Amplifiers 2
At the end of this lesson, you should be able to:
• Draw small signal model for BJT and MOSFET
• Calculate the small signal parameters of BJT and MOSFET
• Construct AC equivalent circuit of BJT and MOSFET inverting amplifiers
• Calculate the following performance characteristics of C-E and C-S amplifiers
- Voltage gain
- Input resistance
- Output resistance
EE2002 Analog Electronics
Hybrid-Pi Model of BJT
AC Analysis of BJT and MOSFET Inverting Amplifiers 3
• This hybrid-pi small-signal model is
the intrinsic low-frequency
representation of the BJT.
• Small-signal parameters are
controlled by the Q-point and are
independent of geometry of BJT.
Transconductance:
Output resistance:
ic
gmvbe
ib
E
CB
vbe rp ro vce
40C
m C
T
I
g I
V
 
where 25 mV@25 CT
kT
V
q
  
Input resistance:
m
r
g
p


–
+
–
+
A CE A
o
C C
V V V
r
I I

  if
EE2002 Analog Electronics
Equivalent Forms of Small-Signal
Model for BJT
AC Analysis of BJT and MOSFET Inverting Amplifiers 4
Voltage-controlled current source gmvbe can be transformed into current-controlled
current source.
gmvbe
ib
E
CB
vbe
rp ro
ic
 ib
ib
E
CB
vbe rp ro
ic
Basic relationship ic=  ib is useful
for both dc and ac analysis when
BJT is in forward-active region.
be b
m be m b b
ce
c b b
o
v i r
g v g i r i
v
i i i
r
p
p 
 

 
  
–
+
–
+
EE2002 Analog Electronics
VCC
E
Q
RC
RE
C1  
C2  
C3  
B
C
RB
RC
RI
vi
+
-
RL
RI
RB2
RB1
vi
+
-
Q
C
B
E RL
+
vo
-
1 2B B BR R R
• The ac equivalent circuit is
constructed by assuming that all
capacitances have zero
impedance at signal frequency and
dc voltage source is ac ground.
• Assume that Q-point has already
been calculated from DC analysis.
Hence, gm, rp , and ro of BJT can
be calculated.
AC Analysis of BJT and MOSFET Inverting Amplifiers 5
Small Signal Analysis of C-E Amplifier
with Fully Bypass RE
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 6
C-E Amplifier with Fully Bypass RE:
Voltage Gain
c o m be L
vt m L
b be be
v v g v R
A g R
v v v
-
    -
Terminal voltage gain between
base and collector is:
Overall voltage gain from source vi
to output voltage vo across RL is:
o c b b
v vt
i b i i
v v v v
A A
v v v v
    
RI
vi
+
- RB ro RL
+
vo
-
B C
E
rp
RC
gmvbe
+
vbe
-
0L C L
R r R R 
B
v m L
I B
R r
A g R
R R r
p
p
   -   
B
R rp
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 7
C-E Amplifier with Fully Bypass RE:
Input Resistance
C1  
VCC
1 2B B BR R R
RI
RB2
vi
Q
RB1
RC C2  
C3  
+
-
RL
Rin
vx
+
-
ix
RE
 x x B
x
in
x
B
v i R r
v
R
i
R r
p
p



RB rp ro RL
+
vo
-
B C
E
RC
gmvbe
+
vbe
-
B
R rp
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 8
C-E Amplifier with Fully Bypass RE:
Output Resistance
C1  
0
x x
x m be
C o
x x
be x
C o
v v
i g v
R r
v v
v i
R r
  
   
VCC
RI
Q
RB2
RB1 C2  
C3  
RL
Rour
1 2B B BR R R
vx
+
-
RB ro
RI
B C
E
rp RC
gmvbe
+
vbe
-
ix
RE
RC
vi
+
-
1
1 1x
out C o
x C o
out C
v
R R r
i R r
R R
-
 
    
 
 if
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 9
C-E Amplifier with Fully Bypass RE:
Example
Problem: Find voltage gain, input
and output resistances.
Given:  = 65, VA = 50 V
Find the Q-point from dc equivalent circuit:
 3 3
100 10 0.7 66 16 10 ( 5) 0B BI I     - 
3.71 μA
65 241 μA
66 245 μA
B
C B
E B
I
I I
I I

 
 
 5 10 k 241 μ 16 k 245 μ 5 0CEV-  - -  - - 
+5 V
10 k
voRout
RC
C3  
C1  
Rin
vi
+
–
220 k
RB
100 k
RE
16 k
330 
RI
–5 V
RL
C2  
+5 V
10 k
IC
VCE
+
–
IE
16 k
–5 V
100 k VBE
+
–
IB
3.67 VCEV 
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 10
AC Analysis of C-E Amplifier with
Fully Bypass RE
9.64 (223 10 220) 6.23 k
84.0
330 6.23 k
bein
v vt
I in be
vR
A A
R R v
 -     
   -   
     
40 241 μ 9.64 mSmg   
65
6.64 k
9.64 m
rp   
50 3.67
223 k
241 μ
or

  
100 k 6.64 k 6.23 kinR   
223 k 10 k 9.57 koutR   
330 
vi
+
-
100k
ro
220k
+
vo
-
B C
E
rp
10k
gmvbe
+
vbe
-
Rin
100k
223k
220k
B C
E
6.64k
10k
9.64mvbe9.64mvbe
Rout
100k
223k
330
B C
E
6.64k
10k
Small signal equivalent
EE2002 Analog Electronics
Small Signal Parameters of MOSFET
AC Analysis of BJT and MOSFET Inverting Amplifiers 11
• Since gate is insulated from channel
by gate-oxide, input resistance = .
• Small-signal parameters are
controlled by the Q-point.
• MOSFET transconductance is
geometry dependent.
   
  
2
1
2
1
n
D GS TN DS
D
m n GS TN DS
GS Q pt
K
i v V v
i
g K V V V
v


-
 - 

  - 

n n OX
W
K C
L
    
 
where
+
gmvgs
ig
S
DG
vgs ro
id
vds
–
+
–
2
2
1
1
D
m n D
GS TN
DS
o
D D
I
g K I
V V
V
r
I I


 
-

  if
1

EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 12
Small Signal Analysis of C-S
Amplifier with Fully Bypass RS
VDD
S
• AC equivalent circuit is constructed by
assuming that all capacitances have zero
impedance at signal frequency and dc
voltage sources represent ac grounds. M
RD
RS
C1  
C2  
C3  
G
D
RG
RD
RI
vi
+
-
RL
RI
RG2
RG1
vi
+
-
D
MG
S RL
+
vo
-
1 2G G GR R R
vo
• The small signal parameters, gm and ro of
the MOSFET is calculated at the Q-point,
ID and VDS.
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 13
C-S Amplifier with Fully Bypass RS:
Voltage Gain
Terminal voltage gain between
gate and drain is:
Overall voltage gain from source vi
to output voltage vo across RL is:
RI
vi
+
-
RG ro RL
+
vo
-
G D
S
RD
gmvgs
+
vgs
-
0L D LR r R R 
m gs Ld
vt m L
g gs
g v Rv
A g R
v v
-
   -
g go o
v vt
i g i i
G
m L
I G
v vv v
A A
v v v v
R
g R
R R
    
 
 -  
 
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 14
C-S Amplifier with Fully Bypass RS:
Input Resistance
RG2
M
RG1
RD
RS
C2  
C3  
RL
Rin
C1  
RG ro RL
+
vo
-
G D
S
RD
gmvgs
+
vgs
-
1 2G G GR R R
VDD
RI
vi
+
-
x x G
x
in
x
G
v i R
v
R
i
R



vx
+
-
ix
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 15
C-S Amplifier with Fully Bypass RS:
Output Resistance
vi
+
-
VDD
C1  
Since vgs = 0, gmvgs = 0.
RI
RG2
RG1 C2  
Rour
RS
RD
RL
M
gmvgs
vx
+
-
RG ro
RI
G D
S
RD
+
vgs
-
ix
C3  
21 GG G
R R R
 x x D o
x
out D o
x
out D
v i R r
v
R R r
i
R R

 
 if
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 16
C-S Amplifier with Fully Bypass RS:
Example
Problem: Find voltage gain, input and
output resistances.
Given: Kn = 500 A/V2, VTN = 1V,
 = 0.0167 V–1
DC Analysis:
61
1
6
5 10
2 10
0.4
S
GS
D
DS
I
IV
V
V


  

5 , 2 , 250 μADS GS DV V V V I   
 
2500 μ
1
2
GSD VI  - (1)
 3
110 20 10DS D IV I -   (2)
+10 V
20 k
C3  
vo
+
–
Rout
100 k
RL
RD
RG3
1 M
RG2 2 M
RG1 2 M
C2  
C1  
10 k
RI
Rin
vi
+10 V
20 k
RG3
1 M
RG2
2 M
RG1
2 M
vDS
+
–
+
–
vGS
I1
ID
IG
RD
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 17
C-S Amplifier with Fully Bypass RS:
Example
   0.5 260 k 1 M 20 k 100 k 1 M
7.93
10 k 1 M
gsin
v vt
I in gs
m vR
A A
R R v
-   
   -     
2 2 1inR M M M  
2 500 μ 250 μ 0.5 mSm
g    
1
5
0.0167 260 k
250 μ
or

  
10k 
vi
+
-
2M
ro
100k
+
vo
-
G
D
S
1M
gmvgs
+
vgs
-
2M
20k
Rin
1M
260k
16.4k
G D
S
0.5mvgs
0.5mvgs
S
Rout
1M
260k
10k
G D
1M
20k
260 1 20 18.2outR k M k k  
16.4 k1 M
Small Signal Equivalent Circuit:
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 18
C-E Amplifier with Unbypass RE:
Terminal Voltage Gain
VCC
RB2
RB1
RC
RE2 C3  
RI
vi
+
-
RL
C2  
C1  
vo
RE1
Q
vi
+
- RB
RL
+
vo
-
RC
ro
B C
E
rp
gmvbe
+
vbe
-
RI
RE1
ic
L C LR R R 
1 1 11
c c L m be L m L
vt
b be e E be m be E m E
v i R g v R g R
A
v v i R v g v R g R
  - - -
   
  
ie
1 2B B BR R R
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 19
C-E Amplifier with Unbypass RE:
Input Resistance
vi
+
-
RI
 
 
1
1
1
1
b b b E
b
in E
b
v i r i R
v
R r R
i
p
p


  
    
Rin
vb
+
- RB
RL
+
vo
-
RC
ro
B C
E
rp
ib
+
vbe
-
RE1
(+1)ib
ib
Rin
in in BR R R
inR
RB
RL
+
vo
-
RC
ro
B C
E
rp
gmvbe
+
vbe
-
RE1
1 2B B BR R R
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 20
C-E Amplifier with Unbypass RE:
Overall Voltage Gain
+
vb
-
Rin
RB = RB1 || RB2
RL
+
vo
-
RC
ro
B C
E
rp
gmvbe
+
vbe
-
RE1
Rin
VCC
RB2
RB1
RC
RE2 C3  
RI
vi
+
-
RL
C2  
C1  
vo
RE1
Q
o o b in
v vt
i b i I in
v v v R
A A
v v v R R
    

vi
+
-
RI
Rin
vi
+
-
RI
+
vb
-
Rin
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 21
C-E Amplifier with Unbypass RE:
Output Resistance
Rth
RC
ro
B C
E
rp
gmvbe
+
vbe
-
RE1
vx
+
-
ix
Rout
1 2th I B BR R R R
ix
  1
be e
th
x th E
th
r
v v
r R
r
i r R R
r R
p
p
p
p
p
 
 -  
 
 -   
  1e x th Ev i r R Rp 
 x m bex o ev g v r vi -
current through ro
outR
RL
+
vo
-
RI
RB
RC
ro
B C
E
rp
gmvbe
+
vbe
-
RE1
1 2B B BR R R
Rout
vi
+
-
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 22
C-E Amplifier with Unbypass RE:
Output Resistance
 
  
  
1
1
x x m be o e
x m x th E o
th
x th E
v i g v r v
r
i g i r R R r
r R
i r R R
p
p
p
p
 - 
  
    
  
 Rth
ro
B C
E
rp
gmvbe
+
vbe
-
RE1
vx
+
-
ix
ix
out out CR R R
RC
Rout
  11x
out m th E o
x th
v r
R g r R R r
i r R
p
p
p
          
outR
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 23
C-S Amplifier with Unbypass RS:
Terminal Voltage Gain
VDD
RG2
RG1
RD
RS2 C3  
RI
vi
+
-
RL
C2  
C1  
vo
RS1
vi
+
- RG
RL
+
vo
-
RD
RI
RS1
id  gmvgs
L D LR R R 
1 1 11
m gs Ld d L m L
vt
g gs d S gs m gs S m S
g v Rv i R g R
A
v v i R v g v R g R
- - -
   
  
id
1 2G G GR R R
ro
G D
S
gmvgs
+
vgs
-
M
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 24
C-S Amplifier with Unbypass RS:
Input Resistance
RG = RG1 || RG1
RL
+
vo
-
RD
RS1
Rin
in GR R
ro
G D
S
gmvgs
+
vgs
-vi
+
-
RI
Rin
RG
RL
+
vo
-
RD
RS1
1 2G G GR R R
ro
G D
gmvgs
+
vgs
-
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 25
C-S Amplifier with Unbypass RS:
Overall Voltage Gain
VDD
RG = RG1 || RG2
RL
+
vo
-
RD
ro
G D
S
gmvgs
+
vgs
-
RS1
Rin
RG2
RG1
RD
RS2 C3  
RI
vi
+
-
RL
C2  
C1  
vo
RS1
vi
+
-
RI
go o in
v vt
i g i I in
vv v R
A A
v v v R R
    

Rin
vi
+
-
RI
+
vg
-
Rin
M
EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 26
C-S Amplifier with Unbypass RS:
Output Resistance
Rth
RD
RS1
vx
+
-
ix
Rout
1 2th I G GR R R R
ix
1
10,
s x S
g gs s x S
v i R
v v v i R

  -  -
ro
G D
S
gmvgs
+
vgs
-
 1 1x x m x S o x Sv i g i R r i R  
 11x
out m S o
x
v
R g R r
i
   
out out DR R R
 x m gsx o sv g v r vi -
current through ro
outR
RL
+
vo
-
RI
RG
RD
RS1
1 2G G GR R R
ro
G D
gmvgs
+
vgs
-
Rout
vi
+
-

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Ac analysis of bjt and mosfet inverting amplifiers

  • 1. AC Analysis of BJT and MOSFET Inverting Amplifiers Assoc Prof Chang Chip Hong email: echchang@ntu.edu.sg EE2002 Analog Electronics
  • 2. EE2002 Analog Electronics Lesson Objectives AC Analysis of BJT and MOSFET Inverting Amplifiers 2 At the end of this lesson, you should be able to: • Draw small signal model for BJT and MOSFET • Calculate the small signal parameters of BJT and MOSFET • Construct AC equivalent circuit of BJT and MOSFET inverting amplifiers • Calculate the following performance characteristics of C-E and C-S amplifiers - Voltage gain - Input resistance - Output resistance
  • 3. EE2002 Analog Electronics Hybrid-Pi Model of BJT AC Analysis of BJT and MOSFET Inverting Amplifiers 3 • This hybrid-pi small-signal model is the intrinsic low-frequency representation of the BJT. • Small-signal parameters are controlled by the Q-point and are independent of geometry of BJT. Transconductance: Output resistance: ic gmvbe ib E CB vbe rp ro vce 40C m C T I g I V   where 25 mV@25 CT kT V q    Input resistance: m r g p   – + – + A CE A o C C V V V r I I    if
  • 4. EE2002 Analog Electronics Equivalent Forms of Small-Signal Model for BJT AC Analysis of BJT and MOSFET Inverting Amplifiers 4 Voltage-controlled current source gmvbe can be transformed into current-controlled current source. gmvbe ib E CB vbe rp ro ic  ib ib E CB vbe rp ro ic Basic relationship ic=  ib is useful for both dc and ac analysis when BJT is in forward-active region. be b m be m b b ce c b b o v i r g v g i r i v i i i r p p          – + – +
  • 5. EE2002 Analog Electronics VCC E Q RC RE C1   C2   C3   B C RB RC RI vi + - RL RI RB2 RB1 vi + - Q C B E RL + vo - 1 2B B BR R R • The ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage source is ac ground. • Assume that Q-point has already been calculated from DC analysis. Hence, gm, rp , and ro of BJT can be calculated. AC Analysis of BJT and MOSFET Inverting Amplifiers 5 Small Signal Analysis of C-E Amplifier with Fully Bypass RE
  • 6. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 6 C-E Amplifier with Fully Bypass RE: Voltage Gain c o m be L vt m L b be be v v g v R A g R v v v -     - Terminal voltage gain between base and collector is: Overall voltage gain from source vi to output voltage vo across RL is: o c b b v vt i b i i v v v v A A v v v v      RI vi + - RB ro RL + vo - B C E rp RC gmvbe + vbe - 0L C L R r R R  B v m L I B R r A g R R R r p p    -    B R rp
  • 7. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 7 C-E Amplifier with Fully Bypass RE: Input Resistance C1   VCC 1 2B B BR R R RI RB2 vi Q RB1 RC C2   C3   + - RL Rin vx + - ix RE  x x B x in x B v i R r v R i R r p p    RB rp ro RL + vo - B C E RC gmvbe + vbe - B R rp
  • 8. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 8 C-E Amplifier with Fully Bypass RE: Output Resistance C1   0 x x x m be C o x x be x C o v v i g v R r v v v i R r        VCC RI Q RB2 RB1 C2   C3   RL Rour 1 2B B BR R R vx + - RB ro RI B C E rp RC gmvbe + vbe - ix RE RC vi + - 1 1 1x out C o x C o out C v R R r i R r R R -           if
  • 9. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 9 C-E Amplifier with Fully Bypass RE: Example Problem: Find voltage gain, input and output resistances. Given:  = 65, VA = 50 V Find the Q-point from dc equivalent circuit:  3 3 100 10 0.7 66 16 10 ( 5) 0B BI I     -  3.71 μA 65 241 μA 66 245 μA B C B E B I I I I I       5 10 k 241 μ 16 k 245 μ 5 0CEV-  - -  - -  +5 V 10 k voRout RC C3   C1   Rin vi + – 220 k RB 100 k RE 16 k 330  RI –5 V RL C2   +5 V 10 k IC VCE + – IE 16 k –5 V 100 k VBE + – IB 3.67 VCEV 
  • 10. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 10 AC Analysis of C-E Amplifier with Fully Bypass RE 9.64 (223 10 220) 6.23 k 84.0 330 6.23 k bein v vt I in be vR A A R R v  -         -          40 241 μ 9.64 mSmg    65 6.64 k 9.64 m rp    50 3.67 223 k 241 μ or     100 k 6.64 k 6.23 kinR    223 k 10 k 9.57 koutR    330  vi + - 100k ro 220k + vo - B C E rp 10k gmvbe + vbe - Rin 100k 223k 220k B C E 6.64k 10k 9.64mvbe9.64mvbe Rout 100k 223k 330 B C E 6.64k 10k Small signal equivalent
  • 11. EE2002 Analog Electronics Small Signal Parameters of MOSFET AC Analysis of BJT and MOSFET Inverting Amplifiers 11 • Since gate is insulated from channel by gate-oxide, input resistance = . • Small-signal parameters are controlled by the Q-point. • MOSFET transconductance is geometry dependent.        2 1 2 1 n D GS TN DS D m n GS TN DS GS Q pt K i v V v i g K V V V v   -  -     -   n n OX W K C L        where + gmvgs ig S DG vgs ro id vds – + – 2 2 1 1 D m n D GS TN DS o D D I g K I V V V r I I     -    if 1 
  • 12. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 12 Small Signal Analysis of C-S Amplifier with Fully Bypass RS VDD S • AC equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources represent ac grounds. M RD RS C1   C2   C3   G D RG RD RI vi + - RL RI RG2 RG1 vi + - D MG S RL + vo - 1 2G G GR R R vo • The small signal parameters, gm and ro of the MOSFET is calculated at the Q-point, ID and VDS.
  • 13. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 13 C-S Amplifier with Fully Bypass RS: Voltage Gain Terminal voltage gain between gate and drain is: Overall voltage gain from source vi to output voltage vo across RL is: RI vi + - RG ro RL + vo - G D S RD gmvgs + vgs - 0L D LR r R R  m gs Ld vt m L g gs g v Rv A g R v v -    - g go o v vt i g i i G m L I G v vv v A A v v v v R g R R R         -    
  • 14. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 14 C-S Amplifier with Fully Bypass RS: Input Resistance RG2 M RG1 RD RS C2   C3   RL Rin C1   RG ro RL + vo - G D S RD gmvgs + vgs - 1 2G G GR R R VDD RI vi + - x x G x in x G v i R v R i R    vx + - ix
  • 15. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 15 C-S Amplifier with Fully Bypass RS: Output Resistance vi + - VDD C1   Since vgs = 0, gmvgs = 0. RI RG2 RG1 C2   Rour RS RD RL M gmvgs vx + - RG ro RI G D S RD + vgs - ix C3   21 GG G R R R  x x D o x out D o x out D v i R r v R R r i R R     if
  • 16. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 16 C-S Amplifier with Fully Bypass RS: Example Problem: Find voltage gain, input and output resistances. Given: Kn = 500 A/V2, VTN = 1V,  = 0.0167 V–1 DC Analysis: 61 1 6 5 10 2 10 0.4 S GS D DS I IV V V       5 , 2 , 250 μADS GS DV V V V I      2500 μ 1 2 GSD VI  - (1)  3 110 20 10DS D IV I -   (2) +10 V 20 k C3   vo + – Rout 100 k RL RD RG3 1 M RG2 2 M RG1 2 M C2   C1   10 k RI Rin vi +10 V 20 k RG3 1 M RG2 2 M RG1 2 M vDS + – + – vGS I1 ID IG RD
  • 17. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 17 C-S Amplifier with Fully Bypass RS: Example    0.5 260 k 1 M 20 k 100 k 1 M 7.93 10 k 1 M gsin v vt I in gs m vR A A R R v -       -      2 2 1inR M M M   2 500 μ 250 μ 0.5 mSm g     1 5 0.0167 260 k 250 μ or     10k  vi + - 2M ro 100k + vo - G D S 1M gmvgs + vgs - 2M 20k Rin 1M 260k 16.4k G D S 0.5mvgs 0.5mvgs S Rout 1M 260k 10k G D 1M 20k 260 1 20 18.2outR k M k k   16.4 k1 M Small Signal Equivalent Circuit:
  • 18. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 18 C-E Amplifier with Unbypass RE: Terminal Voltage Gain VCC RB2 RB1 RC RE2 C3   RI vi + - RL C2   C1   vo RE1 Q vi + - RB RL + vo - RC ro B C E rp gmvbe + vbe - RI RE1 ic L C LR R R  1 1 11 c c L m be L m L vt b be e E be m be E m E v i R g v R g R A v v i R v g v R g R   - - -        ie 1 2B B BR R R
  • 19. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 19 C-E Amplifier with Unbypass RE: Input Resistance vi + - RI     1 1 1 1 b b b E b in E b v i r i R v R r R i p p           Rin vb + - RB RL + vo - RC ro B C E rp ib + vbe - RE1 (+1)ib ib Rin in in BR R R inR RB RL + vo - RC ro B C E rp gmvbe + vbe - RE1 1 2B B BR R R
  • 20. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 20 C-E Amplifier with Unbypass RE: Overall Voltage Gain + vb - Rin RB = RB1 || RB2 RL + vo - RC ro B C E rp gmvbe + vbe - RE1 Rin VCC RB2 RB1 RC RE2 C3   RI vi + - RL C2   C1   vo RE1 Q o o b in v vt i b i I in v v v R A A v v v R R       vi + - RI Rin vi + - RI + vb - Rin
  • 21. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 21 C-E Amplifier with Unbypass RE: Output Resistance Rth RC ro B C E rp gmvbe + vbe - RE1 vx + - ix Rout 1 2th I B BR R R R ix   1 be e th x th E th r v v r R r i r R R r R p p p p p    -      -      1e x th Ev i r R Rp   x m bex o ev g v r vi - current through ro outR RL + vo - RI RB RC ro B C E rp gmvbe + vbe - RE1 1 2B B BR R R Rout vi + -
  • 22. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 22 C-E Amplifier with Unbypass RE: Output Resistance         1 1 x x m be o e x m x th E o th x th E v i g v r v r i g i r R R r r R i r R R p p p p  -              Rth ro B C E rp gmvbe + vbe - RE1 vx + - ix ix out out CR R R RC Rout   11x out m th E o x th v r R g r R R r i r R p p p            outR
  • 23. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 23 C-S Amplifier with Unbypass RS: Terminal Voltage Gain VDD RG2 RG1 RD RS2 C3   RI vi + - RL C2   C1   vo RS1 vi + - RG RL + vo - RD RI RS1 id  gmvgs L D LR R R  1 1 11 m gs Ld d L m L vt g gs d S gs m gs S m S g v Rv i R g R A v v i R v g v R g R - - -        id 1 2G G GR R R ro G D S gmvgs + vgs - M
  • 24. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 24 C-S Amplifier with Unbypass RS: Input Resistance RG = RG1 || RG1 RL + vo - RD RS1 Rin in GR R ro G D S gmvgs + vgs -vi + - RI Rin RG RL + vo - RD RS1 1 2G G GR R R ro G D gmvgs + vgs -
  • 25. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 25 C-S Amplifier with Unbypass RS: Overall Voltage Gain VDD RG = RG1 || RG2 RL + vo - RD ro G D S gmvgs + vgs - RS1 Rin RG2 RG1 RD RS2 C3   RI vi + - RL C2   C1   vo RS1 vi + - RI go o in v vt i g i I in vv v R A A v v v R R       Rin vi + - RI + vg - Rin M
  • 26. EE2002 Analog Electronics AC Analysis of BJT and MOSFET Inverting Amplifiers 26 C-S Amplifier with Unbypass RS: Output Resistance Rth RD RS1 vx + - ix Rout 1 2th I G GR R R R ix 1 10, s x S g gs s x S v i R v v v i R    -  - ro G D S gmvgs + vgs -  1 1x x m x S o x Sv i g i R r i R    11x out m S o x v R g R r i     out out DR R R  x m gsx o sv g v r vi - current through ro outR RL + vo - RI RG RD RS1 1 2G G GR R R ro G D gmvgs + vgs - Rout vi + -