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1
Small Signal ModelMOS Field-Effect Transistors (MOSFETs)
Quiz No 3 DE 27 (CE)
(a) Draw small signal model (4)(b) Find expression for Rout (2) (c) Prove vo/vsig = (β1α2RC)/(Rsig+rπ) (4).
Rout.
20-03-07
Figure 4.2 The enhancement-type NMOS transistor with a positive voltage applied to the gate. An n channel is induced at the top of the substrate beneath the gate.
Enhancement-type NMOS transistor:
MOSFET Analysis
iD = iS, iG = 0
Large-signal equivalent-circuit model of an n-channel MOSFET : Operating in the saturation region.
Large-signal equivalent-circuit model of an p-channel MOSFET : Operating in the saturation region.
Large Signal Model : MOSFET
Transfer characteristic of an amplifier
Conceptual circuit utilized to study the operation of the MOSFET as a small-signal amplifier.
The DC BIAS POINT
To Ensure Saturation-region Operation
Signal Current in Drain Terminal
Total instantaneous voltages vGS and vD
Small-signal ‘π’ models for the MOSFET
Common Source amplifier circuitExample 4-10
Small Signal ‘T’ Model : NMOSFET
Small Signal Models
‘T’ Model
Single Stage MOS Amplifier
Amplifiers Configurations
Common Source Amplifier (CS) :Configuration
Common Source Amplifier (CS)• Most widely used
• Signal ground or an ac earth is at the source through a bypass capacitor
• Not to disturb dc bias current & voltages coupling capacitors are used to pass the signal voltages to the input terminal of the amplifier or to the Load Resistance
• CS circuit is unilateral – – Rin does not depend on RL and vice versa
Small Signal Hybrid “π” Model (CS)
Small Signal Hybrid “π” Model : (CS)
Gin RR sig
sigG
Ggs v
RR
Rv
LDogsmo RRrvgv ||||
sigG
GLDom
gs
ov RR
RRRrg
v
vG ||||
Doo Rr ||R
sig
gs
gs
o
sig
ov v
v
v
v
v
vG
Small-signal analysis performed directly on the amplifier circuit with the MOSFET model implicitly utilized.
Gin RR
sigG
GLDom
gs
o
RR
RRRrg
v
v||||
Doo Rr ||R
• Input Resistance is infinite (Ri=∞)
• Output Resistance = RD
• Voltage Gain is substantial
Common Source Amplifier (CS) Summary
Gin RR
sigG
GLDom
gs
o
RR
RRRrg
v
v||||
Doo Rr ||R
Common-source amplifier with a resistance RS in the source lead
The Common Source Amplifier with a Source Resistance
• The ‘T’ Model is preferred, whenever a resistance is connected to the source terminal.
• ro (output resistance due to Early Effect) is not included, as it would make the amplifier non unilateral & effect of using ro in model would be studied in Chapter ‘6’
Small-signal equivalent circuit with ro neglected.
Sm
g
Rg
vi
1
Do
Gin
RR
RR
Small-signal Analysis.
sig
i
i
gs
gs
o
sig
ov v
v
v
v
v
v
v
vG
Sm
LDm
sigG
Gv
sig
o
sigsigG
Gi
Sm
ii
Sm
mgs
LDgsmo
Rg
RRg
RR
RG
v
v
vRR
Rv
Rg
vv
Rg
gv
RRvgv
1
||
11
1
||
Voltage Gain : CS with RS
Common Source Configuration with Rs
• Rs causes a negative feedback thus improving the stability of drain current of the circuit but at the cost of voltage gain
• Rs reduces id by the factor
– (1+gmRs) = Amount of feedback
• Rs is called Source degeneration resistance as it reduces the gain
Small-signal equivalent circuit directly on Circuit
A common-gate amplifier based on the circuit
Common Gate (CG) Amplifier• The input signal is applied to the source
• Output is taken from the drain
• The gate is formed as a common input & output port.
• ‘T’ Model is more Convenient
• ro is neglected
A small-signal equivalent circuit
A small-signal Analusis : CG
mim
i
i
iin gvg
v
i
vR
1
Dout RR
A small-signal Analusis : CG
sigm
LDm
sig
ov
sigm
sigsig
sigm
msig
sigin
ini
LDimo
sig
i
i
o
sig
ov
Rg
RRg
v
vG
Rg
vv
Rg
gv
RR
Rv
RRvgv
v
v
v
v
v
vG
1
||
11
1
||
Small signal analysis directly on circuit
The common-gate amplifier fed with a current-signal input.
Summary : CG
4. CG has much higher output Resistance5. CG is unity current Gain amplifier or a Current Buffer6. CG has superior High Frequency Response.
A common-drain or source-follower amplifier.
Small-signal equivalent-circuit model
Small-signal Analysis : CD
(a) A common-drain or source-follower amplifier :output resistance Rout of the source follower.
mmoout gg
rR11
||
(a) A common-drain or source-follower amplifier. : Small-signal analysis performed directly on the circuit.
Common Source Circuit (CS)
Common Source Circuit (CS) With RS
Common Gate Circuit (CG)Current Follower
Common Drain Circuit (CD) Source Follower
Summary & Comparison
Quiz No 4• Draw/Write the Following:
27-03-07
BJT MOSFET
Types npn pnp nMOS pMOS
Symbols
‘π’ Model
T Model
gm
Re/rs
rπ/rg
Problem 5-44
SOLUTION : DC Analysis
SOLUTION : DC Analysis
IE
Check for Active Mode
mAI
II
II
E
EE
BE
1
101100
3.3
7.05
0100)1(
7.03.35
01007.03.35
250.1
25
E
te I
Vr
IB
Solution Small Signal Analysis
Solution Small Signal Analysis
Solution Small Signal Analysis : Input Resistance
Rin
ib
LCee
b
b
bin RRr
i
v
i
vR ||)1(
)1(
+
vb
-
Solution Small Signal Analysis : Output ResistanceItest
IE
IE/(1+ß)
IRC
Rout
test
testout I
VR
)1(||
)1(
)1(
)1(
sigeC
sigeC
sigeC
sige
test
C
test
testout
RrR
RrR
RrR
Rr
VR
VV
R
ERtest IIIC
)1(
sig
e
testE R
r
VI
C
testR R
VI
C
Solution Small Signal Analysis : Voltage Gain
+
-
LCmeb
o RRgv
v||+
-
vi
+
-
veb
sig
i
i
eb
eb
o
sig
o
v
v
v
v
v
v
v
v
Vo
Solution Small Signal Analysis : Voltage gain
sig
i
i
eb
eb
o
sig
o
v
v
v
v
v
v
v
v
LCmeb
o RRgv
v||+
-
vi
+
-
veb
LCe
e
i
eb
RRr
r
v
v
||
Solution Small Signal Analysis : Voltage Gain
sig
i
i
eb
eb
o
sig
o
v
v
v
v
v
v
v
v
LCmeb
o RRgv
v||
+
-
vi LCe
e
i
eb
RRr
r
v
v
||
LCein RRrR ||)1(
sigLCe
LCe
sigin
in
sig
i
RRRr
RRr
RR
R
v
v
||)1(
||)1(
Solution Small Signal Analysis : Voltage Gain
sig
i
i
eb
eb
o
sig
o
v
v
v
v
v
v
v
v LCm
eb
o RRgv
v||
LCe
e
i
eb
RRr
r
v
v
||
sigin
in
LCe
eLCm
sig
o
RR
R
)R(Rr
r)||R(Rg
v
v
||
sigin
in
sig
i
RR
R
v
v
sigin
in
LCe
LC
sig
o
RR
R
)R(Rr
)||R(R
v
v
||
sigin
in
LCe
LCem
sig
o
RR
R
)R(Rr
)||R(Rrg
v
v
||
Solution Small Signal Analysis : Voltage Gain
+
-
LCe
LC
i
o
RRr
RR
v
v
||
||
+
-
vi
sig
i
i
o
sig
o
v
v
v
v
v
v
sigin
in
sig
i
RR
R
v
v
sigin
in
LCe
LC
sig
o
RR
R
)R(Rr
)||R(R
v
v
||
Vo
Problem
Small Signal Model MOSFET : CD
Solution Small Signal Analysis
1/gm
gmvsg
D
1/gm
gmvsg
D
Solution Small Signal Analysis : Input Resistance
Rin
Ig=0
inR
1/gm
gmvsg
D
Solution Small Signal Analysis : Output ResistanceItest
ID
IG=0
IRD
Rout
test
testout I
VR
mD
m
test
D
test
testout g
R
gV
RV
VR
1||
/1
DRtest IIIC
m
testD
g
VI
1
D
testR R
VI
D
Vtest
1/gm
gmvsg
D
Solution Small Signal Analysis : Voltage Gain
+
-
LDmsg
o RRgv
v||+
-
vi
+
-
vsg
sig
i
i
sg
sg
o
sig
o
v
v
v
v
v
v
v
v
1/gm
gmvsg
D
Solution Small Signal Analysis : Voltage gain
+
-
vi
+
-
vsg
LDm
m
i
sg
RRg
g
v
v
||1
1
LDmsg
o RRgv
v||
sig
i
i
sg
sg
o
sig
o
v
v
v
v
v
v
v
v
Solution Small Signal Analysis : Voltage Gain
+
-
vi
inR
sigi vv
LDmsg
o RRgv
v||
sig
i
i
sg
sg
o
sig
o
v
v
v
v
v
v
v
v
LDm
m
i
sg
RRg
g
v
v
||1
1
Solution Small Signal Analysis : Voltage Gain
)R(Rg
g)||R(Rg
v
v
LDm
mLDm
sig
o
||1
1
sigi vv LDmsg
o RRgv
v||
sig
i
i
sg
sg
o
sig
o
v
v
v
v
v
v
v
v
LDm
m
i
sg
RRg
g
v
v
||1
1
)R(Rg
)||R(R
v
v
LDm
LD
sig
o
||1
Solution Small Signal Analysis : Voltage Gain
+
- LC
m
LD
i
o
RRg
RR
v
v
||1||
+
-
vi
sig
i
i
o
sig
o
v
v
v
v
v
v
sigi vv
LC
m
LD
sig
o
RRg
RR
v
v
||1||
Solution Small Signal Analysis
LCein RRrR ||)1(
)1(
||
sigeCout
RrRR
sigin
in
LCe
LC
sig
o
RR
R
)R(Rr
)||R(R
v
v
||
1
inR
mDout g
RR1
||
LC
m
LD
sig
o
RRg
RR
v
v
||1||
Problem 6-127(e)
DC Analysis 6-127(e)
mAI
AI
mAI
C
B
E
5.0
05101/5.0
5.0
100
1
1
1
mAI
AI
mAI
C
B
E
5.0
05101/5.0
5.0
100
2
2
2
eActiveinQ
VVVVV
VV
BC
C
mod
6.44.054.0
5105.010
2
22
2
eActiveinQ
VVVV
VV
BC
C
mod
4.04.0)10(5104.0
3.47.05
1
311
1
Small Signal Model
Small Signal Model
Small Signal Model
Rin
1rRin Rout
0 sigCout VRR
sigsig
be
Rr
r
v
v
1
11
sig
be
be
eb
eb
o
sig
o
v
v
v
v
v
v
v
v 1
1
2
2
21
1
2em
be
eb rgv
vCm
eb
o Rgv
v2
2
11211212
rR
R
rR
rrgRg
v
v
sig
C
sig
emCm
sig
o
Problem6-127(f)Replacing BJT with MOSFET
Small Signal Model
Small Signal Model
Small Signal Model
Rin
inR Rout
0 sigDout VRR
sigsg vv 1sig
gs
gs
sg
sg
o
sig
o
v
v
v
v
v
v
v
v 1
1
2
2
2
1
1
2
m
m
gs
sg
g
g
v
vDm
sg
o Rgv
v2
2
Dmm
mDm
sig
o Rgg
gRg
v
v1
2
12
1rRin
Cout RR
121
rR
R
v
v
sig
C
sig
o
1
inR
Dout RR
Dmsig
o Rgv
v1
11
2
1
m
sig
C
sig
o
g
RR
v
v
Problem 6-127(f)
Solution P6-127(f)
+
+
-
-
vbe2
veb1
+
+
-
-
vbe2
veb1
+
vi
-
Solution P6-127(f)))(1( 211
1
1ee
b
bin rr
i
vR
Lout RR
sig
i
i
be
be
O
sig
O
v
v
v
v
v
v
v
v 2
2
Lmbe
O Rgv
v
2
sige
L
sigee
Lem
sigeeee
eeeLm
sig
o
Rr
R
Rrr
Rrg
Rrrrr
rrrRg
v
v
)2)(1(
)1(
))(1(
)1(
))(1(
))(1(
1
21
211
122
21121
21122
21
22
ee
e
i
be
rr
r
v
v
sigee
ee
sigin
in
sig
i
Rrr
rr
RR
R
v
v
))(1(
))(1(
211
211
Problem 6-127(f) with MOSFET
-
-
vgs2
vsg1
+
+
Solution P6-127(f)
-
-
vgs2
vsg1
+
+
Solution P6-127(f)
+vi
-
1g
iin i
vR
Lout RR
sig
i
i
gs
gs
O
sig
O
v
v
v
v
v
v
v
v 2
2
Lmgs
O Rgv
v
2
221
12 Lm
mm
Lmm
sig
o Rg
gg
Rgg
v
v
21
1
21
22
11
1
mm
m
mm
m
i
gs
gg
g
gg
g
v
v
sigi vv
ig1=0
Comparison BJT/MOSFET Cct
inR
Lout RR
2Lm
sig
o Rg
v
v
))(1( 211 eein rrR
Lout RR
sige
L
sig
o
Rr
R
v
v
)2)(1(
)1(
1
21
1
Small Signal Model
Figure P6.123
Problem 6-123
VBE=0.7 Vβ =200K’n(W/L)=2mA/V2
Vt=1V
Figure P6.123
DC Analysis
DC AnalysisVBE=0.7 Vβ =200K’n(W/L)=2mA/V2
Vt1=1VVt2=25mV
0.7V
I=0.7/6.8=0.1mA
0 ,1. 211 BSD ImAoII
VVV GSGS 316.112211.0 2
21 '2
1tGSnD VV
L
WKI
IG=0
2V
VVVV BEGSC 22
1mA
mAII C 13
252
VmAV
Ig
VOV
Dm /63.0
2 11
kg
rVmAV
Ig
mt
Cm 5,/40
22
22
Small Signal Model
Small Signal Model
Small Signal Model : Voltage Gain
ig=0+
vi
-
+vbe2
-
1
2
2 gs
i
i
be
be
o
sig
o
v
v
v
v
v
v
v
v
MRofeffectgNegelectin
RRgv
v
G
C Lmbe
o
10
-30V/V )||(22
VVrR
g
rR
v
v
Sm
S
i
be /64.0)||(
1)||(
211
212
VVRR
R
v
v
sigin
in
sig
i /83.0
VVgRR
R
rRg
rRRRg
v
v
siin
ni
Sm
SCLm
sig
/16)||(
1)||(
)||(
211
212
0
Small Signal Model : Input Resistance
ig=0
Rin
+
vi
-
ii
k
vv
R
Rvv
v
i
vR
i
o
G
Goi
i
i
iin 495
1/
VVrR
g
rRRRg
v
v
Sm
SCLm
i
/2.19)||(
1)||(
)||(
211
212
0