Sedra/Smith Microelectronic Circuits 5/e

Sedra/Smith
Microelectronic Circuits 5/e
Chapter 5B MOS Field-Effect Transistors
(MOSFETs)
S. C. Lin, EE National Chin-Yi University of Technology
1
5.5 Small-Signal operation and models
5.5.1 The DC Bias Point
1
1
2
2
I D = kn (VGS − Vt ) = knVOV
2
2
VDS = VDD − I D RD
To ensure saturation region operation.
We must have VD > VOV
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5.5.2 The Signal Current in the Drain Terminal
vGS = vgs + VGS
1
∴ iD = kn (vgs + VGS − Vt ) 2
N 2
i +I
d
D
1
1
2
= kn (VGS − Vt ) + kn (VGS − Vt )vgs + kn vgs 2 (5.43)
2
2
DC bias current I D
nonlinear distotion
The Current component that is directly
proportional to the input signal vgs
To reduce the nonlinear distotion, the input signal shonld be kept small
1 'W 2
' W
so that kn vgs << kn (VGS − Vt )vgs ⇒ vgs <<2(VGS − Vt )
L
2 L
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Example :
For the amplifier in Fig.A, let the MOSFET is specified to have Vt = 2V
and kn' W / L = 2mA/V 2 ,VGS = 5V, assume λ =0. if vgs = 0.5sin ωt volts
(a) Determine the various components. (b)Find the second harmonic
component as a percentage of the amplitude of fundamental
Sol:
(a ) iD = 1mA/V 2 × (5 - 2) 2 V 2 + 2mA/V 2 × 0.5sin ωt × (5 - 2)V 2
+ 0.25sin 2 ωt ×1mA
1 1
= 9 + 3sin ωt + 0.25( − cos 2ωt ) ( mA )
2 2
= 9.125 + 3sin ωt − 0.125cos 2ωt ( mA ) ▲
(b) Harmonic distotion percentage
0.125
× 100% = 4.16% ▲
D% =
3
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Small-signal operation of the enhancement MOSFET amplifier.
id
' W
gm ≡
= kn (VGS − Vt )(5.47)
vgs
L
∂I D
gm ≡
∂VGS
(5.49)
vgs =VGS
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5.5.3 The voltage gain
vD = VDD − iD RD
= VDD − (id + I D ) RD
= VDD − I D RD − id RD
The biasing component is
VD = VDD − I D RD
The signal component is
vds = −id RD = − g m vgs RD
vd
∴ Av ≡
= − g m RD
vgs
The minus sign, indicates that
the vd is 1800 out of phase
with respct to the vgs
(5.50)
(5.51)
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vi = vgs
small signal condition
Total instantaneous voltages vGS and vD for the circuit in Fig. 4.34.
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5.5.5 Small-signal Equivalent-Circuit Models
(a) neglecting the dependence of iD
on vDS in saturation (the channel
-length modulation effect);
(b) including the effect of channel
-length modulation, modeled by
output resistance ro = |VA| /ID.
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5.5.6 The Transconductance gm
gm =
id
= kn ( vGS − Vt )
vgs
W
' W
v
V
k
−
=
( GS t ) n VOV (5.55)
L
L
1 'W
2
' W
I D = kn VOV ⇒ VOV = 2 I D / kn
2 L
L
Subtituting for VOV in (5.55) by
= kn'
VOV = 2 I D / k (W / L ) , we obtain
'
n
g m = 2kn'
(W / L ) I D
(5.56)
iD
ID
Q
Slop ≡ g m =
0
1
2
VOV VOV
2I D
( vGS − Vt )
, we obtain g m =
2
VOV
Figure 5.38
Yet another useful expression for g m , Subtituting for kn'
in (5.55) by
ID
1
V
2 OV
W
L
2I D
ID
=
vGS − Vt VOV / 2
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(5.57)
9
Example 5.10 For the amplifier in Fig.a, let the MOSFET is specified to
have Vt = 1.5V, VA = 50Vand kn = 0.25mA/V 2 . Assume the coupling
capacitors to be sufficiently large so as to act as short circuit at the
signal frequencies of intrrest.(a) Find the voltage gain (b)Find the input
resistance (c) Find the largest allowable input signal.
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(b)ii = (Vi − Vo ) / RG
1 W
Sol:(a ) I D = kn'
(VGS − Vt ) 2
2 L
1
= × 0.25 × (VGS − 1.5) 2
N
2
V
DS
=
Vi
V
(1 − o )
RG
Vi
=
Vi
(1 − Av )
RG
=
4.3
Vi
RG
VDS = 15 − I D RD = 15 − 10 I D
⇒ I D = 1.06mA ▲ VD = 4.4V ▲
W
gm = k
(VGS − Vt )
L
= 0.25 × (4.4 − 1.5) = 0.724mA/V ▲
V
50V
ro = A =
= 47kΩ ▲
I D 1.06mA
'
n
⇒ vo = − g m vgs ( RD // RL // ro )
v
Av = o = − g m ( RD // RL // ro )
vi
= −0.725(10 //10 // 47) = −3.3V/V ▲
Thus : Rin =
Vi RG
=
ii 4.3
= 10 / 4.3 = 2.33MΩ ▲
(c)vDS (min) = vGS (max) − Vt
vDS − Av vˆi = vGS + vˆi − Vt
4.4 − 3.3vˆi = 4.4 + vˆi − 1.5
Vt
1.5
=
= 0.35V ▲
vˆi =
Av + 1 3.3 + 1
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5.5.6 The T Equvalent-Circuit Model
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Figure 5.41 (a) The T model of the MOSFET augmented with the
drain-to-source resistance ro. (b) An alternative representation of the
T model.
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5.6 Basic MOSFET Amplifier Configurations
5.6.1 The Three Basic Configurations
+
vi +
RD
−
+
RD
vo
vi +
−
−
vo
−
(a) Common-Source (CS)
(b) Common-Gate(CG)
vi −+
RD
+
vo (c) Common-Drain(CD)
−
or Source Follower
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5.6.2 Characterizing Amplifier
Rin
vi =
vsig ,
Rin + Rsig
RL
vo =
Avo vi
Ro + RL
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vi
“ Input resistance: Rin ≡
ii
vo
“ Open-circuit voltage gain: Avo ≡
vi
RL =∞
vo
RL
“ Voltage gain with RL : Av ≡ = Avo
vi
RL + Ro
io
“ Short-circuit current gain: Ais ≡
ii
(5.67)
RL = 0
io
“ Current gain: Ai ≡
ii
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vx
“ Output resistance : Ro ≡
ix
vi = 0, RL =∞
vo
vi vo
Rin
“ overall voltage gain: G v ≡
=
⋅ =
Av
vsig vsig vi Rin + Rsig
Rin
RL
=
Avo
Rin + Rsig
RL + Ro
S. C. Lin, EE National Chin-Yi University of Technology
(5.68)
17
5.6.3 The Common-Source (CS) Amplifier
Characterizing Parameters of CS Amplifier
Rin = ∞
+
Rsig
vsig +
−
Ri
RD
+
vi
−
vo = −( g m vgs ) ( RD // ro )
Avo = − g m ( RD // ro )
vo
Ro
≈ − g m RD
−
Rsig
vsig
Rin = ∞
−
(5.71)
Ro = RD // ro
≈ RD
+
vi = vgs
(5.69)
(5.73)
+
g m vgs
ro
RD
vo
−
Ro = RD // ro
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Concludes:
X The input resistance is ideally infinite.
Y The Ro ≈ RC is moderate to high in value (typically, in the
kilohms to tens kilohms range). Reducing RD to lower Ro
is not a viable proportional, because the Av is also reduced.
Z The open circuit voltage gain Avo can be hig, making the
CS configration the work-hourse in MOS amplifier design.
However the bandwidth of the CS amplifier is severely limited.
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Overall Voltage Gain
v
Av = o = − g m ( RD // RL // ro )
vi
(5.75)
vi = vsig
(5.74)
vo
vi vo
=
⋅ = − g m ( RD // RL // ro )
Gv ≡
vsig vsig vi
(4.76)
Performing the Analysis Directly on the Circuit
g mvgs
Rsig
vsig +
−
+
0
+
vi = vgs
Rin = ∞
−
ro
RD
vo
Ro = RD // ro
−
vo = − g mvgs ( RD // ro )
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4.6.4 The Common-Source (CS) Amplifier with an
Source Resistance
+
Rsig
+
vsig +
−
RD
vi
Rin
−
0
++ i
vi
Rin = ∞
Ro
Rs
−
Rsig
vsig
vo
−
vgs
1
gm
+
i
RD
−
Rs
vo
−
Ro = RD
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21
1/ g m )
(
1
vgs =
vi =
vi
1 + g m Rs
(1/ g m ) + Rs
vi
gm
=
i=
vi
(1/ g m ) + Rs 1 + g m Rs
(5.78)
g m RD
vo
g m RD
=
vo = −iRD =
vi ⇒ Avo =
1 + g m Rs
vi
1 + g m Rs
(5.80)
Total resistance in drain
Av (from gate to drain ) = −
(5.81)
Total resistance in source
Ro = RD
Alternatively, Av can be obtained by simply replacing RD
in (5.80) by ( RD //RL )
g m ( RD //RL )
vo
RD //RL
Av =
=
=
1/ g m + Rs
1 + g m Rs
vi
S. C. Lin, EE National Chin-Yi University of Technology
(5.83)
22
5.6.5 The Common-Gate (CG) Amplifier
RD
Rsig
+
vi +
Rin = 1/ g m
vo
vi
vo = −iRD , i = −
1/ g m
−
vi
−
+
Ro
−
Rin
vo
Avo ≡ = g m RD
vi
(5.85)
Ro = RD
(5.86)
1/ g m
Rsig
+
vsig
(5.84)
vi
−
Rin = 1/ g m
i
+
i
RD
vo
−
Ro = RD
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Overall Voltage Gain
1/ g m )
(
vi
Rin
=
=
vsig Rin + Rsig (1/ g m ) + Rsig
(5.87)
vo
Av = = g m ( RD // RL )
vi
1/ g m )
(
vo
vi vo
=
⋅ =
g m ( RD // RL )
Gv ≡
vsig vsig vi (1/ g m ) + Rsig
RD // RL
=
1 / g m + Rsig
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(5.88)
24
5.6.6 The CD Amplifier or Source Follower
The Need for Voltage Buffers
Rsig = 1 MΩ
Rsig = 1MΩ
vsig = 1V
RL = 1 kΩ
+
1 kΩ vo 1mV
1V
−
Ro = 100Ω
Rsig = 1 MΩ
1V
Avo = 1
vsig = 1 V
+
RL = 1 kΩ
Rin very large
vo 0.9V
−
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D
i
Rsig
Rsig G 0
+
vsig +
−
+
vi
Rin
−
RL
Ro
+
vo
−
vsig
vi
−
vsig
+ i
1/ g m
vi
Rin = ∞
−
S
RL
i
0
Rsig
1/ g m
i
RL
ro
+
vo
−
Ro = 1/ g m
+
vo
−
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Rin = ∞
vo
RL
Av ≡ =
(5.89),
vi RL + (1/ g m )
Ro = 1/ g m
vo
Avo ≡
vi
1 (5.90)
RL =∞
(5.91)
Overall Voltage Gain
vo
vi vo
Gv ≡
=
⋅ = 1⋅ Av
vsig vsig vi
RL
=
≈1
RL + (1/ g m )
(5.92)
G v will be lower than unity,but close to unity.
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5.7 Biasing in MOS amplifier circuits
iD =
1
W
⎫
µn Cox (vI − Vt ) 2 ⎪
2
L
⎬ Biasing by Fixed VGS
⎪⎭
VG = VGS + I D RS
Figure 5.51 The use of fixed bias (constant VGS) can result in a
large variability in the value of ID. Devices 1 and 2 represent
extremes among units of the same type.
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Figure 5.52 Biasing using a fixed voltage at the gate, VG, and
a resistance in the source lead, RS: (a) basic arrangement;
(b) reduced variability in ID;
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(c) practical implementation using a single supply;
(d) coupling of a signal source to the gate using a capacitor CC1;
(e) practical implementation using two supplies.
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Example 5.12 Design the circuit of Fig.5.53 to establish a I D = 0.5mA.
The MOSFET is specified to have Vt = 1V and kn = 1mA/V 2 .
For simplicity, assume λ = 0. Calculate the percentage change in value
of I D obtained when the MOSFET is replaced with another unit having
the same kn' W / L but Vt = 1.5V.
Solution
VDD − VD 15 − 10
=
= 10kΩ
(a ) RD =
ID
0.5
VS
5
=
= 10kΩ
RS =
I D 0.5
▲
▲
1 '
1
2
I D = kn (W / L)VOV = ⋅1⋅ VOV 2
N 2
2
0.5mA
⇒ VOV = 1V
▲
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Thus. VGS = Vt + VOV = 2V
VG = VS + VGS = 5V+2V = 7V
we may select RG1 = 8MΩ, and RG2 = 7MΩ
▲
(b) If the NMOSFET is replaced with another having
Vt = 1.5V
1
2
I D = ⋅1 ⋅ (VGS − Vt )
2
VG = VGS + I D RS ⇒ VGS = 7 − 10 I D
2I D = ( 7 − 10 I D − 1.5 ) ⇒ I D = 0.455mA
2
∆I D = 0.455 − 0.5 = −0.045mA
−0.045mA
×100% = −9%
which
0.5mA
▲
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5.7.3 Biasing Using a Drain-to-Gate Feedback Resistor
Here the large feedback resistance RG
(Usually in the MΩ range) ⇒ I G = 0
We can written
VGS = VDS = VDD − I D RD
or
VDD = VGS + I D RD
Figure 5.54 Biasing the MOSFET using a large drainto-gate feedback resistance, RG.
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5.7.4 Biasing Using a Constant-Current Source.
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1
2
I D1 = ( µnCox )(W / L )1 ( vGS − Vt )
2
VDD + VSS − VGS
I D1 = I ref =
R
1
2
I = I D2 = ( µnCox ) (W / L )2 ( vGS − Vt )
2
1
( µnCox ) (W / L )2 (vGS − Vt ) 2
I D2
I
2
=
=
I D1 I ref
1
( µn Cox ) (W / L )1 (vGS − Vt ) 2
2
I = I ref
(W / L )2
(W / L )1
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5.8 Discrete-Circuit MOS Amplifier.
5.8.2 The Common-Source Amplifier
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Analysis of the circuit is straight ward and proceeds in step-by
step manner, from the signal source to the amplifier load.
ig = 0,
Rin = RG
vo = − g m vgs ( ro // RD // RL )
vo
Av ≡ = − g m ( ro // RD // RL )
vi
The overall voltage gain from the signal-source to the load will be
RG
Rin
Gv =
Av = −
g m ( ro // RD // RL ) (5.101)
Rin + Rsig
RG + Rsig
Rout = ro // RD
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5.8.3 The Common-Source Amplifier with a Source Resistance
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40
Small-signal equivalent circuit with ro neglected.
ig = 0,
Rin = RG
vo
g m ( RD // RL )
Av = = −
vi
1 + g m Rs
Avo = Av
RL =∞
g m RD
=−
1 + g m Rs
RG
g m ( RD // RL )
⋅
Gv = −
1 + g m Rs
RG + Rsig
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(a) A common-gate amplifier based on the circuit of Fig. 5.56
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(b) A small-signal equivalent circuit of the amplifier in (a).
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(c) The common-gate amplifier fed with a current-signal input.
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5.8.5 The Common- Drain or Source-Follower Amplifier.
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Small-signal equivalent-circuit model.
vo
Av =
vi
=
RL // ro
1
( RL // ro ) +
gm
Avo = Av
=
Gv =
,
RL =∞
ro
1
ro +
gm
Ro = (1/ g m ) // ro
RG
RL // ro
⋅
RG + Rsig R // r + 1
( L o)
gm
1
which approaches unity for RG >> Rsig , ro >>
and ro >> RL
gm
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5.8.6 The Amplifier Frequency Response
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5.9 The Body Effect and other Topics
5.9.1 The role of the substrate-The body effect
(5.107)
Vt = Vto + γ ⎡⎣ 2φ f + vSB − 2φ f ⎤⎦
where Vto is the threshold voltage for vSB = 0
φ f is a physical parameter with typically 0.3V
γ is a fabrication-process parameter
2qN A εs
, typically γ = 0.4V 1/ 2 (5.108)
given by γ =
Cox
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Modeling the body effect
The body effect occurs in a MOSFET when the source is
not tied to the substrate. We consider the body effect
{
)}
(
2
1 'W
iD = kn
VGS − ⎡Vt0 + γ 2φ f + VSB − 2φ f ⎤
⎣
⎦
2 L
∂iD
' W
= kn
gm ≡
VGS − ⎡Vt0 + γ 2φ f + VSB − 2φ f ⎤
⎣
⎦
∂VGS
L
{
g mb
∂iD
∂iD
≡
=−
∂VBS
∂VBS
{
)}
(
vGS , vDS = constant
)}
(
1
−
1
W
= kn'
VGS − ⎡Vt0 + γ 2φ f + VSB − 2φ f ⎤ × γ ( 2φ f + VSB ) 2
⎣
⎦
L
2
gm
= gm
∂Vt
γ
γ
= χ g m where χ ≡
=
(5.111)
∂VSB 2 2φ f + VSB
2 2φ f + VSB
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Figure 5.62 Small-signal equivalent-circuit model of a MOSFET in
which the source is not connected to the body.
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5.9.3 Temperature effects
(1)T ↑ Vt ↓ ( −2mV / o C ) ⇒ iD ↑
(2)T ↑ k ' ↓
⇒ iD ↓ (Dominant)
5.9.4 Breakdown and input protection
(1)As the vDS ↑,a value is reached at which the PN junction
between the drain region and substrate suffers avalanche
breakdown. Usually occurs at voltages of 20V ↔ 150V
(2)Punch-through in drain to source
(3) When the vGS exceeds about 30V the oxide breakdown.
The protection mechanism invariably makes use of clamping
diodes.
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