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Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
References B. Razavi, Design of Analog CMOS Integrated Circuits, McGraw-Hill, 2001.
H. Aboushady
University of Paris VI
1
Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
Basic Concepts I Amplification is an essential function in most analog circuits ! Why do we amplify a signal ? The signal is too small to drive a load To overcome the noise of a subsequent stage Amplification plays a critical role in feedback systems In this lecture: Low frequency behavior of single stage CMOS amplifiers: Common Source, Common Gate, Source Follower, ... Large and small signal analysis. We begin with a simple model and gradually add 2nd order effects Understand basic building blocks for more complex systems.H. Aboushady University of Paris VI
2
Approximation of a nonlinear systemInput-Output Characteristic of a nonlinear system
y (t ) 0 + 1 x (t ) + 2 x 2 (t ) + ... + n x n (t )In a sufficiently narrow range:
x1 x x2
y (t ) 0 + 1 x(t )
where 0 can be considered the operating (bias) point and 1 the small signal gain
H. Aboushady
University of Paris VI
Analog Design Octagon
H. Aboushady
University of Paris VI
3
Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
Common Source Stage with Resistive Load
Vout = VDD RD I DM1 in the saturation region:
Vout = VDD RD
nCox W
M1 in limit of saturation: Vin1 VTH M1 in the linear region:H. Aboushady
2 L C W (Vin1 VTH ) 2 = VDD RD n ox 2 L
(Vin VTH ) 2
Vout = VDD RD nCox
W L
V2 (Vin VTH )Vout out 2 University of Paris VI
4
Common Source Stage with Resistive Load
M1 in deep linear region:
Vout = VDD
Ron VDD = W Ron + RD 1 + C RD (Vin VTH ) n ox LUniversity of Paris VI
H. Aboushady
Common Source Stage with Resistive LoadM1 in the saturation region:
Vout = VDD RD
nCox W2 L
(Vin VTH ) 2
Small signal gain:
Av =
Vout W = RD nCox (Vin VTH ) Vin L = g m RDSmall signal model for the saturation region
Same relation can be derived from the small signal equivalent circuit
To minimize nonlinearity, the gain equation must be a weak function of signal dependent parameters such as gm !
H. Aboushady
University of Paris VI
5
Example 1Sketch ID and gm of M1 as a function of the Vin:
M1 in the saturation region:
M1 in the linear region:2 W Vout I D = nCox (Vin1 VTH )Vout 2 L I W g m = D = nCox Vout VGS L VDD Vout = W 1 + nCox RD (Vin VTH ) L University of Paris VI
(Vin VTH ) 2 L I W g m = D = nCox (Vin VTH ) VGS L
ID =
nCox W
2
H. Aboushady
Voltage Gain of a Common Source StageAv = g m RDAv = 2 nCoxAv = 2 nCox
V W I D RD ID LW VRD L ID
How to increase Av ? Trade-offs: Increase W/L Increase VRD Reduce IDH. Aboushady
Greater device capacitances. Limits Vout swing. Greater Time Constant.University of Paris VI
6
Taking Channel Length Modulation into accountCalculating Av starting from the Large Signal Equations:
Vout = VDD RD
nCox W2 L
(Vin VTH ) 2 (1 + Vout )
Av =
Vout W = RD nCox (Vin VTH )(1 + Vout ) Vin L RD
nCox W2 L
(Vin VTH ) 2
Vout Vin
Av = RD g m RD I D AvAv = RD g m 1 + RD I D
I D = 1 / rO
Av = g m
rO RD rO + RDUniversity of Paris VI
H. Aboushady
Taking Channel Length Modulation into accountCalculating Av starting from the Small Signal model:
g mV1 (rO // RD ) = Vout V1 = Vin
Av =
Vout = g m (rO // RD ) Vin
H. Aboushady
University of Paris VI
7
Example 2Assuming M1 biased in saturation, calculate the small signal voltage gain : I1 : Ideal current source Infinite Impedance
Av = g m rO Intrinsic gain of a transistor: This quantity represents the maximum voltage gain that can be achieved using a single device.
I D1 =
nCox W2 L
(Vin VTH ) 2 (1 + Vout ) = I1
Constant Current: As Vin increases, Vout must decrease such that the product remains constantH. Aboushady University of Paris VI
CS Stage with Current-Source Load Both transistors operate in the saturation region:
Av = g m (rO1 // rO 2 ) The output impedance and the minimum required VDS of M2 are less strongly coupled than the value and voltage drop of a resistor.
VDS 2,min = VGS 2 VTH 2 This value can be reduced to a few hundred millivolts by simply increasing the width of M2. If rO2 is not sufficiently high, the length and width of M2 can be increased to achieve a smaller while maintaining the same overdrive voltage. The penalty is the large capacitance introduced by M2 at the output node. Increasing L2 while keeping W2 constant increases rO2 and hence the voltage gain, but at the cost of higher |VDS2| required to maintain M2 in saturationH. Aboushady University of Paris VI
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CS with Source DegenerationLarge Signal model: Small Signal model:
I I VGS Gm = D = D Vin VGS Vin
VGS = Vin I D RSVGS I = 1 D RS Vin Vin
Gm =
I 1 RS D Vin Gm = g m (1 RS Gm ) I D VGSGm = gm 1 + g m RS
gm =
I D VGS
Gm =
ID g mV1 = Vin V1 + g mV1RS gm 1 + g m RS
Av = Gm RDAv = g m RD 1 + g m RS
Gm =
H. Aboushady
University of Paris VI
CS with Source DegenerationGm = gm 1 = 1 + g m RS 1 / g m + RS
for
RS >> 1 / g m
Gm 1 / RS
ID is linearized at the cost of lower gain. Small Signal model including body effect and channel length modulation:
I out = g mV1 g mbVX
VX rO I out RS rO
= g m (Vin I out RS ) + g mb ( I out RS ) Gm =
I out g m rO = Vin RS + [1 + ( g m + g mb ) RS ]rOUniversity of Paris VI
H. Aboushady
9
With and Without Source DegenerationGm = g m rO 1 + [1 + ( g m + g mb ) RS ]rO
RS = 0
RS 0
H. Aboushady
University of Paris VI
Estimating Gain by InspectionAv = g m RD RD = 1 + g m RS 1 / g m + RS
Gain =
Resistance seen at the Drain Total Resistance in the Source Path
Example:
Av =
RD 1 / g m1 + 1 / g m 2
H. Aboushady
University of Paris VI
10
Output Resistance of Degenerated CSV1 = I X RSThe current flowing in rO :
I X ( g m + g mb )V1
= I X + ( g m + g mb ) RS I X VX = rO [ I X + ( g m + g mb ) RS I X ] + I X RSRout = VX = rO [1 + ( g m + g mb ) RS ] + RS IX
Rout = [1 + ( g m + g mb )rO ]RS + rO Rout ( g m + g mb )rO RS + rOH. Aboushady
Rout = [1 + ( g m + g mb ) RS ]rOUniversity of Paris VI
Voltage Gain of Degenerated CSThe current through RS must equal that through RD:
I R D = I RS = VS = Vout
Vout RD RS RD I rO =
VS
Vout ( g mV1 + g mbVBS ) RD R V R V Vout = I rO rO out RS I rO = out [ g m (Vin + Vout S ) + g mbVout S ] RD RD RD RD R R R V Vout = out rO [ g m (Vin + Vout S ) + g mbVout S ]rO Vout S RD RD RD RDThe current through rO :
H. Aboushady
Vout g m rO RD = Vin RD + RS + rO + ( g m + g mb ) RS rO
University of Paris VI
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Voltage Gain of Degenerated CSVout g m rO RD = Vin RD + RS + rO + ( g m + g mb ) RS rO Vout g m rO RD [ RS + rO + ( g m + g mb ) RS rO ] = Vin RS + [1 + ( g m + g mb ) RS ]rO RD + RS + rO + ( g m + g mb ) RS rO Vout = Gm (Rout // RD ) VinThe output resistance of a degenerated CS stage: The Transconductance of a degenerated CS stage:H. Aboushady
Rout = [1 + ( g m + g mb ) RS ]rO Gm = I out g m rO = Vin RS + [1 + ( g m + g mb ) RS ]rOUniversity of Paris VI
General expression to calculate Av by inspectionLemma:
Av = Gm RoutGm : the transconductance of the circuit when the output is shorted to grounded. Rout : the output resistance of the circuit when the input voltage is set to zero.
For high voltage gain the output resistance must be high! A buffer is needed to drive a low-impedance load. The source follower can operate as a voltage buffer.H. Aboushady University of Paris VI
12
Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
Source Follower (Common Drain)Large Signal Behavior M1 turns on in saturation:
Vout = I D RSVout =
nCox W2 L
(Vin Vout VTH ) 2 RS
To calculate gm :
W V V Vout = nCox (Vin Vout VTH )(1 out TH ) RS L Vin Vin Vin
VTH = VTH 0 + 2F + VSB 2F VSB VTH VTH VSB = = Vin VSB Vin 2 2F + VSB Vin Vout = Vin H. AboushadySince,
(
)
University of Paris VI
13
Source Follower Voltage GainW V V Vout = nCox (Vin Vout VTH )(1 out TH ) RS L Vin Vin Vin W V V Vout = nCox (Vin Vout VTH )(1 out out ) RS L Vin Vin Vin
W nCox (Vin Vout VTH ) RS Vout L = Vin 1 + C W (V V V ) R (1 + ) n ox in out TH S LWe also have,
g m = nCox Av =
W (Vin Vout VTH ) L
g m RS 1 + ( g m + g mb ) RSUniversity of Paris VI
H. Aboushady
Source Follower Voltage GainSmall Signal Equivalent Circuit
Vout = [g mV1 + g mbVBS ]RS
= [g m (Vin Vout ) g mbVout ]RSVout g m RS = Vin 1 + ( g m + g mb ) RS
Av =Since: And for :
g mb = g m g m RS >> 1 Av 1 (1 + )University of Paris VI
H. Aboushady
14
Source Follower Output ResistanceRout : the output resistance when the input voltage is set to zero.
V1 = VBS = VX I X g mVX g mbVX = 0Rout =
VX 1 = I X g m + g mb
Body Effect decreases the output resistance of source followers.
VX VGS and VTH ID H. Aboushady University of Paris VI
Source Follower body effectRout : the output resistance when the input voltage is set to zero. Small Signal Model Simplification
Note that the value of the current source gmbVbs is linearly proportional to the voltage across it.
Rout =H. Aboushady
1 1 1 // = g m g mb g m + g mbUniversity of Paris VI
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Source Follower Thvenin Equivalent
Av =
1 g mb 1 1 + g m g mb
=
gm g m + g mb
H. Aboushady
University of Paris VI
Channel Length Modulation in M1 and M2
Av =
1 // rO1 // rO 2 // RL g mb 1 1 // rO1 // rO 2 // RL + g mb gm
H. Aboushady
University of Paris VI
16
Source Follower Characteristics+ High input impedance and Moderate output impedance - Nonlinearity - Limited voltage swing Example:
VTH VSBPMOS source follower with VSB=0
Without the source follower stage:
VX > VGS 1 VTH 1
p < n g mp < g mn Routp > RoutnH. Aboushady
With the source follower stage:
VX > VGS 2 + (VGS 3 VTH 3 )University of Paris VI
Low Load Impedance: CS vs SFSource Follower Amplifier Common Source Amplifier
AvSF
RL RL + 1 / g m
AvCS g m RL
Assuming RL=1/gm
AvSF 1 / 2
AvCS 1
Source Followers are not necessarily efficient drivers.H. Aboushady University of Paris VI
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Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
Common Gate StageLarge Signal Behavior
Vout = VDD I D RDAssuming M1 in saturation:
Vout = VDD
nCox W (Vb Vin VTH ) 2 RD 2 L RD
V W Vout = n Cox (Vb Vin VTH ) 1 TH L Vin Vin W Vout = n Cox (Vb Vin VTH )(1 + )RD L Vin
Av = g m (1 + ) RDH. Aboushady University of Paris VI
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Common Gate Stage Input ResistanceSame as Output Resistance of Source Follower:
Rin =
1 g m + g mb
Body Effect: increases Av decreases Rin
H. Aboushady
University of Paris VI
Common Gate GainSmall Signal Signal Equivalent Circuit
The current through RS is equal to -Vout / RD :
V1
Vout RS + Vin = 0 RD
The current through rO is equal to -Vout / RD - gmV1 - gmbV1 :
V rO out RDH. Aboushady
V V rO out g mV1 g mbV1 out RS + Vin = Vout R R D D V R ( g m + g mb )Vout S Vin out RS + Vin = Vout RD RDUniversity of Paris VI
19
Common Gate GainCommon Gate Amplifier:
AvCG =
( g m + g mb )rO + 1 RD RD + RS + rO + ( g m + g mb )rO RS
Degenerated Common Source Amplifier:
AvCS =
g m rO RD RD + RS + rO + ( g m + g mb )rO RS
H. Aboushady
University of Paris VI
Common Gate Stage Input ResistanceSince V1 = -VX :
VX = RD I X + rO [I X ( g m + g mb )VX ]VX RD + rO = I X 1 + ( g m + g mb )rO Rin RD 1 + ( g m + g mb )rO ( g m + g mb ) Replace RD by ideal current source:
Assume RD = 0 :
Rin =
1 1 / rO + ( g m + g mb )
Rin =
Rin of a common gate stage is low only if RD is small.H. Aboushady University of Paris VI
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Common Gate Stage Output ImpedanceSimilar to Output Impedance of a Degenerated Common Source Stage
Rout = ([1 + ( g m + g mb )rO ]RS + rO ) // RD
H. Aboushady
University of Paris VI
Single Stage Amplifiers
Basic Concepts Common Source Stage Source Follower Common Gate Stage Cascode Stage
Hassan Aboushady University of Paris VI
21
Biasing of a Cascode StageThe cascade of CS stage and a CG stage is called cascode. M1 : the input device M2 : the cascode device Biasing conditions: M1 in saturation:
VX = Vb VGS 2 Vb VGS 2 Vin VTH 1 Vb Vin + VGS 2 VTH 1 M2 in saturation:
Vout VX Vb VX VTH 2 Vout Vin VTH 1 + VGS 2 VTH 2H. Aboushady University of Paris VI
Cascode Stage CharacteristicsLarge signal behavior: As Vin goes from zero to VDD For Vin < VTH M1 and M2 are OFF
Vout =VDD
Output Resistance: Same common source stage with a degeneration resistor equal to rO1
Rout = [1 + ( g m 2 + g mb 2 )rO 2 ]rO1 + rO 2 Rout ( g m 2 + g mb 2 )rO 2 rO1 M2 boosts the output impedance of M1 by a factor of gmr02 Triple cascode Rout difficult biasing at low supply voltage.H. Aboushady University of Paris VI
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Cascode Stage Voltage GainAv = Gm Rout Gm g m1Ideal Current Source:
Rout ( g m 2 + g mb 2 )rO 2 rO1 Av ( g m 2 + g mb 2 )rO 2 g m1rO1Cascode Current Source:
Rout g m 2 rO 2 rO1 // g m3rO 3rO 4 Av g m1 ( g m 2 rO 2 rO1 // g m 3rO 3 rO 4 )
H. Aboushady
University of Paris VI
Shielding PropertyAssume VX is higher than VY by V. Calculate the resulting difference between ID1 and ID2 (with 0 ).
I D1 I D 2 = I D1 I D 2
nCox 2 C = n ox 2
W (Vb VTH ) 2 (VDS 1 VDS 2 ) L W (Vb VTH ) 2 ( VDS ) L
VPQ = V
rO1 [1 + ( g m 3 + g mb3 )rO 3 ]rO1 + rO 3
V ( g m3 + g mb 3 )rO 3
I D1 I D 2 =H. Aboushady
nCox W V (Vb VTH ) 2 2 L ( g m3 + g mb 3 )rO 3University of Paris VI
23
Folded Cascode
Simple Folded Cascode
Folded Cascode with biasing
Folded Cascode with NMOS input
Large Signal Characteristics:
H. Aboushady
University of Paris VI
Output Resistance of Folded CascodeDegenerated Common Source Stage:
Rout = [1 + ( g m1 + g mb1 )rO1 ]RS + rO1
Folded Cascode Stage:
M1 RS
M2 rO1 // rO3
Rout = [1 + ( g m 2 + g mb 2 )rO 2 ](rO1 // rO 3 ) + rO 2H. Aboushady University of Paris VI
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