SINGLE-STAGE AMPLIFIERS EMT451/4. AMPLIFIER : DEFINITIONS Amplification the process of increasing...
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Transcript of SINGLE-STAGE AMPLIFIERS EMT451/4. AMPLIFIER : DEFINITIONS Amplification the process of increasing...
SINGLE-STAGE AMPLIFIERS
EMT451/4
AMPLIFIER : DEFINITIONS
Amplification the process of increasing an ac signal’s power
Amplifier The circuit that amplifies
Two Amplifier properties that are of interest to us: Gain Impedance
AMPLIFICATION?
Essential function in integrated circuits Signal too small:
to drive load Overcome noise of subsequent stage Provide the right level to the subsequent circuit (e.g. logic level)
Analyze large-signal & small-signal characteristics of single-stage amplifiers (CS, CG, CD, cascode) Goal: develop intuitive technique and fundamental models
Need to be able to simplify analysis yet maintain acceptable accuracy
Approach: simple model -> add second order phenomena (e.g. body effect, channel-length modulation)
TRADE-OFFS in ANALOG DESIGN
AMPLIFIER GAIN
Amplifier Ratio of output signal to input signal
Ratio < 1: attenuator Ratio = 1: buffer Ratio > 1: amplifier
3 types of gains associated with an amplifier Voltage gain Current gain Power gain
VOLTAGE GAIN
Defined as the ratio of ac output voltage to ac input voltage
Or, the mathematical expression:
In
OutV V
VA
CURRENT GAIN
Defined as the ratio of ac output current to ac input current
Mathematically, expressed as:
In
OutI I
IA
AMPLIFIER IMPEDANCE
When a signal (current or voltage) is fed into the input, a portion of it will not get through the amplifier. This is due to external resistance effects.
2 types of impedances associated with an amplifier:
Input impedance
Output impedance
SINGLE-STAGE AMPLIFIER
Single-stage amplifiers are used in virtually every op-amp design
By replacing a passive load (resistor) with a MOS transistor (called an active load), minimize chip area
Active load : produce higher values of resistance higher gain
Types of active load : G-D load and current source load.
COMMON-SOURCE STAGEwith resistive load
Common-source stage
Equivalent circuit in
deep triode region
Input-output
characteristic
Small-signal
model for saturation
region
ANALYSIS
If Vin is low (below threshold) transistor M1 is OFF
For Vin not-too-much-above threshold, transistor M1 is in Saturation, and Vout decreases.
For (Vin>Vin1) – M1 is in Triode Mode.
If Vin is high enough to drive M1 into deep triode region, Vout << 2(Vin – VTH)
2
2
1THinoxnDDDout VV
L
WCRVV
222
1outoutTHinoxnDDDout VVVV
L
WCRVV
Don
onDDout RR
RVV
THinDoxn
DDout
VVRL
WC
VV
1
Voltage gain
DmTHinoxnDin
outv RgVV
L
WCR
V
VA
Av 2nCoxW
L
VRD
ID
Design tradeoffs
Gain is determined by 3 factors: W/L, RD , DC voltage and ID
If current and RD are kept constant, an increase of W/L increases the gain, but it also increases the gate capacitance – lower bandwidth
If ID and W/L are kept constant, and we increase RD, then VDS becomes smaller. Operating Point gets closer to the borderline of Triode Mode.
It means – less “swing” (room for the amplified signal)
If ID decreases and W/L and VRD are kept constant, then we must increase RD
Large RD consumes too much space, increases the noise level and slows the amplifier down (time constant with input capacitance of next stage).
COMMON-SOURCE tradeoffs
Av gm ro || RD
Larger RD values increase the influence of channel-length modulation (ro term begins to strongly affect the gain)
Common Source Maximum Gain
Av gm ro || RD Av gmro
" intrinsic gain"
“Diode-Connected” MOSFET is only a name. In BJT if Base and Collector are short-circuited, then the BJT acts exactly as a diode.
COMMON-SOURCE STAGEwith diode-connected load
We want to replace RD with a MOSFET that will operate like a small-signal resistor.
mo
md
Xmo
XXX
gr
gR
Vgr
VIVV
1||
1
1
Common Source with Diode-Connected NMOS Load (Body Effect included)
Vx
Ix
1
gm gmb|| ro
1
gm gmb
(gm gmb)Vx Vx
ro Ix
CS with Diode-Connected NMOS Load Voltage Gain Calculation
Av gm1 1
gm2 gmb2
gm1
gm2
1
1
1
1
)/(2
)/(2
22
11
DOXn
DOXnv
ILWC
ILWCA
Av (W / L)1
(W / L)2
1
1
CS with Diode-Connected NMOS Load Voltage Gain Calculation
Av (W / L)1
(W / L)2
1
1
If variations of η with the output voltage are neglected, the gain is independent of the bias currents and voltages (as long as M1 stays in Saturation)
The point: Gain does not depend on Vin
Amplifier is relatively linear (even for large signal analysis!)
22
2
2
1
)(5.0)(5.0 THoutDDCnTH1inCn VVVL
WVV
L
WOXOX
)()( 221
THoutDDTH1in VVVL
WVV
L
W
Assumptions made along the way:
)()( 221
THoutDDTH1in VVVL
WVV
L
W
We neglected channel-length modulation effect We neglected VTH voltage dependent variations
We assumed that two transistors are matching in terms of COX.
Comment about “cutting off”
If I1 is made smaller and smaller, what happens to Vout?
It should be equal to VDD at the end of the current reduction process, or is it?
Comment about “cutting off” (cont’d)
If I1 is made smaller and smaller, VGS gets closer and closer to VTH.
Very near I1=0, if we neglect sub-threshold conduction, we should have VGS≈VTH2, and therefore Vout≈VDD-VTH2 !
Cutoff conflict resolved:
In reality, sub-threshold conduction, at a very low current, eventually brings Vout to the value VDD.
Output node capacitance slows down this transition. In high-speed switching, sometimes indeed Vout doesn’t
make it to VDD
Back to large-signal behavior of the CS amplifier with diode-connected NMOS load:
)()( 221
THoutDDTH1in VVVL
WVV
L
W
Large signal behavior of CS amplifier with diode-connected load
When Vin<VTH1 (but near), we have the above “imperfect cutoff” effect.
Then we have a, more or less, linear region. When Vin exceeds Vout+VTH1 amplifier enters the
Triode mode, and becomes nonlinear.
CS with Diode-Connected PMOS Load Voltage Gain
2
1
)/(
)/(
LW
LWA
p
nv
No body effect!
Numerical Example
Say that we want the voltage gain to be 10. Then:
100)/(
)/(
10)/(
)/(
2
1
2
1
LW
LW
LW
LWA
p
n
p
nv
Example continued
100)/(
)/(
2
1 LW
LW
p
n
Typically pn 2
Example continued
Need a “strong” input device, and a “weak” load device.
Large dimension ratios lead to either a larger input capacitance (if we make input device very wide (W/L)1>>1) or to a larger output capacitance (if we make the load device very narrow (W/L)2<<1).
Latter option (narrow load) is preferred from bandwidth considerations.
50)/(
)/(
2
1 LW
LW
CS with Diode-Connected Load Swing Issues
)(
||
)/(
)/(
2
1
TH1GS1
TH2GS2
p
nv
VV
VV
LW
LWA
,21 DD II
This implies substantial voltage swing constraint. Why?
2
2
2
1
)()( TH2GS2pTH1GS1n VVL
WVV
L
W
Numerical Example to illustrate the swing problems
Assume for instance VDD=3V Let’s assume that VGS1-VTH1=200mV (arbitrary
selection, consistent with current selection) Assume also that |VTH2|=0.7V (for PMOS load
VTH2=-0.7V) For a gain of 10, we now need |VGS2|=2.7V (for
PMOS load VGS2<-2.7V) Therefore because VGD2=0: VDS2=VGS2=-2.7V). Now VDS1=VDD-VSD2<3-2.7=0.3V, and recall that
VDS1>VGS1-VTH1=0.2V. Not much room left for the amplified signal vds1.
DC Q-Point of the Amplifier
VG1 determines the current and the voltage VGS2
If we neglect the effect of the transistors’ λ, the error in predicting the Q-point solution may be large!
CS with Diode-Connected Load –How do we take into account λ?
)||||1
21( oomb2m2
m1v rrgg
gA
)||||1
21(1 oo
m2v rr
ggA
m
Example as Introduction to Current Source Load
M1 is biased to be in Saturation and have a current of I1.
A current source of IS=0.75I1 is hooked up in parallel to the load – does this addition ease up the amplifier’s swing problems?
Example as Introduction to Current Source Load
Now ID2=I1/4. Therefore (from the ratio of the two transconductances with different currents):
2
1
)/(
)/(4
LW
LWA
p
nv
Example: Swing Issues
)(
||4
TH1GS1
TH2GS2A
VV
VVv
,4 21 DD II
It sure helps in terms of swing.
2
2
2
1
)(4)( TH2GS2pTH1GS1n VVL
WVV
L
W
COMMON-SOURCE STAGEwith current-source load
How can Vin change the current of M1 if I1 is constant?
As Vin increases, Vout must decrease
12
11 )1()(5.0 IVVV
L
WI outTH1inCnD OX
Simple Implementation: Current Source obtained from M2 in Saturation
CS with Current Source Load
)||( o2o1mv rrgA
DC Conditions
{(W/L)1,VG1} and {(W/L)2,Vb} need to be more or less consistent, if we wish to avoid too much dependence on λ values.
Need DC feedback to fix better the DC Vout
])[1()(5.0
)1()(5.0
22
22
212
11
DDoutTHDDbCp
DoutTH1inCnD
VVVVVL
W
IVVVL
WI
OX
OX
Swing Considerations
We can make |VDS2|>|VGS2-VTH2| small (say a few hundreds of mV), if we compensate by making W2 wider.
])[1()(5.0
)1()(5.0
22
22
212
11
DDoutTHDDbCp
DoutTH1inCnD
VVVVVL
W
IVVVL
WI
OX
OX
How do we make the gain large?
)||( o2o1mv rrgA
Recall: λ is inversely proportional to the channel length L. To make λ values smaller (so that ro be larger) need to
increase L. In order to keep the same current, need to increase W by the same proportion as the L increase.
CS Amplifier with Current-Source Load Gains
Typical gains that such an amplifier can achieve are in the range of -10 to -100.
To achieve similar gains with a RD load would require much larger VDD values.
For low-gain and high-frequency applications, RD load may be preferred because of its smaller parasitic capacitance (compared to a MOSFET load)
Numerical Example
Let W/L for both transistors be W/L = 100µm / 1.6µm
Let µnCox=90µA/V2, µpCox=30µA/V2
Bias current is ID=100µA
Numerical Example (Cont’d)
Let ro1=8000L/ID and ro2=12000L/ID where L is in µm and ID is in mA.
What is the gain of this stage?
Numerical Example (Cont’d)
4.81)||(
1921.0/6.112000
1281.0/6.18000
/06.1)/(2
211
2
1
11
oomV
o
o
DOXnm
rrgA
Kr
Kr
VmAILWCg
How does L influence the gain?
Av gm ro1 || ro2
Assuming ro2 large,
DDoxnomv IL
WICrgA
1
21
1
How does L influence the gain?
DDoxnomv IL
WICrgA
11
1
12
As L1 increases the gain increases, because λ1 depends on L1 more strongly than gm1 does!
As ID increases the gain decreases. Increasing L2 while keeping W2 constant increases ro2
and the gain, but |VDS2| necessary to keep M2 in Saturation increases.
21 ONmv RgA
)(
1
22
2THbDDL
Woxp
ONVVVC
R
COMMON-SOURCE STAGEwith triode load
How should Vb be chosen?
M2 must conduct: Vb-VDD≤VTH2, or Vb≤VDD+VTH2
M2 must be in Triode Mode: Vout-VDD≤Vb-VDD-VTH2, or Vb≥Vout+VTH2
M2 must be “deep inside” Triode Mode: 2(Vb-VDD-VTH2)>>Vout-VDD, or: Vb>>VDD/2+VTH2+Vout/2
Also Vb and (W/L)2 determine the desired value of Ron2
It’s not easy at all to determine (and implement) a working value
for Vb
CS Amplifiers with Triode Region load is rarely used
COMMON-SOURCE STAGEwith source degeneration
Av GmRD
Sm
Dmv Rg
RgA
1
Sm
mm Rg
gG
1
Summary of key formulas, neglecting channel-length modulation and body effect
CS Amplifier with Source Degeneration – Formula Explained
Sm
Dmv Rg
RgA
1
Dmout
Sm
in
SminSDin
RVgV
Rg
VV
RVgVRIVV
1
1
11
1
Idea is the same as that of adding emitter resistor to a BJT CE amplifier
With RS=0, the gain strongly depends on gm, and as we recall gm depends on the input’s amplitude, temperature and other effects.
If gmRS>>1, then AV≈-RD/RS
Sm
DmV Rg
RgA
1
CS Amplifier with Source Degeneration Key Formulas with λ and Body-Effect Included
Gm gm
RS
ro
[1 (gm gmb )RS ]
)||( OUTDmv RRGA
ROUT [1 (gm gmb )ro ]RS ro
Derivation is similar to that of the simplified case – generalized transconductance
oSmbmS
om
in
Dm
o
SDSDmbSDinm
o
SDXmbm
robsmbmD
rRggR
rg
V
IG
r
RIRIgRIVg
r
RIVgVg
IVgVgI
])(1[
)()(
1
1
IIt doesn’t matter what RD is because we treat ID as “input”
Linearizing effect of RS:
Av gm RD
1 gm RS
RD
1/ gm RS
Linearizing effect of RS:
For low current levels 1/gm>>RS and therefore Gm≈gm.
For very large Vin, if transistor is still in Saturation, Gm approaches 1/RS.
Av gm RD
1 gm RS
RD
1/ gm RS
Estimating Gain by Inspection
Denominator: Resistance seen the Source path, “looking up” from ground towards Source.
Numerator: Resistance seen at Drain.
Av gm RD
1 gm RS
RD
1/ gm RS
Example to demonstrate method:
Note that M2 is “diode-connected”, thus acting like a resistor 1/gm2
AV=-RD/(1/gm1+1/gm2)
RS effect on CS Output Resistance
roSXmbSXm
robsmbmX
IRIgRIg
IVgVgI
)(1
SXXSmbmXoX RIIRggIrV ))((
CS Output Resistance
ROUT [1 (gm gmb )ro ]RS ro
ROUT ro' ro [1 (gm gmb )RS ]
RS causes a significant increase in the output resistance of the amplifier
CS Amplifier with Source Degeneration Gain Formula with λ and Body-Effect Included
ROUT [1 (gm gmb )ro ]RS ro
Gm gm
RS
ro
[1 (gm gmb )RS ]
)||( OUTDmv RRGA
Derivation of Gain Formula
))([
)( 1
D
Soutmb
D
Soutinm
D
out
bsmbmD
outro
R
RVg
R
RVVg
R
V
VgVgR
VI
Derivation of Gain Formula
D
Souto
D
Soutmb
D
Soutinmo
D
out
SD
outoroout
R
RVr
R
RVg
R
RVVgr
R
V
RR
VrIV
])([
Now we can relate Vout to Vin - formula results
General Networks Result
Voltage gain in any linear circuit equals –GmRout
Gm is circuit transconductance when output is shorted to ground
Rout is output resistance when circuit’s input voltage is set to zero.
General Networks Result
–GmRout formula is useful if Gm and Rout can be determined by inspection.
Proof: Rout is Norton equivalent. Vout=-IoutRout
Gm=Iout/Vin, which leads to the result.