Transcript ANALOG CMOS SUBCIRCUITS

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:
VOut
AV 
V In
CURRENT GAIN
 Defined as the ratio of ac output current to ac
input current
Mathematically, expressed as:
I Out
AI 
I In
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 STAGE
with resistive load
Commonsource 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.
Vout
1
W
2
 VDD  RD  nCox Vin  VTH 
2
L
 For (Vin>Vin1) – M1 is in Triode Mode.
Vout  VDD  RD

1
W
2
 nCox
2Vin  VTH Vout  Vout
2
L

 If Vin is high enough to drive M1 into deep triode region, Vout
VDD
<< 2(Vin – VTH)
V 
Vout  VDD
Ron
Ron  RD
out
1   nCox
W
RD Vin  VTH 
L
Voltage gain
Vout
W
Av 
  RD  nCox Vin  VTH    g m RD
Vin
L
Av   2n C
W
ox
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
" intri nsi c gain"
COMMON-SOURCE STAGE
with diode-connected load
 “Diode-Connected” MOSFET is only a name. In
BJT if Base and Collector are short-circuited,
then the BJT acts exactly as a diode.
 We want to replace RD with a MOSFET that will
operate like a small-signal resistor.
V1  V X  I X 
 Rd 
VX
 g mVX
ro
1
1
|| ro 
gm
gm
Common Source with Diode-Connected NMOS
Load (Body Effect included)
Vx
(gm  gmb)V x   Ix
ro
Vx
1
1

|| ro 
Ix gm  gmb
gm  gmb
CS with Diode-Connected NMOS
Load Voltage Gain Calculation
Av   gm1
1
gm1 1

gm2  gmb2
gm2 1  
2nCOX (W / L)1 I D1 1
Av  
2nCOX (W / L) 2 I D 2 1  
(W / L)1 1
Av  
(W / L)2 1  
CS with Diode-Connected NMOS Load Voltage
Gain Calculation
(W / L)1 1
Av  
(W / L)2 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!)
W 
W 
0.5n COX   (Vin  VTH1 ) 2  0.5n COX   (VDD  Vout  VTH 2 ) 2
 L 1
 L 2
W 
W 
  (Vin  VTH1 )    (VDD  Vout  VTH 2 )
 L 1
 L 2
Assumptions made along the way:
W 
W 
  (Vin  VTH1 )    (VDD  Vout  VTH 2 )
 L 1
 L 2
 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:
W 
W 
  (Vin  VTH1 )    (VDD  Vout  VTH 2 )
 L 1
 L 2
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
 n (W / L)1
Av  
 p (W / L) 2
No body effect!
Numerical Example
 Say that we want the voltage gain to be 10.
Then:
Av  
 n (W / L)1
 10
 p (W / L) 2
 n (W / L)1

 100
 p (W / L) 2
Example continued
n (W / L)1
 100
 p (W / L) 2
 Typically
 n  2 p
Example continued
(W / L )1
 50
(W / L) 2
 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.
CS with Diode-Connected Load Swing Issues
I D1  I D2 , 
W 
W 
2
n   (VGS1  VTH1)   p   (VGS2  VTH2) 2
 L 1
 L 2
 n (W / L)1 | VGS2  VTH2 |
Av  

 p (W / L) 2 (VGS1  VTH1)
This implies substantial voltage swing constraint. Why?
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
Av   gm1 (
|| ro1 || ro 2 )
gm2  gmb2
1
Av   g m1 (
|| ro1 || ro 2 )
gm2
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):
4 n (W / L)1
Av  
 p (W / L) 2
Example: Swing Issues
I D1  4I D 2 , 
W 
W 
2
n   (VGS1  VTH1)  4 p   (VGS2  VTH2) 2
 L 1
 L 2
Av
4
| VGS2  VTH2 |

(VGS1  VTH1)
It sure helps in terms of swing.
COMMON-SOURCE STAGE
with current-source load
How can Vin change the current of M1 if I1
is constant?
W 
I D1  0.5n COX   (Vin  VTH1 ) 2 (1  Vout )  I1
 L 1
 As Vin increases, Vout must decrease
Simple Implementation: Current Source obtained from
M2 in Saturation
CS with Current Source Load
Av   g m ( ro1 || ro2)
DC Conditions
W 
I D1  0.5 n C OX   (Vin  VTH1 ) 2 (1  1Vout )  I D 2
 L 1
W 
 0.5 p C OX   (Vb  VDD  VTH 2 ) 2 (1  2 [Vout  VDD ])
 L 2
 {(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
Swing Considerations
W 
I D1  0.5 n C OX   (Vin  VTH1 ) 2 (1  1Vout )  I D 2
 L 1
W 
 0.5 p C OX   (Vb  VDD  VTH 2 ) 2 (1  2 [Vout  VDD ])
 L 2
 We can make |VDS2|>|VGS2-VTH2| small (say a
few hundreds of mV), if we compensate by
making W2 wider.
How do we make the gain large?
Av   g m ( ro1 || ro2)
 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)
g m1  2  nCOX (W / L)1 I D  1.06mA / V
ro1  8000 1.6 / 0.1  128 K
ro 2  12000 1.6 / 0.1  192 K
AV   g m1 (ro1 || ro 2 )  81.4
How does L influence the gain?
Av  gm ro1 || ro2
Assuming ro2 large,


ox D 

W  1
Av   g m ro1   2 nC I

L  1 I D
How does L influence the gain?


ox D 

W  1
Av   g m ro1   2 nC I

L  1 1 I D
 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.
COMMON-SOURCE STAGE
with triode load
Av   g m1RON 2
RON 2 
1
 p Cox WL  (VDD  Vb  VTH 2)
2
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-VDDVTH2)>>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 STAGE
with source degeneration
gm
Gm 
1  g m RS
Av  Gm RD
Summary of key formulas,
neglecting channel-length
modulation and body
effect
 g m RD
Av 
1  g m RS
CS Amplifier with Source Degeneration –
Formula Explained
V1  Vin  I D RS  Vin  g mV1 RS
Vin
 V1 

1  g m RS
 Vout   g mV1 RD
 g m RD
Av 
1  g m RS
Idea is the same as that of adding emitter resistor to a
BJT CE amplifier
 g m RD
AV 
1  g m RS
 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
CS Amplifier with Source Degeneration Key Formulas
with λ and Body-Effect Included
gm
Gm  R
S
 [1 (gm  gmb )RS ]
ro
ROUT  [1  (gm  gmb )ro ]RS  ro
Av  Gm ( RD || ROUT )
Derivation is similar to that of the simplified case –
generalized transconductance
I
It doesn’t matter what RD is because we treat ID as “input”
I D  g mV1  g mbVbs  I ro
 g mV1  g mbVX 
I D RS
ro
I D RS
 g m (Vin  I D RS )  g mb ( I D RS ) 
ro
 Gm 
g m ro
ID

Vin RS  [1  ( g m  g mb ) RS ]ro
Linearizing effect of RS:
gm RD
RD
Av 
 
1 gm RS
1/ gm  RS
Linearizing effect of RS:
gm RD
RD
Av 
 
1 gm RS
1/ gm  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.
Estimating Gain by Inspection
gm RD
RD
Av 
 
1 gm RS
1/ gm  RS
 Denominator: Resistance seen the
Source path, “looking up” from
ground towards Source.
 Numerator: Resistance seen at
Drain.
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
I X  g mV1  g mbVbs  I ro
 g m ( I X RS )  g mb I X RS  I ro
VX  ro ( I X  ( g m  g mb ) RS I X )  I X RS
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
gm
Gm  R
S
 [1 (gm  gmb )RS ]
ro
ROUT  [1  (gm  gmb )ro ]RS  ro
Av  Gm ( RD || ROUT )
Derivation of Gain Formula
Vout
I ro  
 ( g mV1  g mbVbs )
RD
Vout
RS
RS

 [ g m (Vin  Vout
)  g mbVout
)
RD
RD
RD
Derivation of Gain Formula
Vout  I ro ro 
Vout
RS 
RD
Vout
RS
RS
RS

ro  [ g m (Vin  Vout
)  g mbVout
]ro  Vout
RD
RD
RD
RD
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.