cmos lecture5-1.pdf
TRANSCRIPT
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Zou Zhige VLSI, EST 3
Last Class Exercise
Calculate the
Output voltage swing
(Vx, Vy, Vout=Vx-Vy)
low frequency Av
1in
out
2
DD
SS
b 3 4
( )//V m on opA g r r=
, , ,
, 1
,
in com thn x y DD od mp
in com GS P
od mn P
V V V V V
V V V
V V
< <
= +
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Introduction
We have learned current source:Current source can act as a large resistorwithout consuming excessivevoltage headroom.
MOS in saturation can act as current source.All the amplifier need current source or resistor for load current.
Current source has many applications:
current source load for common source amplifiers
tail current source for differential pairsbias currents for folded cascode amplifierDAC
Current count and so on.This Chapter we will learn current mirrorforbias elements and signalprocessing components.
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Table of Contents
Introduction
Basic Current Mirrors
Cascode Current Mirrors
Active Current Mirrors
Introduction
Large-signal analyses
Small-signal analyses Common response
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Current Source in Differential amplifier
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Basic Current Mirrors
The simplest current source is as follows:
2 ,1
( )2 n oxout TGS
WI C V V
L=
2
1 2DD
GS
RV V
R R=
+
Is this current source
always constant ?
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The problem of Basic CM
Influence of: power supply, process, temperature
Vgs and Vth are not constant.
Even if the Vgs is precisely defined, the Id is not constant.
, Vth will change due to different process or temperature
We must seek other method of biasing MOS currentsources.
Current copy from a constant current reference!
Why?
But how to
copy?
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Copying Current
Current
Mirror
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Basic Current Mirror
Both in saturation and Neglecting the Lambda, We get:
It allows precise copying of the current with no dependence on processand temperature !
2 1( ) /( )out ref W W
I I
L L
=
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Current Source & Current Sink
current source
current sink
out1
REF
out2
DD
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Amplifier Bias Example
All the MOS should have the same channel length!
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Another Example
2
31
)/()/(
LWLWRgA LmV=
Current Transfer!
)||()||1
||( 3322
11 Lomo
m
omV Rrgrg
rgA =
Av=?
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Table of Contents
Introduction
Basic Current Mirrors
Cascode Current Mirrors
Active Current Mirrors
Introduction
Large-signal analyses
Small-signal analyses Common response
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Drawback of Basic CM
Consider the channel length modulation effect:
Because the load is not constant, so:
We should reduce the channel length modulation. Or there isother method?
)1()(
2
1 2DSTHGSoxnOUT VVV
L
WCuI +=
)1(
)1(
1)/(
2)/(
1
2
1
2
DS
DS
D
D
V
V
LW
LW
I
I
+
+=
21 DSDS VV
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How about this current mirror ?
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Simple cascode amplifier
Large signal behavior (Vin fixed to VG1, Vout (VDS) sweeping from 0 to 3V)
Which one is better for the ideal current source?
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Cascode Current Mirror
Choose , let ,then
Output resistance:
Trade-off: accuracy and voltage headroom
3 3 2pout m o oR g r r=
YX VV =bV REFout II
2
3 3 2 3 3
1oy p p
m o o m o
rV V V
g r r g r = =
2oY rR =
How to get Vb ?
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Cascode Current Mirror
To ensure:
We should have :
XGSb VVV += 3
YX VV =
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Cascode Current Mirror
Add M0 for Vb
Choose proper dimension of M0 and M3 for VGS3=VGS0
If Then:3 1 2( ) /( ) ( ) /( )oW W W W
L L L L=
YX VV =
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Region of M2 and M3
M2 and M3 both are in triode
M2 in sat, and M3 in triode
M2 and M3 both are in sat
When Vout
if M2
triode first, but Vgs2
constant
so Vds2 and Id2,3
Vb constant, Vgs3 its impossible for M3 still in saturation.
Conclusion: M3 enter triode zone first, then M2
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Voltage Headroom 1
min 2 2 21y GS T odV V V V = =
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Voltage Headroom 2
min 3 2 3
1 3
2
p GS od TH
od od
od
V V V V V V
V
= = +
=
Notice: YX VV
REFout II So:
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Voltage Headroom 3
min 3 2
3 2 2 2
2
p od GS
od GS TH TH
od TH
V V V
V V V V
V V
= +
= + += +
To ensure:YX VV =
REFout II So:
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Low voltage Cascode CM
In Sub-micro process, the ro isvery small. Cascode is useful.
In sub-micro process, the voltage
supply is very low. So, we should reduce the voltage
headroom.
Here is A good example
out
1
2
3
4b
DD
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Low voltage Cascode CM
M1 in satu
M2 in satu
Range of Vb
The over-drive voltage of M2 is lower than Vth1
2 2 1GS TH TH V V V
2x b THV V V
2 2 1x T H b G S x T HV V V V V V + +
112 HTGSGSbA VVVVV =
1GSX VV =
out
1
2
3
4
b
DD
122 THXGSAGSb VVVVVV ++=
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Vb of Low voltage Cascode CM M1 at the edge of satu:
Vout reaches the lowest level (headroom)
min 4 2 1 4p b TH GS x TH THV V V V V V V = = +
4 4 3 4 3GS TH x TH od od V V V V V V = + = +
)( 112 THGSGSb VVVV +=
out
1
2
3
4b
DD
How to generate Vb?
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How to generate Vb?
The left figure:
I*R=VTH1
VGS2=VGS5
The right figure:
We need big (W/L)7 for VGS7=VTH1
VGS2=VGS5
)( 112 THGSGSb VVVV +=
Are the above result always correct?
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General Advantages of MOS Current Sources
Effective current gain
No dc loading of slave stages on the master stage
( Unlike the BJT multi-stage current mirrors)Current ratioMOS channel geometric ratio
Iout can be as small as several nA
The ratio is not constant any more
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Design Considerations
Work with integer ratios and unit devices as much as possible.Using a unit device of size 1
Keep mirror ratio (IOUT/IREF) reasonably small
Typically no larger than 1020 Typically, we'll only have one single reference current
generator on a chip
Can generate/distribute currents across chip in two different
ways Distribute gate voltage
Can cause big problems due to IR drop
Usually limited to local distribution
Distribute currents Have one global bias cell close to reference that sends currents into local
biasing sub-circuits
Disadvantage: Consumes additional current
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Class Exercise
Ignore the outputvoltage of IREF
All W/L, except M4:W/4L
Please calculate :
Iout And the min Vpand VDD
2
12
34
DD
REF
out
5
6
A
B
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Thanks!