Chapter 5

Passive and Active Current Mirrors

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Basic Current Mirrors

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Dependence of Iout on supply, process, and temperature.

- Overdrive voltage is the function of VDD and VTH.

(VTH might vary by 100mV from wafer to wafer.)

- µn and VTH have temperature dependence.

Basic Current Mirrors

Copying from

precisely-defined current source

“Current Mirror”

If channel-length modulation

is neglected,

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� “no dependence on process & temperature”

Basic Current Mirrors

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Mirror pair should have the same L.

1. if Ldrawn is doubled, Leff (=Ldrawn-∆L) is not.

2. VTH has dependence on L.Gain :

Cascode Current Mirrors

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Channel-length modulation

results in current mismatch.

� Cascode can improve it.

(higher Rout = less C.L.M.)

For VY=VX � Vb - VGS3 = VX � Vb = VGS3 + VX

“worse voltage headroom”

VP,min = 2 * overdrive voltage + VTH

Q) L3 should be equal to L1?

Cascode Current Mirrors : Low Voltage

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For M1 & M2 to be in saturation,

RbI1 is not well-controlled.Similar errors due to body effect.

Current Mirrors : Summary

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Active Current Mirrors

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< Thevenin Equivalence >

1. Output short

� equivalent output current

2. Source disabled

� equivalent output resistance 2

2

1

1

m

in

outm

in

mout

g

V

IG

VgI

==

=( )[ ]

( ) 42

410222

||2

||/1

oo

ommoout

rr

rgrgrR

=

+≈

( )

( )∞→→

≈=

421

421

if

||22

oomv

oom

outmv

rrgA

rrg

RGA

Active Current Mirrors

“M1 drain current is wasted” differential pair with active current mirror

� Differential to single-ended converter

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Active Current Mirrors : Large-Signal

M1 : off � M3 & M4 : off

� no current from VDD

� Vout = 0

M1 : on � M3 & M4 : on

�Vout depends on

the difference between ID4 & ID2.

ID1, ID3, ID4 : ↑ , ID2 : ↓

� M2 : triode region

� If input is too large,

M2 : off (Vout = VDD)

If Vin1 > VF+VTH,

M1� triode region.

@Vin1=Vin2 � Vout = VF = VDD - |VGS3|

But, if not symmetric (ID1≠ID2) � channel-length modulation gives large variation of Vout.

� This circuit is rarely used in an open-loop configuration.

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Active Current Mirrors : Small-Signal

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ID1 = ID3 = ID 4 = gm1,2Vin / 2 ID2 = −gm1,2Vin / 2

2,12,142 mminmDDout gGVgIII =⇒−=−=

IX = 2VX

2ro1, 2 +1/ gm 3

+VX

ro4

Rout ≈ ro2 || ro4 , (2ro1,2 >> [1/ gm 3] || ro3 )

Active Current Mirrors : CM Response

4,3

2,1

2,1

2,1

4,3

4,3

21

1

2

1

2||

2

1

m

m

SSmSS

m

o

m

CMg

g

RgR

g

r

gA

+

−≈

+

−

=

CMRR =ADM

ACM

= gm1, 2(ro1,2 || ro3,4 )gm 3,4 (1 + 2gm1,2 RSS )

gm1,2

= gm 3,4 (ro1, 2 || ro3,4 )(1 + 2gm1, 2RSS)

Even with perfect symmetry,

output is corrupted by input CM variations.

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