Download - Current Sources Slide
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8/3/2019 Current Sources Slide
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A simple two-transistor current source.
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Ic1 = Ic2
Summing currents at the collector of Q1
yields
021
1!
F
Ccref
III
Fand thus
212
1
C
F
ref
C I
I
I !
!
F
ifFF is large, the collector current of Q2 is nearly equal to the
reference current:
R
VVII
onBECC
refC
)(
2
!$
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Simple current source with current gain.
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The emitter current of transistor Q3 is equal to
2
21
3
2
C
FF
C
F
CE I
III
FFF!!
The base current of transistor Q3 is equal to
2
3
3
1
2
1C
FFF
EB I
II
!
!
FFFFinally, summing currents at the collector of Q1, we obtain
0
1
221
!
CFF
Cref IIIFF
Since IC1 and IC2 are equal,
FF
ref
Co
III
FF
!!
2
22
1
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Widlar current source
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Summing voltages around the emitter-base loop, assuming that
VA is infinite, and neglecting base currents results in
022
2
2
1
1 ! RI
I
IInV
I
IInV C
S
CT
S
CT
For identical transistors, IS1 and IS2 are equal, the equation
becomes
22
2
1RI
IIInV CS
CT !
02221 ! RIVV CBEBE
and thus
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8/3/2019 Current Sources Slide
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Cascode current source with bipolar transistors.
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Thus, cascode current sources can boost the output resistance
and equivalent open-circuit voltage approximatelyF0 / 2.
ForF0 = 100 and VA =130V, we have
VVIr
V AooThev 650022
00 !!!FF
At an output current Io = 1mA, we find that
;!! M
mA
VRo 5.6
1
6500
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N-Channel mirror
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In the most general case, the ration of iO to iI is
-
!
1
2
1
2
2
1
2
21
21
1
1
K
K
vv
VV
VV
LW
WL
i
i
DS
DS
TGS
TGS
I
O
PP
-
!1
2
21
21
1
1
DS
DS
I
O
v
v
LW
WL
i
i
P
P
If vDS2 = vGS1, then the ratio of iO/iIbecomes
!
21
21
LWWL
ii
I
O
-
!
1
2
11
DS
DS
I
O
vv
ii
PP
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The Differential Amplifier
The emitter-coupled or differential pair stage
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CmCo
DM
oDM Rg
r
R
V
VA !!!
T
F1
Small-signal models for(a) the differential mode and (b) the common mode.
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Common-Mode Gain ACM
TFF
rR
R
V
V
A Eo
Co
CM
o
CM
!! 12
1
with Fo >> 1 and division by rT, reduces to
E
C
Em
CmCM
R
R
Rg
RgA
221
}
!
The 2gmRE >> 1. Because the same signal is applied to Q1 and Q2,
both Vo1 and Vo2 are 180o out of phase with VCM.
The Common-Mode Rejection Ratio:
CM
DM
A
ACMRR !
CMRR = 1 + 2gmRE } 2gmRE
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Output for Arbitrary Input signals
22
21 d
DM
VVV
V !
! 2
21VV
VCM
!
The output voltage Vo1 is
CMCMDMDMO VAVAV !1
!CMRR
VVA CMDMDM
! CMRR
VVV
AV d
DMo
21
1
2
!
CMRR
VVV
AV
d
DMo
21
22
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A source-coupled pair
For rd >> RD and Q >>1,
SgmSgm RRCMRR 221 }!
FETDifferential Amplifier
Dd
S
Rr
RCMRR
!
Q121