CMOS,Current source, Differential Amplifier
21
EE2603-04 1 4. CMOS,Current source, Differential Amplifier (12.6 - 12.9 ) • CMOS logic circuit • Current Source circuits • Differential Amplifier circuits
description
CMOS,Current source, Differential Amplifier - CMOS logic circuit- Current Source circuits- Differential Amplifier circuits
Transcript of CMOS,Current source, Differential Amplifier
EE2603-04 1
4. CMOS,Current source, Differential Amplifier (12.6 - 12.9 )
• CMOS logic circuit • Current Source circuits • Differential Amplifier circuits
EE2603-04 2
nMOS and pMOS ID
VGS01234567
62 4
ID
VDS0Vt=2V
VGS= 6V
VGS= 4VVGS= 2V
VGS= 8V
VDS
VGS
ID
IG=0
IDON , VGSON ,Vt
VGSON
IDON
nMOS characteristics
ID ID
VDS
VGS= -6V
VGS= -4VVGS= -2V
VGS= -8V
VDS
-VGS
-ID
IG=0
IDON , VGSON ,Vt01234567
6 24 0
IDON
Vt=-2V-VGSON
-VGS
pMOS characteristics
nMOS will conduct when VGS > 0
pMOS will conduct when VGS < 0
nMOS and pMOS, how they works
nMOS will conduct or Qn (on) when VGS > 0 (high)
pMOS will open or Qp (off) when VGS ≥ 0 (zero or positive)
nMOS will open or Qn (off) when VGS ≤ 0 (zero or negative)
pMOS will conduct or Qp (on) when VGS ‹ 0 (negative)
5V
0
-5V
QP
Qn (on)
(off)VGS > 0
V
G
G
S
VGS=+5V
0V
S
D
D
S
D
(off)QP
S
D
Qn (on)
+5
0
0
5V
+5
0
QP
Qn (on)
(off)VGS=0
V
G
G
S
VGS=+5V
5V
S
D
D
0V
+5
0
QP
Qn(off)
(on)VGS=-
5V
G
G
S
VGS=0V
0V
D
D
S
S
D
(off)QnS
D
Qp(on)
+5
+5-5V
0V
5V
QP
Qn
VGS=-5V
G
G
S
VGS ≤ 0V
0V
D
D
S
(on)
(off)
When input Vin = 5V, Vo = 0V
CMOS inverter, how it works
inverter
A 0V 5V
Q=A 5V 0V
When input Vin = 0V, Vo = 5V
Vo
+5
0
=5VQP
Qn(off)
(on)
Vin=0V Vo
+5
0
=5VQP
Qn(off)
(on)VGS
=-5V
VGS=-5
V
0VGS=0V
G
G
S
SVGS=0V
G
G
S
S
Vo
+5
0
=0VQP
Qn (on)
(off)
Vin=5V Vo
+5
0
=0VQP
Qn (on)
(off)VGS
=0V
5G
G
S
SVGS=+5V
G
G
S
S
VGS=0V
VGS=+5V
Vin VoVGS(QP) VGS(Qn)Qp QnLogic(in)
Logic(out)
5VHi 0V off5V on 0V Lo
0VLo -5V on0V off 5V Hi
When Control terminal is High TG becomes inactive making
Output Disconnected or high impedance whatever the input is
CMOS Transmission Gate Logic (Active low control)
C Input OutputC
Lo (0V) X
Hi (5V)
Hi (5V)
Lo (0V)
Lo (0V)
High impd
Hi (5V)Lo (0V)Lo (0V)
Hi (5V)
Hi (5V)
C
TG
C
C
Input Output
G
S D
G
S D
Qn
Qp VDD
Input
C
C (Control) Output
TG
C=5V(Hi)
C=0V(Lo)
Input OutputFloating(high Impedance)
TG
C=0V(Lo)
C=5V(Hi)
Input Output
When Control terminal is Low TG becomes active making
Output connected to the input whatever the input is
G
S D
G
S D
Qn
QpC (Control) OutputInput
C=5V Input=0V(Lo)Input=5V(Hi)
High Impedance (Floating)
How the Output of TG is always High Impedance (Floating) if Control “C” is 5V(Hi)
5V
C=0V
Qp gate is always positive
C=5V
Therefore Qp is always open
Qp(off)
Qn(off)
Qn gate is always zero voltTherefore Qn is always open
C=0V
G
S
G
S
CMOS Transmission Gate,(Active low control gate) how it works
When Control input terminal is High and whatever the Input voltage is When Control C = 5V (Hi) Qp Gate is +5V (positive) so whatever the Source voltage is, VGS of Qp will be positve or zero and therefore Qp will be always open (off) When Control C = 5V (Hi) Qn Gate is 0V(zero) so whatever the Source voltage is, VGS of Qn will be negative or zero so Qn will be always open (off) Therefore as both Qpand Qn are open, output terminal is High Impedance or Floating and the Input is not transmitted through the TG gate to the Output
C=G(Qn) Input=S OutputC=G(Qp)
Lo (0V) XHi (5V) High impd
VGS(Qp)
VGS ≥ 0Qp
off
Qn
off
VGS(Qn)
VGS ≤ 0
When Control terminal is Low and Input voltage is High When Control C = 0V (Lo) Qp Gate is 0V(zero) so when the input (Source voltage) is 5V (Hi), VGS of Qp will be -5V making Qp to conduct (on) When Control C = 0V (Lo) Qn Gate is 5V so when the input (Source voltage) is 5V (Hi), VGS of Qn will be zero making Qn to be open (off) Therefore Input 5V(Hi) is transmitted through conducting Qp to the Output
G
S D
G
S D
Qnopen
QpconductC (Control) Output
Therefore Qp is conductingInput
C=0V Input=5V(Hi)
How the Input 5V (Hi) is transmitted to the Output through Qp if Control “C” is 0V(Lo)
0V
C=5V
Output=5V(Hi)
VGS of Qp gate is negative
C=0V Qp(on)
Qn(off)
VGS of Qn gate is 0 voltTherefore Qn off
G
G
S=5V
C=5V
S=5V
C=G(Qn) Input=S OutputC=G(Qp)
Hi (5V)Lo (0V) Hi (5V)Hi (5V)
VGS(Qp)
VGS = -5VQp Qn
on off
VGS(Qn)VGS = 0V
When Control terminal is Low and Input voltage is Low When Control C = 0V (Lo) Qp Gate is 0V(zero) so when the input (Source voltage) is 0V (Lo), VGS of Qp will be 0V making Qp to “off” When Control C = 0V (Lo) Qn Gate is 5V so when the input (Source voltage) is 0V (Lo), VGS of Qn will be 5V making Qn to “on” Therefore Input 0V(Lo) is transmitted through conducting Qn to the Output
G
S D
G
S
Qp(off)
Qn(on)C (Control) OutputInputC=0V Input=0V(Lo)
0V
C=5V
DOutput=0V(Lo)
How Input=0V(Lo) is transmitted to the Output through Qn if Control “C” is 0V(Lo)
VGS of Qp gate is zero
C=0V
Therefore Qp is off
Qp(off)
Qn(on)
VGS of Qn gate is 5 voltTherefore Qn is on
G
G
S=0V
C=5V
S=0V
C=G(Qn) Input=S OutputC=G(Qp) VGS(Qp) Qp QnVGS(Qn)
Lo (0V) Lo (0V)Lo (0V) Hi (5V) onoff VGS = 5VVGS = 0V
EE2603-04 9
Current Source circuits
0W
1000W
1V1mA
1V/1000W=1mA1V/1001W=1mA
1V/1010W=1mA
1W
1mA
10W
1mA1mA
10W
1mA
1W
1mA1mA
0W
Current source of 1mA
VGS=0
ID=IDSS
IDSS
Current source of IDSS
EE2603-04 10
BJT Current source
BJT-Zener Current source
other circuits using IC
IB
IE
IC
RE
-VEE
0.7
other circuits using IC
IB
IE
IC
RE+VCC
0.7R2R1R2R1
++
+ -
--+-
++-
-
+
-
-
+
-VEE
IC IC
⎥⎦
⎤⎢⎣
⎡−
+
×=∴
+=
+
×=×=
+=
+
−−=
7.0RRRV
R1I
RI7.0VRRRV
RIV
RRV
RR)V(0
I
21
2EE
EC
EC2R
21
2EE22R2R
21
EE
21
EE2R
[ ]7.0VR1I
RI7.0V
ZE
C
ECZ
−=∴
+=
other circuits using IC
IB
IE
IC
RE
-VEE
0.7
other circuits using IC
IB
IE
IC
RE+VCC
0.7R1VZR1
++
+ -
--+-
++-
-
-
+
-VEE
IC IC + +
-VZ
EE2603-04 11
BJT Mirror Current sources
other circuits using IC
IC
0.7+
-
IX
RX
+-0.7
VCC
2IC/bIC
[ ]
X
BECCX
CCC
CXBEC
B
RVV
I
II2I2II7.0V
II
−=
≈β
+β=
β+=⇒=
β=
[ ]
X
BECCX
CCC
CXBEC
B
RVV
I
II3I3
II7.0VI
I
−=
≈β
+β=
β+=⇒=
β=
IC for circuit 1 and 2
IC1
0.7+
-
IX
RX
+-
0.7
VCC
3IC/b
IC2IC1
IC2
0.7+
-
EE2603-04 12
other circuits using ICIC
0.7+
-
IX
RX
+-0.7
VCC
IC/b
IC
[ ]
X
BECCX
CCC
CXBEC
B
RVV
I
II1I
II7.0VI
I
−=
≈β
+β=
β+=⇒=
β=
other circuits using IC
IC
0.7+
-
IDSSVGS=0
+-0.7
+V
IC2IC/bIC
[ ]
CCC
CDSS
CBE
II2I2IIbut
equalareIboth7.0V
≈β
+β=
β+=
⇒=
EE2603-04 13
Differential Amplifier circuits • In differential amplifier, there are two inputs Vi1 and Vi2 and there are three outputs (1) at Vo1, (2) at Vo2, and (3) across between Vo1 and Vo2
• Dual supply +VCC and –VEE are used so that Vi1 and Vi2 can be connected to the BJT directly without coupling capacitor
• IC1 + IC2 = 2Ic flows through RE then VRE = 2Icx RE
Vi1
RC
-VEE
+VCC
Vo1 Vo2
RC
RE
Vi2
Q1 Q2
AC
AC
VE = -0.7V
VB = 0V dc
+ -
( ) ( )
Ce
E
EEC
EEEEREEC
ImV26r
R27.0V
I
7.0VV7.0VRI2
=⇒−
=∴
−=−−−==×
CCCC2C1C RIVVV −==
DC analysis
EE2603-04 14
Example
In differential amplifier shown, find (1) VB1= VB2 (2) VE1= VE2 (3) IC1 = IC2 and (4) VC1= VC2
VB = 0V dc
VE = -0.7V
Vi1
-9V
+9V
Vo1 Vo2
RC
3.3kWVi2
Q1 Q2AC
AC
3.9kW
+ -
VC
IC
( ) ( )
mA26.1k3.327.09
R27.0V
I
7.0VV7.0VRI2
E
EEC
EEEEEEC
=×
−=
−=∴
−=−−−==×
V1.4k9.3mA26.19RIVV CCCCC =×−=−=
V7.0V,V0V EB −==
EE2603-04 15
AC analysis
Vi
Vo
RC
Q1 Q2
AC
RE
RC
AC ground
Vi
VoRC
Vi
AC
RE>>bre
RC
bre
ib
bib bre
then ib2=ib
bib+
++-
-
- IE<<ib2
Vo = -bibRC
AC equivalent circuit
Vi1=Vi
-VEE
+VCC
Vo
RC
Vi2=0
Q1 Q2
AC
RC
RE
Circuit
1. Single-ended input and single output
( ) ( )
e
C
e
Cv
i
o
e
iCCbo
e
ibebebebi
EeE
r2R
r2R
AVV
r2VRRiV
r2Viri2ririVthen
openRtakerRLet
−=β
β−==⇒
β×β−=β−=
β=⇒β×=β×+β×=
⇒β>>
For Single-ended input and single output
EE2603-04 16
Vi1
Vo
RC
Q1 Q2
AC
RE
RC
ACVi2
AC ground
Vi1
VoRC
Vi2
AC
RE>>bre
RC
bre
ib
bib brebib+
+ +-
-
- IE<<ib2
Vo = -bibRC
AC+-
AC equivalent circuit
( ) ( )
i
C
e
C
e
Cd
d
o
2i1i
o
e
2i1iCCbo
e
2i1ibeb2i1i
ebeb2i1i
EeE
r2R
r2R
r2R
AVV
VVV
r2VVRRiV
r2VViri2VV
0ririVVgiveslooptheofKVLthenopenRtakerRLet
β−=−=
β
β−===
−⇒
β
−×β−=β−=
β
−=⇒β×=−
=β×−β×−−
⇒β>>
For Differential input and single output
Vi1
-VEE
+VCC
Vo
RC
Vi2
Q1 Q2
AC
RC
RE
Vd=Vi1-Vi2
AC
Circuit
2. Differential-mode input and single output
EE2603-04 17
Vi
-VEE
+VCC
Vo
RC
Q1 Q2
AC
RC
RE
Circuit
3. Common-mode input and single output
Vo
RC
Q1 Q2
RE
RC
Vi
AC
AC ground
Vi
VoRC
AC
RC
bre
ib
bib brebib+
+- -
-
Vo = -bibRC
+
RE
ib
2(b+1)ib
ib ib
AC equivalent circuit
( ) ( )( ) ( ) ( )
( ) ( )
E
C
E
Cv
i
o
E
iCCbo
E
i
Ee
ib
EebEbebi
R2R
R2R
AVV
R2VRRiV
R2V
R2rVi
R2riRi12riV
−=β
β−==⇒
β×β−=β−=
β=
+β=⇒
+β=×+β+β×=
For Common input and single output
EE2603-04 18
Example
In differential amplifier shown, find Vo when (1) Vi1= 1mV and Vi2= 0 (2) Vi1= 1mV and Vi2= -1mV (3) Vi1= 1mV and Vi2= 1mV
Vi1
-9V
+9V
Vo
Vi2
Q1 Q2
AC
3.9kW
AC
3.9kW
3.3kW
( ) ( )
Ω==⇒=×
−=
−=∴
−=−−−==×
63.20mA26.1mV26rmA26.1
k3.327.09
R27.0V
I
7.0VV7.0VRI2
eE
EEC
EEEEEEC
DC analysis
1. Vi1= 1mV and Vi2= 0 = Single-ended input and single output
mV52.94mV152.94V52.9463.202
3900r2R
mV1V
VV
oe
Co
i
o −=×−=⇒−=×
−=
−==
2. Vi1= 1mV and Vi2= -1mV = Differential-mode input and single output
mV189mV252.94V52.9463.202
3900r2R
)mV1(mV1V
VVV
oe
Co
2i1i
o −=×−=⇒−=×
−=
−=
−−=
−
3. Vi1= 1mV and Vi2= 1mV = Common-mode input and single output mV59.0mV159.0V59.0
330023900
R2R
mV1V
VV
oE
Co
i
o −=×−⇒−=×
−=
−==
EE2603-04 19
4. Use of Constant Current Source
Vi1
-VEE
+VCC
Vo
RC
Vi2
Q1 Q2
AC
RC
RE
AC
ICIC
I=2ICVi1
-VEE
+VCC
Vo
RC
Vi2
Q1 Q2
AC
RC
AC
I=2IC ro
ICIC
IE
RE
-VEE
0.7R2R1
+ -
--+
-
Vi1
+VCC
Vo
RC
Vi2
Q1 Q2AC
RCAC
I=2IC
++
ICIC
• Use of current source in-place-of RE is to get I=2IC and creating a very high resistance (equivalent current source resistance) at the emitter circuit • If RE is very high common-mode gain becomes very low (see common-mode gain equation) • For the same Differential gain, low common-mode gain makes Common Mode Rejection Ratio (CMRR) very high. • Good Differential amplifier has a very high CMRR. Thus if an equivalent constant current source is used in Differential amplifier, it becomes better as CMRR is high.
⎥⎦
⎤⎢⎣
⎡−
+
×=∴
+=
+
×=×=
+=
+
−−=
7.0RRRV
R1I
IR7.0VRRRV
RIV
RRV
RR)V(0
I
21
2EE
E
E2R
21
2EE22R2R
21
EE
21
EE2R
EE2603-04 20
IE
5.1kW
-9
0.7
+ -
--+
-
Vi1
+9
Vo
Vi2
Q1 Q2AC
10kW
AC
I=2IC
++
ICIC
10kW
b1=b2=b=75
b3=75ro=200kW
8.2kW1kW
Example
In differential amplifier shown, find (1) I= 2IC (2) re1= re2= re (3) Common-mode gain Ac , (4) Differential-mode gain Ad , and (5) CMRR
3. Common-mode input and single output
025.0k2002
k10r2R
R2R
VVA
o
C
E
C
i
oC −=
×
−=
−=
−==
mA44.11.57.08Ik1.5I7.0IR7.08V
V8k2.8mA98.0RIVmA98.0k2.9
9k2.8k1)9(0I
E2R
22R2R2R
=−
=⇒∴×+=+==
=×=×=⇒==+
−−=
1. Value of I (=2IC)
Ω=×=β=
Ω====⇒==−
=
k72.222.3675rr
22.36mA72.0mV26rrrmA72.0
2mA44.1
1.57.08I
ei
e2e1eC
2. Value of re
4. Differential-mode input and single output
13822.362
000,10r2R
AVV
Ve
Cd
2i1i
o −=×
−=
−==
−
5521025.0138
AA
CMRRc
d =−
−==
5. CMRR
EE2603-04 21
Vi1
-V
+V
Vo
Vi2
AC
RD
AC
Vi1
-V
+V
Vo
Vi2
AC
AC
Vi1
Vo
Vi2
AC
AC
+V
+V +V
5. FET Differential amplifier circuits
BiFET Differential amplifier BiMOS Differential amplifier
CMOS Differential amplifier
