EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-1
Two Stages of Analysis
University of California
Berkeley
STEP 1 ) DC ANALYSIS
a) Region of Operation of Transistor
b) DC Currents and Voltages
College of Engineering
Department of Electrical Engineering
and Computer Science
Robert W. Brodersen
EECS140
STEP 2 ) AC ANALYSIS
a) Calculate Small Signal parameters
g m, χ, r o
b) Perform 2 Port analysis
R in, R out, Av, Gm
(A v = G m R out )
Analog Circuit Design
Lectures
on
SINGLE TRANSISTOR CIRCUITS
ROBERT W. BRODERSEN
LECTURE 5
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
INTRODUCTION
Two Port Analysis
i in
EECS140 ANALOG CIRCUIT DESIGN
S-1
νi n
R IN
-
R out
+A ν ⋅ ν in
R IN , R OUT , a υ
Rin
R out
ν in
= ----iin
iO U T = 0
ν out
= -------------( –i out )
S-2
V DD
+
-
INTRODUCTION
Common Source (Inverter, Gain Stage)
(High Gain, Moderate R out )
i out
+
LECTURE 5
Small Signal
Analysis
ID
νo u t
k’ = 60µA/V
λ = 0.01
νo,VO
Set independent sources to zero
ν out
A υ = ------νi n
VT = 1V
RD
Vin
W/L = 5
νi
io u t = 0
ν in at input
VB +-
υi n = 0
Vout
VDD = 5V
2
R D = 1kΩ
γ = 0.1
Cutoff Saturation
Linear
V DD
Edge of
Saturation
Problem :
Find VB
for EOS
(EOS)
VB-VT
VT
ROBERT W. BRODERSEN
LECTURE 1
ROBERT W. BRODERSEN
LECTURE 1
VB
EECS140 ANALOG CIRCUIT DESIGN
INTRODUCTION
S-3
Common Source (Cont.)
Step 1) DC Analysis :
EECS140 ANALOG CIRCUIT DESIGN
– V T ) 2 ⋅ RD
νin
G



V GS
VB
V B(A)
3.00
4.47
3.19
:
k' W
2
VO ,E O S = V D D – --- ⋅ ----- ⋅ ( V B – V T ) ⋅ R D
2 L
k' W
2
– --- ⋅ ----- ⋅ ( V B – V T ) ⋅ R D
2 L
V B = V T + V DD
V
(A )
B
=
VDD – V (BB ) + VT
--------------------------------k' W
--- ⋅ ----- ⋅ R D
2 L
R D = 1k Ω
1
-2
+ VT
ROBERT W. BRODERSEN
VB(B)
4.47
3.19
4.33
:
4.12
gmbν bs
Lets say
V OUT = 2.5 V
,then
I D = 2.5 m A
1
--
ν
-----o = – g m ⋅ ( 1k Ω || 40 k Ω ) ≈ – g m ⋅ ( 1 k Ω ) = – 1.2
ν in
S-5
LECTURE 1
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-6
Rout Calculation :
VDD
I DS = ------------2 ⋅ RD
RD //ro
νo
W VDD 
A ν = ----- = – g m ⋅ ( R D || r o ) = –  2k ' ⋅ ----- ⋅ ------⋅ ( R D || r o )

νi n
L 2 
-12
R out
Vt
ROUT = R D || r o
g m ⋅ νi n = 0
νin
gmν in
2
W
ν o = – g m ⋅ ( RD || r o ) ⋅ ν i n = –  2 k' ⋅ ----- ⋅ I DS  ⋅ ( RD || r o ) ⋅ νi n


L
1
1
r o = ----------- = ------------------------------------–3 = 40 k Ω
λ ⋅ ID
0.01 ⋅ 2.5 ⋅ 10
LECTURES ON SINGLE TRANSISTOR CIRCUITS
1
-2
RIN
4
+-
ROUT
a νν in
νoνi
νt
Ro u t = -it
υin=0
it
it = test current
ν t = ( RD || r o ) ⋅ it
ν
Ro u t = -- t = R D || r o
it
ROBERT W. BRODERSEN
LECTURE 6
νo
ν in
ro
ROBERT W. BRODERSEN
Common Source (Cont.)
If we bias the output at VDD /2 , then the equation for gain is,
RD || ro
νo
S
LECTURE 1
EECS140 ANALOG CIRCUIT DESIGN
RIN = ∞
since vbs = 0
D
gm νgs
VO ≥ V D S A T = V G S – VT = V B – V T
S-4
Common Source (Cont.)
Step 2) Small Signal Analysis :
VO = V D D – ID RD
k' W
V O = V DD – --- ⋅ ----- ⋅ (
2 L
INTRODUCTION
ROBERT W. BRODERSEN
LECTURE 6
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-7
Circuit Transconductance GM :
For Voltage Two Port, set iout = 0
S-8
V DD
i out
G M = ----ν in
i out
νi n
RL
ν in
R IN
+-
-
Vout = 0
ν o u t = A ν ⋅ νi n
+
+
RD || r o
(GM = gm )
R out
ν out
A ν ⋅ ν in
ν out = G M ⋅ ν in ⋅ R out
ν out
------- = a ν = GM ⋅ R o u t
νi n
-
For Current Two Port, set V out = 0
+
νi n
R out
R IN
-
ν out
GM ⋅ ν in
ROBERT W. BRODERSEN
R out
ν in
+
+
νo u t = GM ⋅ νi n ⋅ R o u t
R IN
-
iout
GM ⋅ ν in
i o u t = GM ⋅ ν i n
i out
GM = ----νi n
-
LECTURE 6
EECS140 ANALOG CIRCUIT DESIGN
ROBERT W. BRODERSEN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
LECTURE 6
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-9
Calculating GM
S-10
A ν = G M ⋅ R out
1
--
g m ∝ I 2DS , GM = – g m
A ν = – g m ⋅ ( RD || r o )
1
--
GM ∝ I2DS
RD » r o
R D || r o
-12
io u t = – g m ⋅ ν in
v in
g m ⋅ ν in
For Bipolar
IC
g m = ----------------V THERMAL
VEARLY
r o = -----------IC
ROBERT W. BRODERSEN
LECTURE 6
More gain with
Lower Current
I DS 1
a ν, m a x = g m ⋅ r o ∝ ----∼ -----1
I DS
2
IDS
i out
G M = ----- = – g m
ν in
i out
1
r 0 = ------------λ ⋅ I DS
ROBERT W. BRODERSEN
V EARLY
( g m ⋅ r o ) M A X = ----------------V THERMAL
(Not dependent on
Current)
V thermal = kT
LECTURE 6
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-11
Common Source with Source Degeneration
DS
io u t
GM = ----νi n
i out
RD
νin
gmν gs
,Vo
+
VDS
νVmin
-
gmbνbs
RD
is
RS
Vo = 0
is
= – ----ν in
νs
is = g m ⋅ ( ν in – ν s ) + g mb ⋅ ( – ν s ) – ---ro
1
i s = ν i n ⋅ g m – νs  g m + g mb + --- 

ro 
ro
+
-
RS
Vo = 0 for GM Calc.
(Next Page)
small compared to gm
ν s = is ⋅ R S
– is
– gm
– gm
GM = ------ = --------------------------------------------- = ---------------------------------------------υ in
1 + R S ( g m + g m b)
1 + R S ⋅ gm ⋅ ( 1 + χ )
mb 
 χ = g-----
gm
ROUT = RD // R’OUT
LECTURE 6
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Source with Source Degeneration (Cont.)
S-13
‘ROUT,
g mbνbs
= ( it – g m ⋅ ν gs – g mb ⋅ ν bs ) ⋅ r o + i t ⋅ RS
RD
it
Test Current
For R’ OUT
ROUT = RD // R’OUT
i t ⋅ RS
→
νt = it ⋅ [ R S + r o ⋅ { 1 + ( g m + g mb ) ⋅ R S }]
νt
R OUT = -- = { R S + r o ⋅ [ 1 + ( g m + g mb ) ⋅ RS ] } || R D
it
gm ⋅ R O U T
a ν = GM ⋅ R OUT = – -----------------------------------------1 + R S ⋅ ( g m + g mb )
If
LECTURE 6
νg = νb = 0
= ( it – ( g m + g mb ) ⋅ νs ) ⋅ r o + i t ⋅ R S
ro
s
ROBERT W. BRODERSEN
ν s = i t ⋅ RS
ν t = i r o ⋅ ro + ν s
νout
RS
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Source with Source Degeneration (Cont.)
i s = it
R OUT
( GroundVνinm )
is
EECS140 ANALOG CIRCUIT DESIGN
R O U T = R' O U T || R D
Rout Calculation :
g mνgs
LECTURE 6
{
ROBERT W. BRODERSEN
S-12
{
νo
R OUT
Common Source with Source Degeneration (Cont.)
Calculate circuit transconductance GM :
V DS = VDD – I DS ( R D + RS ) > V D S A T
R’ OUT
LECTURES ON SINGLE TRANSISTOR CIRCUITS





DC Analysis : I
V DD – Vo
= ------------------RD
EECS140 ANALOG CIRCUIT DESIGN
ROBERT W. BRODERSEN
R D → ∞ a = –g r
v
m
o
LECTURE 6
S-14
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-15
S-16
SUMMARY
VDD
RD
SUMMARY
VDD
i out
GM = ----- = – g m
ν in
RD
R OUT = R D || r o
ν in
RIN = ∞
R out
ro
g mb
χ = -----gm
R O U T'
R out
ν in
A ν = GM ⋅ R OUT
R O U T' = R S + r o ⋅ { 1 + ( 1 + χ ) ⋅ g m ⋅ R S }
R o u t = R D || R OUT '
RS
gm ⋅ R s » 1
ROUT ' ≈ r o ⋅ ( 1 + χ ) ⋅ g m ⋅ R S
If substrate is tied to the source, then it is the same equation with
χ = 0 (since vbs = 0)
ROBERT W. BRODERSEN
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
ROBERT W. BRODERSEN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
νt
v bs = v b – v s = – R S ⋅ i t
vt
+
g m b v bs
ro
-
ROBERT W. BRODERSEN
More current goes
into ro because of polarity of
vgs and vbs due to vRs.
Increasing vro and thus Vt
1
--
νS +
V
-
For Saturation :
νout
VOUT
+ V
- G
VOUT > V G – V S – V T
W
g m = k ' ⋅ ----- ⋅ ( V GS – VT )
L
k' W
I DS = --- ⋅ ----- ⋅ ( V GS – V T ) 2
2 L
v gs = vb s = – v R s
LECTURE 7
ROBERT W. BRODERSEN
1
--
VT = V T O + γ ⋅ [ ( 2 ⋅ φ f + V B S )2 – ( 2 ⋅ φ f ) 2 ]
VS
νin
it
Rs
VDD
v gs = v g – v s = – R S ⋅ i t
-
S-18
γ
χ = ----------------------------2 ( 2φ f + V B S)
v R = RS ⋅ i t
RS
g m vgs
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Gate (High Gain, Non-Inverting,
Low input resistance)
νt
R o u t' = -it
+
v Rs
EECS140 ANALOG CIRCUIT DESIGN
S-17
it
LECTURE 7
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-19
Common Gate (Cont.)
Vs
ro
EECS140 ANALOG CIRCUIT DESIGN
νin
m
g
νout
-gmbν in
ν bs = 0 – ν i n = v – v
b
νtT
it
s
νt
ro + R D
---- = R i n = --------------------------------------------it
1 + (1 + χ ) ⋅ g m ⋅ r o
ν OUT = i t ⋅ R D
νo
[ 1 + ( g m + g m b) ⋅ r o ] ⋅ R D
A ν = ----- = -------------------------------------------------------νi n
ro + R D
1
R i n → -------------------------(1 + x ) ⋅ g m
R D < ro
ro → ∞
R D > ro
RD → ∞
Aν = G m ⋅ R O U T
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
1 +χ
(HX)g
m νt
νt – ν O U T
i t = – ------------------- – g m ⋅ ( 1 + χ ) ⋅ νt = 0
ro
νi n – ν O U T
ν o = ( g m + g mb ) ⋅ R D ⋅ νi n + -------------------⋅ RD
ro
ROBERT W. BRODERSEN
RD
Test
Current
-gmνin
R O U T = r o || R D
νout
ro
s
RD
1
G m = --- + g m ⋅ ( 1 + χ )
ro
S-20
Common Gate (Cont.)
Rin for common gate :
vg = vb = 0
ν gs = 0 – ν i n = v – v
LECTURES ON SINGLE TRANSISTOR CIRCUITS
RD
R in → ----------------------------------( 1 + x ) ⋅ gm ⋅ r o
ROBERT W. BRODERSEN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
EECS140 ANALOG CIRCUIT DESIGN
S-21
Common Gate with RS
LECTURE 7
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-22
Common Gate with RS (Cont.)
GM Calculation
R’ OUT
R OUT
νs
V in
RS
RS
νin
+- V
G
ro
+
RD
iout
νs
-
(1 + χ ) ⋅ g m ⋅ νs







VDD
ROUT = R’OUT // RD
R O U T = ( RS + r o ⋅ [ 1 + ( g m + g mb ) ⋅ R S ] ) || R D
ROBERT W. BRODERSEN
LECTURE 7
1
RE Q = --------------------------( 1 + χ ) ⋅ gm
ROBERT W. BRODERSEN
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Gate with RS (Cont.)
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-23
ν in
i O U T = --------------------------------------1
R S + --------------------------(1 + χ ) ⋅ g m
Gm
EECS140 ANALOG CIRCUIT DESIGN
S-24
Source Follower
DC Analysis :
VDD
iO U T
1
= ------- = --------------------------------------νi n
1
R S + --------------------------( 1 + χ) ⋅ g m
Good for Buffering &
Impedance Transformation
Vin
IDS
V
νsS
( 1 + χ) ⋅ g m
= ---------------------------------------------1 + (1 + χ ) ⋅ g m ⋅ R S
VOUT
Voltage GAIN ∼ 1
High
Ri n = ∞
RS
A ν = G A I N = G M ⋅ R OUT
R OUT ∼ 10 – 1k Ω
Low
For RD >> ro , Rs << ro
A ν = g m ⋅ (1 + χ ) ⋅ r o
ROBERT W. BRODERSEN
LECTURE 7
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-25
Source Follower (Cont.)
Vνinin
V OUT = I DS ⋅ R S
V OUT
LECTURE 7
g mνgs
VνOUT
out
νs
RS
OUT
= VIN – VT – VD S A T
Need to solve iteratively if VIN is given and want to find VOUT
ROBERT W. BRODERSEN
ν min
IDS
νs
k' W
=  R S ⋅ --- ⋅ -----  ⋅ ( V IN – V O U T – V T ) 2

2 L
1

 2-V OUT ⋅ 2 

+ V T + ----------------------= V IN
OR V

W
 R S ⋅ k' ⋅ ----- 
L
Calculate Av :
ν gs = ν i n – νs
-gmbν s
ro
Vo
RS
νO U T = νs
ν s = ( g m ⋅ ν gs – g mb ⋅ ν s) ⋅ ( r o || R S )
= g m ⋅ R S ⋅ νi n – ( 1 + χ ) ⋅ g m ⋅ R S
ro » R S
g m ⋅ RS ⋅ ν i n
ν s = ---------------------------------------------- = ν OUT
1 + (1 + χ ) ⋅ g m ⋅ R S
ROBERT W. BRODERSEN
S-26
Source Follower (Cont.)
VDD
k' W
I DS = --- ⋅ ----- ⋅ ( V IN – VO U T – V T) 2
2 L
V OUT
Small Signal :
LECTURE 7
vOUT
g mR s
Av = -------- = ------------------------------------v IN
1 + ( 1 + χ )g m R s
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-27
Source Follower (Cont.)
ν gs= - ν s
Calc. ROUT :
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Single Transistor Circuits Summary
Common Source :
νin =0
S-28
Common Source
with Source Degeneration :
ro
1/gmb
1/gm
EECS140 ANALOG CIRCUIT DESIGN
RD
νo u t
ROUT
ν in
ν in
RS
RD
ν out
RS
GM = – g m
RO U T
ROUT = r o || R D
1
1
= ----- || ------ || r o || R S
gm g m b
1
≈ ----------------------Rs
gm ( 1 + χ )
νs
ROBERT W. BRODERSEN
g mνgs
νs
g mνs
νs
1/gm
LECTURE 7
EECS140 ANALOG CIRCUIT DESIGN
ROBERT W. BRODERSEN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Single Transistor Circuits Summary
Common Drain
(Source Follower) :
νo u t
S-29
gm ⋅ R S
A ν = ---------------------------------------------( 1 + χ ) ⋅ g m ⋅ RS + 1
RO U T
1
= --------------------------- || R S
( 1 + χ ) ⋅ gm
LECTURE 8
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Source Av,max Calculation
Aν , M A X = – g m ⋅ R O U T R
D
» ro
= – gm ⋅ r o
W
k' ⋅ ----- ⋅ ( VGS – V T )
L
W
1
= k' ⋅ ----- ⋅ ( V GS – V T ) ⋅ ------------- = ---------------------------------------------------k' W
2
L
λ ⋅ I DS
λ ⋅ --- ⋅ ----- ⋅ ( V G S – VT )
2 L
V DD
GM = g m
A ν = GM ⋅ R o u t
EECS140 ANALOG CIRCUIT DESIGN
RD
ν out
RS
R O U T ≈ r o ⋅ [ 1 + g m ⋅ ( 1 + χ ) ⋅ R S ] || R D
Common Gate :
ν in
ν in
– gm
GM = ---------------------------------------------1 + g m ⋅ (1 + χ ) ⋅ R S
A ν = GM ⋅ R o u t
2
1
Aν , m a x = – ---------- ⋅ --- ; ( R D » r o )
V DSAT λ
GM = g m
A ν = g m ⋅ ( 1 + χ ) ⋅ ROUT
2
2
1
= – ----------------------- ⋅ --- = – -----------------λ ⋅ V DSAT
( V G S – VT ) λ
R OUT = R D || r o
ro + R D
Ri n = --------------------------------------------1 + ( 1 + χ ) ⋅ g m ⋅ ro
1
1
A ν , m a x = ------------ ; R S » ----- 
g m
1+χ
ROBERT W. BRODERSEN
LECTURE 8
ROBERT W. BRODERSEN
LECTURE 8
S-30
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-31
Differential Pair
R D « ro
DC Analysis :
S-32
Differential Pair (Cont.)
SLOPE ~ gm R D (to be shown)
Vo d
I SS R D
V od = V o1 – Vo2
RD
slope = -g m RD
I SS
----2
VDD
V DSAT = V GS - VT
RD
VoO1
1

 -2I SS 

V G S = V T + ------------
W
 k ' ⋅ ----- 
L
I
Vo1 = V o2 = V D D – ----s s R D
2
V s = Vi 1 – V T – VD S A T
= V i 2 – V T – V DSAT
Vi1i1
+
- VGS2
+
vs
Assume:
V i1 = Vi2
Vii22
-I S SR D
Solve eqns :
V i 1 – V GS1 + VGS 2 – V i 2 = 0
I DS1 + I DS2 = I SS
Tail Current Source
LECTURE 8
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
LECTURE 8
EECS140 ANALOG CIRCUIT DESIGN
S-33
Differential Pair (Cont.)
If we have a lot of gain, low current, then the transition is very abrupt
I SS
= ----21
2
W
g m =  2 ⋅ k' ⋅ ----- ⋅ I D S


L


2 ⋅ I DS1
V G S1 = V T +  -------------
W
 k' ⋅ ----- 
L
1

 2-I SS ⋅ RD
I SS ⋅ RD  I SS 
also
∆V i d ≈ ------------------- = --------------- = ------------S L O P E g m ⋅ RD 
W
 k' ⋅ ----- 
L
1
-2
I SS
ROBERT W. BRODERSEN
Vid
1
VoO2
2
VGS1 -
V id = V i1 – V i2
∆Vid
-∆Vid
}
I SS
----2
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Differential Pair (Cont.)
Define New Variables for Differential 2-Port, small signal model :
Vo d
Vid
ν in1
νout1
ν in2
νout2
To make sure your output voltage is at zero when your input voltage is zero,
Differential Mode :
do a DC sweep in SPICE, and find the exact value of V id to set the output V DD = 0.
ν od = ν OUT 1 – ν OUT 2
Vod
Small offset
-34mV
ROBERT W. BRODERSEN
LECTURE 8
Vid
Common Mode :
ν O U T 1 + ν O U T2
ν o c = --------------------------2
ν in 1 + ν i n 2
ν i d = νi n 1 – ν i n2
ν i c = -------------------2
ν od
AD M = -----Then the differential mode gain is
νi d
ν oc
and the common mode gain is
A C M = -----νi c
ROBERT W. BRODERSEN
LECTURE 8
S-34
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-35
Differential Pair (Cont.)
Inverting the equations,
νod
νo1 = νo c + -----2
ν id
ν i2 = νi c – ----2
ν od
νo2 = νo c – -----2
νi d
ν i1 = νi c + ----2
νo1
νi d = νi n – 0 = ν i n
νo2
ν in + 0
ν in
νi c = --------------- = ----2
2
ν id
LECTURES ON SINGLE TRANSISTOR CIRCUITS
iic
+-
iic
+-
+A ocνic
νic
A CM
ROBERT W. BRODERSEN
R ic
ν od
Rod = -----i od
Differential Pair - Half Circuits
VDD
Roc
ν id/2
+-
ν
= -----in
i ic
LECTURE 8
R oc
ν
= -----i n
i ic
ROBERT W. BRODERSEN
I SS
----- – ∆ i ss
2
RD
νod/2
-νod/2
vs
Rs
ν
= ------o c
νi c
ν id
R i d = ----ii d
LECTURES ON SINGLE TRANSISTOR CIRCUITS
RD
Ao cνic
IDC
Ric
i od
EECS140 ANALOG CIRCUIT DESIGN
-I---SS- + ∆ i s s
2
Ro c
Ric
RL
Differential Mode Operation :
(Vid = 0)
νic
ADMADM νid
LECTURE 8
S-37
Differential Pair (Cont.)
iod
+
-
Rid
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
+-
+
-
ν od
A D M = -----νi d
LECTURE 8
Common Mode 2-Port :
Rod
iid
ν in A DM
AC M A D M
ν o1 = A C M ⋅ ----- + -------⋅ ν i n = νi n ⋅  -------- + -------- 
 2
2
2 
2
ROBERT W. BRODERSEN
S-36
Differential Pair (Cont.)
(Vic = 0)
Differential Mode 2-Port :
ν od
ADM
= ν o c + ------ = AC M ⋅ ν i n + -------⋅ νi n
2
2
So for our example,
LECTURES ON SINGLE TRANSISTOR CIRCUITS
i id
ν in
ν o1
EECS140 ANALOG CIRCUIT DESIGN
- νid/2
V s acts like a small signal ground
because there isn’t any small signal
current (∆ iss ) flowing into Rs
+-
v i r t u a l ground
LECTURE 8
S-38
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-39
Differential Pair - Half Circuits (Cont.)
Half Circuit :
RD
νod/2
iod
νid/2
ν od
R od = ------ = 2 ⋅ ( R D || r o )
i id
R D || r o
S-40
Differential Pair - Half Circuits (Cont.)
VDD
RD
VDD
RD
νoc
RD
RD
νoc
νoc
ν id/2
νt
+-
νο1
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Common Mode Operation :
ν od
-----2
νod
A υ = ------ = ------ = – g m ( R D || r o )
ν id
ν id
----2
ν od
-----2
------ = R D || r o
Ri d = ∞
i od
it
EECS140 ANALOG CIRCUIT DESIGN
ν oc
ν ic
νic
νic
+-
ν ο2
2RS
R D || r o
2RS
RS
Remove since no current flow
through it.
ROBERT W. BRODERSEN
LECTURE 8
EECS140 ANALOG CIRCUIT DESIGN
ROBERT W. BRODERSEN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
Differential Pair - Half Circuits (Cont.)
Half circuit :
EECS140 ANALOG CIRCUIT DESIGN
S-41
νoc
A CM
ioc
S-42
A CM
C M R R ≡ Common Mode Rejection Ratio (dB) = -20 ⋅ log -------A DM
νin
garbage gets
coupled in
noise gets amplified
along with signal









RD
gm
-----------------------------------------------------⋅ [ R O U T]
1 + 2 ⋅ RS ⋅ ( 1 + χ ) ⋅ g m
LECTURES ON SINGLE TRANSISTOR CIRCUITS
CMRR and Single vs Double Ended Diff. Pairs
VDD
ν
= ------o c =
νi c
LECTURE 8
GM
ν ic
2R S
R OUT ≈ r o ⋅ [ 1 + g m ⋅ ( 1 + χ ) ⋅ RS ] || RD ≈ R O C
νin
since the noise is common,and the common mode gain
is much smaller than the differential mode gain, the
noise doesn’t get amplified as much as the signal
ROBERT W. BRODERSEN
LECTURE 8
ROBERT W. BRODERSEN
LECTURE 9
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON SINGLE TRANSISTOR CIRCUITS
S-43
Single Ended Differential Pair
VDD
VDD
RD
RD
νOUT
ν OUT
+νid/2
-νi d/2
i OUT
-νid/2
RS
still approximately
a virtual ground
ν OUT
--------- = – g m ⋅ ( R D || r o )
ν id
– ----2
ν OUT
g m ⋅ ( RD || r o )
--------- = ----------------------------νi d
2
νOUT
R OUT = --------- = RD || r o
i OUT
ROBERT W. BRODERSEN
LECTURE 9