7. Differential Amplifiers
Sedra & Smith Sec. 2.1.3 and Sec. 8 (MOS Portion)
(S&S 5th Ed: Sec. 2.1.3 and Sec. 7 MOS Portion &
ignore frequency-response)
ECE 102, Fall 2012, F. Najmabadi
Common-Mode and Differential-Mode
Signals & Gain
F. Najmabadi, ECE102, Fall 2012 (2/33)
Differential and Common-Mode Signals/Gain
Consider a linear circuit
with TWO inputs
By superposition:
vo = A1 ⋅ v1 + A2 ⋅ v2
Define:
vd = v2 − v1
vc =
v1 + v2
2
Difference (or differential) Mode
Common Mode
Substituting for v1 = vc −
vd
2
v
v2 = vc + d
2
v1 = vc −
vd
v
and v2 = vc + d in the expression for vo:
2
2
v 
v 


 A − A1 
vo = A1 ⋅  vc − d  + A2 ⋅  vc + d  = ( A1 + A2 ) ⋅ vc +  2
 ⋅ vd
2
2
 2 


vo = Ac ⋅ vc + Ad ⋅ vd
F. Najmabadi, ECE102, Fall 2012 (3/33)
Differential and common-mode signal/gain is an
alternative way of finding the system response
vd = v2 − v1
vc =
v1 + v2
2
Ac = A1 + A2
Ad =
A2 − A1
2
vo = Ac ⋅ vc + Ad ⋅ vd
vo = A1 ⋅ v1 + A2 ⋅ v2
vd
2
v
v2 = vc + d
2
v1 = vc −
Ac
− Ad
2
A
A2 = c + Ad
2
A1 =
Differential Gain:
Common Mode Gain:
Common Mode Rejection Ratio (CMRR)*:
F. Najmabadi, ECE102, Fall 2012 (4/33)
Ad
Ac
|Ad|/|Ac|
* CMRR is usually given in dB: CMRR(dB) = 20 log (|Ad|/|Ac|)
To find vo , we can calculate/measure
either A1 A2 pair or Ac Ad pair
Superposition (finding A1 and A2 ):
Difference Method (finding Ad and Ac ):
2. Set v1 = 0, compute A2 from
vo = A2 v2
2. Set vd = 0 (or set v1 = + vc & v2 = + vc )
compute Ac from vo = Ac vc
1. Set v2 = 0, compute A1 from
vo = A1 v1
3. For any v1 and v2 :
vo = A1 v1 + A2 v2
1. Set vc = 0 (or set v1 = − 0.5 vd & v2 = + 0.5 vd)
compute Ad from vo = Ad vd
3. For any v1 and v2 :
vd = v2 − v1
vo = Ad vd + Ac vc
vc = 0.5(v1 + v2 )
 Both methods give the same answer for vo (or Av ).
 The choice of the method is driven by application:
o Easier solution
o More relevant parameters
F. Najmabadi, ECE102, Fall 2012 (5/33)
Caution
 In Chapter 2.1.3, Sedra & Smith defines vd = v2 − v1
v
v1 = vc − d
2
v 2 = vc +
vd
2
 But in Chapter 8, Sedra & Smith uses vd = v1 − v2
v1 = vc +
vd
2
v2 = vc −
vd
2
While keeping vo = vo2 − vo1 as before (this is inconsistent)
 Here we use vd = v2 − v1 and vo = vo2 − vo1 throughout
vd
v1 = vc −
2
v 2 = vc +
vd
2
 Therefore, Ad (lecture slides) = −Ad (Sedra & Smith) for
difference Amplifiers.
 Use Lecture Slides Notation!
F. Najmabadi, ECE102, Fall 2012 (6/33)
Differential Amplifiers:
Fundamental Properties
F. Najmabadi, ECE102, Fall 2012 (7/33)
Differential Amplifier




Identical transistors.
Circuit elements are symmetric about the mid-plane.
Identical bias voltages at Q1 & Q2 gates (VG1 = VG2 ).
Signal voltages & currents are different because v1 ≠ v2.
Load RD: resistor, currentmirror, active load, …
Q1 & Q2 are in CS-like
configuration (input at
the gate, output at the
drain) but with sources
connected to each other.
RSS: Bias resistor, current
source (current-mirror)
o For now, we keep track of “two” output, vo1 and vo2 , because there
are several ways to configure “one” output from this circuit.
F. Najmabadi, ECE102, Fall 2012 (8/33)
Differential Amplifier – Bias
Since VG1 = VG 2 = VG
and
ID
VS 1 = VS 2 = VS
ID
VGS 1 = VGS 2 = VGS
VOV 1 = VOV 2 = VOV
I D1 = I D 2 = I D
VDS 1 = VDS 2 = VDS
Also:
ID
ID
2ID
g m1 = g m 2 = g m
ro1 = ro 2 = ro
This is correct even if channel-length
modulation is included because
I D1 RD + VDS1 = I D 2 RD + VDS 2
F. Najmabadi, ECE102, Fall 2012 (9/33)
Differential Amplifier – Gain
 Signal voltages & currents are
different because v1 ≠ v2
 We cannot use fundamental
amplifier configuration for
arbitrary values of v1 and v2.
 We have to replace each NMOS
with its small-signal model.
F. Najmabadi, ECE102, Fall 2012 (10/33)
Differential Amplifier – Gain
v gs1 = v1 − v3
v gs 2 = v2 − v3
Node Voltage Method:
Node vo1: vo1 + vo1 − v3 + g (v − v ) = 0
m 1
3
RD
ro
Node vo2: vo 2 + vo 2 − v3 + g m (v2 − v3 ) = 0
RD
ro
v3 v3 − vo 2 v3 − vo1
+
+
− g m (v1 − v3 ) − g m (v2 − v3 ) = 0
RSS
ro
ro
Above three equations should be solved to find vo1 , vo2 and v3 (lengthy calculations)
Node v3:
 Because the circuit is symmetric, differential/common-mode
method is the preferred method to solve this circuit (and we
can use fundamental configuration formulas).
F. Najmabadi, ECE102, Fall 2012 (11/33)
Differential Amplifier – Common Mode (1)
Common Mode: Set vd = 0 (or set v1 = + vc and v2 = + vc )
id
Because of summery of
the circuit and input signals*:
vo1 = vo 2 and id 1 = id 2 = id
We can solve for vo1 by node voltage method
but there is a simpler and more elegant way.
id
2id
id
* If you do not see this, set v1 = v2 = vc in node equations of the previous slide, subtract the
first two equations to get vo1 = vo2 . Ohm’s law on RD then gives id1 = id2 = id
F. Najmabadi, ECE102, Fall 2012 (12/33)
Differential Amplifier – Common Mode (2)
id
id
0
id
2id
id
id
v3 = 2id RSS *
 Because of the symmetry, the common-mode circuit breaks into two
identical “half-circuits”.
F. Najmabadi, ECE102, Fall 2012 (13/33)
* Vss is grounded for signal
Differential Amplifier – Common Mode (3)
 The common-mode circuit breaks into two identical half-circuits.
0
CS Amplifiers with Rs
vo1 vo 2
gm R D
=
=−
vc
vc
1 + 2 g m R SS + R D / ro
F. Najmabadi, ECE102, Fall 2012 (14/33)
Differential Amplifier – Differential Mode (1)
Differential Mode: Set vc = 0 (or set v1 = − vd /2 and v2 = + vd /2 )
v gs1 = −0.5vd − v3
v gs 2 = +0.5vd − v3
Node Voltage Method:
Node vo1: vo1 + vo1 − v3 + g (−0.5v − v ) = 0
m
d
3
RD
ro
Node vo2: vo 2 + vo 2 − v3 + g (+0.5v − v ) = 0
m
d
3
RD
Node v3:
ro
v3 v3 − vo 2 v3 − vo1
+
+
− g m (−0.5vd − v3 ) − g m (+0.5vd − v3 ) = 0
RSS
ro
ro
 1 1
2

 + (vo1 + vo 2 ) −  + 2 g m v3 = 0
 RD ro 
 ro

 1 2

1
Node v3: − (vo1 + vo 2 ) + 
+ − 2 g m v3 = 0

ro
 RSS ro

F. Najmabadi, ECE102, Fall 2012 (15/33)
Node vo1 + Node vo2 :
Only possible solution:
vo1 + vo 2 = 0 ⇒ vo1 = −vo 2
v3 = 0
Differential Amplifier – Differential Mode (2)
v3 = 0 and
id
vo1 = −vo 2 ⇒ id 1 = −id 2
id
id
v3 = 0
id
v3 = 0
id
id
0
CS Amplifier
vo1
vo 2
= − g m (ro ||R D ) ,
= − g m (ro ||R D )
− 0.5vd
+ 0.5vd
 Because of the symmetry, the differential-mode circuit also breaks into two
identical half-circuits.
F. Najmabadi, ECE102, Fall 2012 (16/33)
Concept of “Half Circuit”
 For a symmetric circuit, differential- and common-mode
analysis can be performed using “half-circuits.”
Common Mode
F. Najmabadi, ECE102, Fall 2012 (17/33)
Differential Mode
Common-Mode “Half Circuit”
Common Mode circuit
id
id
vo1 = vo 2
0
id
vs1 = vs 2
id
Common Mode Half-circuit
1. Currents about symmetry line are equal.
2. Voltages about the symmetry line are equal (e.g., vo1 = vo2)
3. No current crosses the symmetry line.
F. Najmabadi, ECE102, Fall 2012 (18/33)
Differential-Mode “Half Circuit”
Differential Mode circuit
id
id
vo1 = −vo 2
id
id
vs1 = vs 2 = 0
Differential Mode Half-circuit
1. Currents about the symmetry line are equal in value and opposite in sign.
2. Voltages about the symmetry line are equal in value and opposite in sign.
3. Voltage at the summery line is zero
F. Najmabadi, ECE102, Fall 2012 (19/33)
Constructing “Half Circuits”
Step 1:
Divide ALL elements that cross the symmetry line (e.g., RL) and/or
are located on the symmetry line (current source) such that we
have a symmetric circuit (only wires should cross the symmetry
line, nothing should be located on the symmetry line!)
F. Najmabadi, ECE102, Fall 2012 (20/33)
Constructing “Half Circuit”– Common Mode
Step 2: Common Mode Half-circuit
1. Currents about symmetry line are equal (e.g., id1 = id2).
2. Voltages about the symmetry line are equal (e.g., vo1 = vo2).
3. No current crosses the symmetry line.
0
vo1,c = vo 2,c
F. Najmabadi, ECE102, Fall 2012 (21/33)
Constructing “Half Circuit”– Differential Mode
Step 3: Differential Mode Half-Circuit
1. Currents about symmetry line are equal but opposite sign (e.g., id1 = − id2)
2. Voltages about the symmetry line are equal but opposite sign (e.g., vo1 = − vo2)
3. Voltage on the symmetry line is zero.
vo1,d = −vo 2,d
F. Najmabadi, ECE102, Fall 2012 (22/33)
“Half-Circuit” works only if the circuit is
symmetric!
 Half circuits for common-mode and differential mode are different.
 Bias circuit is similar to Half circuit for common mode.
 Not all difference amplifiers are symmetric. Look at the load
carefully!
 We can still use half circuit concept if the deviation from prefect
symmetry is small (i.e., if one transistor has RD and the other RD
+ ∆RD with ∆RD << RD).
o However, we need to solve BOTH half-circuits (see slide 30)
F. Najmabadi, ECE102, Fall 2012 (23/33)
Why are Differential Amplifiers popular?
 They are much less sensitive to noise (CMRR >>1).
 Biasing: Relatively easy direct coupling of stages:
o Biasing resistor (RSS) does not affect the differential gain
(and does not need a by-pass capacitor).
o No need for precise biasing of the gate in ICs
o DC amplifiers (no coupling/bypass capacitors).
 …
F. Najmabadi, ECE102, Fall 2012 (24/33)
Why is a large CMRR useful?
 A major goal in circuit design is to minimize the noise level (or improve
signal-to-noise ratio). Noise comes from many sources (thermal, EM, …)
 A regular amplifier “amplifies” both signal and noise.
v1 = vsig + vnoise
vo = A ⋅ v1 = A ⋅ vsig + A ⋅ vnoise
 However, if the signal is applied between two inputs and we use a
difference amplifier with a large CMRR, the signal is amplified a lot more
than the noise which improves the signal to noise ratio.*
v1 = −0.5vsig + vnoise & v2 = +0.5vsig + vnoise
vd = v2 − v1 = vsig
& vc = vnoise
vo = Ad ⋅ vd + Ac ⋅ vc = Ad ⋅ vsig +
F. Najmabadi, ECE102, Fall 2012 (25/33)
Ad
⋅ vnoise
CMRR
* Assuming that noise levels are similar to both inputs.
Comparing a differential amplifier
two identical CS amplifiers (perfectly matched)
Differential Amplifier
F. Najmabadi, ECE102, Fall 2012 (26/33)
Two CS Amplifiers
Comparison of a differential amplifier with two
identical CS amplifiers – Differential Mode
Differential amplifier
Two CS amplifiers
Identical
Half-Circuits
vo1,d = − g m (ro ||R D ) (−0.5vd )
vo 2,d = − g m (ro ||R D ) (+0.5vd )
vod = vo 2,d − vo1,d = − g m (ro ||R D )vd
Ad = vod / vd = − g m (ro ||R D )
 vo1,d , vo2,d , vod, and differential gain, Ad, are identical.
F. Najmabadi, ECE102, Fall 2012 (27/33)
Comparison of a differential amplifier with two
identical CS amplifiers – Common Mode
Differential amplifier
Two CS amplifiers
NOT Identical
Half-Circuits
vo1,c = vo 2,c = −
gm R D
vc
1 + 2 g m R SS + RD / ro
voc = vo 2,c − vo1,c = 0
Ac = voc / vc = 0
vo1,c = vo 2,c = − g m (ro ||R D )vc
voc = vo 2,c − vo1,c = 0
Ac = voc / vc = 0
 vo1,c & vo2,c are different! But voc = 0 and CMMR = ∞.
F. Najmabadi, ECE102, Fall 2012 (28/33)
Comparison of a differential amplifier with two
identical CS amplifiers – Summary
Differential Amplifier
Ad =
vod
v
= − g m (ro ||R D ) , Ac = oc = 0
vd
vc
CMRR = ∞
Two CS Amplifiers
Ad =
vod
v
= − g m (ro ||R D ) , Ac = oc = 0
vd
vc
CMRR = ∞
 For perfectly matched circuits, there is no difference between
a differential amplifier and two identical CS amplifiers.
o But one can never make perfectly matched circuits!
F. Najmabadi, ECE102, Fall 2012 (29/33)
Consider a “slight” mis-match in the load resistors
 We will ignore ro in the this analysis (to make equations simpler)
F. Najmabadi, ECE102, Fall 2012 (30/33)
“Slightly” mis-matched loads – Differential Mode
Differential amplifier
Two CS amplifiers
Identical
Half-Circuits
vo1,d = − g m ( R D ) (−0.5vd )
vo 2,d = − g m ( R D + ∆R D ) (+0.5vd )
vod = vo 2,d − vo1,d = − g m ( R D +0.5∆R D )vd
Ad = vod / vd = − g m ( R D +0.5∆R D )
 vo1, vo2, vod, and differential gain, Ad, are identical.
F. Najmabadi, ECE102, Fall 2012 (31/33)
“Slightly” mis-matched loads – Common Mode
Differential amplifier
Two CS amplifiers
NOT Identical
Half-Circuits
vo1,c = −
gm R D
g ( R + ∆R D )
vc , vo 2,c = − m D
vc
1 + 2 g m R SS
1 + 2 g m R SS
voc = vo 2,c − vo1,c = −
Ac =
g m ∆R D
vc
1 + 2 g m R SS
voc
g ∆R
=− m D
1 + 2 g m R SS
vc
vo1,c = − g m R D vc
vo 2,c = − g m ( R D + ∆R D )vc
voc = vo 2,c − vo1,c = + g m ∆R D vc
Ac =
voc
= + g m ∆R D
vc
 vo1 and vo2 are different. In addition, voc ≠ 0 and CMMR ≠ ∞.
F. Najmabadi, ECE102, Fall 2012 (32/33)
A differential amplifier increases CMRR
substantially for a slight mis-match (∆RD ≠0)
Two CS Amplifiers
Differential Amplifier
Ad = − g m ( R D +0.5∆R D )
Ad = − g m ( R D +0.5∆R D )
Ac = + g m ∆R D
Ac = −
CMRR ≈
1
∆R D / R D
g m ∆R D
1+ 2 g m R SS
CMRR ≈
1 + 2 g m R SS
∆R D / R D
 Differential amplifier reduces Ac and increases CMRR substantially
(by a factor of: 1 + 2 gmRSS).
 The common-mode half-circuits for a differential amplifier are
CS amplifiers with RS (thus common mode gain is much smaller
than two CS amplifiers).
 We should use a large RSS in a differential amplifier!
F. Najmabadi, ECE102, Fall 2012 (33/33)
* Exercise: Compare a differential amplifier and
two CS amplifiers with a mis-match in gm