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7. Differential Amplifiers
ECE 102, Fall 2012, F. Najmabadi
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)
F. Najmabadi, ECE102, Fall 2012 (2/33)
Common-Mode and Differential-Mode Signals & Gain
Differential and Common-Mode Signals/Gain
F. Najmabadi, ECE102, Fall 2012 (3/33)
Consider a linear circuit with TWO inputs
2211 vAvAvo ⋅+⋅=
By superposition:
Define: 12 vvvd −=
2
21 vvvc+
=
Difference (or differential) Mode
Common Mode
Substituting for and in the expression for vo: 2
1d
cvvv −=
2 2
dc
vvv +=
( ) dcd
cd
co vAAvAAvvAvvAv ⋅
−
+⋅+=
+⋅+
−⋅=
22212
2121
ddcco vAvAv ⋅+⋅=
2
2
2
1
dc
dc
vvv
vvv
+=
−=
Differential and common-mode signal/gain is an alternative way of finding the system response
F. Najmabadi, ECE102, Fall 2012 (4/33)
2
2
1221
2112
AAAvvv
AAAvvv
dc
cd
−=
+=
+=−=
2211 vAvAvo ⋅+⋅= ddcco vAvAv ⋅+⋅=
Differential Gain: Ad Common Mode Gain: Ac Common Mode Rejection Ratio (CMRR)*: |Ad|/|Ac|
* CMRR is usually given in dB: CMRR(dB) = 20 log (|Ad|/|Ac|)
2
2
2
2
22
11
dcd
c
dcd
c
AAAvvv
AAAvvv
+=+=
−=−=
To find vo , we can calculate/measure either A1 A2 pair or Ac Ad pair
F. Najmabadi, ECE102, Fall 2012 (5/33)
Difference Method (finding Ad and Ac ): 1. Set vc = 0 (or set v1 = − 0.5 vd & v2 = + 0.5 vd)
compute Ad from vo = Ad vd 2. Set vd = 0 (or set v1 = + vc & v2 = + vc )
compute Ac from vo = Ac vc 3. For any v1 and v2 : vo = Ad vd + Ac vc
Superposition (finding A1 and A2 ): 1. Set v2 = 0, compute A1 from
vo = A1 v1 2. Set v1 = 0, compute A2 from
vo = A2 v2
3. For any v1 and v2 : vo = A1 v1 + A2 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
)(5.0 2112 vvvvvv cd +=−=
Caution
F. Najmabadi, ECE102, Fall 2012 (6/33)
In Chapter 2.1.3, Sedra & Smith defines vd = v2 − v1 But in Chapter 8, Sedra & Smith uses vd = v1 − v2 While keeping vo = vo2 − vo1 as before (this is inconsistent)
21d
cvvv −=
22d
cvvv +=
21d
cvvv +=
22d
cvvv −=
Here we use vd = v2 − v1 and vo = vo2 − vo1 throughout
Therefore, Ad (lecture slides) = −Ad (Sedra & Smith) for difference Amplifiers.
Use Lecture Slides Notation!
21d
cvvv −=
22d
cvvv +=
F. Najmabadi, ECE102, Fall 2012 (7/33)
Differential Amplifiers: Fundamental Properties
Differential Amplifier
F. Najmabadi, ECE102, Fall 2012 (8/33)
o For now, we keep track of “two” output, vo1 and vo2 , because there are several ways to configure “one” output from this circuit.
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, current-mirror, active load, …
RSS: Bias resistor, current source (current-mirror)
Q1 & Q2 are in CS-like configuration (input at the gate, output at the drain) but with sources connected to each other.
Differential Amplifier – Bias
F. Najmabadi, ECE102, Fall 2012 (9/33)
ID ID
ID ID
2ID
21
21
21
21
DSDSDS
DDD
OVOVOV
GSGSGS
VVVIII
VVVVVV
====
====
SSS
GGG
VVVVVV
====
21
21
and Since
ooo
mmm
rrrggg
====
21
21Also:
This is correct even if channel-length modulation is included because 2211 DSDDDSDD VRIVRI +=+
Differential Amplifier – Gain
F. Najmabadi, ECE102, Fall 2012 (10/33)
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.
Differential Amplifier – Gain
F. Najmabadi, ECE102, Fall 2012 (11/33)
0)()(
0)(
0)(
323113233
32322
31311
=−−−−−
+−
+
=−+−
+
=−+−
+
vvgvvgr
vvr
vvRv
vvgr
vvRv
vvgr
vvRv
mmo
o
o
o
SS
mo
o
D
o
mo
o
D
o
Node Voltage Method:
Node vo1:
Node vo2:
Node v3:
Above three equations should be solved to find vo1 , vo2 and v3 (lengthy calculations)
322
311
vvvvvv
gs
gs
−=
−=
Because the circuit is symmetric, differential/common-mode method is the preferred method to solve this circuit (and we can use fundamental configuration formulas).
Differential Amplifier – Common Mode (1)
F. Najmabadi, ECE102, Fall 2012 (12/33)
Because of summery of the circuit and input signals*:
Common Mode: Set vd = 0 (or set v1 = + vc and v2 = + vc )
dddoo iiivv === 2121 and
We can solve for vo1 by node voltage method but there is a simpler and more elegant way.
id
id id 2id
* 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
Differential Amplifier – Common Mode (2)
F. Najmabadi, ECE102, Fall 2012 (13/33)
Because of the symmetry, the common-mode circuit breaks into two identical “half-circuits”.
id
id id 2id
id
0
id
* 23 SSd Riv =
* Vss is grounded for signal
Differential Amplifier – Common Mode (3)
F. Najmabadi, ECE102, Fall 2012 (14/33)
oDSSm
Dm
c
o
c
o
rRRgRg
vv
vv
/2121
++−==
CS Amplifiers with Rs
0
The common-mode circuit breaks into two identical half-circuits.
Differential Amplifier – Differential Mode (1)
F. Najmabadi, ECE102, Fall 2012 (15/33)
32
31
5.0
5.0
vvvvvv
dgs
dgs
−+=
−−=
0)5.0()5.0(
0)5.0(
0)5.0(
3313233
3322
3311
=−+−−−−−
+−
+
=−++−
+
=−−+−
+
vvgvvgr
vvr
vvRv
vvgr
vvRv
vvgr
vvRv
dmdmo
o
o
o
SS
dmo
o
D
o
dmo
o
D
o
Node Voltage Method:
Node vo1:
Node vo2:
Node v3:
Differential Mode: Set vc = 0 (or set v1 = − vd /2 and v2 = + vd /2 )
022)(11321 =
+−+
+ vg
rvv
rR mo
oooD
( ) 02211321 =
−+++− vg
rRvv
r moSS
ooo
Node vo1 + Node vo2 :
Node v3: 0 0
3
2121
=−=⇒=+
vvvvv oooo
Only possible solution:
Differential Amplifier – Differential Mode (2)
F. Najmabadi, ECE102, Fall 2012 (16/33)
Because of the symmetry, the differential-mode circuit also breaks into two identical half-circuits.
21213 and 0 ddoo iivvv −=⇒−==
v3 = 0
id id
v3 = 0
CS Amplifier
)||(5.0
, )||(5.0
21Dom
d
oDom
d
o Rrgv
vRrgv
v−=
+−=
−
id id
0
id id
Concept of “Half Circuit”
F. Najmabadi, ECE102, Fall 2012 (17/33)
Common Mode Differential Mode
For a symmetric circuit, differential- and common-mode analysis can be performed using “half-circuits.”
Common-Mode “Half Circuit”
F. Najmabadi, ECE102, Fall 2012 (18/33)
id id
id id
0
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.
21 oo vv =
Common Mode circuit
21 ss vv =
Differential-Mode “Half Circuit”
F. Najmabadi, ECE102, Fall 2012 (19/33)
Differential Mode circuit
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
21 oo vv −=
021 == ss vv
id id
id id
Constructing “Half Circuits”
F. Najmabadi, ECE102, Fall 2012 (20/33)
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!)
Constructing “Half Circuit”– Common Mode
F. Najmabadi, ECE102, Fall 2012 (21/33)
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.
coco vv ,2,1 =
0
Constructing “Half Circuit”– Differential Mode
F. Najmabadi, ECE102, Fall 2012 (22/33)
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.
dodo vv ,2,1 −=
“Half-Circuit” works only if the circuit is symmetric!
F. Najmabadi, ECE102, Fall 2012 (23/33)
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 (24/33)
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).
…
Why is a large CMRR useful?
F. Najmabadi, ECE102, Fall 2012 (25/33)
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, …)
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.*
A regular amplifier “amplifies” both signal and noise.
noised
sigdccddo vCMRR
AvAvAvAv ⋅+⋅=⋅+⋅=
1 noisesig vvv +=
1 noisesigo vAvAvAv ⋅+⋅=⋅=
noisecsigd
noisesignoisesig
vvvvvvvvvvvv
==−=
++=+−=
&
5.0 & 5.0
12
21
* Assuming that noise levels are similar to both inputs.
Comparing a differential amplifier two identical CS amplifiers (perfectly matched)
F. Najmabadi, ECE102, Fall 2012 (26/33)
Differential Amplifier Two CS Amplifiers
Comparison of a differential amplifier with two identical CS amplifiers – Differential Mode
F. Najmabadi, ECE102, Fall 2012 (27/33)
)||(/ )||(
)5.0( )||( )5.0( )||(
,1,2
,2
,1
Domdodd
dDomdodood
dDomdo
dDomdo
RrgvvAvRrgvvv
vRrgvvRrgv
−==
−=−=
+−=
−−=
vo1,d , vo2,d , vod, and differential gain, Ad, are identical.
Half-Circuits
Identical
Differential amplifier Two CS amplifiers
Comparison of a differential amplifier with two identical CS amplifiers – Common Mode
F. Najmabadi, ECE102, Fall 2012 (28/33)
Half-Circuits
NOT Identical
vo1,c & vo2,c are different! But voc = 0 and CMMR = ∞.
0/ 0
/21
,1,2
,2,1
==
=−=++
−==
cocc
cocooc
coDSSm
Dmcoco
vvAvvv
vrRRg
Rgvv
0/ 0
)||(
,1,2
,2,1
==
=−=
−==
cocc
cocooc
cDomcoco
vvAvvv
vRrgvv
Differential amplifier Two CS amplifiers
Comparison of a differential amplifier with two identical CS amplifiers – Summary
F. Najmabadi, ECE102, Fall 2012 (29/33)
∞=
==−==
CMRR
0 , )||( c
occDom
d
odd v
vARrgvvA
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!
∞=
==−==
CMRR
0 , )||( c
occDom
d
odd v
vARrgvvA
Differential Amplifier Two CS Amplifiers
Consider a “slight” mis-match in the load resistors
F. Najmabadi, ECE102, Fall 2012 (30/33)
We will ignore ro in the this analysis (to make equations simpler)
“Slightly” mis-matched loads – Differential Mode
F. Najmabadi, ECE102, Fall 2012 (31/33)
)5.0(/ )5.0(
)5.0( )( )5.0( )(
,1,2
,2
,1
DDmdodd
dDDmdodood
dDDmdo
dDmdo
RRgvvAvRRgvvv
vRRgvvRgv
∆+−==
∆+−=−=
+∆+−=
−−=
vo1, vo2, vod, and differential gain, Ad, are identical.
Half-Circuits
Identical
Differential amplifier Two CS amplifiers
“Slightly” mis-matched loads – Common Mode
F. Najmabadi, ECE102, Fall 2012 (32/33)
Half-Circuits
NOT Identical
Differential amplifier Two CS amplifiers
vo1 and vo2 are different. In addition, voc ≠ 0 and CMMR ≠ ∞.
SSm
Dm
c
occ
cSSm
Dmcocooc
cSSm
DDmcoc
SSm
Dmco
RgRg
vvA
vRg
Rgvvv
vRg
RRgvvRg
Rgv
21
21
21
)( , 21
,1,2
,2,1
+∆
−==
+∆
−=−=
+∆+
−=+
−=
Dmc
occ
cDmcocooc
cDDmco
cDmco
RgvvA
vRgvvvvRRgv
vRgv
∆+==
∆+=−=
∆+−=
−=
)(
,1,2
,2
,1
A differential amplifier increases CMRR substantially for a slight mis-match (∆RD ≠0)
F. Najmabadi, ECE102, Fall 2012 (33/33)
Dmc RgA ∆+=
)5.0( DDmd RRgA ∆+−=
DD RR /1CMRR
∆≈
Two CS Amplifiers
)5.0( DDmd RRgA ∆+−=
SSm
Dmc Rg
RgA21+
∆−=
DD
SSm
RRRg
/21CMRR
∆+
≈
Differential Amplifier
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!
* Exercise: Compare a differential amplifier and two CS amplifiers with a mis-match in gm