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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 5. th. Ed: Sec. 2.1.3 and Sec. 7 MOS Portion &
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

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