7. Differential Amplifiers

7. Differential Amplifiers

ECE 102, Fall 2012, F. Najmabadi

Sedra & Smith Sec. 2.1.3 and Sec. 8 (MOS Portion)

(S&S 5

_{Ed: Sec. 2.1.3 and Sec. 7 MOS Portion &}

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

2 2 1 1

v

A

v

A

v

=

⋅

+

⋅

v

=

v

−

v

2

v

v

v

=

+

Difference (or differential) Mode Common Mode

Substituting for and in the expression for v_o:

2

v

v

v

=

−

2

v

v

v

=

+

(

)

v

A

A

v

A

A

v

v

A

v

v

A

v



⋅











−

+

⋅

+

=











 +

⋅

+











 −

⋅

=

2

2

2

A

v

A

v

v

=

⋅

+

⋅

2

2

v

v

v

v

v

v

+

=

−

=

Differential and common-mode signal/gain is an

alternative way of finding the system response

2 2 1 2 2 1 2 1 1 2 A A A v v v A A A v v v d c c d − = + = + = − = 2 2 1 1

v

A

v

A

v

=

⋅

+

⋅

v

=

A

⋅

v

+

A

⋅

v

|A |/|A |) 2 2 2 2 2 2 1 1 d c d c d c d c A A A v v v A A A v v v + = + = − = − =

To find

v

_o

, we can calculate/measure

either

A

₁

A

₂

pair or

A

_c

A

_d

pair

F. Najmabadi, ECE102, Fall 2012 (5/33)

Difference Method (finding A_d and A_c ):

1. Set v_c = 0 (or set v₁ = − _{0.5 vd} & v₂ = + _{0.5 vd}) compute A_d from v_o = A_d v_d

2. Set v_d = 0 (or set v₁ = + _vc & v₂ = + _vc )

compute A_c from v_o = A_c v_c

3. For any v₁ and v₂ : v_o = A_d v_d + A_c v_c

Superposition (finding A₁ and A₂ ):

1. Set v₂ = 0, compute A₁ from

v_o = A₁ v₁

2. Set v₁ = 0, compute A₂ from

v_o = A₂ v₂

3. For any v₁ and v₂ :

v_o = A₁ v₁ + A₂ v₂

 _{Both methods give the same answer for} v_{o (or} A_{v ).}  The choice of the method is driven by application:

o Easier solution

o More relevant parameters

) ( 5 . 0 _{1 2} 1 2 v v v v v v_d = − _c = +

Caution

 In Chapter 2.1.3, Sedra & Smith defines v_d = v₂ − _v1

 But in Chapter 8, Sedra & Smith uses v_d = v₁ − _v2

While keeping v_o = v_o2 − _{vo1 as before (this is inconsistent)}

2 1 d c v v v = − 2 2 d c v v v = + 2 1 d c v v v = + 2 2 d c v v v = −

 Here we use v_d = v₂ − _v1 and v_o = v_o2 − _vo1throughout

 Therefore, A_d(lecture slides) = −_Ad(Sedra & Smith) for difference Amplifiers.

 Use Lecture Slides Notation!

2 1 d c v v v = − 2 2 d c v v v = +

F. Najmabadi, ECE102, Fall 2012 (7/33)

Differential Amplifiers:

Fundamental Properties

Differential Amplifier

o For now, we keep track of “two” output, v_o1 and v_o2 , 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 (}V_{G1 =} V_{G2 ).}  _{Signal voltages & currents are different because} v₁ ≠ v_2.

Load R_D: resistor, current-mirror, active load, …

R_SS: 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)

I_D I_D I_D I_D 2I_D 2 1 2 1 2 1 2 1 DS DS DS D D D OV OV OV GS GS GS

V

V

V

I

I

I

V

V

V

V

V

V

=

=

=

=

=

=

=

=

V

V

V

V

V

V

=

=

=

=

and

Since

r

r

r

g

g

g

=

=

=

=

This is correct even if channel-length modulation is included because

2 2 1 1 D DS D D DS D

R

V

I

R

V

I

+

=

+

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}

Differential Amplifier – Gain

F. Najmabadi, ECE102, Fall 2012 (11/33)

0 ) ( ) ( 0 ) ( 0 ) ( 3 2 3 1 1 3 2 3 3 3 2 3 2 2 3 1 3 1 1 = − − − − − + − + = − + − + = − + − + v v g v v g r v v r v v R v v v g r v v R v v v g r v v R v m m o o o o SS m o o D o m o o D o

Node Voltage Method: Node v_o1:

Node v_o2: Node v₃:

Above three equations should be solved to find v_o1 , v_o2 and v₃ (lengthy calculations) 3 2 2 3 1 1 v v v v v v 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)

Because of summery of

the circuit and input signals*:

Common Mode: Set v_d = 0 (or set v₁ = + v_c and v₂ = + v_c )

d d d o o

v

i

i

i

v

=

and

=

=

We can solve for v_o1 by node voltage method but there is a simpler and more elegant way.

i_d

i_d i_d

2i_d

* If you do not see this, set v₁ = v₂ = v_c in node equations of the previous slide, subtract the first two equations to get v_o1 = v_o2 . Ohm’s law on R_D then gives i_d1 = i_d2 = i_d

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”. i_d i_d i_d 2i_d id 0 i_d * 2 3 idRSS v =

Differential Amplifier – Common Mode (3)

Differential Amplifier – Differential Mode (1)

F. Najmabadi, ECE102, Fall 2012 (15/33)

3 2 3 1 5 . 0 5 . 0 v v v v v v d gs d gs − + = − − = 0 ) 5 . 0 ( ) 5 . 0 ( 0 ) 5 . 0 ( 0 ) 5 . 0 ( 3 3 1 3 2 3 3 3 3 2 2 3 3 1 1 = − + − − − − − + − + = − + + − + = − − + − + v v g v v g r v v r v v R v v v g r v v R v v v g r v v R v d m d m o o o o SS d m o o D o d m o o D o

Node Voltage Method:

Node v_o1: Node v_o2: Node v₃:

Differential Mode: Set v_c = 0 (or set v₁ = − _{vd /2 and v2 =} + _{vd /2 )}

0

2

2

)

(

1

1



=











+

−

+













+

g

v

r

v

v

r

R

(

)

1

2

2

0

1



=











−

+

+

+

−

g

v

r

R

v

v

r

Differential Amplifier – Differential Mode (2)

 Because of the symmetry, the differential-mode circuit also breaks into two

identical half-circuits. 2 1 2 1 3

0

and

v

v

i

i

v

=

=

−

⇒

=

−

Concept of “Half Circuit”

F. Najmabadi, ECE102, Fall 2012 (17/33)

Common Mode Differential Mode



For a symmetric circuit, differential- and common-mode

Common-Mode “Half Circuit”

i_d i_d

i_d i_d

0

Common Mode Half-circuit

1. Currents about symmetry line are equal.

2. Voltages about the symmetry line are equal (e.g., v_o1 = v_o2) 3. No current crosses the symmetry line.

2 1 o

o v

v =

Common Mode circuit

2 1 s

s v

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

2 1 o o v v = − 0 2 1 = s = s v v i_d i_d i_d i_d

Constructing “Half Circuits”

Step 1:

Divide ALL elements that cross the symmetry line (e.g., R_L) 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., i_d1 = i_d2). 2. Voltages about the symmetry line are equal (e.g., v_o1 = v_o2).

3. No current crosses the symmetry line.

c o c o

v

v

=

Constructing “Half Circuit”– Differential Mode

Step 3: Differential Mode Half-Circuit

1. Currents about symmetry line are equal but opposite sign (e.g., i_d1 =

−

2. Voltages about the symmetry line are equal but opposite sign (e.g., v_o1 =

−

3. Voltage on the symmetry line is zero.

d o d

o

v

“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

R

and the other

R

+

∆

R

with

∆

_R

_<<

_R

_).

Why are Differential Amplifiers popular?

 _{Biasing: Relatively easy direct coupling of stages:}

o Biasing resistor (R_SS) 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.}

noise d sig d c c d d o v CMRR A v A v A v A v = ⋅ + ⋅ = ⋅ + ⋅ 1 vsig vnoise v = + 1 sig noise o A v A v A v v = ⋅ = ⋅ + ⋅ noise c sig d noise sig noise sig v v v v v v v v v v v v = = − = + + = + − = & 5 . 0 & 5 . 0 1 2 2 1

Comparing a differential amplifier

two identical CS amplifiers (perfectly matched)

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 D o m d od d d D o m d o d o od d D o m d o d D o m d o R r g v v A v R r g v v v v R r g v v R r g v − = = − = − = + − = − − =

 v_{o1,d ,} v_{o2,d ,} v_{od, and differential gain,} A_{d, are identical.}

Half-Circuits

Identical

Comparison of a differential amplifier with two

identical CS amplifiers – Common Mode

Half-Circuits

NOT Identical

 v_{o1,c &} v_{o2,c are different! But} v_oc = 0 _{and CMMR =} ∞_. 0 / 0 / 2 1 , 1 , 2 , 2 , 1 = = = − = + + − = = c oc c c o c o oc c o D SS m D m c o c o v v A v v v v r R R g R g v v

0

/

0

)

||

(

=

=

=

−

=

−

=

=

v

v

A

v

v

v

v

R

r

g

v

v

Comparison of a differential amplifier with two

identical CS amplifiers – Summary

F. Najmabadi, ECE102, Fall 2012 (29/33)

∞ = = = − = = CMRR 0 , ) || ( c oc c D o m d od d v v A R r g v v A

 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 oc c D o m d od d v v A R r g v v A

Consider a “slight” mis-match in the load resistors

“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 D D m d od d d D D m d o d o od d D D m d o d D m d o R R g v v A v R R g v v v v R R g v v R g v ∆ + − = = ∆ + − = − = + ∆ + − = − − =

 v_o1, v_o2, v_{od, and differential gain,} A_{d, are identical.}

Half-Circuits

Identical

“Slightly” mis-matched loads – Common Mode

Half-Circuits

NOT Identical

Differential amplifier Two CS amplifiers

 v_{o1 and} v_{o2 are different. In addition,} v_oc ≠ 0 _{and CMMR} ≠ ∞_.

SS m D m c oc c c SS m D m c o c o oc c SS m D D m c o c SS m D m c o R g R g v v A v R g R g v v v v R g R R g v v R g R g v 2 1 2 1 2 1 ) ( , 2 1 , 1 , 2 , 2 , 1 + ∆ − = = + ∆ − = − = + ∆ + − = + − = D m c oc c c D m c o c o oc c D D m c o c D m c o R g v v A v R g v v v v R R g v v R g v ∆ + = = ∆ + = − = ∆ + − = − = ) ( , 1 , 2 , 2 , 1

A differential amplifier increases CMRR

substantially for a slight mis-match (

∆

_R

_D

≠

₀

₎

F. Najmabadi, ECE102, Fall 2012 (33/33)

D m c g R A = + ∆ ) 5 . 0 ( _{D D} m d g R R A = − + ∆ D D R R / 1 CMRR ∆ ≈ Two CS Amplifiers ) 5 . 0 ( _{D D} m d g R R A = − + ∆ SS m D m c R g R g A 2 1+ ∆ − = D D SS m R R R g / 2 1 CMRR ∆ + ≈ Differential Amplifier

 _{Differential amplifier reduces} A_{c and increases CMRR substantially}

(by a factor of: 1 + 2 g_mR_SS).

 _{The common-mode half-circuits for a differential amplifier are}

CS amplifiers with R_S (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 g_m