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1

Common Mode Rejection Ratio

(CMRR) in Differential AmplifiersPete Semig

Analog Applications Engineer-Precision Linear

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2

Outline

Definitions

Differential-input amplifier

Common-mode voltage

Common-mode rejection ratio (CMRR)

Common-mode rejection (CMR)

CMRR in Operational Amplifiers

CMRR in Difference Amplifiers

CMRR in Instrumentation Amplifiers

CMRR in Hybrid Amplifiers

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3

Differential Input Amplifier

Differential input amplifiers are devices/circuits that can input and

amplify differential signals while suppressing common-mode signals This includes operational amplifiers, instrumentation amplifiers, and

difference amplifiers

Difference

Amplifier

InstrumentationAmplifiersOperational

Amplifier

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Common-Mode Voltage (Alternate defn.)

For a differential amplifier, common-mode voltage is defined as the average of

the two input voltages. [2]

gainmode-Common

gainmode-alDifferenti

where

cm

dm

cmcmiddmout

A

A

VAVAV

-

+

Vout

Vcm

-

+

Vid

Vid/2Vid/2

-

+

IOP1

Vcm

=

Vp+V

n

2

w here

Vp=V

cm+

Vid

2

Vn=V

cm-

Vid

2

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6

Common-Mode Voltage

Ideally a differential input amplifier only responds to a differential input voltage,

not a common-mode voltage.

V-

V+V-

V+

Vb 0 Va 0

-

+

Vid

Vs+ 5

Vs- 5

Vcm 0

-

++3

2

6

7

4 OP1

-

+

Vo 0V

0V

V-

V+

Vb 0 Va 0

-

+

Vid

Vcm 1

-

++3

2

6

7

4 OP1

-

+

Vo 0V

0V

V-

V+

Vb 0 Va 1m

-

+

Vid

Vcm 1

-

++3

2

6

7

4 OP1

-

+

Vo

3.826745V

1000uV

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CMRR and CMR

Common-Mode Rejection Ratio is defined as the ratio of the differential gain to

the common-mode gain

CMR is defined as follows [2]:

CMR and CMRR are often used interchangeably

cm

dm

A

ACMRR

CMRRdBCMR

10

log20

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Ideal Differential Amplifier CMRR

What is the CMRR of an ideal differential input amplifier (e.g. op-amp)?

Recall that the ideal common-mode gain of a differential input amplifier is 0.

Voltage Amplifier Model [1]

Also recall the differential gain of an ideal op-amp is infinity.

So

cm

dm

cm

dmOAideal

A

A

A

ACMRR

Vs

Rs

Ri

-

+

-

+

VCVS

-

+

Vi Vi

Ro

Rload-

+

Vo

Source Amplifier

Adm

->Infinity

Load

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Real Op-Amp CMRR

In an operational amplifier, the differential gain is known as the open-

loop gain.

The open-loop gain of an operational amplifier is fixed and determinedby its design

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Real Op-Amp CMRR

However, there will be a common-mode gain due to the following

Asymmetry in the circuit

Mismatched source and drain resistors

Signal source resistances

Gate-drain capacitances

Forward transconductances

Gate leakage currents Output impedance of the tail current source

Changes with frequency due to tail current sources shunt capacitance

These issues will manifest themselves through converting common-mode variations to differential components at the output and variation of

the output common-mode level. [4]

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Resistor Mismatch

Lets look at the case of a slight

mismatch in drain resistances [4] in theinput stage (diff-in, diff-out) of an op-amp

What happens to Vx and Vy as Vin,cmchanges?

Assuming M1 and M2 are identical, Vx

and Vy will change by different amounts:

This imbalance will introduce adifferential component at the output

So changes in the input common-modecan corrupt the output signal

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Transistor Mismatch

What about mismatches with respect to M1

and M2? Threshold mismatches

Dimension mismatches

These mismatches will cause the transistorsto conduct slightly different currents and

have unequal transconductances.

We find the conversion of input commonmode variations to a differential error by thefollowing factor [4]

121

SSmm

DmDMCM

Rgg

RgA

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Tail Current Source Capacitance

As the frequency of the CM disturbance increases the capacitance shunting

the tail current source will introduce larger current variations. [4]

OPA333

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Modeling CMRR

Now that we understand what CMRR is and what affects it in operational

amplifiers, lets see how it can affect a circuit.

First, however, we need to understand the model

To be useful, CMRR needs to be referred-to-input (RTI)

We can therefore represent it as a voltage source (aka offset voltage) in series

with an input. The magnitude (RTI) is Vcm/CMRR [2]

-

+Vo

Vcm/CMRR

+

Vn

+

Vp

-

+

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OA CMRR Error

Example: non-inverting buffer

CMRRV

V

AA

CMRRA

V

V

CMRRAVAV

CMRR

AVAVAVV

CMRR

VVVAV

VV

CMRR

VVV

VnVpAV

p

O

p

O

pO

p

OpO

p

OpO

Ocm

cmOn

O

11

As1

11

111

thatNote

Vn

-

+

Vo

Vcm/CMRR

+

Vp

-

+

A

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Real CMRR Example

To understand the effects CMRR can have at the output of a device,

lets look at an example.

OPA376 PDS

Notice the Vcm is specified at the top of the page

Deviation from this value will induce an offset error

Remember CMRR is RTI

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Real CMRR Example

Remember

In reality, CMRR is measured by changing the input common-mode voltageand observing the output change.

For an operational amplifier, this is usually done with a composite amplifier

It is then referred-to-input by dividing by the gain and can be though of as an

offset voltage From reference [3], in TI datasheets CMRR is defined as follows so that the

value is positive

)(log20)( 10 CMRRdBCMR

os

cm

V

VCMRR

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Real CMRR Example

For the OPA376, CMRR(min)=76dB. Note this is really CMR!

uVV

V

V

V

V

VdB

CMRRdBCMR

os

os

cm

os

cm

5.1585.6309

1

modecommoninchange1VaFor

5.630910

log2076

)(log20)(

20

76

10

10

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19

CMRR of Difference Amplifiers

A difference amplifier is made up of a differential amplifier (operational

amplifier) and a resistor network as shown below.

The circuitmeets our definition of a differential amplifier

The output is proportional to the difference between the input signals

-

+

R3 R4

R1 R2

+

V1

+

V2

-

+

Vo

Ri1

Ri2

Ro

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Difference Amplifier Impedances

2432

111

21

432

11

then

andsresistanceoutputhavesourcesinputIfequal.

ynecessarilnotandfiniteare

sourcesby theseensResistance

si

si

ss

i

i

RRRR

RRR

RR

RRRRR

-

+

R3 R4

R1 R2

+

V1

+

V2

-

+

Vo

Ri1

Ri2

Ro

2

2

:)and(Assuming

followsassresistanceinputmode-commonand

mode-aldifferentithedefinealsocanWe

21

1

4231

RRR

RR

RRRR

ic

id

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DA Derivation

Use superposition to determine Vo

as a function of V1 and V2

V1 off

Non-inverting amplifier

1

2

43

422

43

42

1

22

1

So,

1

R

R

RR

RVV

RR

RVV

RRVV

o

p

poVp

-

+

R3 R4

R1 R2

+

V2

-

+

Vo2

Ri1

Ri2

Ro

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DA Derivation

V2 off

Inverting amplifier

1

211

R

R

VVo

-

+

R3 R4

R1 R2

+

V1

-

+

Vo1

Ri1

Ri2

Ro

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DA Derivation

Combining the contributions of V1 and V2

Re-arranging yields

1

2

43

42

1

21

21

1R

R

RR

RV

R

RVV

VVV

o

ooo

121

2

4

3

2

1

4

3

2

1

12

1

2

thenIf

1

1

VVR

RV

RR

RR

R

R

R

R

VVR

RV

o

o

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DA CMRR

Lets replace V1 and V2 with our alternate definition of the inputs (in terms of

differential-mode and common-mode signals)

It is readily observed that an ideal difference amplifiers output shouldonlyamplify the differential-mode signalnot the common-mode signal.

dmo

dmcm

dmcmo

o

dmcm

dmcm

V

R

RV

VV

VV

R

RV

VVR

RV

VVV

VVV

1

2

1

2

12

1

2

2

1

22

2

2

-

+

R1 R2

R1 R2

-

+

VoVcm

+

Vdm/2

+Vdm/2

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DA CMRR

This assumes that the operational amplifier is ideal and that the resistors are

balanced.

Keeping the assumption that the operational amplifier is ideal, lets see what

happens when an imbalance factor () is introduced.

-

+

R1 R2

R1

-

+

Vo

Vcm+

Vdm/2

+Vdm/2 R2(1- )

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DA CMRR

Using superposition we find that

After some algebra we find that [1]

As expected, an imbalance affects the differential and common-mode gains, which willaffect CMRR!

As the error->0, Adm->R2/R1 and Acm->0.

1

11

2

1

2 21

2

21

2

1

2

RR

R

RR

RVV

R

RVVV dmcm

dmcmo

21

2

21

21

1

2

2

21

where

RR

RA

RR

RR

R

RA

VAVAV

cm

dm

cmcmdmdmo

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DA CMRR

Since we have equations for Acm and Adm, lets look at CMR

If the imbalance is sufficiently small we can neglect its effect on Adm

With that and some algebra we find [1]

21

2

21

21

1

2

1010

2

21

log20log20)(

RR

R

RR

RR

R

R

A

AdBCMR

cm

dm

1

2

10

1

log20)( R

R

dBCMR

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DA CMRR

This equation shows two very important relationships

As the gain of a difference amplifier increases (R2/R1), CMR increases

As the mismatch () increases, CMR decreases

Please remember that this just shows the effects of the resistor network andassumes an ideal amplifier

1

2

10

1

log20)( R

R

dBCMR

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-

+

R1 R2

R1 R2

-

+

VoVcm

+

Vdm/2

+Vdm/2

DA CMRR

Another possible source for CMRR degradation is the impedance at the

reference pin.

So far we have connected this pin to low-impedance ground.

Placing and impedance here will disturb the voltage divider we come acrossduring superposition analysis.

This will negatively affect CMR

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Why not make our own DA?

If a DA is simply an operational amplifier and 4 resistors, I can save money

by making my own, right?

Not if you want good CMRR.

They must be well-matched (ratio-metric) They must have good drift with temperature

You can purchase a 4 precision resistor pack with 0.1% relative accuracy for$0.85 (1k pricing). Drift is 25ppm. You still have to buy an op-amp.

-

+

R2 25kR1 25k

R3 25k R4 25k -

+

Vout

+

Vcm

0% 0%

0% 0%

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-319.09

-319.09

-319.09

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Why not make our own DA?

Lets assume an ideal amplifier and

just look at resistor mismatchesusing TINA (only changing R2)

Monte Carlo analysis

Gaussian distribution (6), 100cases

Values are negative due to TINA

Assuming 0% tolerance for R1, R3,and R4 and only 0.1% tolerance forR2 this network can degrade CMRRto 66dB (calculated), 69.16dB(simulated).

-

+

R2 25kR1 25k

R3 25k R4 25k -

+

Vout

+

Vcm

0% 0.1%

0% 0%

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-140.00

-120.00

-100.00

-80.00

-60.00

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Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-100.00

-80.00

-60.00

-40.00

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-120.00

-100.00

-80.00

-60.00

-40.00

Why not make our own DA?0.5%: 52dB (calc), 53.64dB (sim) 1.0%: 46dB (calc), 46.85dB (sim)

5.0%: 32dB (calc), 33.34dB (sim)

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-100.00

-80.00

-60.00

-40.00

-20.00

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-

+

R2 150kR1 150k

R3 150k R4 150k -

+

Vout

+

Vcm

5%5%

5% 5%

-

+

R2 150kR1 150k

R3 150k R4 150k -

+

Vout

+

Vcm

0.1%

0.1%0.1%

0.1%

Frequency (Hz)

10.00 1.00k 100.00k

Ga

in(dB)

-72.72

-49.71

-26.70

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-125.86

-93.35

-60.84

Why not make our own DA?

What if all resistors are 0.1% or 5%?

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Why not make our own DA?

Now that we understand how the resistor matching can affect CMRR,

what about an integrated solution?

TI can trim resistors to within 0.001% relative accuracy (e.g. INA105has a minimum CMR of 80dB, including OA!)

INA105 gain error is 5ppm/C (max)

On-chip resistors will drift together

Less board space

1k price on www.ti.com: $3.50

Includes amplifier!

So, you maysave $ but will lose performance

Customer will require 2 suppliers (1 for OA, 1 for precision resistors)

INA105

http://www.ti.com/http://www.ti.com/

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DA Gain

We learned that the gain of a difference amplifier is set by R2 and R1.

What if we wanted variable gain?

We would have to adjust 2 resistors due to the topology.

To retain good CMR they would have to be tightly matched, too.

This is difficult and expensive

Alternately, you could use an external operational amplifier (with very lowoutput impedance so as not to degrade CMR) to drive the reference pin asshown below [4]

1231

2 vvRRRRv Go

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DA Gain

But, R3 should be a precision resistor. Its error will be seen as a gain error.

You also need to purchase an external operational amplifier and potentiometer.

If you need variable gain, there are better options

Instrumentation amplifiers (IAs) usually have an external resistor that can be used toset the gain

Programmable Gain Amplifiers (PGAs) can be programmed (either with pin settingsor digitally) with a particular gain

In summary, difference amplifiers are typically manufactured with a set gain soas to preserve CMR and since there are alternate (better) solutions for variablegain

Since difference amplifiers come with a fixed gain, you will only see 1 CMRcurve in the datasheet

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Ground Interference-Operational Amplifier

A ground bus is not a perfect conductor

It has an impedance (distributed resistance, capacitance, and inductance)

Typically the input voltage source (e.g. sensor) is located remotely withrespect to the amplifiers location

Currents flowing through the ground bus can create a potential between thesensors ground and the amplifiers ground

If the input signal is particularly small, this potential difference (aka ground-loop interference) may degrade the output signal

gio vv

R

Rv

1

2

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Ground Interference-Difference Amplifier

Solution: replace the operational amplifier with a difference amplifier. An

additional wire needs to be run for the sensors ground.

This eliminates the errors introduced by the ground-loop interference becauseVg is now viewed as a common-mode signal.

io vR

Rv

1

2

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Difference Amplifiers-Summary

Pros:

Difference amplifiers amplify differential signals and reject common-mode signals The common-mode rejection is based mainly resistor matching

Making your own difference amplifier will not yield the same performance

Difference amplifiers can be used to protect against ground disturbances

Cons:

Externally changing the gain of a difference amplifier is not worthwhile The input impedance is finite

This means that a difference amplifier will load the input signals

If the input signal sources impedances are not balanced, CMR could be degraded

Is there a way we can amplify differential signals, change the gain, retain high

CMR, and not load our source? Yes! Buffer the inputsthis creates an Instrumentation Amplifier (IA).

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Instrumentation Amplifier

There are 2 types of

instrumentation amplifiers 2 op-amp (e.g. INA122)

3 op-amp (e.g. INA333)

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Instrumentation Amplifier

Notice both have gain equations so you can vary the gain

Notice the input impedance is that of the non-inverting terminal of anon-inverting amplifier

Difference Amp

High-Z Nodes

High-Z Nodes

Variable Gain

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Instrumentation Amplifier

From the 0V criteria we can determine that the voltage across Rg=Vin1-Vin2

21

dcmin

VVV

-

+

A1

R3 50k

R3 50k

R1 150k R2 150k

R1 150k R2 150k

-

+

Vout-

+

A2

Rg

Vo1

Vo2-

+

A3

+

Vcm

+

Vd/2

Vin1

Vin2

+

Vd/2

-

+

Vd

22

dcmin

VVV

V=0V

V=0V

+

-VRg=Vin1-Vin2

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Instrumentation Amplifier

We can now solve for VRg in terms of Vcm and Vd.

Notice the common-mode voltage cancels!

21

2

1

21

22

2

2

inind

dR

dcm

dcmR

dcmin

dcmin

ininR

VVV

So

VV

VV

VVV

VVV

VVV

VVV

g

g

g

-

+

A1

R3 50k

R3 50k

R1 150k R2 150k

R1 150k R2 150k

-

+

Vout-

+

A2

Rg

Vo1

Vo2-

+

A3

+

Vcm

+

Vd/2

Vin1

Vin2

+

Vd/2

-

+Vd

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Instrumentation Amplifier

From the 0A criteria we know that the current through the R3 resistors must be

the same as the current through Rg.

I

0A

I

0A

-

+

A1

R3 50k

R3 50k

R1 150k R2 150k

R1 150k R2 150k

-

+

Vout-

+

A2

Rg

Vo1

Vo2-

+

A3

+

Vcm

+

Vd/2

Vin1

Vin2

+

Vd/2

-

+

Vd

I i A lifi

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46

Instrumentation Amplifier

From Ohms law we therefore know that

Substituting for Vin1-Vin2 and re-arranging we get

gg

ininOO RR

R

VVVV

3

2121 2

-

+

A1

R3 50k

R3 50k

R1 150k R2 150k

R1 150k R2 150k

-

+

Vout-

+

A2

Rg

Vo1

Vo2-

+

A3

+

Vcm

+

Vd/2

Vin1

Vin2

+

Vd/2

-

+

Vd

g

dOO

g

g

dOO

g

g

dOO

g

g

ininOO

RRVVV

R

RRVVV

RRRVVV

RRR

VVVV

321

3

21

321

321

21

21

2

2

2

Differential Gain

Current through Rg

I t t ti A lifi INA333

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47

Instrumentation Amplifier-INA333

From previous slide:

GR

RGain 3

21

I t t ti A lifi

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48

Instrumentation Amplifier

Input stage also known as:

Diff-in, diff-out stage Gain stage

The difference amplifier stage is also known as:

Output stage

Diff-in, single-ended output stage

From our difference amplifier discussion we know the differential gain of theoutput stage is 1. Therefore,

G

OUT

GOUT

stageoutputstageinputOUT

R

kVVV

VVR

R

V

VVGGV

1001

1

2

1

12

12

3

12

Same asPDS

IA CMRR

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49

IA CMRR

So, what is the CMRR of an instrumentation amplifier?

Instrumentation amplifiers reject common-mode signals (Acm->0)

Recall

CMRR is directly related to differential gain. Since we can change thedifferential gain of an IA, we also change the CMRR.

cm

dm

A

ACMRR

INA826 CMRR M d l V ifi ti

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50

INA826 CMRR Model Verification

+V 15

++

-

Rg

RgRef

U1 INA826

V1 15

-

+

Vout

Rg 1k

+

Vcm

G1

G10

G100

G1000

Frequency (Hz)

10 215 5k 100k

Gain(dB)

0

20

40

60

80

100

120

140

160

G1

G10

G100

G1000

INA826 Eff t f R T l CMRR

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INA826-Effects of Rg Tolerance on CMRR

Now that we see our INA826 model is accurate, lets look at the effects

of Rgs tolerance on CMRR

Set G=100, 6resistors, 100 cases.

Note that due to the number of cases, no post-processing was performed

Normally this would be Gain/Waveform. Therefore we have to mentallysubtract 20dB from this cluster of waveforms.

Frequency (Hz)

10.00 1.00k 100.00k

Gain(dB)

-88.13

-81.16

-74.19

-87.97dB

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52

2-OA Instrumentation Amplifiers

What are the properties of 2-OA

Instrumentation Amplifiers?

Pros

Lower cost (only 2 op-amps), less trimming

High impedance input

Can be placed in a smaller package

Cons

Compare signal path to Vo for Vin+ and Vin-

Vin+ has a shorter path than V-

This delay does not allow the common-modecomponents to cancel each other as well as

frequency increases Therefore CMR degradation occurs earlier in

frequency than the 3-OA designs

Since we can change thedifferential gain, the CMR also

changes.

H b id Diff A lifi

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Hybrid Difference Amplifiers

Some devices have unique topologies (e.g. INA321).

How do we determine whether CMRR will change with the gain of this

device?

2OAInstrumentation

Amp

Op-amp (has

fixed differentialgain)

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H brid Differential Amplifiers

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No. The differential gain of the device is set internally!

If you cant change the differentialgain of the device, the CMRR will not changewith gain.

Remember the differentialgain of an op-amp (A3) is fixed (its the open-loopgain)

Hybrid Differential Amplifiers

Summary

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Summary

A differential amplifier amplifies differential signals, not common-mode

signals Examples include operational amplifiers, difference amplifiers, and

instrumentation amplifiers

CMRR is defined as the ratio of differential gain to common-mode gain

All differential amplifiers have an ideal common-mode gain of 0

To determine if a circuits CMRR is going to change with gain, you must

look at the differentialgain. Remember an op-amps differential gain isfixed.

If you can change the differential gain of the device/circuit, the CMRRwill also change

References

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References

[1] Franco, Design with Operational Amplifiers and Analog Integrated

Circuits, 3rd

Edition, McGraw-Hill, 2002.

[2] Tobey, Graeme, Huelsman, Operational Amplifiers: Design and

Applications, McGraw-Hill, 1971.

[3] Karki, Understanding Operational Amplifier Specifications, White Paper:

SLOA011, Texas Instruments, 1998.

[4] Razavi, Design of Analog CMOS Integrated Circuits, McGraw-Hill, 2001.

Razavi .pdf

Franco.pdf