Transcript CMRR of Difference Amplifiers

Fully Understanding CMRR in
DAs, IAs, and OAs
Pete Semig
Analog Applications Engineer-Precision Linear
1
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
2
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
Instrumentation
Amplifiers
Operational
Amplifier
Difference
Amplifier
3
Common-Mode Voltage
• For a differential input amplifier, common-mode voltage is defined as the
average of the two input voltages. [2]
V p+V n
V cm=
2
-
Vn
+
+
+
+
-
Vo
Vp
4
Common-Mode Voltage (Alternate defn.)
• For a differential amplifier, common-mode voltage is defined as the average of
the two input voltages. [2]
-
IOP1
Vid
+
+
+
-
Vid/2
Vid/2
Vcm
V p+V n
V cm=
2
w here
V id
V p=V cm+
2
V id
V n=V cm2
Vout
Vout  Adm Vid   Acm Vcm 
where
Adm  Differential - mode gain
Acm  Common- mode gain
5
Common-Mode Voltage
• Ideally a differential input amplifier only responds to a differential input voltage,
not a common-mode voltage.
V+
2
V-
6
-
Vid 0V
3
+
V-
+
+
V+
+
-
+
+
V+
Vo 0V
Vb 0
Va 0
+
-
Vo 3.826745V
Va 1m
2
Vcm 0
4
V-
6
-
Vid 0V
3
+
+
+
V+
Vb 0
Vcm 1
OP1
+
7
Vb 0
3
+
7
Vs- 5
-
OP1
6
-
Vid 1000uV
OP1
4
2
-
7
Vs+ 5
4
V-
-
Vo 0V
Va 0
Vcm 1
6
CMRR and CMR
• Common-Mode Rejection Ratio is defined as the ratio of the differential gain to
the common-mode gain
Adm
CMRR 
Acm
• CMR is defined as follows [2]:
CMRdB  20log10 CMRR
• CMR and CMRR are often used interchangeably
7
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]
Amplifier
Source
Load
Ro
Rs
+
Vs
Vi
Ri
VCVS
+
+
+
-
-
Vi
-
-
Vo
Rload
A dm->Infinity
• Also recall the differential gain of an ideal op-amp is infinity.
• So
CMRRidealOA
Adm Adm  



Acm Acm  
8
Real Op-Amp CMRR
• In an operational amplifier, the differential gain is known as the openloop gain.
• The open-loop gain of an operational amplifier is fixed and determined
by its design
9
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 source’s shunt capacitance
• These issues will manifest themselves through converting commonmode variations to differential components at the output and variation of
the output common-mode level. [4]
10
Resistor Mismatch
• Let’s look at the case of a slight
mismatch in drain resistances [4] in the
input stage (diff-in, diff-out) of an op-amp
• What happens to Vx and Vy as Vin,cm
changes?
• Assuming M1 and M2 are identical, Vx
and Vy will change by different amounts:
• This imbalance will introduce a
differential component at the output
• So changes in the input common-mode
can corrupt the output signal
11
Transistor Mismatch
• What about mismatches with respect to M1
and M2?
– Threshold mismatches
– Dimension mismatches
• These mismatches will cause the transistors
to conduct slightly different currents and
have unequal transconductances.
• We find the conversion of input common
mode variations to a differential error by the
following factor [4]
ACM  DM
g m RD

g m1  g m 2 RSS  1
12
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
13
Modeling CMRR
• Now that we understand what CMRR is and what affects it in operational
amplifiers, let’s 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]
Vcm/CMRR
-
Vn
+
-
+
+
+
Vo
Vp
14
VO  AVp  Vn 
OA CMRR Error
Vcm
CMRR
Not e t hatVcm  VO
Vn  VO 
• Example: non-inverting buffer
Vn
Vcm/CMRR
A
+
+
+
Vp
Vo
Vp 


VO  AV p  VO 
CMRR 

AVp
VO  AVp  AVO 
CMRR
1 

VO 1  A  AVp 1 

CMRR


1 

A1 

VO
CMRR 


Vp
1 A
As A  
VO
1
 1
Vp
CMRR
15
Real CMRR Example
• To understand the effects CMRR can have at the output of a device,
let’s 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
16
Real CMRR Example
• Remember
CMR(dB)  20log10 (CMRR)
• In reality, CMRR is measured by changing the input common-mode voltage
and 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
Vcm
CMRR 
Vos
17
Real CMRR Example
• For the OPA376, CMRR(min)=76dB. Note this is really CMR!
CMR(dB)  20 log10 (CMRR)
 Vcm 

76dB  20 log10 
 Vos 
76
20
10  6309.5 
Vcm
Vos
For a 1V changein commonmode
1V
 158.5uV
Vos 
6309.5
18
CMRR of Difference Amplifiers
• A difference amplifier is made up of a differential amplifier (operational
amplifier) and a resistor network as shown below.
• The circuit meets our definition of a differential amplifier
• The output is proportional to the difference between the input signals
R2
R1
+
Ri1
Ro
-
V1
+
Ri2
+
+
R3
Vo
R4
V2
19
DA CMRR
• Let’s replace V1 and V2 with our alternate definition of the inputs (in terms of
differential-mode and common-mode signals)
Vdm
2
V
V2  Vcm  dm
2
R
Vo  2 V2  V1 
R1
V1  Vcm 
Vdm/2
+
R1
R2
+
+
-
Vcm
+
Vdm/2
R1
R2
Vo
Vo 
V  
V 
R2  
 Vcm  dm   Vcm  dm  
R1  
2  
2 
Vo 
R2
Vdm 
R1
• It is readily observed that an ideal difference amplifier’s output should only
amplify the differential-mode signal…not the common-mode signal.
20
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, let’s see what
happens when an imbalance factor (ε) is introduced.
Vdm/2
+
R1
R2(1- )
+
+
-
Vcm
+
Vdm/2
R1
Vo
R2
21
DA CMRR
• Using superposition we find that
V  R 1     
V  R2 
R2 1    

  Vcm  dm 
1 

Vo  Vcm  dm   2
2 
R1  
2  R1  R2  R1  R2 1    

• After some algebra we find that [1]
Vo  AdmVdm  AcmVcm
where
Adm 
R2  R1  2 R2  
1 
 
R1 
R1  R2 2 
Acm 
R2

R1  R2
• As expected, an imbalance affects the differential and common-mode gains, which will
affect CMRR!
• As the error->0, Adm->R2/R1 and Acm->0.
22
DA CMRR
• Since we have equations for Acm and Adm, let’s look at CMR
 R2  R1  2 R2   
 1 
  
R1  R2 2  
 Adm 
 R1 
  20 log10 
CMR(dB)  20 log10 

R
A
2
 cm 





R1  R2


• If the imbalance is sufficiently small we can neglect its effect on Adm
• With that and some algebra we find [1]
R2

1


R1
CMR(dB)  20 log10 
 








23
DA CMRR
• This equation shows two very important relationships
R

1 2
R1
CMR(dB)  20 log10 
 








– As the gain of a difference amplifier increases (R2/R1), CMR increases
– As the mismatch (ε) increases, CMR decreases
• Please remember that this just shows the effects of the resistor network and
assumes an ideal amplifier
24
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.
Vdm/2
+
R1
R2
+
+
-
Vcm
+
Vdm/2
R1
Vo
R2
• Placing and impedance here will disturb the voltage divider we come across
during superposition analysis.
• This will negatively affect CMR
25
Real DA CMRR Example (INA149 PDS)
26
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?
-319.09
R2 25k
0%
0%
Gain (dB)
R1 25k
-
+
+
-319.09
+
R3 25k
R4 25k
0%
0%
-
Vout
Vcm
-319.09
10.00
1.00k
Frequency (Hz)
100.00k
• Should be well-matched
• Should have low temperature drift
27
Why not make our own DA?
• Let’s assume an ideal amplifier and
just look at resistor mismatches
using TINA (only changing R2)
R1 25k
• Monte Carlo analysis
R2 25k
0%
0.1%
-
• Gaussian distribution (6σ), 100
cases
+
+
• Values are negative due to TINA
+
R3 25k
R4 25k
0%
0%
-
Vout
-60.00
Vcm
Gain (dB)
-80.00
-100.00
-120.00
-140.00
10.00
1.00k
Frequency (Hz)
100.00k
Assuming 0% tolerance for R1, R3,
and R4 and only 0.1% tolerance for
R2 this network can degrade CMRR
to 66dB (calculated), 69.16dB
(simulated).
28
Why not make our own DA?
• What if all resistors are 0.01% or 0.1%?
R2 25k
0.01%
0.01%
Worse performance
than all of our DAs
-
+
+
R2 150k
0.1%
0.1%
+
+
R3 25k
R4 25k
0.01%
0.01%
-
Vout
R3 150k
0.1%
+
R4 150k
-
Vout
0.1%
Vcm
Vcm
-60.84
Gain (dB)
-81.93
Gain (dB)
R1 150k
-
+
R1 25k
-100.84
-93.35
-125.86
-119.74
10.00
1.00k
Frequency (Hz)
100.00k
10.00
1.00k
Frequency (Hz)
100.00k
29
Why not make our own DA?
0.5%: 52dB (calc), 53.64dB (sim)
1.0%: 46dB (calc), 46.85dB (sim)
-40.00
-40.00
-60.00
Gain (dB)
-80.00
-80.00
-100.00
-120.00
-100.00
10.00
1.00k
Frequency (Hz)
100.00k
10.00
1.00k
Frequency (Hz)
100.00k
5.0%: 32dB (calc), 33.34dB (sim)
-20.00
-40.00
Gain (dB)
Gain (dB)
-60.00
-60.00
-80.00
-100.00
10.00
1.00k
Frequency (Hz)
100.00k
30
Why not make our own DA?
• 80dB: Lowest cost of one 0.01%, 10ppm/C resistor (1k pricing)
–
–
–
–
1206 package:
0805 package:
0603 package:
0402 package:
$0.45 ($1.80 total cost)
$0.53 ($2.12 total cost)
$0.53 ($2.12 total cost)
$0.50 ($2.00 total cost, 10k pricing!)
• 60dB: Lowest cost 4-pack 0.1%, 25ppm/C resistor (1k pricing)
– SO-8 package: $0.98 ($0.98 total cost)
• Footprint size comparison:
SO-8
1206
0805
0603
1 required
0402
(need op amp)
4 required
31
Why not make our own DA?
• Now that we understand how the resistor matching can affect CMRR and
the related cost, what about an integrated solution?
• TI can trim resistors to within 0.01% relative accuracy
• INA152
–
–
–
–
–
–
CMR(min)=80dB
GE=10ppm/˚C (max)
On-chip resistors will drift together
MSOP-8
1k price on www.ti.com: $1.20
Includes amplifier!
• Some DA’s can give CMR(min)=74dB @ $1.05!
• Customer will require 2 suppliers (1 for OA, 1 for precision resistors)
SO-8
MSOP-8
Op amp included!
32
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 low
output impedance so as not to degrade CMR) to drive the reference pin as
shown below [4]
R2 RG
v2  v1 
vo 
R1R3
33
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 to
set the gain
– Programmable Gain Amplifiers (PGAs) can be programmed (either with pin settings
or digitally) with a particular gain
• In summary, difference amplifiers are typically manufactured with a set gain so
as to preserve CMR and since there are alternate (better) solutions for variable
gain
• Since difference amplifiers come with a fixed gain, you will only see 1 CMR
curve in the datasheet
34
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 source’s 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 inputs…this creates an Instrumentation Amplifier (IA).
35
Instrumentation Amplifier
• There are 2 common types of
instrumentation amplifiers
– 2 op-amp (e.g. INA122)
– 3 op-amp (e.g. INA333)
36
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 a
non-inverting amplifier
High-Z Nodes
Difference Amp
High-Z Nodes
Variable Gain
37
IA CMRR
• So, what is the CMRR of an instrumentation amplifier?
• Instrumentation amplifiers reject common-mode signals (Acm->0)
• Recall
Adm
CMRR 
Acm
• CMRR is directly related to differential gain. Since we can change the
differential gain of an IA, we also change the CMRR.
38
INA826 CMRR Model Verification
V1 15
Rg
Rg 1k
Ref
+
Rg
+
+
U1 INA826
160
+
-
Vout
G1000
140
G100
Vcm
120
G10
G1
+V 15
Gain (dB)
100
80
60
40
20
0
10
215
5k
100k
Frequency (Hz)
39
INA826-Effects of Rg Tolerance on CMRR
• Now that we see our INA826 model is accurate, let’s look at the effects
of Rg’s tolerance on CMRR
• Set G=100, 6σ resistors, 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 mentally
subtract 20dB from this cluster of waveforms.
-74.19
-74.19
1% Resistor
-81.16
Gain (dB)
-87.97dB<CMR<-88.13dB
Adjusted for gain:
-107.97dB<CMR<-108.13dB
-74.19
-88.04dB<CMR<-88.07dB
Adjusted for gain:
-108.04dB<CMR<-108.07dB
-81.13
-88.08
-88.13
10.00
10.00
1.00k
Frequency (Hz)
100.00k
1.00k
Frequency (Hz)
100.00k
0.1% Resistor
Gain (dB)
Gain (dB)
5% Resistor
-88.065531dB<CMR<-88.06869dB
Adjusted for gain:
-108.065531dB<CMR<-108.06869dB
-81.13
-88.07
10.00
1.00k
Frequency (Hz)
100.00k
Notice the gain setting
resistor tolerance does
not significantly affect the
CMR.
40
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-mode
components 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 the
differential gain, the CMR also
changes.
41
‘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?
Op-amp (has
fixed differential
gain)
2OA
Instrumentation
Amp
42
‘Hybrid’ Difference Amplifiers
• Depends on what ‘gain’ you’re talking about.
• With respect to CMRR, it’s all about the differential gain since the
common-mode gain of all differential amplifiers is ideally 0.
• When you place resistors for R1 and R2, are you changing the
differential gain?
43
‘Hybrid’ Differential Amplifiers
• No. The differential gain of the device is set internally!
• If you can’t change the differential gain of the device, the CMRR will not change
with gain.
• Remember the differential gain of an op-amp (A3) is fixed (it’s the open-loop
gain)
44
Real IA CMR Competitive Analysis
45
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 circuit’s CMRR is going to change with gain, you must
look at the differential gain. Remember an op-amp’s differential gain is
fixed.
• If you can change the differential gain of the device/circuit, the CMRR
will also change
46
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.
47
Questions?
48