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- 1. Fully Understanding CMRR inDAs, IAs, and OAsPete SemigAnalog Applications Engineer-Precision Linear 1
- 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 2
- 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 Instrumentation Operational Amplifiers Amplifier Difference Amplifier 3
- 4. 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 - + + Vo + + - Vn Vp 4
- 5. 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 + + + Vout - V p+V n Vid/2 Vid/2 V cm= 2 Vout = Adm (Vid ) + Acm (Vcm ) w here where V id Vcm V p=V cm+ Adm = Differential - mode gain 2 V id Acm = Common - mode gain V n=V cm- 2 5
- 6. Common-Mode Voltage• Ideally a differential input amplifier only responds to a differential input voltage,V+ not a common-mode voltage. V- OP1 4 2 Vs+ 5 V- - 6 - 4 OP1 Vid 1000uV 3 + 2 - + + + 7 6 Vs- 5 - Vo 3.826745V Vid 0V 3 + + + - + V+ 7 Vo 0VV- - Va 1m V+ Vb 0 Vb 0 Va 0 V- OP1 Vcm 1 4 2 - Vcm 0 6 - Vid 0V 3 + + + + 7 Vo 0V - V+ Vb 0 Va 0 Vcm 1 6
- 7. 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 ) = 20 log10 ( CMRR )• CMR and CMRR are often used interchangeably 7
- 8. 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] Source Amplifier Load Rs Ro VCVS + + + + Vs Vi Ri Vi Vo Rload - - - - A dm->Infinity• Also recall the differential gain of an ideal op-amp is infinity.• So Adm Adm → ∞ CMRRideal −OA = = →∞ Acm Acm → 0 8
- 9. 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 determined by its design 9
- 10. 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 common- mode variations to differential components at the output and variation of the output common-mode level. [4] 10
- 11. 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
- 12. 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] ∆g m RD ACM − DM =− ( g m1 + g m 2 ) RSS + 1 12
- 13. 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
- 14. 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 - + + Vo - + + Vn Vp 14
- 15. VO = A(Vp − Vn )OA CMRR Error Vcm Vn = VO ± CMRR• Example: non-inverting buffer Note that Vcm ≈ VO Vp VO = AV p − VO ± CMRR AV p Vn VO = AV p − AVO ± CMRR - 1 Vcm/CMRR A VO (1 + A) = AV p 1 ± + + CMRR Vo - 1 A1 ± + VO = Vp CMRR Vp 1+ A As A → ∞ VO 1 → 1± Vp CMRR 15
- 16. 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
- 17. Real CMRR Example• Remember CMR(dB) = 20 log10 (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
- 18. Real CMRR Example• For the OPA376, CMRR(min)=76dB. Note this is really CMR! CMR (dB ) = 20 log10 (CMRR) ∆Vcm 76dB = 20 log10 ∆V os 76 ∆Vcm 10 20 = 6309.5 = ∆Vos For a 1V change in common mode 1V ∆Vos = = 158.5uV 6309.5 18
- 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 circuit meets our definition of a differential amplifier• The output is proportional to the difference between the input signals Ri1 R1 R2 + Ro V1 - + + Ri2 Vo - R3 R4 + V2 19
- 20. 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 V1 = Vcm − 2 V + Vdm/2 R1 R2 V2 = Vcm + dm 2 - R + + Vo = 2 (V2 − V1 )Vcm - Vo R1 + Vdm/2 R1 R2 R2 V V Vo = Vcm + dm − Vcm − dm R1 2 2 R2 Vo = (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
- 21. 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-ε) + - + + Vo Vcm - + R1 R2 Vdm/2 21
- 22. DA CMRR• Using superposition we find that V R (1 − ε ) V R2 R2 (1 − ε ) Vo = Vcm − dm − 2 + Vcm + dm R + R R + R (1 − ε ) 1 + 2 R1 2 1 2 1 2 • After some algebra we find that [1] Vo = AdmVdm + AcmVcm where R2 R1 + 2 R2 ε Adm = 1 − × R1 R1 + R2 2 R2 Acm = ×ε 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
- 23. DA CMRR• Since we have equations for Acm and Adm, let’s look at CMR R2 R1 + 2 R2 ε 1 − × Adm R1 R1 + R2 2 CMR (dB ) = 20 log10 = 20 log10 A R2 cm ×ε R1 + R2 • If the imbalance is sufficiently small we can neglect its effect on Adm• With that and some algebra we find [1] R 1+ 2 R1 CMR(dB ) ≅ 20 log10 ε 23
- 24. DA CMRR• This equation shows two very important relationships R 1+ 2 R1 CMR(dB ) ≅ 20 log10 ε – 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
- 25. 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 + - + + Vo Vcm - + R1 R2 Vdm/2• Placing and impedance here will disturb the voltage divider we come across during superposition analysis.• This will negatively affect CMR 25
- 26. Real DA CMRR Example (INA149 PDS) 26
- 27. 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? T -319.09 R1 25k R2 25k 0% 0% Gain (dB) - -319.09 + + R3 25k R4 25k Vout - 0% 0% + Vcm -319.09 10.00 1.00k 100.00k Frequency (Hz)• Should be well-matched• Should have low temperature drift 27
- 28. 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 R2 25k • Monte Carlo analysis 0% 0.1% - • Gaussian distribution (6σ), 100 + cases + R3 25k R4 25k Vout - • Values are negative due to TINA 0% 0% + T -60.00 Vcm -80.00Gain (dB) -100.00 Assuming 0% tolerance for R1, R3, -120.00 and R4 and only 0.1% tolerance for R2 this network can degrade CMRR -140.00 10.00 1.00k 100.00k to 66dB (calculated), 69.16dB Frequency (Hz) (simulated). 28
- 29. Why not make our own DA?• What if all resistors are 0.01% or 0.1%? R1 25k R2 25k R1 150k R2 150k 0.01% 0.01% 0.1% 0.1% Worse performance - than all of our DAs - + + + + Vout R3 150k R4 150k Vout R3 25k R4 25k - - 0.01% 0.01% 0.1% 0.1% + + Vcm Vcm T -81.93 T -60.84 Gain (dB)Gain (dB) -100.84 -93.35 -119.74 -125.86 10.00 1.00k 100.00k 10.00 1.00k 100.00k Frequency (Hz) Frequency (Hz) 29
- 30. Why not make our own DA? 0.5%: 52dB (calc), 53.64dB (sim) 1.0%: 46dB (calc), 46.85dB (sim) T -40.00 T -40.00 -60.00 -60.00Gain (dB) Gain (dB) -80.00 -80.00 -100.00 -120.00 -100.00 10.00 1.00k 100.00k 10.00 1.00k 100.00k Frequency (Hz) Frequency (Hz) 5.0%: 32dB (calc), 33.34dB (sim) T -20.00 -40.00 Gain (dB) -60.00 -80.00 -100.00 10.00 1.00k 100.00k Frequency (Hz) 30
- 31. Why not make our own DA?• 80dB: Lowest cost of one 0.01%, 10ppm/C resistor (1k pricing) – 1206 package: $0.45 ($1.80 total cost) – 0805 package: $0.53 ($2.12 total cost) – 0603 package: $0.53 ($2.12 total cost) – 0402 package: $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: 1 required (need op amp) 4 required 31
- 32. 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) Op amp included! 32
- 33. 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 vo = ( v2 − v1 ) R1 R3 33
- 34. 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
- 35. 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
- 36. Instrumentation Amplifier• There are 2 common types of instrumentation amplifiers – 2 op-amp (e.g. INA122) – 3 op-amp (e.g. INA333) 36
- 37. 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
- 38. 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
- 39. INA826 CMRR Model Verification V1 15 - Rg Rg 1k Ref U1 INA826 T Rg + 160 + + Vout G1000 + - 140 G100 Vcm 120 G10 100 G1 +V 15 Gain (dB) 80 60 40 20 0 10 215 5k 100k Frequency (Hz) 39
- 40. 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. T -74.19 T -74.19 1% Resistor 5% Resistor -88.04dB<CMR<-88.07dB -87.97dB<CMR<-88.13dB Gain (dB) Adjusted for gain: Gain (dB) Adjusted for gain: -81.13 -108.04dB<CMR<-108.07dB -81.16 -107.97dB<CMR<-108.13dB T -74.19 -88.08 -88.13 10.00 1.00k 100.00k 10.00 1.00k 100.00k Frequency (Hz) Frequency (Hz) 0.1% Resistor -88.065531dB<CMR<-88.06869dB Adjusted for gain: Notice the gain setting Gain (dB) -81.13 -108.065531dB<CMR<-108.06869dB resistor tolerance does -88.07 not significantly affect the 10.00 1.00k Frequency (Hz) 100.00k CMR. 40
- 41. 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
- 42. ‘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) 2OAInstrumentation Amp 42
- 43. ‘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
- 44. ‘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
- 45. Real IA CMR Competitive Analysis 45
- 46. 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
- 47. 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
- 48. Questions? 48

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