Lecture Block 1: Operational Amplifier ECEN303:Analogue ...
Transcript of Lecture Block 1: Operational Amplifier ECEN303:Analogue ...
Topics
β’ Basic characteristics and properties.
β’ Device imperfections.
β’ Circuit limitations.
β’ Design problems and solutions.
OperationalAmplifiers
β’ Voltage relationship:
π£π = π π£π β π£π
β’ Golden rules of op amp circuits:
β’ The output tries to force π£π = π£π (with
negative feedback) .
β’ The inputs draw no current.
β’ The output has no impedance.
vNvP vO
Impedance
Black box model: Op-amp consists of:
β’ Internal input impedance (ππ).
β’ Voltage source (ππ ).
β’ Internal output impedance (ππ).
vIZ i
vOZovS
Impedance (Input Side)
β’ This internal input impedance reduces the
input voltage from the source to op-amp
(input loading).
β’ Want πππ = β for ideal op-amp.
Z s
vs
Device 1 Op-Amp
vi
Z i π£π =ππ
ππ + ππ π£π
Impedance (Output Side)
β’ Output internal impedance reduces the output
voltage driven to the load from op-amp
(output loading) β reduce efficiency.
β’ Want πππ’π‘ = 0 for ideal op-amp.
Zo
avi
Op-Amp Device 2
vo
Z L π£π =ππ
ππ + ππππ£π
Impedances of Op-Amp Loading
β’ Inclusion of internal impedance and
measurements are required for precise op-amp
design consideration.
β’ For ideal op-amp, we want πππ = β and πππ’π‘ = 0.
viZ i
voZ o
Aocvi
Zs
vs
Device 1 Op-Amp
vi
Z i
Zo
ZLAocvi
Op-Amp Device 2
vo
Impedances of Op-Amp Loading
viZ i
vo
Z o
Aocvi
Zs
vs
Input
Source Op-Amp
vi
Z i
Zo
ZLAocvi
Op-Amp Load
vo
Put equation (1) into equation (2):
ππππ
=ππ
ππ + ππ
ππππ + ππ
π΄ππππ
ππ =ππ
ππ + ππππ (πΈπ. 1)
ππ =ππ
ππ + πππ΄ππππ (πΈπ. 2)
Impedance: Inverting Amplifier
R1
R2
vOvI
Inverting Amplifier:
π΄ =ππππ
= βπ 2π 1
1
1 + (1 + π 2/π 1)/π
Note: finite open loop gain (π) of the op amp -> reduce overall gain of the amplifier.
Impedance: Non-Inverting Amplifier
Non-Inverting Amplifier:
π΄ =ππππ
= 1 +π 2π 1
1
1 + (1 + π 2/π 1)/π
R1
R2
vOvI
Note: finite open loop gain (π) of the op amp -> reduce overall gain of the amplifier.
Impedance: SummingAmplifier
Summing Amplifier:
ππ =π π
π 1π1 +
π π
π 2π2 +β―
v2 vO
R 2
R fR 1v1
Impedance: DifferenceAmplifier
R1 R2
vO
v1
R3R4
v2
Difference Amplifier:
ππ =π 2π 1
1 + π 1/π 21 + π 3/π 4
π2 β π1
Common and Differential Modes
Differential Mode:
ππ·π = π2 β π1
π ππ = 2π 1
Common Mode:
ππΆπ =π2 + π1
2
π ππ =π 1 + π 2
2
R1 R2
vO
R2R1
vD M
2
vD M
2
vCM
vDM
Analysis of difference amplifier circuit.
Non-Ideal Difference Amplifier
β’ Resistor mismatch: due to tolerance of the resistors used in the circuit.
β’ Note that β = imbalance factor.
R1 R2(1 - β)
vO
R2R1
vD M
2
vD M
2
vCM
vDMπ΄πΆπ =
π 2
π 1 + π 2β
π΄π·π =π 2
π 11 β
π 1 + 2π 2
π 1 + π 2
β
2
(Eq.1)
(Eq.2)
Difference Amplifier Calibration
R1 R2
vO
R3R4
βvcalib
R calib
+vcalib
Circuit calibration:
β’ Trimming using Howland circuit.
β’ Downside: increase cost of production.
Common Mode Rejection Ratio
β’ Common Mode Rejection Ratio (CMRR) is one of the factors used for determining signal quality in a given electronic circuit.
β’ Other factors: Signal gain, signal to noise ratio (SNR), total harmonic distortion (THD), etc.
β’ CMRR is ability of the device to reject common-mode signals, i.e. those that appear simultaneously and in-phase on both inputs.
Common Mode Rejection Ratio
β’ An ideal differential amplifier would have infinite
CMRR.
β’ However this is not achievable in practice.
β’ Gain equations for common & differential modes:
ACM = [R2 / (R1 + R2)] Ο΅ (Eq. 1)
ADM = R2/R1 [1 - (R1 + 2R2)/(R1 + R2)Ο΅/2] (Eq. 2)
β’ For the given difference amplifier:
CMRR = 20 log10 |ADM/ACM|
CMRR β 20 log10 |(1 + R2/R1)/Ο΅| (Eq. 3)
Differential Mode Signalling
β’ Problem: GND β Earth.
β’ This is due to design of the circuit, analogue/ digital sections, grounding via chassis.
vO
Zg
vI
R 1 R 2
Differential Mode Signalling
β’ Solution: Make π£πΊ common.
β’ This is to ensure that both of the of the top and bottom paths have common reference.
vOvI
R3R4
Zg
R1 R2
Differential Mode Signalling
Problem: Ground loop
β’ As source and amplifier are often far apart β voltage drop due to ground bus impedance
vO
Zg
vI
R 1 R 2
Differential Mode Signalling
β’ Solution: isolate source from GND through useof twisted pair cabling.
β’ The common mode noises will be cancelled out by each others at matching opposite pairs.
β’ Reduces the natural self inductance property of wire and eliminate field interferences.
R1 R2
vO
R1R2
vI
Differential Mode Signalling
β’ Solution: isolate source from GND through useof RF choke.
β’ The common mode noises will be cancelled out by each others (through out of phase pair of signals).
R1 R2
vO
R1R2
vI
T1
Design Problem
vref
R1
R(1 + )
v1
β’ Want to measure very small v.
β’ From big background voltage.
β’ Where R, R1 are all BIG.
β’ Some gain would be nice, too!
For electronic circuits used in the sensor applications, it is quite common to encounter these requirements and issues:
β’ Want to measure very small voltage (v) -> voltage
divisions and differential signal amplification.
β’ Accuracy of voltage divider depends the tolerance
of the resistors used in the circuit.
vref
R1
v1
R(1 + )
R3R2
Design Problem
β’ From big background voltage -> resistor bridge.
β’ Resistor bridge has inherent sensitivity for detecting
small voltage differences.
vref
R1
R R(1 + )
R1
v2 v1
R3
R3R2
R2
Design Problem
vref
R1
R R(1 + )
R1
v2 v1
R2 R3
v1
R3R2v2
R4
R4
β’ Where R, R1 are all BIG -> input buffers isolation
through voltage followers on both ends.
Design Problem
Design Problem
β’ Some gain would be nice, too! -> gain adjustment
trimmed/tailored towards design requirements.
β’ Adjustable gain for tuning the accuracy of the
measurement.
vref
R1
R R(1 + )
R1
v2 v1
R2 R3v1
R3R2v2
R4
RG
R4
Instrumentation Amplifiers
For even better performance than differenceamplifiers:
β’ Very high Zin.
β’ Very high CMRR.
R1 R2
vO
v1
R2R1v2
R3
RG
R3
vo2
vo1
Instrumentation Amplifiers
Components:
β’ Difference amplifier:
A2 = R2/R1
β’ Input buffers:
A1 = (1+2R3/ RG)
R1 R2
vO
v1
R2R1v2
R3
RG
R3
vo2
vo1
Instrumentation Amplifier Connections
β’ Discrete package integrated circuit.
β’ Sense pin e.g. remote or local.
β’ External reference voltage.
β’ External adjustable gain.
RG
Sense
Ref
R1 R2v1
R1v2
R3
RG
R3
R2Amplifier
Instrument Amplifier Application (Transducer Bridge)
β’ Encapsulate the stated solutions: small measurement (), imbalance impedance, imbalance voltage, and adjustable trimmed gain.
vO
vref
R1
RG
Sense
Ref
R1
R R(1 +)
v2 v1
PracticalTransducer Bridges
β’ Due to resistors tolerance, most of the occasions, require trimming using variable resistors.
vref
R3
R2
vO
R1
RG
Sense
Ref
R1
R R(1 +)
Two Op-Amp Instrumentation Amplifier
β’ Just two precision amplifiers can be configured
to create a differential to single ended
instrumentation amplifier.
π£π = 1 + π 2/π 1 π£2 β π£1
β’ These op amp instrumentation amplifiers:
β’ Cheap.
β’ Asymmetric.
v1 v2
R1 R2
vO
R1R2
Two Op-Amp Instrumentation Amplifier
β’ Gain tuning: realised with feedback path added
to two op amp instrumentation amplifier.
π£π = 1 + π 2/π 1 + 2π 2/π πΊ π£2 β π£1
β’ Gain accuracy is tuned by π πΊ .
v1 v2 vO
RG
R2 R1 R1 R2