35
Operational Amplifier XMUT303 Analogue Electronics

Lecture Block 1: Operational Amplifier ECEN303:Analogue

  • Upload
    others

  • View
    0

  • Download
    0

Embed Size (px)

Citation preview

Operational Amplifier

XMUT303 Analogue Electronics

Topics

• Basic characteristics and properties.

• Device imperfections.

• Circuit limitations.

• Design problems and solutions.

OperationalAmplifiers

• Voltage relationship:

𝑣𝑂 = 𝑎 𝑣𝑃 − 𝑣𝑁

• Golden rules of op amp circuits:

• The output tries to force 𝑣𝑃 = 𝑣𝑁 (with

negative feedback) .

• The inputs draw no current.

• The output has no impedance.

vNvP vO

Impedance

Black box model: Op-amp consists of:

• Internal input impedance (𝑍𝑖).

• Voltage source (𝑉𝑠).

• Internal output impedance (𝑍𝑜).

vIZ i

vOZovS

Impedance (Input Side)

• This internal input impedance reduces the

input voltage from the source to op-amp

(input loading).

• Want 𝑍𝑖𝑛 = ∞ for ideal op-amp.

Z s

vs

Device 1 Op-Amp

vi

Z i 𝑣𝑖 =𝑍𝑖

𝑍𝑖 + 𝑍𝑠𝑣𝑠

Impedance (Output Side)

• Output internal impedance reduces the output

voltage driven to the load from op-amp

(output loading) – reduce efficiency.

• Want 𝑍𝑜𝑢𝑡 = 0 for ideal op-amp.

Zo

avi

Op-Amp Device 2

vo

Z L 𝑣𝑜 =𝑍𝑙

𝑍𝑙 + 𝑍𝑜𝑎𝑣𝑖

Impedances of Op-Amp Loading

• Inclusion of internal impedance and

measurements are required for precise op-amp

design consideration.

• For ideal op-amp, we want 𝑍𝑖𝑛 = ∞ and 𝑍𝑜𝑢𝑡 = 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):

𝑉𝑜𝑉𝑖

=𝑍𝑖

𝑍𝑠 + 𝑍𝑖

𝑍𝑙𝑍𝑜 + 𝑍𝑙

𝐴𝑜𝑐𝑉𝑠

𝑉𝑖 =𝑍𝑖

𝑍𝑠 + 𝑍𝑖𝑉𝑠 (𝐸𝑞. 1)

𝑉𝑜 =𝑍𝑙

𝑍𝑜 + 𝑍𝑙𝐴𝑜𝑐𝑉𝑖 (𝐸𝑞. 2)

Impedance: Inverting Amplifier

R1

R2

vOvI

Inverting Amplifier:

𝐴 =𝑉𝑜𝑉𝑖

= −𝑅2𝑅1

1

1 + (1 + 𝑅2/𝑅1)/𝑎

Note: finite open loop gain (𝑎) of the op amp -> reduce overall gain of the amplifier.

Impedance: Non-Inverting Amplifier

Non-Inverting Amplifier:

𝐴 =𝑉𝑜𝑉𝑖

= 1 +𝑅2𝑅1

1

1 + (1 + 𝑅2/𝑅1)/𝑎

R1

R2

vOvI

Note: finite open loop gain (𝑎) of the op amp -> reduce overall gain of the amplifier.

Impedance: SummingAmplifier

Summing Amplifier:

𝑉𝑜 =𝑅𝑓

𝑅1𝑉1 +

𝑅𝑓

𝑅2𝑉2 +⋯

v2 vO

R 2

R fR 1v1

Impedance: DifferenceAmplifier

R1 R2

vO

v1

R3R4

v2

Difference Amplifier:

𝑉𝑜 =𝑅2𝑅1

1 + 𝑅1/𝑅21 + 𝑅3/𝑅4

𝑉2 − 𝑉1

Common and Differential Modes

Differential Mode:

𝑉𝐷𝑀 = 𝑉2 − 𝑉1

𝑅𝑖𝑑 = 2𝑅1

Common Mode:

𝑉𝐶𝑀 =𝑉2 + 𝑉1

2

𝑅𝑖𝑐 =𝑅1 + 𝑅2

2

R1 R2

vO

R2R1

vD M

2

vD M

2

vCM

vDM

Analysis of difference amplifier circuit.

Non-Ideal Difference Amplifier

• Resistor mismatch: due to tolerance of the resistors used in the circuit.

• Note that ∈ = imbalance factor.

R1 R2(1 - ∈)

vO

R2R1

vD M

2

vD M

2

vCM

vDM𝐴𝐶𝑀 =

𝑅2

𝑅1 + 𝑅2∈

𝐴𝐷𝑀 =𝑅2

𝑅11 −

𝑅1 + 2𝑅2

𝑅1 + 𝑅2

2

(Eq.1)

(Eq.2)

Difference Amplifier Calibration

R1 R2

vO

R3R4

—vcalib

R calib

+vcalib

Circuit calibration:

• Trimming using Howland circuit.

• Downside: increase cost of production.

Common Mode Rejection Ratio

• Common Mode Rejection Ratio (CMRR) is one of the factors used for determining signal quality in a given electronic circuit.

• Other factors: Signal gain, signal to noise ratio (SNR), total harmonic distortion (THD), etc.

• 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

• An ideal differential amplifier would have infinite

CMRR.

• However this is not achievable in practice.

• Gain equations for common & differential modes:

ACM = [R2 / (R1 + R2)] ϵ (Eq. 1)

ADM = R2/R1 [1 - (R1 + 2R2)/(R1 + R2)ϵ/2] (Eq. 2)

• For the given difference amplifier:

CMRR = 20 log10 |ADM/ACM|

CMRR ≈ 20 log10 |(1 + R2/R1)/ϵ| (Eq. 3)

Differential Mode Signalling

• Problem: GND ≠ Earth.

• This is due to design of the circuit, analogue/ digital sections, grounding via chassis.

vO

Zg

vI

R 1 R 2

Differential Mode Signalling

• Solution: Make 𝑣𝐺 common.

• 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

• As source and amplifier are often far apart – voltage drop due to ground bus impedance

vO

Zg

vI

R 1 R 2

Differential Mode Signalling

• Solution: isolate source from GND through useof twisted pair cabling.

• The common mode noises will be cancelled out by each others at matching opposite pairs.

• Reduces the natural self inductance property of wire and eliminate field interferences.

R1 R2

vO

R1R2

vI

Differential Mode Signalling

• Solution: isolate source from GND through useof RF choke.

• 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

• Want to measure very small v.

• From big background voltage.

• Where R, R1 are all BIG.

• Some gain would be nice, too!

For electronic circuits used in the sensor applications, it is quite common to encounter these requirements and issues:

• Want to measure very small voltage (v) -> voltage

divisions and differential signal amplification.

• Accuracy of voltage divider depends the tolerance

of the resistors used in the circuit.

vref

R1

v1

R(1 + )

R3R2

Design Problem

• From big background voltage -> resistor bridge.

• 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

• Where R, R1 are all BIG -> input buffers isolation

through voltage followers on both ends.

Design Problem

Design Problem

• Some gain would be nice, too! -> gain adjustment

trimmed/tailored towards design requirements.

• 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:

• Very high Zin.

• Very high CMRR.

R1 R2

vO

v1

R2R1v2

R3

RG

R3

vo2

vo1

Instrumentation Amplifiers

Components:

• Difference amplifier:

A2 = R2/R1

• Input buffers:

A1 = (1+2R3/ RG)

R1 R2

vO

v1

R2R1v2

R3

RG

R3

vo2

vo1

Instrumentation Amplifier Connections

• Discrete package integrated circuit.

• Sense pin e.g. remote or local.

• External reference voltage.

• External adjustable gain.

RG

Sense

Ref

R1 R2v1

R1v2

R3

RG

R3

R2Amplifier

Instrument Amplifier Application (Transducer Bridge)

• 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

• 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

• Just two precision amplifiers can be configured

to create a differential to single ended

instrumentation amplifier.

𝑣𝑜 = 1 + 𝑅2/𝑅1 𝑣2 − 𝑣1

• These op amp instrumentation amplifiers:

• Cheap.

• Asymmetric.

v1 v2

R1 R2

vO

R1R2

Two Op-Amp Instrumentation Amplifier

• Gain tuning: realised with feedback path added

to two op amp instrumentation amplifier.

𝑣𝑜 = 1 + 𝑅2/𝑅1 + 2𝑅2/𝑅𝐺 𝑣2 − 𝑣1

• Gain accuracy is tuned by 𝑅𝐺 .

v1 v2 vO

RG

R2 R1 R1 R2

Instrument AmplifierApplication (Active Guard)

• Use of double ended transmission, noise appears as common-mode component, hence rejected by op-amp.

• CMRR degradation with frequency due to RC imbalance.

R1 R2

vO

R2R1

R3

RG

R3 R s

R s

vI vCM