Conditioning Circuits

Conditioning Circuits

Dr. Ashraf Saleem

Dr. Ashraf Saleem 2

Structure of Mechatronic Systems

Dr. Ashraf Saleem 3

Signal Amplifiers

Designed to amplify input signals to a right level to be noticeable for further uses.

Typical input signals are:thermocouple, RTD, pressure, strain, flow, etc.

Typical outputs include:high level dc voltages (0 to 5 or 0 to 10 volts), process current (0 to 20 mA or 4 to 20 mA)

There are commercial signal conditioners with computer interface ready.

Dr. Ashraf Saleem 4

Operational Amplifier (Op Amp)

An operational amplifier (Op Amp) is an integrated circuit of a complete amplifier circuit.

Op amps have an extremely high gain (A=10A=1055 typicallytypically).

Op amps also have a high input impedance (R=4 MR=4 MΩΩ , , typicallytypically) and a low output impedance (in order of 100 in order of 100 ΩΩ , , typicallytypically) .

-

+

Vi1 VoutA

BVi2

( )AVVV iiout 21 −=

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Characters of Operational Amplifiers

high open loop gain

high input impedance

low output impedance

low input bias current

wide bandwidth

large common mode rejection ratio (CMRR)

11 22 33 44

88 77 66 55

Offset nullOffset null

Offset nullOffset nullNot usedNot used

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Input Impedance

WHY HIGH?

For an instrument the ZINshould be very high (ideally infinity) so it does not divert any current from the input to itself even if the input has very high resistance.

e.g. an op-amp taking input from a microelectrode.

Input Circuit Output

Impedance between input terminals = input impedance

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Output Impedance

Input Circuit Output

Impedance between output terminals = output impedance

WHY LOW?

For an instrument the ZOUTshould be very low (ideally zero) so it can supply output even to very low resistive loads and not expend most of it on itself.

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Voltage Output from an Amplifier

The linear range of an amplifier is finite, and limited by the supply voltage and the characteristics of the amplifier.

If an amplifier is driven beyond the linear range (overdrivenoverdriven), serious errors can result if the gain is treated as a constant.

AA

LinearLinearregionregion

NonNon--linearlinearregionregion

VVoutout

VVinin

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Analysis of Op-Amp Circuits

The following rules can be applied to almost all op-amp circuits with external feedback:No current can enter op amp input terminals, because of infinite input impedanceThe +ve and –ve (non-inverting and inverting) inputs are forced to be at the same potential.

-

+

Vi1 VoutA

BVi2

( )−+ −= VVAV OGout

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Voltage Follower

V+ = VIN.

By virtual ground, V- = V+

Thus Vout = V- = V+ = VIN !!!!

•Due to the infinite input impedance of an op amp, no current at all can be drawn from the circuit before VIN. Thus this part is effectively isolated.

•Very useful for interfacing to high impedance sensors such as microelectrode, microphone…


Inverting Amplifier

Point B is grounded, so does point A (very small).

Voltage across R1 is Vin, and across RF is Vout.

The output node voltage determined by Kirchhoff'sCurrent Law (KCL).

Circuit voltage gaindetermined by the ratio of R1 and RF.

Vout

1RR

VVG F

in

out −==

-

+

Vin

R1

RF

A

B

F

F

RRRRR+

=1

13


Analysis of Inverting Amplifier

Ideal transfer characteristics:Ideal transfer characteristics:

-

+

Vin

Vout

R1

RF

A

BR

ii++VV++

iiFF

ii11ii--

VV--

0== +− ii

+− = VV

FF iiii =+= −1

F

outF

IN

RVViand

RVVi −

=−

= −− 1

1

000 =→=→= −++ VVi

F

outIN

RV

RV

−=1

oror1R

RVV F

in

out −=


Op-amp circuit is a voltage divider.

Noninverting Amplifier

-

+VinVout

R1

RF

A

BF

outA RRRVV+

⋅=1

1

1

1RR

VVG F

in

out +==

Circuit voltage gain determined by the ratio of R1 and RF.

Point VA equals to Vin .


Differential AmplifierDifferential Amplifier

Point B is grounded, so does point A (very small).

Voltage across R1 is V1, and across R2 is V2.

Normally: R1 = R2, and RF= R3.

Commonly used as a single op-amp instrumentation amplifier.)( 12

1

VVRRV F

out −=

RF

-

+

V1

Vout

R1

A

BR3

V2R2


Design an Instrumentation Amplifier

Design a single op-amp instrumentation amplifier.

R1 = R2, RF = R3Determine the instrumentation gain.-

+

V1

Vout

R1

RF

A

BR3

V2

R2

AF

OUTAA iR

VVR

VV−

−=

−

1

1

32

2

RVi

RVV B

BB =+

−

0→= BA ii

2

2

31

1

RVV

RV

RVV

RVV BB

F

OUTAA −+−

−=

−

( ) ( ) ( )1

12

RVVVV

RVVV BA

F

BAOUT −−−=

−−)( 121

VVRRV F

out −=

BA VV ≈


Instrumentation Amplifier with isolators

In order to avoid high current driving to the circuit, V1 and V2 input lines are connected straight to the inputs of two voltage-follower op-amps, giving very high impedance. The two op-amps on the left now handle the driving of current through the resistors instead of letting the input voltage sources (whatever they may be) do it.


Example 1:

A Sensor outputs a voltage range of 20 to 250 mV. The sensor output has to feed computer based controller that work at voltage range of 0 to 5 V. Design the required conditioning circuit in order to get the demand voltage range.


Instrumentation Amplifier with Adjustable Gain

Inverting amplifier

Non-inverting amplifier

Differential amplifier but with very high input impedance

- So, you can connect to sensors

Differential amplifier -> it rejects common-mode interference -> so you can reject noise

Gain in the multiple stages: i.e. High Gain – so, you can amplify small signals

Put some lowpassand high pass filters!


Instrumentation Amplifier: Stage 1

I1

Recall virtual ground of opamps

I1 = (V1 – V2)/R1

Recall no current can enter opamps and Kirchoff’s current law

I2 = I3 = I1Recall Kirchoff’s voltage law

VOUT = (R1 + 2R2)(V1 – V2)/R1

= (V1 – V2)(1+2R2/R1)

I2I3


Instrumentation Amplifier: Stage 2

I1

Recall virtual ground of opampsand voltage divider

V- = V+ = V2R4/(R3 + R4)

Recall no current can enter opamps

(V1 – V-)/R3 = (V- – VOUT)/R4

Solving,

VOUT = (V2 – V1)R4/R3

I2I3


VOUT = (V2 – V1)(1 + 2R2/R1)(R4/R3)Gain from Stage I and Stage II

Instrumentation Amplifier: CompleteInstrumentation Amplifier: Complete


Example 2:

Re-solve example 1 by using the adjustable instrumentation amplifier.


Example 3:

A sensor outputs a voltage ranging -2.4 to -1.1 V. For interface to ADC, this needs to be 0 to 2.5 V. Develop the required signal conditioning.


Example 4:

Temperature is to be measured in the range of 250oC to 450oC. The sensor resistance that varies linearly from 280Ω to 1060Ω for the temperature range. Power dissipated in the sensor must be kept below 5mW. Develop analog signal conditioning that provides a voltage varying linearly from -5 to 5 V for the temperature range. The load is high impedance recorder.


Sources of Amplifier ErrorsSources of Amplifier Errors

Temperature drift: a drift in the output signal per unity change in the temperature (e.g., µv/ oC) Offset current: present at the input leads due to bias currents that are needed to operate the solid-state circuitry. Offset voltage that might be present at the output even when the input leads are open. Common mode output voltageThe inverting gain is not equal to the non-inverting gain.Internal noiseGround Loop Noise: which can enter the signal leads because of the possible potential difference between the two ground points.


Outlines of FiltersOutlines of Filters

Filterinput output

Filtering: Certain desirable features are retainedOther undesirable features are suppressed


Classification of FiltersClassification of Filters

Signal Filter

Analog Filter Digital Filter

Element Type Frequency Band

Active Passive Low-Pass

High-Pass

Band-Pass

Band-Reject

All-Pass


Filter classification according to implementationActive filters include RC networks and op-amps

Suitable for low frequency, small signalActive filters are preferred since avoid the bulk and non-linearity of inductors. However, active filters require a power supply

Passive filters consist of RCL networksSimple, more suitable for frequencies above audio range, where active filters are limited by the op-amp bandwidth

Digital filters



Filter classification according to frequency responseLow-pass filterHigh-pass filterBand-pass filterBand-stop (Notch) filter



StateState--variable filtersvariable filters

Also known as a Universal Active FilterConsists of one amplifier and two integratorsHigh-pass, low-pass and band-pass in the same ICExample below: Burr Brown UAF42


Data Acquisition SystemData Acquisition System

Mechatronic systems use digital data acquisition for a variety of purposes such as:process condition monitoring and performance evaluation,fault detection and diagnosis, product quality assessment, dynamic testing, system identification (i.e., experimental modeling), and feedback control.


Data Acquisition SystemBlock Diagram


A/D Converter: Input Signal

AnalogSignal is continuousExample: strain gage. Most transducers

produce analog signals

DigitalSignal is either ON or OFFExample: light switch.


DigitalDigital--Analog Converter (DAC)Analog Converter (DAC)Weighted Resistor DACWeighted Resistor DAC

22...

2 102

1ref

nn

n

vbbbv ⎥⎦⎤

⎢⎣⎡ +++= −

−−


DigitalDigital--Analog Converter (DAC)Analog Converter (DAC)Ladder DAC (2RLadder DAC (2R--R)R)

22...

2 102

1ref

nn

n

vbbbv ⎥⎦⎤

⎢⎣⎡ +++= −

−−


Example 1

What is the output voltage of a 10-bit Ladder DAC with a 10 V reference voltage if the input is 00101101012.


Conversion Resolution

The conversion resolution is a function of the reference voltage and the number of bits in the word. The more bits, the smaller the change in analog output for a 1-bit change in binary word.

nRout VV −=∆ 2

Where ∆Vout = smallest output change VR = reference voltage n = number of bits in the word


Example 2:

Determine how many bits a D/A converter must have to provide output increments of 0.04V or less. The reference is 10V


Example 3:

A control valve has a linear variation of opening as the input voltage varies from 0 to 10V. A microcomputer outputs an 8-bit word to control the valve opening using an 8-bit DAC to generate the valve voltage.

a- Find the reference voltage required to obtain a full open valve (10V).

b- Find the percentage of valve opening for a 1-bit change in the input word.


DAC Error SourcesDAC Error Sources

Code Ambiguity. In many digital codes (e.g., in the straight binary code), incrementing a number by an LSB will involve more than one bit switching. If the speed of switching from 0-1 is different from that for 1-0, and if switching pulses are not applied to the switching circuit simultaneously, the switching of the bits will not take place simultaneously. Settling Time. The circuit hardware in a DAC unit will have some dynamics, with associated time constants and perhaps oscillations (underdamped response). Hence, the output voltage cannot instantaneously settle to its ideal value upon switching.The time required for the analog output to settle within a certain band (say :t2% of the final value or :t resolution), following the application of the digital data, is termed settling time.


DAC Error SourcesDAC Error Sources

Parametric Errors. resistor elements in a DAC might not be very precise, particularly when resistors within a wide range of magnitudes are employed, as in the case of weighted-resistor DAC. These errors appear at the analog output. Furthermore, aging and environmental changes (primarily, change in temperature) will change the values of circuit parameters, resistance in partic-ular. This also will result in DAC error. Reference Voltage Variation: Since the analog output of a DAC is proportional to the reference voltage vref, any variations in the voltage supply will directly appear as an error. This problem can be overcome by using stabilized voltage sources with sufficiently low output impedance.


Analog to Digital Converters (ADC)Analog to Digital Converters (ADC)

r

innn V

Vbbb =+++ −−− 2.........22 22

11

The formulae of analog to digital conversion is as following:

Where b1b2b3….bn = n-bit digital output

Vin = analog input voltage

Vr = analog reference voltage

The output uncertainty/resolution can be given by:

∆V=Vr2-n


Example 1:

Temperature is measured by a sensor with an output of 0.02 V/oC. Determine the required ADC reference and word size to measure 0o to 100oC with 0.1oCresolution


Example 2:

Find the digital word that results from a 3.127 V input to a 5-bit ADC with a 5V reference?


Successive Approximation ADCSuccessive Approximation ADC


Example 3:

Find the successive approximation ADC output for a 4-bit converter to a 3.217V input if the reference is 5V.


DualDual--Slope ADC Slope ADC


Example 4:

A dual-slope ADC as shown previously has R=100 kΩ and C=0.01µF. The reference is 10V, and the fixed integration time is 10 ms. Find the conversion time for a 6.8V input.


Bridge Circuits

Wheatstone Bridge: The primary application of this bridge in signal conditioning is to convert variations of resistance into variation of voltage


Wheatstone Bridge

VO will equal zero. In this situation, the bridge is said to beBALANCED. Any change in any resistive element will result in anUNBALANCED condition, hence VO will be non-zero.


Wheatstone Bridge

If we replace R3 with an active strain gauge, RG, which will vary by ∆R when pressure is applied (the arrow through the resistor shows that it has a variable value), then


Example 1:

The resistors in a bridge are given by R1=R2=R3=120Ω and R4=121Ω. If the supply is 10V. Find the voltage offset?


AC Bridges

AC bridges are used mainly to detect changes in inductance and/or capacitance

At balance condition


Example 2:

An ac bridge is in balance with thefollowing constants: arm AB, R = 200 Ωin series with L = 15.9 mH ; arm BC, R =300 Ω in series with C = 0.265 µF; armCD,unknown; arm DA, = 450 Ω. Theoscillator frequency is 1 kHz. Find theconstants of arm CD.


Comparison Bridge: Capacitance

Measure an unknown capacitance by comparing it with a Known capacitance.


Comparison Bridge: Inductance

Measure an unknown inductance by comparing it with a Known inductance.


Maxwell Bridge

Measure an unknown inductance in terms of a known capacitance

Conditioning Circuits

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