et zc344-l5 part2

BITS PilaniPilani Campus

Instrumentation and controlET ZC 341

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Swapna KulkarniLecturer

BITS PilaniPilani Campus

Operational Amplifiers

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BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Voltage Follower Figure : The Op amp voltage follower. This circuit

has unity gain but very high input impedance • Figure shows an opamp circuit with unity gain and very high input impedance. The input impedance is essentially the input impedance of the op amp itself, which can be greater than 100M . The voltage output tracks the input over a range defined by the plus and minus saturation voltage outputs.

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• Current output is limited to the short circuit ofop amp, and output impedance is typicallymuch less than 100 . In many cases, amanufacturer will market an opamp voltage

follower whose feedback is provided internally.

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• Such a unit is usually specifically designed forvery high input impedance. The unity gainvoltage follower is essentially an impedancetransformer in the sense of converting avoltage at high impedance to the samevoltage at low impedance.

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This connection forces the op-amp to adjust its output voltage simply equal to the input voltage (Vout follows Vin so the circuit is named op-amp voltage follower). The importance of this circuit does not come from any change in voltage, but from the input and output impedances of the op-amp.


Inverting Amplifier

• Equation shows that this circuit inverts theinput signal and may have either attenuationor gain, depending on the ratio of inputresistance, R1 and feedback resistance R2.

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inout VR

RV

1

2


• It is important to note that the input impedance ofthis circuit is essentially equal to R1, the inputresistance. In general, this resistance is not large, andhence the input impedance is not large. The outputimpedance is low.

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Figure: The Op Amp summing Amplifier



Summing Amplifier


Summing Amplifier

• A common modification of the invertingamplifier is an amplifier that sums or adds twoor more applied voltages.

• This circuit is shown in figure for the case ofsumming two input voltages. The transferfunction of this amplifier is given by

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2

3

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1

2 VR

RV

R

RVout

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The sum can be scaled by proper selection ofresistors. For example, if we make R1 = R2=R3,then the output is simply the (inverted) sumof V1 and V2. The average can be found bymaking R1 = R3 and R2 = R1/2.


Develop an op amp circuit that can provide an output voltage related to the input voltage by Vout=3.4 Vin+5

• Solution: One way is to use a summing amplifier with Vin on

one input and 5 V on the other. The gains will be selected to be 3.4 and 1.0,respectively. The summing amplifier is also an inverter, however, so the sign will be wrong.

Thus, a second amplifier will be used with a gain of -1 to make the sign correct. The result is shown in next slide. Selection of the values of resistors is based on the general notion of keeping the currents in mA.

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Figure : The op amp circuit for example.




Noninverting Amplifier

• A noninverting amplifier may be constructed from an op amp, as shown in figure.

• The gain of this circuit is found by summing the currents at the summing point, S, and using the fact that the summing point voltage is Vin so that no voltage difference appears across the input terminals.

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Figure A noninverting amplifier

I1 + I2 = 0 Where I1 = Current through R1

I2 = Current through R2




• But these currents can be found from Ohm’slaw such that this equation becomes

021

R

VV

R

V outinin

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Solving this equation for Vout we find

inout VR

RV

1

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Differential Instrumentation Amplifier

• An ideal differential amplifier provides anoutput voltage with respect to ground that issome gain times the difference between twoinput voltages.

Vout = A(Va-Vb)

Where A is the differential gain and both Va andVb are voltages with respect to ground.

• Such an amplifier plays an important role ininstrumentation and measurement.

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• To define the degree to which a differentialamplifier approaches the ideal, we use thefollowing definition.

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• The common-mode input voltage is theaverage applied to the two input terminals,

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bacm

VVV


• An ideal differential amplifier will not have anyoutput that depends on the value of thecommon-mode voltage; that is the circuit gainfor common–mode voltage, Acm will be zero.

• The common-mode rejection ration (CMRR) ofa differential amplifier is defined as the ratioof the differential gain to the common-modegain.

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• The common – mode rejection (CMR) is the CMRR expressed in dB.

)(log20 10 CMRRCMR

A

ACMRR

cm

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Figure : The basic differential amplifier configuration



Differential Amplifier


Differential Amplifier

• There are a number of op amp circuit fordifferential amplifiers.

• Notice that the circuit uses two pairs ofmatched resistors, R1 and R2. When thematching is perfect and the op amp is ideal, thetransfer function for this amplifier is given by

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12

1

2 VVR

RVout


• If the resistors are not well matched, the CMR willbe poor. The circuit of figure has a disadvantage inthat its input impedance is not very high and,further is not the same for the two inputs. For thisreason, voltage followers are often used on theinput to provide high input impedance. The result iscalled an instrumentation amplifier.

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Instrumentation Amplifier

• Figure shows one type of instrumentation amplifierin common use, voltage followers are simply placedon each input line. The transfer function is stillgiven by equation.

• One disadvantages of this circuit is that changinggain requires changing two resistors and havingthem carefully matched in value.

• Input offset compensation can be provided usingonly the differential amplifier op amp tocompensate for overall offsets of all three op amps.

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12

1

2 VVR

RVout


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Figure : An instrumentation amplifier includes voltagefollowers for input isolation.




• A more common configuration of instrumentationamplifier, however, is the circuit shown in figure. Thiscircuit allows for selection of gain, within certainlimits, by adjustment of a single resistor, RG. It can beshown that the CMR of this circuit, although stilldependent on careful matching of the differentialamplifier resistors, does not depend on matching ofthe two R1s. The transfer function of this amplifier isgiven by

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12

2

3121 VV

R

R

R

RV

G

out


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Figure: This instrumentation amplifier allows the gain to be changed using a single resistor




• The input impedance is very high, and theoutput impedance very low. Many ICmanufacturers provide this circuit with fixeddifferential gain and R1 but allow the user toinsert external RG so the desired gain can beselected. They can thus ensure a high CMR.

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Voltage-to-Current Converter

• Because signals in process-control are mostoften transmitted as a current, specifically 4 to20 mA, it is often necessary to employ a linearvoltage-to-current converter.

• Such a circuit must be capable of sinking acurrent into a number of different loadswithout changing the voltage-to-currenttransfer characteristics.

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Figure : A voltage-to-current using an op amp




• An op amp circuit that provides this function isshown in figure. An analysis of this circuitshows that the relationship between currentand voltage is given by

inVRR

RI

31

2

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• Provided that the resistance are selected sothat

R1(R3+R5)=R2R4

• The circuit can deliver current in eitherdirection, as required by a particularapplication.

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• The maximum load resistance and maximum currentare related and determined by the condition that theamplifier output saturates in voltage. Analysis of thecircuit shows that when the op amp output voltagesaturates, the maximum load resistance andmaximum current are related by

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543

354

RRR

RI

VRR

Rm

sat

ml


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WhereRml = maximum load resistanceVsat= op amp saturation on voltage Im = maximum current

•A study of equation shows that the maximum load resistance is always less than Vsat/Im. The minimum load resistance is zero.


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Figure : A current-to-voltage converter using an op amp. Care must be taken that the current output capability of the op amp is not exceeded.

Current-to-Voltage Converter




• At the receiving end of the process-control signaltransmission system, we often need to convert thecurrent back into a voltage. This can be done mosteasily with the circuit shown in figure. This circuitprovides an output voltage given by

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Vout = -IR

• Provided the op amp saturation voltage has not beenreached. The resistor, R, in the non-inverting terminals isemployed to provide temperature stability to theconfiguration.


Integrator

Figure An integrator circuit using anop amp.

• Another op amp circuit to be considered is the integrator.

• Integrator is also called as Active low pass filter.

• Using the ideal analysis, we can sum the currents at the summing point as

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0dt

dVC

R

V outin



BITS Pilani, Pilani Campus

• Rearranging the equation and integrating from0 to t, we get


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• This result shows that the output voltagevaries as an integral of the input voltage witha scale factor of -1/RC. This circuit is employedin many cases where integration of atransducer output is desired.

dtVRC

V inout

1


• Other function also can be implemented, such as ahighly linear ramp voltage. If the input voltage isconstant Vin=K, equation reduces to

tRC

KVout

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• Which is linear ramp, a negative slope of K/RC. Somemechanism of reset through discharge of thecapacitor must be provided, because otherwise Vout

will rise to the output saturation value and remainfixed there in time.


Differentiator

• It is also possible to construct an op ampcircuit with an output proportional to thederivative of the input voltage.

0R

V

dt

dVC outin

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Figure : This circuit takes the time derivative of theinput voltage



It is also called as Active high pass filter.


• Solving for the output voltage shows that the circuit response is

dt

dVRCV in

out

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Therefore, the output voltage varies as thederivative of the input voltage

Practically speaking, this circuit exhibitserratic or even unstable response and canbe used only in combination with othercircuitry to depress this instability.


Linearization

• The op amp can also implement linearization.

• Generally, this is achieved by placing a non-linear element in the feedback loop of the op amp (shown in fig.)

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F(Vout)

+

VoutRVin




Linearization

• The summarization of currents provides

Vin/R+I(Vout)=0

where Vin=input voltage

R= input resistance

I(Vout)= nonlinear variation of current with voltage

Eq. is solved (in principle) for Vout, we get

Vout=G(Vin/R)

where Vout=output voltage

G(Vin/R)= a nonlinear function of the input voltage[actually the inverse function of I(Vout)]

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Summary

• To present a complete picture of analog signalconditioning, the following points wereconsidered:

1. The need for analog signal conditioning wasreviewed and resolved into the requirementsof signal-level changes, linearization, signalconversions, and filtering and impedancematching.

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2.Bridge circuits are a common example of aconversion process where a changingresistance is measured either by a current orby a voltage signal. Many modifications of thebridge are used, including electronic balancingand techniques of lead compensation.

3. The high and low pass RC filters are passivecircuits used to block undesired free quenciesfrom data signals.

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• Operational amplifiers (op amp) are a specialsignal-conditioning building block aroundwhich many special – function circuits can bedeveloped. The device was demonstrated inapplications involving amplifiers, converters,linearization circuits, integrators, and severalother functions.

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Digital signal conditioningIntroductionDigital information PLCs Computer interfacing ComparatorsDigital to analog converters Data acquisition systems

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