chap1 indus electronics

EE3532 Industrial ElectronicsCourse Notes

Mohamed A. Tahac© PSUT, October 5, 2008

1 Process Control

Process control is the automatic control of an output variable by sensing theamplitude of the output parameter from the process and comparing it tothe desired or set level and feeding an error signal back to control an inputvariable.

1.1 Elements of Process Control

A process control can be described as shown in the block diagram in Figure1. The block diagram shown in Figure 1 consists of the following elements:

Figure 1: Block diagram of a process control

1

1.1 Elements of Process Control 2

1. Controlled or Measured variable: Monitored output of the process(variable to be controlled) which is usually held within given limits.Examples in a process control system may include temperature, pres-sure and flow rate.

2. Manipulated variable: It is the amount of energy that is physicallyaltered by the actuator to adjust the controlled variable. Example inthe level control system is the flow rate where the controlled variableis the level.

3. Set point: It is the desired value of the output (controlled) variable.

4. Sensor: Is a device that can detect physical variables such as temper-ature, lights, and motion, etc.

5. Transducer: Converts one form of energy to another. Example is thepressure to current transducers, temperature to voltage transducers.

6. Controller: Is a device that monitors signals from sensors and take thenecessary action to keep the process within specified limits.

7. Actuator: Is a device that is used to control an input variable in re-sponse to a signal from the controller. Example in the level control isthe valve that controls the flow by the amount it closes and opens.

Example 1.1 Consider the system shown in Figure 2. Identify the processcontrol elements.

In this example, pressure is used to control the rate of fluid flows. This means,the controlled variable is the fluid flow rate and the manipulated variable isthe pressure developed by the fluid.

Sensor is the pressure cell. Transducer is the P/I and I/P convertors. Ac-tuator is the valve that changes the pressure applied to the pipes. Controlleris the processor, memory and summing junction. This controller is a PLC(Programmable Logic Controller) type (more in CH 8).

Important Notes:

• Programmable logic controllers (PLC) are used in process-control ap-plications, and are microprocessor-based systems. Small systems have

c©PSUT, First Semester 2008-2009c©PSUT, First Semester 2008-2009c©PSUT, First Semester 2008-2009

1.1 Elements of Process Control 3

Figure 2: Process control for example 1.1

the ability to monitor several variables and control several actuators,with the capability of being expanded to monitor 60 or 70 variablesand control a corresponding number of actuators, as may be requiredin a petrochemical refinery. PLCs, which have the ability to use ana-log or digital input information and output analog or digital controlsignals, can communicate globally with other controllers, are easilyprogrammed on line or off line, and supply an unprecedented amountof data and information to the operator. Ladder networks are normallyused to program the controllers.

• An error signal is the difference between the set point and the amplitudeof the measured variable.

Example 1.2 The process control shown in Figure 3 is a temperature controlsystem. Identify the elements of the process control for that system. Inthis example, the temperature of the flow is controlled using an input steamadjusted using a valve. The controlled variable is the temperature, and themanipulated variable is the steam. Sensor is attached to the output flow pipes.The controller is shown in the figure.


2 Signal Conditioning and Interfacing circuits 4

Figure 3: Process control for example 1.2

2 Signal Conditioning and Interfacing circuits

When a sensor is used to respond to a change in physical parameter, its out-put is usually very small, which makes it sensitive to noise and interference.Therefore, we need a mechanism to amplify the signal level and attenuatesnoise. This can be performed using signal conditioning which is a combi-nation of amplification and filtering. Signal conditioning are needed for thefollowing reasons:

2.1 Signal Level Changes

This includes amplification and attenuation. The process can be performedusing amplifiers. Amplifiers must satisfy

• Amplifier must be chosen carefully to match the frequency response ofthe measured variable.

• Amplifiers must have input impedance that does not load the outputof the sensor.

Usually signal level changes can be performed easily using Operational Am-plifier (OP-AMP)


2.2 Linearization 5

2.2 Linearization

When the relation between the physical parameter (input of the sensor) andthe output variable (measured variable) is non linear, lineralization is needed.Linearization can be performed using non linear circuits to undo the effectof nonlinearity coming from the sensor I/O relation, or it can be performedusing software.

2.3 Conversions

Conversions are necessary to convert the output of the sensor to anotherform that can be processed easily. For example, if the sensor output is aresistance change, it is more useful to convert such change into voltage orcurrent change. This can be done using bridge circuits and sometimes usingOP-AMP circuits.

2.4 Filtering and Impedance Matching

Filtering is necessary in order to amplify the output of the sensor withoutamplifying the noise, in this case filtering is applied to attenuate the theunwanted signals prior to amplification.

Impedance matching is necessary in order to preserve the signal level forthe next stages. Incorrect impedance matching may reduce the signal whenapplied to the next stages.

2.5 Loading Effect

Consider a sensor with internal resistance Rx and output voltage vx as shownin Figure 4. When the sensor is not loaded (RL = ∞), vout = vx. When thesensor equivalent is connected to the load RL the output vout will be reduced,

vout = vx

(1− Rx

RL + Rx

)= vx

RL

RL + Rx

The reduction in vout is by the amount of the voltage drop across Rx that isvx

Rx

Rx+RL. It is obvious then that we need a mechanism to isolate the sensor

from the load, this can be done using a buffer amplifier (will be shown later).

Example 2.1 An amplifier outputs a voltage that is ten times the voltageon its input terminals. It has an input resistance 10 KΩ. A sensor outputs a


2.5 Loading Effect 6

R x

V x R L v

out

Figure 4: Thevenin’s equivalent of a sensor x

voltage proportional to temperature with a transfer function 20 mvo/C. Thesensor has an output resistance of 5 KΩ. If the temperature is 50oC, find theamplifier output.

The circuit for this example is shown in Figure 5. The voltage from the

T 20mv/ o C

v out

R x

R L

Figure 5: Thevenin’s equivalent of a sensor and the amplifier

sensor at 50oC is

vx(50oC) = 20mv/oC × 50oC = 1 V

The voltage at the input of the amplifier is

vi(50oC) = 1× 10

5 + 10= 0.67 V

The amplifier amplifies the input signal by a factor of 10

vout = 0.67× 10 = 6.7 V

It should be noted here that if the parameter of interest is the frequency,then loading is not very important since although the signal is attenuated


3 Passive Interfacing Circuits 7

because of loading, we can still figure out the frequency of the output signal,this also applies if the signal is a digital signal since we can decide the zerosand ones of the signal even if the level is small.

3 Passive Interfacing Circuits

The most important passive analog passive signal conditioning are the Bridgeand the Divider circuits. The bridge circuits is used when the change in thephysical variable results in a small change in the impedance, in this case,this small change in the impedance is converted into voltage difference bythe bridge.

Another application for passive circuits is the Filtering in which a certainfrequency or band of frequencies are passed and other are rejected. This canbe performed by combination of R and C elements.

3.1 Divider Circuits

When the change in the physical variable results in a considerable change inresistance, divider circuit is used. In this case the change in the impedance isconverted into change in voltage. A typical divider circuit is shown in Figure6. In Figure 6, either R1 or R2 can be a sensor whose remittance (R1 or R2)is changeling with some physical parameter. The divider voltage VD is givenby

VD =R2Vs

R1 + R2

(1)

Important Notes:

• The variation of VD is nonlinear with R1 or R2 even in the case of linearrelation between R1 or R2 and the physical variable.

• The equivalent impedance of the divider circuit is the parallel combi-nation of R1 and R2, i.e., RTH = R1||R2. The value of RTH is notnecessarily high, so loading effect must be considered.

• Same current will flow in R1 and R2 so power dissipation in both re-sistors must be considered.

Example 3.1 In the circuit shown in Figure 6, let R1 = 10.0 KΩ and Vs =5.00 V. Suppose R2 is a sensor whose resistance varies from 4.00 to 12.0 KΩ


3.1 Divider Circuits 8

R 1

R 2

V D

V s

Figure 6: A divider circuit

as some dynamic variable varies over a range. Find (a) the minimum andmaximum of VD. (b) the range of output impedance. (c) the range of powerdissipated in R2.

(a) VD can be found using equation (1). The minimum VD results whenR2 = 4.00 KΩ

VDmin =4× 5

4 + 10= 1.43 V

The maximum VD results when R2 = 12.0 KΩ

VDmax =12× 5

12 + 10= 2.73 V

(b) RTH = R1||R2

When R2 = 4.00 KΩ, RTH = 10||4 = 2.86 KΩ. When R2 = 12.0 KΩ,RTH = 10||12 = 5.45 KΩ.

(c) The power dissipated in R2

R2 = 4.00 KΩ


3.2 Bridge Circuits 9

P =V 2

D

R2

=1.432

4= 0.511 m W

R2 = 12.00 KΩ

P =V 2

D

R2

=2.732

12= 0.621 m W

3.2 Bridge Circuits

Bridge circuit has the advantage of converting small variation of impedanceinto varying voltage around zero. An amplifier can then be used to increasethe sensitivity (produce higher voltages). A typical bridge circuit is calledWheatstone bridge as shown in Figure 7.

In the circuit in Figure 7, a voltage detector (D) is used to detect thevoltage difference between the points a and b. This detector is usually adifferential amplifier with high input impedance. For calibration purposes,D is a low impedance Galvanometer. The detector voltage reading is ∆V =Va − Vb, where Va is the voltage at point a with respect to ground (point c)and Vb is the voltage at point b with respect to ground (point c).

The voltage Va is the voltage division of the right arm

Va =V R3

R1 + R3

similarly

Vb =V R4

R2 + R4



V

R 1 R 2

R 3 R 4

D a b

c

Figure 7: Wheatstone bridge circuit

The voltage difference ∆V

∆V = Va − Vb

=V R3

R1 + R3

− V R4

R2 + R4

= VR3R2 −R1R4

(R1 + R3)(R2 + R4)(2)

The null condition results from ∆V = 0, in this case

R3R2 = R1R4 (3)

Example 3.2 In the bridge circuit shown in Figure 7, the bridge is nulledwith R1 = 1000 Ω, R2 = 842 Ω, and R3 = 500 Ω, find the value of R4.

Since the bridge is nulled we can use equation (3)

R4 =R3R2

R1

=500× 842

1000= 421 Ω

Example 3.3 The resistors in a bridge shown in Figure 7 are given by R1 =R2 = R3 = 120 Ω and R4 = 121 Ω. If the supply voltage V = 10.0 V, findthe voltage offset.



The offset voltage ∆V can be found using equation (2)

∆V = VR3R2 −R1R4

(R1 + R3)(R2 + R4)

= VR3R2 −R1R4

(R1 + R3)(R2 + R4)

= 10120× 120− 120× 121

(120 + 120)(120 + 121)

= −20.8 m V

If the Galvanometer is used to detect the voltage difference, Galvanometerresistance RG must be included in the analysis. If the Galvanometer resis-tance is finite, current will flow through this resistance. To find the value ofthis current, we first assume open bridge (RG = ∞) and find the Thevenin’sequivalent between the points a and b. The equivalent Thevenin’s circuitbetween a and b is shown in Figure 8. In this Figure RTH is the equivalent

V TH

R TH

R G

Figure 8: Wheatstone bridge Thevenin’s equivalent circuit

resistance between a and b, it can be found by short circuit the voltage V ,in this case

RTH = R1||R3 + R2||R4

=R1R3

R1 + R3

+R2R4

R2 + R4

(4)

The open circuit voltage VTH is found using equation (2). The Gal-vanometer current can be found easily as

IG =VTH

RTH + RG

(5)



Example 3.4 A bridge circuit has resistance of R1 = R2 = R3 = 2.00 KΩ,and R4 = 2.05 KΩ, and a 5.00 V supply. If a galvanometer with a 50.0 Ωinternal resistance is used for a detector, find the offset current.

We first find VTH using equation (2)

VTH = VR3R2 −R1R4

(R1 + R3)(R2 + R4)

= 52.00× 2.00− 2.00× 2.05

(2.00 + 2.00)(2.00 + 2.05)

= −30.9 mV

To find RTH we use equation (4)

RTH = R1||R3 + R2||R4

=R1R3

R1 + R3

+R2R4

R2 + R4

=2.00× 2.00

2.00 + 2.00+

2.00× 2.05

2.00 + 2.05= 2.01 KΩ

The offset current can be found using equation (5)

IG =VTH

RTH + RG

=−30.9× 10−3

2.01× 103 + 50= −15 µA

Bridge Resolution: Offset voltage resolution is the minimum voltagethat can be detected by the detector. Resolution of the bridge is the minimumvalue of the resistance change that produces the minimum detected offsetvoltage. For example, if the detector can measure an offset voltage of 100 µV ,then the minimum resistance change that can be measured is the change thatproduces offset voltage of 100 µV .

Example 3.5 A bridge circuit has R1 = R2 = R3 = R4 = 120.0 Ω, anda 10.0 V supply voltage. If the detector has a resolution of 10 mV, find theresolution of R4.


3.3 Filtering 13

We need to find the value of R4 that produces an offset voltage of 10 µV .Equation (2) can be used

10 mV = 10120× 120− 120×R4

(120 + 120)(120 + R4)

Solving for R4, givesR4 = 119.52 Ω

Thus, the change in R4 that produces the given offset is

∆R4 = 0.48 Ω

It should be noted in this example that, if the change in R4 is less than 0.48 Ωthe detector will not detect any offset, since the resulting offset in this caseis less than 10 mV, the resolution of the detector.

Finally, we should note that bridge circuit can be used when the sensoroutput for any physical input is a resistive type and the vhange in the re-sistance is small, this ensures nearly linear relation between ∆V and ∆R asshown in Figure 9.

3.3 Filtering

When the sensor captures data from measurements, they are usually equippedwith noise. To reduce the effect of noise, filtering must be performed to re-ject the frequency bands which the noise signals occupy. Filtering is easilyimplemented using RC circuits, these circuits are called filters. According tothe configuration of the R’s and C’s, filters have several types.

3.4 Low Pass Filter

Low pass filter, is the circuit that passes signals with frequencies lower thansome frequency called cutoff frequency fc, and attenuates signals with fre-quencies beyond fc. Low pass filtering can be performed using the circuitshown in Figure 10.

In the circuit in Figure 10, the output voltage Vout can be found as

Vout =VinXC

R + XC

(6)


3.4 Low Pass Filter 14

0 50 100 150 200 250 300 350 400 450 500−4

−2

0

2

4

6

∆ R4

∆ V

∆ R4=100Ω

0 10 20 30 40 50 60 70 80 90 1001.5

2

2.5

3

3.5

4

4.5

5

∆ R4

∆ V

∆ R4=10Ω

Figure 9: Relation of ∆V as a function of ∆R4 for large and small changesin R4

R

C V out V in

Figure 10: Low pass filter RC circuit



XC = 1sC

= 12πfC

, substitute in equation (6) and find the transfer function

denoted by H(f) = Vout

Vin

H(f) =1

RC

j2πf + 1RC

(7)

Define the critical frequency fc as the frequency at which the transfer functionmagnitude reaches 1√

2of it value at DC (H(0)) or equivalently (20 log10(

1√2) =

−3 dB),

|H(fc)| = 1√2|H(0)|

The magnitude of the transfer function |H(f)| is found as follows

|H(f)| =1

RC

(2πf)2 + ( 1RC

)2(8)

To find the critical frequency fc, we must solve for fc the following equation

|H(fc)| = 1√2|H(0)|

(2πfc)2 + (

1

RC)2 = 2(

1

RC)2

which gives

fc =1

2πRC(9)

We can express equation (8) in terms of fc by substituting 1RC

= 2πfc inequation (8)

|H(f)| = 1√1 + ( f

fc)2

(10)

The magnitude of the transfer function (H(f)) can be plotted as a functionof the ratio f

fcas shown in Figure 11.

Example 3.6 A measured signal has a frequency has a maximum frequencycomponent of 1 KHz. The signal is captured with noise located at frequencyof 1 MHz. Design a low pass filter to attenuate the noise to 0.01 of its valueat the input. What is the effect of the designed filter on the signal.



10−2

10−1

100

101

0

0.2

0.4

0.6

0.8

1

f/fc

|H(f

)|

Figure 11: Low pass frequency response


3.5 High Pass Filter 17

We need to determine the cutoff frequency of the filter using the givenspecifications. The noise signal of 1 MHz needs to be attenuated to 0.01 ofits value at the input, we use equation (10) as follows

|H(1 MHz)| = 1√1 + (1 MHz

fc)2

Solving for fc

fc = 10 KHz

Select C = 0.01µF , and using the equation (9) to find the resistor, R =1.59 KΩ.

To find the effect of the filter on the signal we use equation (10)

|H(1 KHz)| =1√

1 + ( 1 KHz

10 KHz)2

= 0.995

The signal is attenuated by 0.5% using this filter.

3.5 High Pass Filter

High pass filter passes high frequencies and attenuates low frequencies, highand low frequencies are realtime to the critical frequency fc. It can be per-formed using the RC circuit by interchanging the elements R and C in thelow pass filter circuit as shown in Figure 12.

V in V out

C

R

Figure 12: High pass filter RC circuit



The transfer function ratio H = Vout

Vinis as folows

H =R

R + XC

=s

s + 1RC

=jf

jf + fc

(11)

The 3 dB frequency is found on the same manner as in low pass filter andequals fc = 1

2πRC. The magnitude frequency response is found by taking the

magnitude of H in equation (11)

|H| =

∣∣∣∣jf

jf + fc

∣∣∣∣

=

∣∣∣∣jf/fc

1 + jf/fc

∣∣∣∣

=f/fc√

1 + (f/fc)2(12)

The frequency response of the RC high pass filter is shown in Figure 13.

Example 3.7 Pulses for stepping motor are being transmitted at 2000 Hz.Design a filter to reduce 60 -Hz noise, but reduce the pulses by no more than3 dB.

Since signals with frequencies below the cutoff frequency will be attenuatedby more than 3 dB, then fc = 2000 Hz. The pulses are attenuated by 0.707times their values at the input, while the noise by

60/2000√1 + (60/2000)2

= 0.03

This means, if the pulses has amplitude of 1 V at the input of the filter,then this amplitude will be 0.707 V at the output. While, if the noise hasamplitude of 1 V at the input of the filter, then this amplitude will be 0.03 Vat the output of the filter, a reduction by 97%.

The design of the filter will be, selecting C = 0.01µF , and find R as

R =1

2πCfc

=1

2π0.01µF2000 Hz= 8 KΩ



10−2

10−1

100

101

0

0.2

0.4

0.6

0.8

1

f/fc

|H(f

/f c)|

Figure 13: RC High pass filter frequency response



Multistage Filters:Often, filters are cascaded in order to have better response (more attenua-

tion in the stop band region). In this case, loading effect has to be considered.If the input impedance of the second stage is much larger than the outputimpedance of the first stage, the output signal of the first stage will not bereduced by the second stage, and the response will be squared if the bothstages have the same cutoff frequency. Figure 14 shows the scenario whenconnecting two stages, in this configuration Rout1 is the Thevenin’s equiva-lent impedance of the first stage, Vout1 is the Thevenin’s open circuit voltageof the first stage, Vin2 is the input voltage to the second stage and Rin2 is thesum of the input impedance and the load of the second stage .

Vin1 =Vout1

Rout1 + Rin1

If Rin1 >> Rout1

Vin1 = Vout1, No reduction

If Rin ≤ Rout1

Vin1 ≤ 1

2Vout1, A reduction of 0.5 or more

First stage Second stage

V out1

R out1

R in2

V in2

Figure 14: Loading effect in multistage filtering

Example 3.8 A 2 KHz data signal is contaminated by a 60 Hz noise. Com-pare the performance of a signal and two stages high pass filter of the signal(data) for 60dB attenuation of the noise signal.

Since the required attenuation is 60dB (10−60/20 = 10−3) which is morethan 3dB (10−3/20 = 0.707), then the cutoff frequency is not 2000Hz. Todetermine fc we use equation (12)

0.001 =60/fc√

1 + (60/fc)2



Solving for fc, we havefc = 60 KHz

Using a single stage, the amplitude of the signal (2KHz) at the output isfound using equation (12)

Vout

Vin

=2, 000/60, 000√

1 + (2, 000/60, 000)2= 0.033

The signal is almost killed with 0.0333 left at the output.For multistage design, assume no loading effect, the magnitude response

is squared

0.001 =(60/fc)

2

1 + (60/fc)2

Solving for fc

fc = 1896 Hz

To find the effect on the signal

Vout

Vin

=(2000/1896)2

1 + (2000/1896)2= 0.53

Which is much better attenuation than the a single stage (0.53 of the signal isleft). Since both stages are identical then each stage will provide attenuationof (

√0.53 = 0.71). If we consider the loading effect, the attenuation will be

increased.

Example 3.9 In the previous example, assume C = 0.001 µF , find R andobtain the performance of the filter considering the loading effect of the mul-tistage filter.

In the previous example, fc = 1896 Hz,

R =1

2πCfc

=1

2π0.001× 10−6 × 1896= 83.9 KΩ

To evaluate the performance of the data using this filter, we compute XC at2000Hz

XC(2000) =1

2π × 2000× 10−6= 79.6 KΩ

The circuit is shown in Figure 15. In this circuit, the second stage is loadingthe first.



C C

R R V in V out

Figure 15: RC High pass filter multistage circuit

To study the loading effect, we find Thevenin’s equivalent of the first stage,in this case the second stage is eliminated since it represents the load for thefirst stage.

RTH = R||XC = 83.9||79.6 = 40.85 KΩ

Voc = Vout1 = 0.73Vin

The equivalent circuit for the firsts stage is shown in Figure 16.

R out1 =R TH =40.85K

V oc =V out1 = 0.73V in R in2

Figure 16: Thevenin’s equivalent of the RC High pass filter multistage circuit

Where Rin2 is the input impedance from the second stage computed as

Rin2 = R + XC = 83.9 + 79.6 = 163.5 KΩ

Now the input voltage to the second stage can be found by voltage division ofRTH and Rin2.

Vin2 =0.73Vin

RTH + Rin2

×Rin2 =0.73Vin

40.85 + 163.5× 163.5 = 0.584Vin

The overall output can be found by noting the relation between Vin2 and Vout,

Vout = 0.73Vin2

Thus,Vout = 0.73× 0.584Vin = 0.4263


4 Operational Amplifier 23

4 Operational Amplifier

4.1 Ideal Characteristics

Operational amplifier (OP-AMP) is represented by the symbol shown in Fig-ure 17. In this circuit, the + terminal is called the non-inverting input termi-nal and the − terminal is called the inverting input terminal. We also denotethe open loop gain of the OP-AMP by A, it is called open loop since thereare no external components connected to the OP-AMP, the typical valuesfor A is in the order of 105. The output vo is given by

-

+

v o

v 2

v 1

Figure 17: Ideal Operational Amplifier (OP-AMP)

vo = A(v1 − v2) (13)

The OP-AMP output is limited by the supply voltage vcc used for biasing,that is, when the output voltage supposed to exceed the supply voltage it islimited (clipped or saturated) to the supply voltage, or precisely by vcc ± 3,this due to the non-ideal characteristics of the OP-AMP (discussed in thenext section).

Using the relation given by equation (13), we have two cases

1. v1 > v2

vo = A ∆v, Possibly larger than the vcc

This large value will be clipped to the saturation voltage, vsat = vcc−3.

2. v1 < v2

vo = −A ∆v, Possibly smaller than the vcc


4.2 Ideal Inverting Amplifier 24

This small value will be clipped to the negative saturation voltage,−vsat = −(vcc − 3).

4.2 Ideal Inverting Amplifier

Usually, open loop amplifier is unstable because of large open loop amplifica-tion, A, this also causes a DC drift of the operating point with temperature.To solve the problem, feedback resistors are used, Figure 18 shows an op-ampwith input restack R1 and feedback resistance R2, this configuration is calledinverting op-amp circuit.

-

+

v o

V in

R 2

R 1

I 2

v 2

I 1

v 1

Figure 18: Ideal Inverting Operational Amplifier (OP-AMP)

Important Notes:

1. The op-amp is assumed to be ideal, I1 = I2 = 0, current flows in bothterminals are zero.

2. The op-amp is assumed to be ideal, v2 = v1, voltage difference betweenthe inverting and non inverting terminals is zero.

Since v1 = v2 and v1 = 0 then v2 = 0. Writing the node equation atterminal 2,

0− vin

R1

+0− vo

R2

+ I2 = 0


4.3 Non-Inverting Configuration 25

Since I2 = 0, the Gain of the amplifier is the ratio vo

vinis obtained as

vo

vin

= −R2

R1

(14)

From this equation we note the 180o phase difference between vin and vo,that’s why this configuration is called inverting configuration. The inputimpedance of this circuit is

Rin = R1

For a certain application, R1 is set to satisfy the impedance matching issue,and R2 is set to satisfy the required gain, accordingly.

Example 4.1 In the circuit shown in Figure 18, R1 = 1.2 KΩ and R2 =150 KΩ. What is the gain of the amplifier. What is the output voltage if theinput signal is 3.5 mV.

Gain = −R2

R1

= −150

1.2= −125

vo = −3.5mv× 125 = −437.5 mV

4.3 Non-Inverting Configuration

The non-inverting op-amp circuit is shown in Figure 19. In this circuit, sincev1 = v2 and v1 = vin, then v2 = vin, nodal equation can be written at node2, assume negligible input currents (I1 = I2 = 0)

vin − 0

R1

+vin − vo

R2

= 0

This gives amplifier gainvo

vin

= 1 +R2

R1

(15)

The input impedance isRin = rin

where rin is the output impedance of the op-amp itself, which is very large.


4.4 Voltage Follower 26

-

+

v o

V in

R 2

R 1

Figure 19: Ideal Non-Inverting Operational Amplifier (OP-AMP)

4.4 Voltage Follower

An impedance matching op-amp is called a buffer amplifier. Such ampli-fiers have feedback to give unity voltage gain, high input impedance (manymegaohms), and low output impedance (< 20Ω), such an amplifier is shownin Figure 20.

Figure 20: Buffer Operational Amplifier (OP-AMP)

Example 4.2 Consider the divider circuit shown in Figure 21. Obtain theunloaded and loaded divider voltage. Use the buffer circuit to reduce the


4.4 Voltage Follower 27

Figure 21: Buffer Operational Amplifier for example 4.2

loading effect, assume buffer input impedance 2 MΩ and output impedance15 Ω.

For unloaded circuit RL = ∞

vdiv. =12

8 + 4× 8 = 8 V

For loaded circuit (RL = 2 KΩ)

vdiv. =12

8||2 + 4× 8||2 = 3.43 V

A reduction of 57.5% of the 8 VThe buffer can be used to match the output impedance of the divider circuit

to the load as shown in Figure 22. The load is 2 MΩ, so the divider voltageat the input of the buffer is

vdiv. =12

8||2000 + 4× 8||2000 = 7.99 V

The output impedance of the buffer is in series with the 2 KΩ load,

vo =7.99

2000 + 15× 2000 = 7.93 V

A reduction of 0.07 V or 0.9% of the 8 V compared to 57.5% in direct loadingwithout buffer.


5 Summing Amplifier 28

Figure 22: Impedance matching Buffer for example 4.2

5 Summing Amplifier

It is an inverting OP-AMP, with several input voltages, the output is thesum of all these voltages with 1800 phase shift between the input and output.Figure 23 shows a summing amplifier with two inputs v1 and v2, the outputvoltage vo is

vo = −[RF

R1

v1 +RF

R2

v2

](16)

+

-

v o

R F

R 1

R 2

v 1

v 2

Figure 23: Summing amplifier with two inputs


6 Differential and Instrumentation Amplifier 29

Example 5.1 Design an OP-AMP circuit to provide the relation

vo = 3.4vin + 5

The circuit can be implemented using the summing amplifier with properselection for the resistors. Using equation (16) and compare the terms in vx

and vo given by the relation.If we neglect the minus sign for now, and we can compensate it later using

an inverting amplifier with gain −1. We have

RF

R1

v1 = 3.4vin andRF

R2

v2 = 5

The easiest way to satisfy both relations to select

RF

R1

= 3.4, v1 = vin andRF

R2

= 1, v2 = 5

select R1 = 10 KΩ gives RF = 3.4× 10 KΩ = 34 KΩ and R2 = RF = 34 KΩThe designed circuit can be compensated the minus sign using inverting am-plifier with gain 1, the overall design is shown in Figure 24.

Figure 24: Circuit for example 5.1

6 Differential and Instrumentation Amplifier

6.1 Differential Amplifier

In several cases, voltage difference needs to be amplified, e.g. Wheatstonebridge. In this case, Differential or Instrumentation Amplifier is used. A


6.1 Differential Amplifier 30

differential amplifier is a dual input amplifier that amplifies the differencebetween two signals, such that the output is the gain multiplied by the mag-nitude of the difference between the two signals. One signal is fed to thenegative input of the op-amp and the other signal is fed to the positive in-put of the op-amp. Hence the signals are subtracted before being amplified.Figure 25 shows a typical differential amplifier.

Figure 25: Differential Amplifier Circuit

The output of this circuit is given by

vo =R2

R1

(v2 − v1)

The above discussion assumes perfect resistance matching and ideal op-amp. In general, the output of the differential amplifier is written as

vo = Ad(v1 − v2) + Acm

(v1 + v2

2

)

= Advd + Acmvcm

Ad is the differential gain, vd is the differential (difference) voltage appliedto the two terminals, Acm is the common mode voltage (amplifier gain when


6.2 Instrumentation Amplifier 31

same voltage is applied to the two terminals, e.g. Noise) and vcm is the aver-age of the common mode voltage. When the amplifier circuit discussed aboveis ideal, Acm is equal zero, and the output is determined by the differentialcomponents, which attenuates all noise and disturbance signal that affectthe circuit. When the amplifier is non ideal Acm will not be zero and thecommon mode amplified component will appear at the output.

Circuit performance can be assessed by the ratio of the Ad and Acm anddefined by the common mode rejection ratio (CMRR) (the ability of theamplifier to attenuate signals applied commonly to both input terminals)

CMRR =Ad

Acm

If the circuit is ideal this quantity will be infinity, if it is not it would befinite. The CMRR can be expressed in dB and is called CMR

CMR = 20 log10 CMRR

Typical value for a good differential amplifier will be in the range of 60 −100 dB.

Important Notes:

• The op-amp can have different input impedances at the two inputs.

• The input impedances can be relatively low and tend to load the sensoroutput.

• Can have different gains at the inverting and non-inverting inputs, andcommon mode noise can be a problem.

For these reasons instrumentation amplifier can be used, which is a combi-nation of the differential amplifier and buffer circuits.

6.2 Instrumentation Amplifier

Instrumentation amplifier gives the required CMRR by amplifying the differ-ence voltage applied to its two input terminals and gives perfect impedancematching for the two input terminals. High impedance matching can beprovided using buffer amplifiers which reduce loading on the sensor output.Typical instrumentation amplifier circuit is shown in Figure 26.



Figure 26: Instrumentation Differential Amplifier Circuit

Example 6.1 A sensor outputs a range of 20.0 to 250 mV as a variablevaries over its range. Develop signal conditioning so that this becomes 0 to5 V. The circuit must have very high input impedance.

From the specs. given it’s clear that we can use the circuit shown in Figure26. To obtain a relation between the input voltage (20.0 to 250 mV) and theoutput voltage of the amplifier circuit ( 0 to 5 V), we assume linear relation

vo = avin + b

a and b are variables that can be determined as

0 = a(0.02) + b

5 = a(0.25) + b

solving the linear system, we have

a = 21.7 and b = −0.434

That is

vo = 21.7vin − 0.434

= 21.7 (vin − 0.02)



If this relation is compared to the instrumentation amplifier output relation

R2

R1

= 21.7

select R1 = 10 KΩ then R2 = 217 KΩv2 = vin and v1 = 0.02 V. 0.02 V is a small value and needs to be

stabilized against power supply voltage ripple, this can be achieved using asimple voltage regulator, typically a zener diode as shown in Figure 27.

680 Ω

6.8 V30.9 KΩ

100 Ω

10 KΩ

10 KΩ

217 KΩ

217 KΩ

Figure 27: Instrumentation Differential Amplifier circuit for example 27

To reduce the dependance of the circuit performance on the components,the circuit in Figure 28 is usually used. In this circuit, the gain is set by thevariable resistor RG, the CMRR does not depend on matching of R′

1s, but itdepends on the R2 and R3 only. For any zero signal offset R3 can be variedto eliminate that offset. The output voltage vo of the circuit is

vo =

(1 +

2R1

RG

)(R3

R2

)(v2 − v1) (17)



v2

v1

RG

R1

R1

R2

R2

R3

R3

Figure 28: Practical Instrumentation Differential Amplifier circuit


7 Design Guidelines 35

Example 6.2 Figure 7 shows a bridge circuit for which R4 varies from100 Ω to 102 Ω, the resitors in the bridge R1 = R2 = R3 = 100 Ω. Show howthe instrumentation amplifier shown in Figure 28 could be used to provide anoutput of 0 to 2.5 V. Assume in Figure 28, R2 = R3 = 1 Ω, R1 = 100 KΩand V = 5 V.

From the question and using equation (3) the bridge is nulled when R4 =R2R3

R1= 100×100

100= 100 Ω

Voltage difference is obtained using equation (2)

∆V = 5

[100

100 + 100− 102

102 + 100

]= −24.75 mV

When R4 = 100 Ω, ∆V = 0 and vo = 0 as required in the question.While, when R4 = 102 Ω, vo = 2.5 V Substitute in equation (17)

2.5 =

(1 +

2× 100× 103

RG

) (1000

1000

)(24.75× 10−3)

Solving for RG

RG = 2000 Ω

Since the voltage difference in the bridge is negative, Va in the bridge mustbe connected to V1 in the amplifier and Vb in the bridge must be connected toV2 in the amplifier

7 Design Guidelines

A process control is usually consists of, physical process, sensor, signal con-ditioning as shown in Figure ??. In the design, each stage has its ownspecifications and hence design.

1. Define the measurement objective

a. Parameter: nature of the measured variable, e.g. temperature,pressure, etc.

b. Range: the range of the input parameter cmin and cmax, e.g. 20−40oC, 10− 60 psi.

c. Accuracy: the accuracy of the measured variable in terms of thenominal value, e.g. 3% reading.



Process

c max

c min

Sensor Signal

Conditioning

y max

y min

b max

b min

Figure 29: Process control stages

d. Linearity: linearity of the measured variable with respect to thephysical variable.

e. Noise: noise level in the measured variable, noise level will decidethe signal conditioning used later.

2. Select a sensor

a. Parameter: what is the nature of the sensor output, e.g. resis-tance, voltage, etc.

b. Transfer function: the relation between measured variable (inputof the sensor) and output of the sensor.

c. Time response: what is the time response of the sensor, e.g. firstorder time constant, second order damping, and frequency.

d. Power: what is the power specs. of the sensor, power dissipationand maximum current.

3. Signal conditioning

a. Parameter: nature of the output, voltage is the most common,but conversion to current is also possible.

b. Range: desired range of the output parameter.

c. Input impedance and output impedance to ensure impedance match-ing.



Example 7.1 A sensor outputs a voltage from −2.4 V to −1.1 V. For inter-face to an ADC, this needs to be 0 V to 2.5 V. Develop a signal conditioning.

The scenario presented here is the last block in Figure 29, therefore, y andb are the given voltages and the signal conditioning must convert the voltagey to the voltage b. For simplicity, assume linear signal conditioning

b = ay + b0

where a and b0 are constants need to be determined as

bmax = aymax + b0

2.5 = −1.1a + b0 (18)

similarly,

bmin = aymin + b0

0 = −2.4a + b0 (19)

Solving the linear system in equations (18) and (19)

a = 1.9231 and b0 = 4.6154

This gives

b = 1.9231y + 4.6154

= 1.9231(y + 2.4)

From this relation, we can use a summing amplifier to add the signals y and2.4, then an inverting amplifier with gain 1.9231 can be used to provide therequired gain and the minus sign developed by the summing amplifier. Toprevent loading the sensor output, a voltage follower can be used which has ahigh input impedance. The input impedance of the ADC is much larger thanthe op-amp output impedance, therefore no loading on the ADC is guaranteed.The voltage signal 2.4 V can be supplied using a divider circuit with R2 =100 Ω, in this way, there will be no loading from the summing amplifierinput impedance. A variable resistor can be used to adjust for 2.4 V suchthat loading from the summing amplifier and the ripple in the power supplycan be adjusted. The complete design is shown in Figure 30.



10 KΩ

10 KΩ

525 Ω

100 Ω

10 KΩ

10 KΩ

19.23 KΩ

1.923(y+2.4)

−(y+2.4)

Figure 30: Signal conditioning for example 7.1



Example 7.2 Temperature is to be measured in the range of 250oC to 450oCwith an accuracy of ±2oC. The sensor is a resistance that varies linearly from280 Ω to 1060 Ω for this temperature range. Power dissipated in the sensormust be kept below 5 mW. Develop analog signal conditioning that provides avoltage varying linearly from −5 to +5 volts for this temperature range. Theload is a high impedance recorder.

Since the resistance and voltage change is linear, linear relationship be-tween resistance and voltage can be assumed. At the sensor side, the powerdissipation Pd produce by the current I that flows in the resistance (sensor).The value of the current will be higher for smaller resistance, to prevent thesensor from overheating, we design according to the highest resistance.

Pd = I2maxRmax

5 mW = I2max1060 Ω

Solving for Imax

Imax = 2.20 mA

The current Imax is maximum from the design point of view, but it’s minimumfor the resistance range. The accuracy of the design is ±2oC over the wholetemperature range, the overall accuracy

∆T =±2

450− 250= 0.01

In this case the measured temperature in this range will have 3 significantfigures, thus, the obtained design values must be approximated using 3 signif-icant figures.

We develop the linear relation between the output of the sensor (resistance,Rx) and the output of the signal conditioning (voltage, vo)

vo = aRx + b

a and b will be determined using the given values for Rx and vo as follows

vomax = aRxmax + b

5 = 1060a + b (20)

similarly,

vomin = aRxmin + b

−5 = 280a + b (21)



Solving the linear system in equations (20) and (21)

a = 0.0128 and b = −8.58

This givesvo = 0.0128Rx − 8.58

The first term of vo can be implemented using a simple inverting OP-AMPcircuit with feedback resistance Rx and input 1 V. It should be noted here thatthe current flows in the senors resistance will also flow in the input resistanceR1 of the inverting OP-AMP. If Imax is selected to be 1 mA, then

R1 =1 V

1 mA= 1 KΩ

The output of the first stage is shown in Figure 31. In this circuit, we alsosupplied 1 V from the +15 V supply via divider circuit with R2 = 100 Ω toprevent loading from the OP-AMP input impedance (1 KΩ). Now the output

100 Ω

1400 Ω

1.0 KΩ

Rx

Figure 31: First stage of the signal conditioning for example 7.2

of the first stage isvFS = −10−3Rx

To add the other term in vo and change the sign of vFS we use a summingamplifier, the 8.58 V will be supplied by a divider circuit as shown in Figure32.


8 Thermal Sensor 41

Figure 32: Second stage of the signal conditioning for example 7.2

8 Thermal Sensor

8.1 Introduction

Similar to our every day needs of temperature control for comfort, almostall industrial processes need accurately controlled temperatures. Physicalparameters and chemical reactions are temperature dependent, and thereforetemperature control is of major importance. Temperature is without doubtthe most measured variable, and for accurate temperature control its precisemeasurement is required. This chapter discusses the various temperaturescales used, their relation to each other, methods of measuring temperature,and the relationship between temperature and heat.

8.2 Temperature definitions

Temperature is a measure of the thermal energy in a body, which is therelative hotness or coldness of a medium and is normally measured in degreesusing one of the following scales; Fahrenheit (F), Celsius or Centigrade (C),Rankine (R), or Kelvin (K). Absolute zero is the temperature at which allmolecular motion ceases or the energy of the molecule is zero.

Fahrenheit scale was the first temperature scale to gain acceptance. Itwas proposed in the early 1700s by Fahrenheit (Dutch). The two points of


8.2 Temperature definitions 42

reference chosen for 0 and 100 were the freezing point of a concentrated saltsolution (at sea level) and the internal temperature of oxen (which was foundto be very consistent between animals). This eventually led to the acceptanceof 32 and 212 (180 range) as the freezing and boiling point, respectively ofpure water at 1 atm (14.7 psi or 101.36 kPa) for the Fahrenheit scale. Thetemperature of the freezing point and boiling point of water changes withpressure.

Celsius or centigrade scale (C) was proposed in mid 1700s by Celsius(Sweden), who proposed the temperature readings of 0 and 100 (giving a100 scale) for the freezing and boiling points of pure water at 1 atm.

Rankine scale (R) was proposed in the mid 1800s by Rankine. It is a tem-perature scale referenced to absolute zero that was based on the Fahrenheitscale, i.e., a change of 1F = a change of 1R. The freezing and boiling pointof pure water are 491.6R and 671.6R , respectively at 1 atm, see Figure 33.Kelvin scale (K) named after Lord Kelvin was proposed in the late 1800s. Itis referenced to absolute zero but based on the Celsius scale, i.e., a changeof 1C = a change of 1 K. The freezing and boiling point of pure water are273.15 K and 373.15 K, respectively, at 1 atm, see Figure 33. The degreesymbol can be dropped when using the Kelvin scale.

Figure 33: Comparison of temperature scales

The relation between different temperature measures can be summarized


8.3 Metal Resistance Versus Temperature Devices 43

by the following relations

T (C) = T (K)− 273.15 (22)

T (F ) = T (R)− 459.6 (23)

T (F ) =9

5T (C) + 32 (24)

8.3 Metal Resistance Versus Temperature Devices

Usually temperature is measured by placing a temperature sensor in theenvironment where the temperature needs to be measured. The resistanceof the sensor changes with temperature. By measuring the resistance of thesenors, temperature can be determined. A typical thermal-metallic resistancesensor is called Resistance Temperature Detector (RTD).

Metallic materials have their valence and conduction electrons movingthrough the material. When heat energy is present, the motion of the elec-trons increases which causes collision of the atoms and molecules in the mate-rials. This collision tends to increase the resistivity and hence the resistanceof the material (it inhibits the motion of the electrons). Therefore, for metal-lic sensors, as the temperature increases the resistance of the sensor increases.The relation is almost linear with a certain slope for a certain metal.

The resistance of metal as a function of temperature can be expressed as

R(T )

R(25)=

ρ(T )`/A

ρ(25)`/A

=ρ(T )

ρ(25)(25)

where R(T ) is the resistance of the metal at temperature T . ρ(T ) is theresistivity of the metal at temperature T . ` is the length of the metal and Ais the cross-sectional area of the metal.

8.4 Resistance - Temperature (R-T) Approximation

When the temperature range of interest is small, linear relation can be as-sumed. While for larger range, an approximate relation should be deduced.


8.4 Resistance - Temperature (R-T) Approximation 44

8.4.1 Linear Approximation

In this case, the relation (R-T) is approximated linearly. Consider a temper-ature range T ∈ [T1 T2] and the midpoint T0 = T1 + T2−T1

2, the relation can

be expressed as

R(T ) = R(T0) [1 + α0∆T ] , T1 < T < T2

where α0 is the scaled slope of the line in the range [T1 T2] and is given by

α0 =1

R(T0)× slope at T0

=1

R(T0)

(R2 −R1

T2 − T1

)

R(T ) is the resistance at temperature T ∈ [T1 T2]. R(T0) is the resistance attemperature T0.

8.4.2 Quadratic Approximation

If a quadratic approximation about some temperature T0 in the range of[T1 T2], the relation R− T can be expressed as

R(T ) = R(T0)[1 + α1∆T + α2(∆T )2

]

the coefficients of approximation alpha1 and α2 can be determined by formingtwo linear equations using the given data values.

Example 8.1 A sample of metal resistance versus temperature has the fol-lowing measured values.

T (F ) R(Ω)60 106.065 107.670 109.175 110.280 111.185 111.790 112.2

The midpoint is T0 = 75, R(75) = 110.2.

α0 =1

110.2

(112.2− 106.0

90− 60

)= 0.001875



Thus the linear approximation gives

R(T ) = 110.2 [1 + 0.001875(T − 75)]

A simple Matlab script can be used to find the approximation

>> T=[60:5:90];

>> R=[106.0 107.6 109.1 110.2 111.1 111.7 112.2];

>> P=polyfit(T,R,1);

>>P= [0.2057 94.2714]

The values in the vector P are the coefficients of the approximation, that is

R(T ) = P (1)T + P (2)

in this caseR(T ) = 0.2057T + 94.2714

The actual relation and the approximated relation can be plotted on the sameplot as follows

>> RT=P(1)*T+P(2);

>> plot(T,R,T,RT,’r--’)

the output is shown in Figure 34.

Example 8.2 Find the quadratic approximation for the R − T relation forthe data in the previous example.

T0 = 75F , R(T0) = 110.2Ω. To determine the coefficients α1 and α2, weform the quadratic relation between T1 = 60F and T2 = 90F

112.2 = 110.2[1 + α1(90− 75) + α2(90− 75)2

]

106.0 = 110.2[1 + α1(60− 75) + α2(60− 75)2

]

which gives [15 225

−15 225

][α1

α2

]=

[0.0181

−0.0363

]

solving the above two equations

α1 = 1.873× 10−3 α1 = −4.444× 10−5

Thus the quadratic approximation is

R(T ) = 110.2[1 + 1.873× 10−3(T − 75)− 4.444× 10−5(T − 75)2

]

A Matlab script to solve the quadratic approximation



60 65 70 75 80 85 90106

107

108

109

110

111

112

113

T (oF)

R(Ω

)

ActualApproximate

Figure 34: R-T actual and approximated linear relation

>> R=[106.0 107.6 109.1 110.2 111.1 111.7 112.2];

>> T=[60:5:90];

>> P=polyfit(T,R,2)

P =[ -0.004952380952381 0.948571428571433 66.909523809523591]

The quadratic approximation is

R(T ) = −0.004952380952381T 2+0.948571428571433T+66.909523809523591

the plot of actual relation and quadratic approximation is shown in Figure35. It is obvious that the quadratic approximation is better than the linearapproximation.



60 65 70 75 80 85 90105

106

107

108

109

110

111

112

113

T (oF)

R(Ω

)

Actual Quadratic Approx.

Figure 35: R-T actual and approximated quadratic relation


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