ec1355 esd ece

8/10/2019 Ec1355 Esd Ece

1/31

Electronic System Design Lab Manu

VI SEMESTER ECE

LIST OF EXPERIMENTSLIST OF EXPERIMENTSLIST OF EXPERIMENTSLIST OF EXPERIMENTS

EC1355EC1355EC1355EC1355 ELECTRONIC SYSTEM DESIGN LABELECTRONIC SYSTEM DESIGN LABELECTRONIC SYSTEM DESIGN LABELECTRONIC SYSTEM DESIGN LAB 0 0 3 1000 0 3 1000 0 3 1000 0 3 100

1. DC power supply design using buck boost converters

Design the buck-boost converter for the given input voltage variation, load current anoutput voltage. Plot the regulation characteristics.

2. DC power supply design using fly back converter (Isolated type)

Design the fly back converter using ferrite core transformer for the given input voltagvariation load current and output voltage.Plot the regulation characteristics.

3. Design of a 4-20mA transmitter for a bridge type transducer.

Design the Instrumentation amplifier with the bridge type transducer (Thermistor or a

resistance variation transducers) and convert the amplified voltage from thinstrumentation amplifier to 4 20 mA current using op-amp. Plot the variation of thtemperature Vs output current.

4. Design of AC/DC voltage regulator using SCR

Design a phase controlled voltage regulator using full wave rectifier and SCR, vary thconduction angle and plot the output voltage.

5. Design of process control timer

Design a sequential timer to switch on & off at least 3 relays in a particular sequencusing timer IC.

6. Design of AM / FM modulator / demodulator

i. Design AM signal using multiplier IC for the given carrier frequency and modulatioindex and demodulate the AM signal using envelope detector.

ii. Design FM signal using VCO IC NE566 for the given carrier frequency and demodulathe same using PLL NE 565.

7. Design of Wireless date modem.

Design a FSK modulator using 555 and convert it to sine wave using filter and transmthe same using IR LED and demodulate the same PLL NE 565.

8. PCB layout design using CAD

Drawing the schematic of simple electronic circuit and design of PCB layout using CAD

9. Microcontroller based systems design

Design of microcontroller based system for simple applications like security systemcombination lock etc. using 89c series flash micro controller.

8/10/2019 Ec1355 Esd Ece

2/31


VI SEMESTER ECE

10. DSP based system design

Design a DSP based system for simple applications like echo generation, etc. using TM320 DSP kit.

8/10/2019 Ec1355 Esd Ece

3/31


VI SEMESTER ECE

1. DC POWER SUPPLY DESIGN USING BUCK1. DC POWER SUPPLY DESIGN USING BUCK1. DC POWER SUPPLY DESIGN USING BUCK1. DC POWER SUPPLY DESIGN USING BUCK BOOST CONVERTERSBOOST CONVERTERSBOOST CONVERTERSBOOST CONVERTERS

AimAimAimAim

Design the buck-boost converter for the given input voltage variation, load current and outp

voltage. Plot the regulation characteristics.Apparatus RequiredApparatus RequiredApparatus RequiredApparatus Required

DCDCDCDC----DC Converter BasicsDC Converter BasicsDC Converter BasicsDC Converter Basics

A DC-to-DC converter is a device that accepts a DC input voltage and produces a Doutput voltage. Typically the output produced is at a different voltage level than the input. addition, DC-to-DC converters are used to provide noise isolation, power bus regulation, etc. This a summary of some of the popular DC-to-DC converter topologies:

1.1.1.1. Buck Converter StepBuck Converter StepBuck Converter StepBuck Converter Step----Down ConverterDown ConverterDown ConverterDown Converter

In this circuit the transistor turning ON will put voltage Vin on one end of the inductor.This voltage will tend to cause the inductor current to rise. When the transistor is OFF, thecurrent will continue flowing through the inductor but now flowing through the diode. We initialassume that the current through the inductor does not reach zero, thus the voltage at Vx willnow be only the voltage across the conducting diode during the full OFF time. The averagevoltage at Vx will depend on the average ON time of the transistor provided the inductor currenis continuous.

Fig. 1: Buck Converter

8/10/2019 Ec1355 Esd Ece

4/31


VI SEMESTER ECE

Fig. 2: Voltage and current changes

To analyse the voltages of this circuit let us consider the changes in the inductor current overone cycle. From the relation

the change of current satisfies

For steady state operation the current at the start and end of a period T will not change. To get simple relation between voltages we assume no voltage drop across transistor or diode whileON and a perfect switch change. Thus during the ON time Vx=Vinand in the OFF Vx=0. Thus

which simplifies to

or

and defining "duty ratio" as

8/10/2019 Ec1355 Esd Ece

5/31


VI SEMESTER ECE

the voltage relationship becomes Vo=D Vin Since the circuit is lossless and the input and outpupowers must match on the average Vo* Io= Vin* Iin. Thus the average input and output current

must satisfy Iin=D IoThese relations are based on the assumption that the inductor current doenot reach zero.

1.1 Transition between continuous and discontinuous1.1 Transition between continuous and discontinuous1.1 Transition between continuous and discontinuous1.1 Transition between continuous and discontinuous

When the current in the inductor L remains always positive then either the transistor T1 or thediode D1 must be conducting. For continuous conduction the voltage Vxis either Vinor 0. If theinductor current ever goes to zero then the output voltage will not be forced to either of theseconditions. At this transition point the current just reaches zero as seen in Figure 3. During theON time Vin-Voutis across the inductor thus

(1)

The average current which must match the output current satisfies

(2)

Fig. 3: Buck Converter at Boundary

If the input voltage is constant the output current at the transition point satisfies

8/10/2019 Ec1355 Esd Ece

6/31


VI SEMESTER ECE

(3)

1.2 Voltage Ratio of Buck Converter (Discontin1.2 Voltage Ratio of Buck Converter (Discontin1.2 Voltage Ratio of Buck Converter (Discontin1.2 Voltage Ratio of Buck Converter (Discontinuous Mode)uous Mode)uous Mode)uous Mode)

As for the continuous conduction analysis we use the fact that the integral of voltage across th

inductor is zero over a cycle of switching T. The transistor OFF time is now divided into segmentof diode conduction ddT and zero conduction doT. The inductor average voltage thus gives

(Vin- Vo) DT + (-Vo) dT = 0 (4)

Fig. 4: Buck Converter - Discontinuous Conduction

(5)

for the case . To resolve the value of consider the output current which is half the

peak when averaged over the conduction times

(6)

Considering the change of current during the diode conduction time

(7)

Thus from (6) and (7) we can get

8/10/2019 Ec1355 Esd Ece

7/31


VI SEMESTER ECE

(8)

using the relationship in (5)

(9)

and solving for the diode conduction

(10)

The output voltage is thus given as

(11)

defining k* = 2L/(VinT), we can see the effect of discontinuous current on the voltage ratio ofthe converter.

Fig. 5: Output Voltage vs Current

8/10/2019 Ec1355 Esd Ece

8/31


VI SEMESTER ECE

As seen in the figure, once the output current is high enough, the voltage ratio depends only onthe duty ratio "d". At low currents the discontinuous operation tends to increase the outputvoltage of the converter towards Vin.

2. BOOST CONVERTER STEP2. BOOST CONVERTER STEP2. BOOST CONVERTER STEP2. BOOST CONVERTER STEP----UP CONVERTERUP CONVERTERUP CONVERTERUP CONVERTER

The schematic in Fig. 6 shows the basic boost converter. This circuit is used when a higheroutput voltage than input is required.

Fig. 6: Boost Converter Circuit

While the transistor is ON Vx=Vin, and the OFF state the inductor current flows through the diodgiving Vx=Vo. For this analysis it is assumed that the inductor current always remains flowing(continuous conduction). The voltage across the inductor is shown in Fig. 7 and the averagemust be zero for the average current to remain in steady state

This can be rearranged as

and for a lossless circuit the power balance ensures

8/10/2019 Ec1355 Esd Ece

9/31

8/10/2019 Ec1355 Esd Ece

10/31


VI SEMESTER ECE

Fig. 9: Waveforms for buck-boost converter

which gives the voltage ratio

and the corresponding current

Since the duty ratio "D" is between 0 and 1 the output voltage can vary between lower or highethan the input voltage in magnitude. The negative sign indicates a reversal of sense of theoutput voltage.

RESULT:RESULT:RESULT:RESULT:

8/10/2019 Ec1355 Esd Ece

11/31


VI SEMESTER ECE

2. DC POWER SUPPLY DESIGN USING FLYBACK CONVERTERS2. DC POWER SUPPLY DESIGN USING FLYBACK CONVERTERS2. DC POWER SUPPLY DESIGN USING FLYBACK CONVERTERS2. DC POWER SUPPLY DESIGN USING FLYBACK CONVERTERS

AimAimAimAim

Design the fly back converter using ferrite core transformer for the given input voltag

variation load current and output voltage.

Apparatus RequiredApparatus RequiredApparatus RequiredApparatus Required

Isolated DCIsolated DCIsolated DCIsolated DC----DC ConvertersDC ConvertersDC ConvertersDC Converters

In many DC-DC applications, multiple outputs are required and output isolation may need to beimplemented depending on the application. In addition, input to output isolation may berequired to meet safety standards and / or provide impedance matching.

1. Flyback Converter1. Flyback Converter1. Flyback Converter1. Flyback Converter

The flyback converter can be developed as an extension of the Buck-Boost converter. Fig 1shows the basic converter; Fig 2 replaces the inductor by a transformer. The buck-boostconverter works by storing energy in the inductor during the ON phase and releasing it to theoutput during the OFF phase. With the transformer the energy storage is in the magnetisation othe transformer core. To increase the stored energy a gapped core is often used.

In Fig 3 the isolated output is clarified by removal of the common reference of the input andoutput circuits.

Fig. 1: Buck-Boost Converter

8/10/2019 Ec1355 Esd Ece

12/31


VI SEMESTER ECE

Fig. 2: Replacing inductor by transformer

Fig. 3: Flyback converter re-configured

2. Forward Converter2. Forward Converter2. Forward Converter2. Forward Converter

The concept behind the forward converter is that of the ideal transformer converting the inputAC voltage to an isolated secondary output voltage. For the circuit in Fig. 4, when the transistoris ON, Vin appears across the primary and then generates

The diode D1 on the secondary ensures that only positive voltages are applied to the outputcircuit while D2 provides a circulating path for inductor current if the transformer voltage is zeror negative.

8/10/2019 Ec1355 Esd Ece

13/31


VI SEMESTER ECE

Fig. 4: Forward Converter

The problem with the operation of the circuit in Fig 4 is that only positive voltage is appliedacross the core, thus flux can only increase with the application of the supply. The flux willincrease until the core saturates when the magnetizing current increases significantly andcircuit failure occurs. The transformer can only sustain operation when there is no significant D

component to the input voltage. While the switch is ON there is positive voltage across the coreand the flux increases. When the switch turns OFF we need to supply negative voltage to resetthe core flux. The circuit in Fig. 5 shows a tertiary winding with a diode connection to permitreverse current. Note that the "dot" convention for the tertiary winding is opposite those of theother windings. When the switch turns OFF current was flowing in a "dot" terminal. The coreinductance act to continue current in a dotted terminal, thus

Fig. 5: Forward converter with tertiary winding


8/10/2019 Ec1355 Esd Ece

14/31


VI SEMESTER ECE

3. DESIGN OF A 43. DESIGN OF A 43. DESIGN OF A 43. DESIGN OF A 4----20mA TRANSMITTER FOR A BRIDGE TYPE TRANSDUCER20mA TRANSMITTER FOR A BRIDGE TYPE TRANSDUCER20mA TRANSMITTER FOR A BRIDGE TYPE TRANSDUCER20mA TRANSMITTER FOR A BRIDGE TYPE TRANSDUCER

AimAimAimAim

Design the Instrumentation amplifier with the bridge type transducer (Thermistor or aDesign the Instrumentation amplifier with the bridge type transducer (Thermistor or aDesign the Instrumentation amplifier with the bridge type transducer (Thermistor or aDesign the Instrumentation amplifier with the bridge type transducer (Thermistor or aresistance variatresistance variatresistance variatresistance variation transducers) and convert the amplified voltage from the instrumentatioion transducers) and convert the amplified voltage from the instrumentatioion transducers) and convert the amplified voltage from the instrumentatioion transducers) and convert the amplified voltage from the instrumentatioamplifier to 4amplifier to 4amplifier to 4amplifier to 4 20 mA current using op20 mA current using op20 mA current using op20 mA current using op----amp. Plot the variation of the temperature Vs outpamp. Plot the variation of the temperature Vs outpamp. Plot the variation of the temperature Vs outpamp. Plot the variation of the temperature Vs outpcurrentcurrentcurrentcurrent

Apparatus required

Sl.NoSl.NoSl.NoSl.No ComponentsComponentsComponentsComponents Type/RangeType/RangeType/RangeType/Range QuantityQuantityQuantityQuantity

1 OpAmp MA741 3

2 Thermister or RTD - 1

3 Ammeter (0-20)mA 1

4

Resister 8k,47k,

470k, 100k,

470, 1meg

1,2, 2, 2,1,4

5 DRB - 1

6 Power supply 5V, 15V,(0-5)V 1,1

A 4A 4A 4A 4----20ma Transmitter For A Bridge Type Transducer20ma Transmitter For A Bridge Type Transducer20ma Transmitter For A Bridge Type Transducer20ma Transmitter For A Bridge Type Transducer

4444----20 mA20 mA20 mA20 mA is an analog electrical transmission standard for industrial instrumentation ancommunication. The symbol "mA" is standard SI notation for milliampere, or 1/1000 of aampere. The signal is a current loop where 4 mA represents zero percent signals and 20 mrepresents the one hundred percent signal.

The reason zero is at 4 mA and not 0 mA is that this "live zero" allows the receivininstrumentation to differentiate between a zero signal and a broken wire or a dead instrumenThis standard was developed in the 1950s and is still widely used in industry today, even thougmany attempts have been made to replace it with digital forms of communication such a

fieldbus and Profibus. Its benefits of being a widely followed standard, low cost, its reliability animmunity to electrical noise keep it in regular use. Current loop is also much easier understand and debug than more complicated digital fieldbuses. Using fieldbuses and solvinrelated problems usually requires much more education and understanding than required simple current loop solutions. Additional digital communication to the device can be added current loop using HART Protocol.

8/10/2019 Ec1355 Esd Ece

15/31


VI SEMESTER ECE

ProcessProcessProcessProcess----control usecontrol usecontrol usecontrol use

For industrial process control instruments, analog 4-20 mA and 10-50 mA current looare commonly used for analog signaling, with 4 mA representing the lowest end of the rangand 20 mA the highest. The key advantages of the current loop are that the accuracy of thsignal is not affected by voltage drop in the interconnecting wiring, and that the loop can suppoperating power to the device. Even if there is significant electrical resistance in the line, thcurrent loop transmitter will maintain the proper current, up to its maximum voltage capabilitThe live-zero represented by 4 mA allows the receiving instrument to detect some failures of thloop, and also allows transmitter devices to be powered by the same current loop (called twwire transmitters). Such instruments are used to measure pressure, temperature, flow, pH other process variables. A current loop can also be used to control a valve positioner or othoutput actuator. An analog current loop can be converted to a voltage input with a precisioresistor. Since input terminals of instruments may have one side of the current loop input tied the chassis ground (earth), analog isolators may be required when connecting several devices series.

Taking the point of view of the source of current for the loop, devices may be classified aactive (supplying power) or passive (relying on loop power). For example, a chart recorder mprovide loop power to a transmitter instrument such as a pressure transmitter. The pressutransmitter modulates the current on the loop to send the signal to the strip chart recorder, bdoes not in itself supply power to the loop and so is passive. Another loop may contain twpassive chart recorders, a passive pressure transmitter, and a 24 V battery. (The battery is thactive device). Panel mount displays and chart recorders are commonly termed 'indicatdevices' or 'process monitors'. Several passive indicator devices may be connected in series, ba loop must have only one transmitter device and only one power source (active device).

The relationship between current value and process variable measurement is set

calibration, which assigns different ranges of engineering units to the span between 4 and 2mA. Occasionally the mapping between engineering units and current was inverted, so thatmA represented the maximum and 20 mA the minimum.

Design ProcedureDesign ProcedureDesign ProcedureDesign Procedure

Problem: Design a (4-20) mA transmitter for a Thermister based Temperature indicator toindicate 30oC to 600C as 4-20 mA.

Design Data: At 30oC RT= 8k, at 60oC RT= 3k

Design of Bridge:

At 30oC RT= 8k, Differential VoltageVD= ((R4/(R4 + R2) - (RT/(RT+ R1) = 0V ----(1)

At 60oC RT= 3k, Differential VoltageVD= ((R4/(R4 + R2) - (RT/(RT+ R1) = 400mV ----(2)

From equation 1 and 2, R1 = R2 = 51.3 k47 k, R4 = 8k.

Design of Differential Amplifier:

8/10/2019 Ec1355 Esd Ece

16/31


VI SEMESTER ECE

Output of Differential amplifier = (0-4)V for 30oC to 600C Temperature variation.

Gain of Differential amplifier = 10, Assume Rf=1 Meg , Ri= Rf/ 10 = 100k

Design of Summing Amplifier:

Output of Summing Amplifier = (2 to 10) V for (0-4) V input.

Assume, Gain of Summing Amplifier = 2, Rf=1 Meg , Ri= Rf/ 2 = 500k 470k

V2 1V

Design of V to I Converter:

Input voltage (2 10) VRequired output Current = (4 - 20) mA

Ri= 500 470 .

4444----20 mA Transmitter Bridge Circuit20 mA Transmitter Bridge Circuit20 mA Transmitter Bridge Circuit20 mA Transmitter Bridge Circuit

HI

1meg

100k

HIR4

8k 1meg

470

100k

AD741

3

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

0

RTD

(4-20) m

HI

AD7413

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

15

0

470k

HI

AD7413

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

LO

R1

47k

15

0

LO

1meg

0

V2

R2

47k

1meg

LO

LO

470k

0

5 V

8/10/2019 Ec1355 Esd Ece

17/31


VI SEMESTER ECE

ObservationObservationObservationObservationFor temperature from 30oC to 600C measure the current and tabulate the same.

Sl. No.Sl. No.Sl. No.Sl. No. Temperature (Temperature (Temperature (Temperature ( ooooC)C)C)C)Resistance ofResistance ofResistance ofResistance of

Thermistor (kThermistor (kThermistor (kThermistor (k))))

OutputOutputOutputOutput

Current(mA)Current(mA)Current(mA)Current(mA)

ResultResultResultResult:

8/10/2019 Ec1355 Esd Ece

18/31


VI SEMESTER ECE

5. DESIGN OF PROCESS CONTROL TIMER5. DESIGN OF PROCESS CONTROL TIMER5. DESIGN OF PROCESS CONTROL TIMER5. DESIGN OF PROCESS CONTROL TIMER

AimAimAimAim

Design a sequential timer to switch on & relays in a particular sequence using timer IC

Apparatus requiredApparatus requiredApparatus requiredApparatus required

Sl.NoSl.NoSl.NoSl.No ComponentsComponentsComponentsComponents Type/RangeType/RangeType/RangeType/Range QuantityQuantityQuantityQuantity

1 Counter IC 7490 1

2 AND gate IC 7408 1

3 OR gate IC 7432 1

4 NOT gate IC 7404 1

5 XOR gate IC 7486 1

6 LED 67 Pulse generator 1

8 Powersupply 5V 1

Process Control TimerProcess Control TimerProcess Control TimerProcess Control Timer


8/10/2019 Ec1355 Esd Ece

19/31


VI SEMESTER ECE

6. DESIGN OF AM / FM MODULATOR / DEMODULATOR6. DESIGN OF AM / FM MODULATOR / DEMODULATOR6. DESIGN OF AM / FM MODULATOR / DEMODULATOR6. DESIGN OF AM / FM MODULATOR / DEMODULATOR

AimAimAimAim

iii. Design AM signal using multiplier IC for the given carrier frequency and modulatioindex and demodulate the AM signal using envelope detector.

iv. Design FM signal using VCO IC NE566 for the given carrier frequency and demodulathe same using PLL NE 565.

Apparatus RequiredApparatus RequiredApparatus RequiredApparatus Required

Am ModulatorAm ModulatorAm ModulatorAm ModulatorThe basic operation performed in Amplitude Modulation is multiplying two signa

together to produce a modulated signal s(t). The first signal is called the carrier signal andmade of a pure sine wave at a high frequency fc. The carrier signal does not contain ainformation and serves mainly to "carry" the contents of the information signal to highfrequencies so that it can be radiated through an antenna. The second signal is called tcomplex envelope signalg(t). The complex envelope signal has particular meaning in AM wheit provides the slow varying envelope to the fast varying modulated signal s(t). The complenvelopeg(t)is strongly related to the modulating or information signal m(t).

Modulating signal (information) ( ) ( )tfAtm mm 2cos= simple case

Complex envelope signal ( ) ( )[ ]tmAtg c += 1 for Amplitude ModulationCarrier signal tfj ce

2

Modulated signal ( ) ( ) tfj cetgts 2Re=

It can be shown that the modulated signals(t)reduces to the expression in Equation 1.

( ) ( ) ( )( ) ( )( )tffAA

tffAA

tfAts mcmc

mcmc

cc +++= 2cos2

2cos2

2cos 1

The Fourier Transform for positive frequencies of the modulated signal is given by Equation 2.

( ) ( ) ( )( ) ( )( )mcmc

mcmc

cc fff

AAfff

AAff

AfS +++=

442 .2

Figure 1 shows the complex envelope signal g(t) and the carrier signal on the left and thmodulated signal for Am = 0.5 on the right. It can be seen that the complex envelope gcorresponds to the slow varying envelope in the modulated signals(t).

8/10/2019 Ec1355 Esd Ece

20/31


VI SEMESTER ECE

0 20 40 60 80 100 120 140 160 180 200-2

-1

0

1

2

volts

complex envelope

0 20 40 60 80 100 120 140 160 180 200-2

-1

0

1

2

time in microseconds

volts

carrier

0 20 40 60 80 100 120 140 160 180 200

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

time in microseconds

volts

modulated signal

Figure 1: Time representation of amplitude modulation of a sine wave

(fc= 100 kHz, fm= 10 kHz,Ac= 1 andAm= 50% modulation)

Figure 2 shows the frequency magnitude spectrum of the AM modulated signal s(t). There a

three frequency components. The strongest component occurs at the frequency of the carrisignal fc. The other two components have the same magnitude and occur at the differenfrequency fc - fm and at the sum frequency fc + fm. One notes that the magnitude of thcomponents at the difference frequency fc- fmand at the sum frequency fc+ fmis proportional AcandAm, while the magnitude of the component at the carrier frequency fcis only proportiontoAc. It is therefore possible to increase or decrease the magnitude of the fc- fmand at the fcfm components without changing the magnitude of the carrier frequency component changingAm. The value ofAmtherefore represents the relative amount of the modulating signm(t) compared to the carrier signal. The value of Am, often expressed in percent is called thmodulation index. More generally, the modulation index is defined as

cA

AA

M

= 2

minmax

whereAmax,AminandAcare the maximum, minimum and average amplitudes of the envelope the modulated signals(t)respectively.

|S(f)|

0 fc-fm fc+fm ffc

Ac------

2

AcAm

--------4

AcAm

--------4

20 40 60 80 100 120 140 160 180

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

frequency in kHz

Figure 2: Frequency representation of amplitude modulation of a sine wave(fc= 100 kHz, fm= 10 kHz,Ac= 1 andAm= 50% modulation)

8/10/2019 Ec1355 Esd Ece

21/31


VI SEMESTER ECE

Frequency ModulatioFrequency ModulatioFrequency ModulatioFrequency Modulationnnn

Frequency modulation (FM) is the standard technique for high-fidelity communications is evident in the received signals of the FM band (88-108 MHz) vs. the AM band (450-1650 KH

The main reason for the improved fidelity is that FM detectors, when properly designed, are nsensitive to random amplitude variations which are the dominant part of electrical noise (heaas static on the AM radio). Frequency modulation is not only used in commercial radbroadcasts, but also in police and hospital communications, emergency channels, TV sounwireless (cellular) telephone systems, and radio amateur bands above 30 MHz.

The basic idea of an FM signal vs. an AM signal is shown in Fig. 3. In an FM signal, thfrequency of the signal is changed by the modulation (baseband) signal while its amplitudremains the same. In an AM signal, we now know that it is the amplitude (or the envelope) the signal that is changed by the modulation signal. The FM signal can be summarized follows:

Figure 3Figure 3Figure 3Figure 3: FM representation vs. AM representation.

1. The amplitude of the modulation signal determines the amount of the frequency change frothe center frequency.

2. The frequency of the modulation signal determines the rate of the frequency change from thcenter frequency.

3. The amplitude of the FM signal is constant at all times and is independent of the modulatiosignal.

8/10/2019 Ec1355 Esd Ece

22/31


VI SEMESTER ECE

Mathematically, an FM signal is written as:

V = Asin ct +m fsinmt( )

where A = the amplitude of the signalc= center frequency (frequency for no modulation signal)

m= modulation frequency

and mf= FM modulation index = /fm

where = maximum frequency shift caused by the modulation signalfm= frequency of the modulation signal.

The spectrum of an FM signal is quite complicated and is dependent on mf. Actually, it follows

Bessel Function (you will study this in a senior-level math course) and is given by:

FM Spectrum = J0 mf( )cos ct( ) Center Frequency c( )

J1 mf( )cos c m( )tcos c +m( )t[ ] Components at c + /m( )+J2 mf( )cos c2m( )t+cos c2m( )t[ ] Components at c + /2m( )

J3 mf( )cos c 3m( )tcos c+3m( )t[ ] Components at c + /3m( )+ .....

where J0 mf( ),J1 mf( ) etc. are in volts, and are the levels of the frequency components of thFM signal for A = 1 V.

Figure 4 gives some spectrum representation for various modulation indices. Notice froTable 1 that for mf = 2.4, there is no power in the center frequency component (J0(2.4) =0). Th

also occurs at mf = 5.5, 8.6, ... . This does not mean that there is no power transmitted in thsignal. All that it means is that for m = 2.4, 5.5, ..., there is no power at the center frequency aall of the power is in the sidebands.The bandwidth of an FM signal depends on the modulation index (mf), and is approximated

the well-known Carson s Rule:

BW = 2 (+ fm(max))

where fm(max) is the maximum frequency of the modulating signal. The factor (2) in th

equation is to account for both the upper and lower sidebands (left and right of the carrier). Thequation gives the bandwidth which contains 98% of the signal power.

8/10/2019 Ec1355 Esd Ece

23/31


VI SEMESTER ECE

Figure 4Figure 4Figure 4Figure 4: Frequency spectrum of FM signals with different mfand same modulating frequency.

The best way to understand FM signals is to consider a real life example. Let us take an An

Arbor station, 107.1 FM, broadcasting at 107.1 MHz with a power of 50 KW. The bandwidth the modulation signal is from 30 Hz to 15 KHz which is excellent for high-fidelity broadcast. Tmaximum deviation set by the FCC, (), is 75 KHz. The range of the modulation index therefore:

mf(min)= /fm(max)= 75 KHz/15 KHz= 5 (for fm= 15 KHz)

tomf(max) = /fm(min)= 75 KHz/30 Hz = 2500! (for fm= 30 Hz)

Notice that the modulation index changes a lot with the modulation frequency (from 2,500 toFor the 15 KHz signal, mf= 5, we see that the frequency components are up high up to J6

drop quickly afterwards. This means that the bandwidth of the signal is 6x15 KHz = 90 KHzeach side of the center frequency (a total bandwidth of 180 KHz). We can also use Carsons rand, BW = 2 ( + fm(max)) = 180 KHz or 90 KHz on each side of the center frequency.

For the 30 Hz signal and mf= 2,500, a huge number of sidebands exist but remember that th

are now spaced at only 30 Hz apart. The total bandwidth of the signal is BW = 2 (+ f m(min150 KHz or 75 KHz on each side of the center frequency. This means that the bandwidth of FM signal changes from +/- 75 KHz to +/- 90 KHz from the carrier depending on the modulafrequency.

Commercial FM stations are therefore spaced 200 KHz apart to avoid interference for amodulating frequencies. In order to even isolate the stations further, FCC only assigns alternastations for a certain area. For example, in the Detroit/Ann Arbor area, the stations are 107.

107.5 (and 93.1, 93.5, 93.9, ...) spaced 400 KHz apart. In adjoining areas, such as Toledo to thsouth (or Lansing to the north, but very far from Toledo), the stations are also centered at 40KHz, but they are 107.3, 107.7, etc... (and 93.3, 93.7, 94.1 etc...). This allows inexpensive radiowith bad-to-acceptable selectivity to receive FM stations without interference from adjoinistations (since they are 400 KHz away and not only 200 KHz away). The 200 KHz-away stationare very far and therefore their signals would appear as noise in the receiver. However, amentioned before, FM receivers have excellent noise rejection and therefore are not affected bthe far-away stations.

8/10/2019 Ec1355 Esd Ece

24/31


VI SEMESTER ECE

FM Generation Using LM 566C VoltageFM Generation Using LM 566C VoltageFM Generation Using LM 566C VoltageFM Generation Using LM 566C Voltage----Controlled Oscillator:Controlled Oscillator:Controlled Oscillator:Controlled Oscillator:

1. Connect the LM 566C as shown below. The center frequency is set by Ro, Co, V5/V+. Set V

(Vcc) to 10 V.

2. Measure the output voltage Vo(pin #4). It should be a triangular wave with a frequency 100-110 KHz and Vppk around 1.8 V.

3. Connect the signal generator to the modulation input (Vs) and set it to a sine wa

with f = 5 KHz and Vppk= 200 mV. Basically, the voltage at pin #5 is now varyiby +/- 0.1 V.The output frequency of the LM 566 will vary by +/- 10 KHz (this is the maximum deviationThis is the case of mf = 2 (since fm=5 KHz). Notice how the amplitude of the FM sign

remains constant. Look at the output in frequency domain and check that this is correct (dnot make any sideband measurements). Choose a center frequency of 244 KHz. You will seother sidebands due to the third harmonic of the triangular wave.

4. Change from a sine wave to a square wave with Vppk = 400 mV. You will have now only tw

frequencies, one around 80 KHz and one around 120 KHz. This is an FSK signal, which generated by the VCO.

5. Disconnect the connection to pin #5.6. Connect the circuit shown below to the LM 566. Notice the 10 K/560 voltage divider

the output of the LM 566, and the 0.1 F capacitor DC block between the LM 566 and thLM 386.

7. Measure the output of the LM 386. It should be a nice triangular wave with Vppk arou1.6 V

8. Connect a 1.6 nF capacitor in parallel with the 560 resistor. This capacitor results in corner frequency of around 180 KHz and therefore passes the 100 KHz components anattenuates the third and fifth harmonic components of the triangular wave. Look at th

output waveform of the LM 386 and notice that it now resembles a sine wave.9. Measure the frequency components of the LM 386 output for the quasi-sine wavefor(Vppk, fundamental and second/third harmonic levels only)

10.Connect Vsto pin #5 of the LM 566 through the 4.7 F DC-block capacitor, and set V sto 20

mVppk, f = 5 KHz sinewave. Measure the output voltage of the LM 386. Look at the resultin

waveform and plot the graph.FM Generation Using LM 566C VoltageFM Generation Using LM 566C VoltageFM Generation Using LM 566C VoltageFM Generation Using LM 566C Voltage----Controlled OsControlled OsControlled OsControlled Oscillator:cillator:cillator:cillator:

8/10/2019 Ec1355 Esd Ece

25/31


VI SEMESTER ECE

FM Demodulation Using LM 565 Phase Locked Loop:FM Demodulation Using LM 565 Phase Locked Loop:FM Demodulation Using LM 565 Phase Locked Loop:FM Demodulation Using LM 565 Phase Locked Loop:


ModulatedOutput

8/10/2019 Ec1355 Esd Ece

26/31


VI SEMESTER ECE

8. PCB LAYOUT DESIGN USING CAD8. PCB LAYOUT DESIGN USING CAD8. PCB LAYOUT DESIGN USING CAD8. PCB LAYOUT DESIGN USING CAD

AimAimAimAim

Drawing the schematic of simple electronic circuit and design of PCB layout using CAD

Pcb Layout Design Using CadPcb Layout Design Using CadPcb Layout Design Using CadPcb Layout Design Using Cad

Preparing a Capture design for Layout is a two-part process. First, you must create a valdesign and then create a netlist in an .MNL format for Layout. After you have prepared yoCapture design, you can create a new Layout design using the .MNL netlist.

Preparing a Capture design for Layout is a two-part process. First, you must create a valdesign. Then, you need to create a netlist in a .MNL format in Capture and read it into Layout.

To prepare your Capture designTo prepare your Capture designTo prepare your Capture designTo prepare your Capture design

If you want to send part or net information from a Capture schematic design, add a user-defineproperty with a name from the property tables and assign a value to the property. The propername must be in uppercase, as given in the tables.

1 On your schematic page, select the objects you want to send to Layout, then right-cliand choose Edit Properties from the pop-up menu. The property editor will appear.

2 Click the down arrow on the Filter by drop-down list to expand the list, and select LayouThe columns in the spreadsheet change to display properties available in Layout. The cebelow each property column heading appear white with hash marks, indicating that theproperties are not yet assigned to the part or net.

3 Assign values to the properties of the selected objects on the schematic by clicking oncell, typing in the value, and pressing ENTER or clicking Apply. The hash marks disappear athe property is applied to the object.

4 Assign PCB footprints to each of your schematic parts. Use only Layout footprint nameselecting from those shown in the OrCAD Layout for Windows Footprint Libraries, or those your custom footprint libraries.

The pin numbers for pin names where no number exists must match the pad names on thLayout footprint.

Layout cannot accept PCB footprint names or part values that include spaces. Check yodesign to eliminate spaces or tabs in these property values. Capture's spreadsheet editor is veuseful for this.

The layer names you use as property values must be among the standard Layout laynames. The via names you use as property values must be among the standard Layout vnames.

8/10/2019 Ec1355 Esd Ece

27/31


VI SEMESTER ECE

To create a netlist for LayoutTo create a netlist for LayoutTo create a netlist for LayoutTo create a netlist for Layout

1 Open the Capture project.

2 Select the design in the project manager and choose Create Netlist from the Tools menThe Create Netlist dialog box appears.

3 Choose the Layout tab. The Layout tab appears.

4 In the PCB Footprint group box, ensure that {PCB Footprint} appears in the Combineproperty string text box.

5 In the Netlist File text box, ensure that the path to the netlist file is correct. The netltakes the name of the Capture project and adds a .MNL extension.

6 Click OK. Capture processes the netlist, then creates an .MNL file and saves it in t

directory specified in step 6.

It is not necessary to run Capture and Layout simultaneously to take advantage of forwaannotation.

If Capture is unable to create a .MNL file, the errors are written to the Capture session loand to the .ERR file in the target directory for the .MNL file.

Preparing PCB layout using AutoroutePreparing PCB layout using AutoroutePreparing PCB layout using AutoroutePreparing PCB layout using AutorouteThis command systematically autoroutes your printed circuit board design from start

finish, automatically sweeping through the design. If you have set any of the fanout command(Fanout Board, Fanout DRC/Route Box

, or Fanout Component) before you choose the Autoroute Board command, Layout executes tappropriate fanout operation before any routing sweeps.

You set routing priorities and characteristics using the Route Sweep, Route Pass, and RouLayer commands.

You can interrupt autorouting at any time by pressing ESC, then restart by choosing the Resume Routing command.

If you want to autoroute a design after you have routed it to completion, you must edit the RouPass spreadsheet and change the passes from Done to Yes.

8/10/2019 Ec1355 Esd Ece

28/31


VI SEMESTER ECE

Circuit DiagramCircuit DiagramCircuit DiagramCircuit Diagram

U1

AD741

3

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

R5

U2

AD741

3

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

R9

R7R4

R1

U3

AD741

3

2

7

4

6

1

5+

-

V+

V-

OUT

OS1

OS2

0

R2

R8

R10

R3

PCB LayoutPCB LayoutPCB LayoutPCB Layout

Top layerTop layerTop layerTop layer

8/10/2019 Ec1355 Esd Ece

29/31


VI SEMESTER ECE

Bottom layerBottom layerBottom layerBottom layer

Top & Bottom layerTop & Bottom layerTop & Bottom layerTop & Bottom layer


8/10/2019 Ec1355 Esd Ece

30/31


VI SEMESTER ECE

9. MICROCONTROLLER BASED SYSTEMS DESIGNMICROCONTROLLER BASED SYSTEMS DESIGNMICROCONTROLLER BASED SYSTEMS DESIGNMICROCONTROLLER BASED SYSTEMS DESIGN

AimAimAimAimDesign of microcontroller based system for simple applications like security

systems combination lock etc.

AlgorithmAlgorithmAlgorithmAlgorithm

Microcontroller based security systems

1. Read number of bytes in the password

2. Initialize the password

3. Initialize the Keyboard Display IC (8279) to get key and Display

4. Blank the display

5. Read the key from user

6. Compare with the initialized password

7. If it is not equal, Display E to indicate Error.

8. Repeat the steps 6 and 7 to read next key

9. If enterer password equal to initialized password, Display O to indicate open.

ProgramProgramProgramProgram

LABELLABELLABELLABEL MNEMONICSMNEMONICSMNEMONICSMNEMONICS OPERANTOPERANTOPERANTOPERANT COMMENTSCOMMENTSCOMMENTSCOMMENTS

MOV 51H,#

MOV 52H,#

MOV 53H,#

MOV 54H,#

MOV R1,#51

MOV R0,#50

MOV R3,#04

MOV R2,#08

MOV DPTR,#FFC2

MOV A,#00

MOVX @DPTR,AMOV A,#CC

MOVX @DPTR,A

MOV A,#90

MOVX @DPTR,A

MOV A,#FF

MOV DPTR,#FFCO

LOOP: MOVX @DPTR,A

8/10/2019 Ec1355 Esd Ece

31/31


DJNZ R2,LOOP

AGAIN: MOV DPTR,#FFC2

WAIT: MOVX A,@DPTR

ANL A,#07

JZ WAIT

MOV A,#40MOVX @DPTR,A

MOV DPTR,#FFCO

MOVX A,@DPTR

MOV @R0,A

MOV A,@R1

CJNE A,50H,NEQ

INC R1

DJNZ R3,AGAIN

MOV DPTR,#FFCO

MOV A,#OC

MOVX @DPTR,AXX: SJMP XX

NEQ: MOV DPTR,#FFCO

MOV A,#68

MOVX @DPTR,A

YY: SJMP YY

ObservationObservationObservationObservation

ResultResultResultResult

ec1355 esd ece

Documents