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Diploma in Electronics - CP04 Instructors’ Practical Manual V semester Industrial Electronics-I

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Page 1: Industrial Electronics

Diploma in Electronics - CP04

Instructors’ Practical Manual

V semester

Industrial Electronics-I

Page 2: Industrial Electronics

LIST OF EXPERIMENTS

PAGE NO

1. FAMILIARISATION OF DIGITAL STORAGE OSCILLOSCOPE 01

2. FAMILIARISATION OF CURRENT PROBE AND AMPLIFIER 05

3. TRANSISTER SWITCHING CHARACTERISTICS 08

4. MOSFET SWITCHING CHARECTERISTICS 13

5. MAGNTISATION OF RL LOAD & DEMAGNETISATION USING

i. DIODE 19

ii. DIODE&RESISTER 23

iii. DIODE&ZENER 27

6. MAGNETISATION OF L LOAD & DEMAGNETISATION USING

a. DIODE 31

b. DIODE & RESISTOR 35

c. DIODE & ZENER 39

7. FAMILIARISATION OF PWM IC SG3524 43

8. INDUCTOR DESIGN 47

9. BUCK CONVERTER 53

Page 3: Industrial Electronics

10. FAMILIARISATION OF IC TCA785 60

FAMILIARIZATION OF DIGITAL

STORAGE OSCILLOSCOPE

OBJECTIVES

To get familiarized with DSO and its front panel controls.

EQUIPMENTS REQUIRED

SlNo Equipment Model Specification Quantity

01 DSO Tektronix 60MHz 1

02 Probe Tektronix 7A 1

HOW TO USE IT?

Tektronix TDS 350 two channel oscilloscope are digital signal processors and are

superb tools for displaying and measuring and measuring waveforms.

Two input channels, each with a record length of 1000 samples an 8-bit vertical

resolution .both channels acquire wave forms simultaneously.

1 Giga samples/second maximum sample rate (TDS500):500 Mega

samples/second maximum sample rate: 200 Mega samples/second (TDS320).

200 MHz analog band width and fastest time base setting of 2.5ns/div (TDS350).

Page 4: Industrial Electronics

A full compliment of advanced functions including on-screen read out.

Auto set cursors and continuously updated automatic measurements

Wave form averaging, wave form enveloping and hardware peak detection

A unique graphical user interface (GUI) and a logical front panel layout which combine

to deliver the standard in usability pioneered by the TDS family of oscilloscopes.

COMPENSATING A PROBE

Use the following procedure:

Attach the probe BNC connector to channel and attach the probe HP to the probe

comp output signal. Press auto reset.

Check that wave form indicates correct compensation. If the wave form indicates over or

under compensation, use alignment tool provided with the probe to adjust the

compensation.

FRONT PANEL

AUTOSET

Feature produces a stable triggered display of almost any signal .Two use auto

set; connect a signal to either ch1 or ch2.

CLEAR MANU

This button clears all menus from the screen.

VERTICAL CONTROL

Page 5: Industrial Electronics

The vertical knob controls the vertical position of the presently selected

waveforms.

The vertical menu button calls up the vertical operation menu.

The vertical volts/div knob controls the vertical scale of the presently

selected waveform.

The waveform OFF button turns off the presently selected waveforms.

HORIZONTAL CONTROL

The horizontal position knob controls the horizontal position of all

waveforms.

The sec/div knob controls the horizontal scale of active waveforms.

TRIGGER

The trigger level knob controls the trigger.

The force trigger button forces the oscilloscope to start acquiring a

waveform regardless of whether a trigger event occurs. The button has no

effect if the acquisition system is stopped.

The set level to 50 % button sets the trigger level at the half way point

between the peaks of the trigger signal.

The trigger status lights indicate the status of the triggering system.

Page 6: Industrial Electronics

The ready light illuminates when the instrument can accept a valid trigger

and is waiting for the event to occur.

MISCELLANEOUS CONTROLS

The general purpose knob controls the many side menu functions

including the cursor. The toggle button switches control from cursor to

cursor.

The rum/stop button starts and stops acquisition.

The Measure button calls up the automated measurement menu.

The acquire button calls up the acquisition menu.

The save/recall setup button calls up the save/recall setup.

LOGIC CONVENTIONS

This refers to digital logic circuits with standard logic symbols and

terms. Unless otherwise started all logic functions are described using the

positive logic convention. The more positive of two logic levels is the high

(1) state and more negative level is the low (0) state signal states may also

be described as “true” meaning their active state or “false” meaning their

non active state. The specific voltages that constitute a high or low state

vary among the electronic devices.

Page 7: Industrial Electronics

FAMILIARISATION OF

CURRENT PROBE AND CURRENT AMPLIFIER

OBJECTIVES

To familiarize with current probe amplifier and its front panel controls.

EQUIPMENTS REQUIRED:

Slno Equipment Model Specification Quantity

01 Current probe Tektronix 20 A AC/ DC 1

02 Current probe amplifier Tektronix 50Mhz 1

03 DSO Tektronix 50Mhz 1

HOW TO USE IT?

The A6302 is DC to 50Mhz current probe designed for use with AM503 family

of current probe amplifier .The A6302 can measure current up to 20A(DC+ peak AC)

and up to 50A peak current.

The slide must be locked to accurately measure current or degauss the probe. If a probe is

unlocked; the probe open indicator on the amplifier will light.

Page 8: Industrial Electronics

DEGAUSSING AND AUTOBALANCING OF CURRENT PROBE

Verify that current probe is connected to the amplifier.

Remove the current probe from the conductor under test.

Lock the probe slide closed.

Press the amplifier probe degauss auto balance button.

NOTE: The degauss procedure will fail if the amplifier is not properly connected to a 50

ohm termination impedance.

Why degauss the current probes:

Degaussing the probe removes any residual magnetization from the probe core such

residual magnetization can induce measurement error. Auto balancing removes unwanted

dc offset in the amplifier circuitry. Failure to degauss the probe is a leading cause of

measurement error.

MAXIMUM CURRENT LIMITS

Exceeding any of these rating can saturate the probe core and cause measurement error.

Maximum continuous current error refers to the maximum current that can be

continuously measured at DC or at a specified AC frequency the maximum

continuous current value is de rated with frequency. As the frequency increase

maximum continuous current value or rating decreases.

Maximum pulsed current refers to the maximum peak value of pulsed current the

probe can accurately measured, regardless of how short the pulse duration is.

Ampere-second product defines the maximum width of pulsed current that you

can measure when the pulse amplitude is between the maximum continuous and

Page 9: Industrial Electronics

maximum pulsed current specifications. The maximum continuous specifications

it itself varies by frequency.

To determine if your measurement exceeds the ampere-second

Product, perform either procedure A or B.

PROCEDURE A

To determine the maximum allowable pulse width, measure the peak current of the

pulse divide the ampere-second specification of your probe by the measured peak current

of the pulse. The quotient is the maximum allowable pulse width; the pulse width at the

50% point of the measured signal must be less than this value.

PROCEDURE B

To determine the maximum allowable pulse amplitude measures the pulse width at

50% points. Divide the ampere-second (or ampere-microsecond) specification of your

probe by the pulse width .The quotient is the maximum allowable current: the peak

amplitude of the measured pulse must be less than this value.

After degaussing the probe connect the probe to current amplifier. Set the DSO volts

per division to 10mV. Adjust the reference of the current amplifier by keeping the

coupling in reference position. Lock the probe and change the coupling to either dc or

ac depending on the current flow in the conductor. Adjust the ampere per division of

current probe amplifier. Measure the current on the DSO.

Page 10: Industrial Electronics

SWITCHING CHARACTERSTICS OF A TRANSISTOR

OBJECTIVE

To design a switch using a transistor.

Find rise time, fall time, delay time and storage time.

BILL OF MATERIALS

Sl no: Component Specification Value Quantity

1 Transistor BD139 Ic=1.5A,60v 1

2 Resistor R1 ½W CFR 200 ohm 1

3 R2 2W CFR 20 ohm 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement

systems

0-30v,2A

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

Page 11: Industrial Electronics

CIRCUIT THEORY

The above circuit is a base bias circuit, which is designed to operate transistor

in saturation and cut off. This is used in switching application.

When base drive is zero, the switch is open. There is no drop across collector

resistor and the entire supply voltage drop across collector-emitter terminals. Since the

beta of the transistor varies commonly to make sure that the circuit will work in the

saturation it is designed in the hard saturation. Even if the beta changes for a long range

the circuit will remain in the hard saturation.

On time of the transistor= Rise time + Delay Time

Off time of the transistor= Fall time + Storage time

Page 12: Industrial Electronics

DESIGN

To design the transistor as a switch for a current of 0.5A.

Icsat =0.5A

Icsat = Vcc / Rc

Let Vcc = 10V

Then Rc = 10/0.5A

= 20 Ohms

Since the circuit should work in the hard saturation the ratio of the Rb verses Rc

should be 10:1

Therefore Rb = 200 Ohms

Page 13: Industrial Electronics

CIRCUIT DIAGRAM

Page 14: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 100µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the Current and Voltage waveforms.

Measure rise time, fall time, delay time and storage time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

Rise time=

Fall time=

Delay time=

Storage time =

Page 15: Industrial Electronics

SWITCHING CHARACTERISTICS OF MOSFET

OBJECTIVE:

To study the switching characteristics of MOSFET.

To measure On time (Ton ) and Off time (Toff ) of the MOSFET.

BILL OF MATERIALS

Slno Component Specification Value Quantity

01 MOSFET IRF 640 18 A,200V 1

02 Resistor R1 ½w CFR 50 ohm 1

03 R2 2w CFR 10 ohm 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement

systems

0-30v,2A

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

Page 16: Industrial Electronics

CIRCUIT THEORY

With gate current Ig gate source capacitor is charged , once this capacitor is charged

to some extend the mosfet will start conducting , even with Ig , Cgs is getting charged Vgs

remains constant , because Cgd is getting discharged , after this Vgs starts increasing.

The above mentioned properties can be seen in the current wave forms using CPA.

Characteristics of MOSFET:

The MOSFET is preferred in most of the switching circuits because of it’s very

less gate current.

It is a voltage controlled device.

The driving circuit is easy to design.

Switching time is faster.

All these characteristics of the MOSFET make it more useful for the high

frequency switching applications.

The working of the MOSFET is mainly related to the capacitors between Cgs and Cgd.

They are available in two types they are enhancement mode and depletion mode, the

enhancement mode is more preferred because of the very less reverse current. In this

type when the switch is off there is no channel existing between drain and source.

Main disadvantage is

It is sensitive to the Electro static discharges and may get damaged easily

The Rds (on) (on state drain to source resistance) is high.

Page 17: Industrial Electronics

NOTE: The MOSFET can be checked using the multimeter by the following steps

Short all the three pins in order to discharge both capacitors.

Check the drain source resistance it should be infinity if the

MOSFET is working.

Keep the multimeter in diode checking mode

Charge the gate - source capacitance by connecting the probes to

gate and source of the MOSFET.

Check for the channel formation between drain and the source.

If the channel is present the resistance will be very less and the

MOSFET is OK

Page 18: Industrial Electronics

DESIGN:

To design the switch for a current of 1A.

Assume Idsat =1A

Vcc =10V

Calculate Rd

Rd = Vdd / Id

= 10V / 1A

= 10 Ohms

Gate current is Ig+= Qc / Ton

Consider Ton= 1us

Ig+= 63nC / 1 uS

= 63mA

Ig = 2 Ig+

= 2 * 63mA

= 126mA

Rg= Vg- Vgsth / Ig

= 10V – 4V / 126mA

Page 19: Industrial Electronics

= 47.619 Ohms

CIRCUIT DIAGRAM

Page 20: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 50us with a frequency of 200 Hz.

Apply the signal to gate of the switch.

Observe the wave form across the switch.

Measure the on time and off time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The On time of the MOSFET is 900ns and Off time is 700 ns.

Page 21: Industrial Electronics

MAGNETIZATION AND DEMAGNETISATION

OF RL LOADS USING DIODE

OBJECTIVES:

To use a diode for demagnetizing the RL loads.

Study the property of current & voltage through & across the inductor.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core 540µH,0.5A 1

2 Resistor 2w CFR

1/2w

20 ohm,

200ohm

1

1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 DSO TEKTRONIX 50Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement 0-30v,2A

Page 22: Industrial Electronics

systems

4 Current probe and amplifier Tektronics 50Mhz,20A

CIRCUIT THEORY

In this circuit the inductor is demagnetized by a freewheeling diode. In most of the

commonly used circuits we can see this type of demagnetization. When the inductor

produces the back EMF then the diode gets forward biased and the inductor will be

demagnetized. This diode is commonly known as the free wheeling diode because of it’s

free wheeling action in the circuit. This diode suppresses the huge Back EMF and

protects the switch from damaging.

The inductor is magnetized with a series resistor; the current in the circuit is limited

through series resistor even if the inductor is allowed to go saturation by increasing the

pulse width. The current and voltage change is exponential. At 5T the inductor losses the

property of opposing the changes in current and goes to saturation. When the switch is

turned OFF the inductor reverses its polarity and demagnetizes through diode.

Voltage across inductor is equal to diode voltage plus resistor drop.

Page 23: Industrial Electronics

CIRCUIT DIAGRAM

Page 24: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 135µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch and inductor.

Take current probe and see the wave form (current waveform) through all the

components.

Find the demagnetization time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 122µs.

The nature of current waveform is Exponential.

Page 25: Industrial Electronics

MAGNETIZATION AND DEMAGNETISATION OF RL

LOADS USING DIODE AND RESISTOR

OBJECTIVES:

To use a diode and resistor for demagnetizing the RL loads.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core 540µH,0.5A 1

2 Resistor 2w CFR

1/2w

20 ohm,

200ohm

2

1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement 0-30v,2A

Page 26: Industrial Electronics

systems

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

CIRCUIT THEORY

In this circuit the inductor in series with the resister is demagnetized through

diode and a series resistor. At 5T the inductor goes to saturation, but due to presence of a

series resistor the current is limited to the finite value.

When the inductor produces the back EMF then the diode gets forward

biased and the inductor will be demagnetized. When the inductor gets demagnetized then

the voltage across the inductor will be diode voltage plus the drop across the resistor (r1)

and resistor (r2). Thus by connecting resistor in series with the diode will reduce the

demagnetization time. The maximum value of the series resistor that can be connected

depends on the breakdown voltage of the switch.

CALCULATIONS

T=L/R

=540µH / 20ohm

=27µs

So, 5T=135µs

Demagnetizing voltage across inductor is

VL=Vd+Vr1+Vr2

= 1 + 10 + 10

= 21v

Page 27: Industrial Electronics

CIRCUIT DIAGRAM

Page 28: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 135µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch.

Take current probe and see the current waveform through all the components.

Find the demagnetization time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 64µs.

Page 29: Industrial Electronics

DEMAGNETISATION OF RL LOADS

USING DIODE AND ZENER DIODE

OBJECTIVES

To use a diode and zener diode for demagnetizing the RL loads.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core 540µH,0.5A 1

2 Resistor 2W,CFR 20 ohm,

200ohm

1

1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

5 Zener diode 15v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement 0-30v,2A

Page 30: Industrial Electronics

systems

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

CIRCUIT THEORY

In this circuit the inductor in series with the resister is demagnetized by using diode

and zener diode. At 5T the inductor goes to saturation, but due to presence of a series

resistor the current is limited to the finite value i.e. 1A.

When the inductor gets demagnetized then the voltage across the inductor will be

diode voltage plus the drop across the resistor (r1) and break down voltage (Vz) of the

zener. Thus by connecting zener diode in series with the diode will reduce the

demagnetization time.

CALCULATIONS

T=L/R

=540µH/20ohm

=27µs

So, 5T=135µs

Demagnetizing voltage across inductor is

VL=Vd+Vr1+Vz

= 1 + 10 + 15

= 26v

Page 31: Industrial Electronics

CIRCUIT DIAGRAM

Page 32: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 135µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch.

Take current probe and see the current waveform through all the components.

Find the demagnetization time from the waveform.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 51µs.

Page 33: Industrial Electronics

MAGNETIZATION AND DEMAGNETIZATION

OF INDUCTOR USING DIODE

OBJECTIVES

To use a diode for demagnetizing the inductor.

To study the property of current & voltage, through & across the inductor.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core 540µH,0.5A 1

2 Resistor ½ W,CFR 200ohm 1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 Digital Storage Oscilloscope Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement 0-30v,2A

Page 34: Industrial Electronics

systems

4 Current probe and amplifier Tektronics 0-20A DC/AC,

60Mhz

CIRCUIT THEORY

In this circuit the inductor is magnetized by turning on the switch for a short

period of time. Here the inductor is not allowed to go to saturation since there is no

resistance to limit the current. The only resistance in the circuit is coil resistance which is

less than 1ohm.If the magnetizing time is increased the current shoots up to a very high

value and burns the inductor.

When the inductor produces the back EMF then the diode gets forward

biased and the inductor will be demagnetized. This diode suppresses the huge Back EMF

and protects the switch from damaging. In this the demagnetizing time of the inductor

will be more. When the inductor gets demagnetized then the voltage across the inductor

will be 0.7V which makes the demagnetizing time more. The magnetizing and

demagnetizing current is linear.

Since the pulse width is less the voltage across the inductor will be constant.

CALCULATION

E= L I / Ton

10v = 540µh X 0.5A / Ton

Ton=27µs

Page 35: Industrial Electronics

CIRCUIT DIAGRAM

Page 36: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 27µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch.

Take current probe and see the current waveform through all the components.

Find the demagnetization time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 270µs.

The nature of current waveform is linear.

Page 37: Industrial Electronics

MAGNETIZATION AND DEMAGNETIZATION

OF INDUCTOR USING DIODE AND RESISTOR

OBJECTIVES

To reduce demagnetization time by using resistor in series with diode.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core, 540µH,0.5A 1

2 Resistor ½ w,CFR

1 w

200ohm

20ohm

1

1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

1 Digital Storage Oscilloscope Tektronics 60Mhz

Page 38: Industrial Electronics

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement

systems

0-30v,2A

4 Current probe and amplifier Tektronics 0-20A DC/AC,

60Mhz

CIRCUIT THEORY

In this circuit the inductor is magnetized by turning on the switch for a short

period of time. Here the inductor is not allowed to go to saturation since there is no

resistance to limit the current.

When the inductor produces the back EMF then the diode gets forward

biased and the inductor will be demagnetized. This diode suppresses the huge Back EMF

and protects the switch from damaging. In this the demagnetizing time of the inductor is

reduced due to the addition of series resistor. When the inductor gets demagnetized then

the voltage across the inductor will be equal to diode drop plus the drop across the

resistor. The magnetizing current is linear and the demagnetizing current is exponential

decay due to the addition of resistor.

CALCULATION

E= L I / Ton

10v = 540µh X 0.5A / Ton

Ton=27µs

Page 39: Industrial Electronics

CIRCUIT DIAGRAM

Page 40: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 27µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch.

Take current probe and see the current waveform through all the components.

Find the demagnetization time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 20µs.

Page 41: Industrial Electronics

DEMAGNATISATION OF INDUCTOR

USING TRANSIENT VOLTAGE SUPRESSOR AND DIODE

OBJECTIVES

To reduce demagnetization time by using transient voltage suppressor.

BILL OF MATERIALS

Sl

No:

Component Specification Value Quantity`

1 Inductor Ferrite core 540µH,0.5A 1

2 Resistor ½w 200ohm 1

3 Diode KHF812 2A 1

4 Transistor BD139 1.5A,60v 1

5 Zener diode 15v 1

EQUIPMENTS REQUIRED

Sl

No:

Equipment Model Specification

Page 42: Industrial Electronics

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement

systems

0-30v,2A

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

CIRCUIT THEORY

Here demagnetization time is reduced using zener diode. This is one of the most

efficient method of demagnetizing the inductor in a very less time. This suppresses the

transients across the inductor when it is getting demagnetized and thus protects the other

components in the circuit.

Here the inductor voltage will be equal to diode voltage plus the breakdown voltage of

the zener. The magnetizing and demagnetizing current is linear and voltage across the

inductor is constant.

Page 43: Industrial Electronics

CIRCUIT DIAGRAM

Page 44: Industrial Electronics

PROCEDURE

Set the circuit as per circuit diagram.

Set the pulse width to 27µs with a frequency of 200 Hz.

Apply the signal to base of the switch.

Observe the wave form across the switch.

Take current probe and see the current waveform through all the components.

Find the demagnetization time.

NOTE:

TYPICAL RESULT FOR THE CIRCUIT USED

The demagnetization time is 18µs.

Page 45: Industrial Electronics

FAMILIARISATION OF

PULSE-WIDTH MODULATOR IC SG3524

OBJECTIVES:

To familiarize with pulse width modulator IC SG3524.

BILL OF MATERIALS

SLNO Components Specification Value Quantity

1 Capacitors Ceramic disc 0.01µF,.1µF 1,1

2 PWM IC SG3524 40v(max) 1

3 Resistors Carbon film 100Ω 2

4 Trim pot 1K 1

10K 2

EQUIMENTS REQUIRED

Slno: Equipment Model Specification

01 DC supply Measurement

systems

0-30V,2A

Page 46: Industrial Electronics

02 CRO Scientific 30MHz

CIRCUIT THEORY

The IC 3524 is a PWM generator which can vary the output pulse width according to

the output of an error amplifier. It operates at a fixed frequency that is programmed by

the resistor Rt and one timing capacitor Ct . Rt establishes a constant current for charging

of Ct. we can vary the frequency by varying any of these two components.

The output of the 3524 is given by two transistors which can be connected in any

fashion.

Frequency = 1.3 / Rt * Ct

Rt is in Kohms

Ct is in uF

And the frequency will be in Khz

Rt = 600ohm

Ct = 0.1uF

F = 1.30 / 520 * 0.1 uF

= 25 Khz

The 3524 is having the feature of current limiting in the output through the pins current

sense and current limit.

Page 47: Industrial Electronics

This IC is having a pin known shut down pin in which if we give a voltage more than

0.8V the IC will shut down by stopping it’s output to zero. This can be used to give a

thermal or short circuit protection.

CIRCUIT DIAGRAM

Page 48: Industrial Electronics

PROCEDURE

Set the circuit as per the circuit diagram.

Provide required input as Vcc.

Observe the output wave form.

Vary the potential at the pin 2 and observe the variation in the pulse width.

Draw the waveforms.

Page 49: Industrial Electronics

INDUCTOR DESIGN

OBJECTIVE:

To design an inductor of L= 1mH and current of 1A.

To compare the practical value & theoretical inductance value.

BILL OF MATERIALS

Slno: Components Specification Quantity

01 Ferrite Core RM-8 1

02 Copper wire gauge SWG-24

Page 50: Industrial Electronics

DESIGN:-

L = 1mh, Irms = 1A, Ip = 1.1A.

Winding factor ( Kw ) = 0.6

Flux Density ( Bmax ) = 0.25

Current Density ( J ) = 4 X 106

Area Product

AcAw = ( L X Ip X Irms ) / (Kw X Bm X J)

= 1mh X 1.1A X 1A X 108

0.6 X 4 X 106 X 0.25

Ap = 0.175cm4

From the core table, selected core is RM - 8

Ac = 0.630

Aw = 0.320

Page 51: Industrial Electronics

AP = 0.195

Number of turns

N = ( L X Ip ) / ( Bm X Ac )

= 1mH X 1.1 X104

0.25 X 0.630

=70

Length of air gap

lg = 4 X pi X 10 -7 X 70 X 1.1

0.25

= 0.387 mm

Wire guage

Area of the wire = Current / Current density

a = Irms / J

= 1 / 4 X 106

= 0.25 mm2

From the wire guage table 24 SWG is selected.

Page 52: Industrial Electronics

DESIGN STEPS

Compute the Area product from the given input

Select the core from the core table with the required Ac and Aw.

For the selected core, find Ac and Aw.

Compute the number of turns.

Calculate area of the wire required.

Select the wire guage from the wire table.

Finally calculate the required air gap.

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DESCRIPTION:

Inductor designing is a very important task in power electronics

while designing the inductor the value ,max current are the important factor.

Inductor is having the characteristics that it opposes the changes in the current through it.

This opposing characteristic is given by the name reactance. The reactance increases with

the frequency, it’s given by the formula

Reactance = 2*pi* frequency * L

Thus the reactance increases also with the increase in the inductor. The

core of the inductor is also selected by the frequency of operation. When the operating

frequency increases then the size of the inductor comes down. The SWG of the winding

wire is selected by the current carrying capacity of the inductor. As the number of the

turns increases then the inductor value also increases. So according to the inductor value

the number of turns will be designed.

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PROCEDURE:-

Get the specification ,and design the inductor using

given the formula.

Collect the material, and start winding the inductor.

After winding measure the value.

Adjust the value of inductance by reducing or increasing

or decreasing the air gap.

RESULT

The Practical value of the inductor is 1.02mH.

Page 55: Industrial Electronics

BUCK CONVERTER

OBJECTIVES:

To study about the basic DC-DC converter.

To construct a buck regulator for 5V & 1A.

BILL OF MATERIALS

SLNO COMPONENTS SPECIFICATION VALUE Quantity

1. Diode BY 229 10A 1

2. Inductor 1mh 1A rms 1

3. Transistor BD140 60v 1

4. Capacitors Electrolytic 100µF 1

Electrolytic 220µF 1

Ceramic disc 0.01µF,.1µF 1,1

5. PWM IC SG3524 40v(max) 1

6. Resistors Carbon film 100Ω 2

Carbon film 5Ω/5W 1

7. Trim pot 1K 1

10K 2

EQUIPMENTS REQUIRED

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Sl

No:

Equipment Model Specification

1 DSO Tektronics 60Mhz

2 Pulse Generator Scientific 20MHz

3 Power Supply Measurement

systems

0-30v,2A

4 Current probe and amplifier Tektronics 0-20A DC/AC,

50Mhz

SPECIFICATIONS:

Input voltage = 10v

Output voltage = 5v

Output current = 1A

Duty cycle = 50%

Ripple voltage = 0.1% of the output voltage

Ripple current = 10 % of the output current

Frequency = 25Khz

CIRCUIT THEORY

Buck converter is a DC to DC converter which can

step down the input voltage given and thus regulates it for the required voltage. It consists an

inductor ,capacitor and a switch as main components. The inductor and the capacitor acts like the

filtering components which stores and supplies the energy to the load. The switch is p-channel

MOSFET.

When the switch is on the inductor and the capacitor gets charged from the supply

and also the load is connected to the input supply. During this time the diode will be reverse biased

and does not conduct any current. When the switch is off then the inductor reverses it’s polarity

and starts discharging this turns the diode to forward bias. When switch is off then the current and

Page 57: Industrial Electronics

the voltage required for the load will be supplied by the inductor. So when the switch is on the

inductor is getting charged and when it is off the inductor gets discharged. The charging current

and the discharging current of the inductor is the same ,this prevents the inductor going to

saturation. Here the output capacitor is not having so much of importance because no time the load

is driven only by the capacitor.

The main disadvantage of the buck converter is not having isolation between the input

and the output. The buck converter is only used for low power applications.

The output voltage of the buck converter is given by the duty cycle. This is given by the

equation

Vout = Vin * dutycycle

In the given circuit the buck converter is connected with the feedback . For this a

PWM IC3524 is used and the feedback is given through a divider network. As the load varies the

feedback voltage changes and this adjusts the duty cycle and the output voltage is always constant

for all the loads. The supply voltage of the 3524 is the input voltage of the buck converter itself.

The inductor and the capacitor of the buck converter is designed according to the specifications

given. The core of the inductor is selected according to the frequency of operation, and the SWG

of the winding wire is selected by the current carrying capacity of the inductor.

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CIRCUIT DIAGRAM

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FORMULA’S:

L = [ ( Vin -Vo)Vo *T ] / Vi ∆I

Ton = (∆I *L) / (Vin-Vo)

Toff = (∆I *L) / Vo

∆Vc = ∆I /8 Fc

C=((Vin-Vo)Vo) / (8Vin*F^2*L*C)

DESIGN:

L=1mH, Vin=10V , Vo= 5V, ∆I=0.1A , Irms =1A , ∆Vc =0.1V

T = (Vin*∆I*L) / ( (Vin-Vo)*Vo)

= (10*0.1*1mH) / (5*5)

= 40 μ sec

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Ton = (∆I *L) / (Vin-Vo)

= (0.1 * 1mH) / 5

=20 μ sec

Toff = (∆I * L) / Vo

= 20 μ sec

∆Vc = ∆I / 8*F*C

0.1 = 0.1 / (8 * 25Khz *C)

C = 100uF

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PROCEDURE:

Set the circuit as per the circuit diagram.

Provide required input.

Check the pulse width and amplitude of the pulse.

Power all the circuits.

Observe the output and measure with DMM.

Observe the waveform on the DSO and plot the waveform.

Calculate line regulation and load regulation.

RESULT

Output voltage = 5.02V

Output current = 1A

Line Regulation = 1%

Load Regulation = 0.5%

Page 62: Industrial Electronics

FAMILIARISATION OF IC TCA 785

OBJECTIVE

To understand the operation IC TCA 785 for driving SCR’s and

TRAIC’s.

To observe the different waveforms of voltage at various outputs

of TCA 785.

BILL OF MATERIALS

Slno: Components Specification Value Quantity

01. TCA 785 Vs 8-18V 1

02. Resistor Carbon film 1/4W,10K 4

03. Capacitor Ceramic 1nF,47nF 1,1

04. Variable resistor 100K,10K 1,1

EQUIPMENTS REQUIRED

Sl no: Equipment Model Specification

01. Dc supply Measurement systems 0-30v

02. DSO Tektronix 200Ms/S

Page 63: Industrial Electronics

DESCRIPTION

The synchronization signal is obtained via a high-ohmic resistance from the line

voltage. A zero voltage detector evaluates the zero passages and transfers them to the

synchronization register.

This synchronization register controls a ramp generator, the capacitor C10 of

which is charged by a constant current (determined by R9).If the ramp voltage V 10

exceeds the control voltage V11, a signal is processed to the logic. Dependent on the

magnitude of the control voltage V11, the triggering angle can be shifted within a phase

angle of 0° to 180°.

For every half wave, a positive pulse of approx. 30uS duration appears at the outputs Q1

and Q2.The pulse duration can be prolonged up to 180 ° via a capacitor C12.If pin 12 is

connected to ground, pulses with a duration between ǿ and 180° will result.

Outputs Q1 and Q2 supply the inverse signals of Q1 and Q2.

A signal which corresponds to the NOR link of Q1 and Q2 is available at output QZ (pin

7).

The inhibit input can be used to disable outputs Q1 ,Q2 and q1,q2.Pin 13 can be used to

extent the outputs q1 and q2 to full pulse length(180° -ǿ).

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CIRCUIT DIAGRAM

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PROCEDURE:

Set the circuit as per the circuit diagram.

Provide required input as Vcc and control voltage.

Observe the output wave form.

Adjust the pot to get the desired output.

RESULT:

Obtained the trigger pulses for triggering the triac circuitry, triggering angle

varying from 0º to 180º with a pulse width of 30µsec.