ee049 electronic components exercises pr inst

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SRI LANKA INSTITUTE of ADVANCED TECHNOLOGICAL EDUCATION

Training Unit

Electronic Components -

Exercises

Practice

No: EE 049

INDUSTRIETECHNIKINDUSTRIETECHNIK

ELECTRICAL and ELECTRONIC

ENGINEERING

Instructor Manual



1

Training Unit

Electronic Components

Practical Part

No.: EE 049

Edition: 2008 Al l Rights Reserved

Editor: MCE Industrietechnik Linz GmbH & CoEducation and Training Systems, DM-1Lunzerst rasse 64 P.O.Box 36, A 4031 Linz / Aus triaTel. (+ 43 / 732) 6987 – 3475Fax (+ 43 / 732) 6980 – 4271Website: www.mcelinz.com



2

ELECTRONIC COMPONENTS - EXERCISE

CONTENTS Page

1

LE 21105-01 - Fundamentals of Electronic Components - Resistors...........................3

2 LE 21106-01 - Fundamentals of Electronic Components - Diodes ............................ 21

3 LE 21107-01 - Fundamentals of Electronic Components - Capacitors ...................... 37

4

LE 21108-01 - Fundamentals of Electronic Components - Transistor Characteristics

62

5 LE 21109-01 - Fundamentals of Electronic Components - Setting the Transistor

Operating Point..................................................................................................................89

6 LE 21110-01 - Fundamentals of Electronic Components - Transistors as Amplifiers

107

7 LE 21111-01 - Fundamentals of Electronic Components - Transistors as Switches133

8 LE 21112-01 - Fundamentals of Electronic Components - Constant Current Sources

148

9 LE 21113-01 - Fundamentals of Electronic Components - Constant Voltage Sources

169

10 LE 21118-01 - Fundamentals of Electronic Components - High, Low, and Band-

pass Filters ......................................................................................................................187

11 LE 21119-01 - Basic Electronics Circuits - Rectification ......................................226



3

Learning Unit: LE 21105-01

1 LE 21105-01 - Fundamentals of Electronic Components - Resistors

Summary of contents

Contents: Resistors

Skills to be acquired: Use of the resistor as a variable resistor, fixed

resistor, PTC thermistor, photoresistor

Projects: Voltage divider circuits with resistors

Exercise 1 Experimental set-up with modules from

and in the electronics trainer

Exercise 2 Experimental set-up in the electronics

experimenter

Exercise 3 PCB configuration

Directions for the use

This Learning Unit contains solution sheets TE 21105.

The electronics trainer, electronics experimenter and accessories contained in the lists of

equipment under Designation and Item No. can be found in the EA catalogue (Electrical

Engineering Training Equipment).



4

Resistors are used

to convert from current to voltage and vice versa, as well as to limit current and to divide

voltage. The energy produced is converted into heat.

Resistors occur in practically all electronic circuits.

Trimmers are a type of variable resistor. They are used as voltage dividers.

Measuring circuit:

Trimmers have the following design:

There is a circular resistance track made of hard carbon that runs about an energizing

axis. A wiper can be moved along the resistance track by using a screwdriver. Thedesired resistance value subdivision is obtained between the initial soldering tab, the

wiper pick-up, and the final soldering tab. The carrier plate shows the ohmic rating of the

entire resistance track.

Structure:

MCEIndustrietechnik

LE 21105-01 Resistors 1



5

Fixed resistors

are resistors with fixed resistance values that cannot change. They can be used

universally as series resistors, load resistors, or, in series circuits for limiting current and

at the same time as voltage dividers.

Measuring circuit:

Due to considerations of economy, fixed resistors are manufactured only in the standard

values recognized internationally, such as 10, 12, 15, etc. as a continuing gradation.

The tolerance is frequently 5 % or 2 %. The resistors are identified by means of coloured

rings.

Four-ring encoding for fixed resistors. Code of four

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Five-ring encoding of fixed resistors: code of five

Fixed resistors are, as a rule, constructed from a ceramic body with a coating of carbon or

metal oxide.

Structure:

Thermistors

are good conductors only when cold. Except for a slight initial drop, their resistance

increases with temperature. Because they have a positive temperature coefficient, they

are also called PTC resistors (Positive Temperature Coefficient).

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Characteristic curve:

Thermistors are used to monitor temperatures. The material used for them is almostexclusively barium titanate. They are frequently drop-shaped.

Structure:

Photoresistors

have a decreasing resistance as the intensity of light increases. They conduct best when

illuminated. They are also called LDRs (Light Dependent Resistor).

Characteristic curve:

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Photoresistors are used to measure light. They are manufactured from a semiconductor

material that is applied in a meander-shape on a ceramic disc. The ceramic disc is

accommodated in a small plastic housing.

Structure:

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Circuit diagram

Structure of module

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LE 21105-01ResistorsExercise 1 6



10

ItemNo. Qty. Designation Detail No. Remarks

1 1 Trimmer, lin. R1 2.2k 0.2 W

2 1 Fixed resistor R2 1.5 k 1 W

3 1 Fixed resistor R3 3.9 k 1 W4 1 Fixed resistor R4 680 1 W

5 1 Thermistor R5 140 = 40°C

6 1 Fixed resistor R6 820 1 W

7 1 Photoresistor R7 LDR 03

8 1 Electronics trainer 2GA5101-3A

The components in

Item Nos. 1 - 7 are

contained in the

accessories for

"Fundamentals of

Electronic 2GA5101-8F

Components"

List comprises sheet(s) 1 Sheet No. 1

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Circuit diagram

Structure of Experimenter

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8 1 Bare wire 0.5mm,silver-plated

9 1 Electronics experimenter 2GA5101-2A

The components in

Item Nos. 1 - 8 are

contained in the

accessories kit. 2GA5101-8A


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Circuit diagram

Structure of circuit board

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9 1 Printed circuit board E100 (half)

10 1 Pin strip 31-pole



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Skills to be acquired

Use of the resistor as a variable resistor, fixed resistor, PTC thermistor and photoresistor

Job plan (electronics trainer)

1. Arrange all the required components on the assembly board in accordance with the

diagram.

2. Complete the wiring in accordance with the circuit diagrams.

3. Perform the measurements in accordance with the measurement circuits and

measurement sheets 1 and 2 and enter the missing values.

Job plan (electronics experimenter)


diagram. The connection wires of the components are not to be shortened.

Component connections which are too short or too wide are prepared by soldering on

pieces of bare wire.

2. Complete the wiring in accordance with the circuit diagram.


measurement sheets 1 and 2 and enter the missing values.

Job plan (printed circuit board arrangement)

1. Arrange all the required components on the printed circuit board in accordance with

the diagram.

2. Complete the wiring in accordance with the circuit diagrams.

3. Plug the completed PCB into the 31-pole connector of the electronics experimenter.


measurement sheets 1 and 2 enter the missing values.

Aids, tools and equipment

PCB holder, 1.0 mm tin-lead solder, 30 W soldering iron, flat-nosed pliers, soldering

tongs, conductor disconnector, side clippers, 1.3 mm twist drill, holder for 1.3 mm twist

drill, desoldering device, steel rule

Accident prevention measuresBe careful when handling the hot soldering iron (risk of burns and danger of fire).

Be careful when cutting components to length; pieces of wire flying around (risk of eye

injuries).

MCEIndustrietechnik

LE 21105-01ResistorsExercise 1 - 3 12



16

Functional description

A variable voltage can be picked up at the wiper on the variable resistor R1.

A voltage divider consists of fixed resistors R2 and R3 arranged in such a way that two

constant voltage drops are obtained.

R4 and R5 also form a voltage divider. When the thermistor R5 warms up, UR5 rises, and

UR4 drops.

The more strongly photoresistor R7 is illuminated, the better a conductor it is. U R7 drops,

UR6 rises.

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Trade: ____________________________Name:________________________________

Measurement sheet 1

1. Set the operating voltage to 5 V. The exact measurement obtained is

UB =

2. The voltage level at the wiper on R1 to ground can be changed

from Umin = to Umax =

3. At voltage divider R2/R3, the voltage values are proportional to the resistances

R2 = R3 =

UR2 = UR3 =

4. The voltage divider consisting of resistor R4 and a thermistor R5 depends on

temperature.

At room temperature,

UR4 = and UR5 =

When the temperature is raised using a soldering iron, the voltage drops change to

UR4 = and UR5 =

5. In daylight, the partial voltages are

UR6 = and UR7 =

When the photo resistor is shaded,

UR7 drops and UR7 rises

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Trade: ____________________________Name:________________________________

Measurement sheet 2

1. A current IR1 of several mA can be calculated from the operating voltage and the

resistance of R1.

The corresponding measurement is

IR1 =

2. The current flowing across R2 and R3 is

IR2 = IR3 =

3. Measure IR4/R5

When the thermistor is cold, the current flowing is

I =

When the thermistor is heated, the current flowing is

I =

4. Measure IR6/R7

When the photo resistor is illuminated, the current flowing is

I =When the photo resistor is shaded, the current flowing is

I =

5. When all circuits are connected simultaneously to voltage, the total current resulting

(with fluctuations due to the effects of temperature and light) is

I =

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Measurement sheet 1

1. Set the operating voltage to 5 V. The exact measurement obtained is

UB = 5 V

2. The voltage level at the wiper on R1 to ground can be changed

from Umin = 0 V to Umax = 5 V

3. At voltage divider R2/R3, the voltage values are proportional to the resistances

R2 = 1.5 k R3 = 3.9 k

UR2 = 1.4 V UR3 = 3.6 V

4. The voltage divider consisting of resistor R4 and a thermistor R5 depends on

temperature.

At room temperature,

UR4 = 4.3 V and UR5 = 0.7 V

When the temperature is raised using a soldering iron, the voltage drops change to

UR4 = 0.4 V and UR5 = 4.6 V

5. In daylight, the partial voltages are

UR6 = 3.1 V and UR7 = 1.9 V

When the photo resistor is shaded,

UR7 drops and UR7 rises

14aMCEIndustrietechnik

LE 21105-01ResistorsSolution Exercise 1 - 3 16

X



20

Measurement sheet 2

1. A current IR1 of several mA can be calculated from the operating voltage and the

resistance of R1.

The corresponding measurement is

IR1 = 2.2 mA

2. The current flowing across R2 and R3 is

IR2 = IR3 = 0.9 mA

3. Measure IR4/R5

When the thermistor is cold, the current flowing is

I = 6.3 mA

When the thermistor is heated, the current flowing is

I = 0.5 mA

4. Measure IR6/R7

When the photo resistor is illuminated, the current flowing is

I = 3.8 mA

When the photo resistor is shaded, the current flowing is

I = 0.8 mA

5. When all circuits are connected simultaneously to voltage, the total current resulting

(with fluctuations due to the effects of temperature and light) is

I = 13.2 mA


LE 21105-01ResistorsSolution Exercise 1 - 3 17



21


2 LE 21106-01 - Fundamentals of Electronic Components - Diodes

Summary of contents

Contents: Diodes

Skills to be acquired: Measurements on a germanium diode, silicon diode,

LED and Zener diode

Drawing of characteristics

Projects: Recording of diode characteristics




experimenter









22

The use of diodes

is just as necessary in power electronics as in electronic controls.

In power electronics, rectifier diodes are used to convert AC current into DC current. DC

current is used, for example, to drive DC machines. In electronic controls, diodes are used

to link control signals together logically. The resulting signals control additional electronic

components.

A characteristic of diodes is their capability to conduct a current in one direction only, and

to block it in the opposite direction. Characteristic curves show how the diodes behave in

the circuit.

Measuring circuit for recording characteristic curves:

MCEIndustrietechnik

LE 21106-01 Diodes 1



23

Diodes (rectifier diodes)

Diodes are made of germanium or silicon. They are semiconductors,

i.e. they do not conduct as well as metals. Directed impurity (doping) creates a P-layer

and an N-layer in the material.

In the P-layer, there is a lack of electrons (holes).

In the N-layer, there is an excess of electrons (-).

Structure:

When a voltage is applied, it can drive a current only from the anode to the cathode. The

PN junction blocks it in the reverse direction.

LEDs

The PN-junction of LEDs emits light. A special semiconductor material is used for this

purpose. LEDs are used as inertialess Signal transmitters. They require very little power.

A series resistor must limit the current. LEDs are operated in the forward direction. They

look like small, transparent beads with two wire connections.

Structure:

Zener diodes

Zener diodes are silicon diodes operated in the blocking direction. Even at a low voltage,

they become conductors in the blocking direction. This voltage is known as the breakdown

voltage or simply as the Zener voltage.

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Measurement circuit for recording the characteristic curve

After the breakdown, the Zener voltage remains constant even when there are fairly large

variations in current. For that reason, Zener diodes are used mainly for voltage

stabilization. A series resistor must limit the Zener current. Zener diodes are classified into

standard series according to standard values. They are available in the gradation 2.4 V,

3.3 V, 3.9 V, 4.7 V and so on. The most commonly found power classes are 0.5 W or 1.5

W. Their structure does not differ from that of other diodes.

Structure:

Semiconductor designations:

First letter = Material

A = Germanium

B = Silicon

C = Semiconductor material for LEDs

Second letter:

A = Diode

B = Radiation-producing semiconductor material

C = Zener diode

The other letters and numerals identify the type in general. Or the designation

1N ... with two to four numerals identifies a diode.

1N = 1 Blocking layer (diode)

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Circuit diagram

Structure of module

MCEIndustrietechnik

LE 21106-01DiodesExercise 1 4



26


1 1 Fixed resistor R2 1 k 1 W


3 1 Fixed resistor R3 1 k 1 W4 1 Fixed resistor R4 1 k 1 W



7 1 Germanium diode V1 AA 118

8 1 Silicon diode V2 1N 4002

9 1 LED B1 CQV 10-5 RED

10 1 Zener diode B2 CQV 15-5 GR

11 1 Zener diode V3 BZX 83C5V6


13 1 Rheostat 100 10 W


The components in

Item Nos. 1 - 13 are

contained in the

accessories for

"Fundamentals of

Electronic 2GA5101-8F

Components."


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Circuit diagram


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10 1 LED B2 CQV 15-5 GR






The components in


contained in the



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Circuit diagram


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10 1 LED B2 CQV 15-5 GR



13 1 Bare wire 0.5mm, silver-plated

14 1 Printed circuit board E 100 (half)





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31


Measurements on a germanium diode, silicon diode, LED and Zener diode Drawing of

characteristics



diagram

2. Complete the wiring in accordance with the circuit diagrams

3. Use the circuit via the variable resistor (voltage divider)


measurement sheet 1 and enter the values in the table 5. Draw the curves on

measurement sheet 2 from the measured values






2. Complete the wiring in accordance with the circuit diagram

3. Use the circuit via the variable resistor (voltage divider)


measurement sheet 1 and enter the values in the table

5. Draw the curves on measurement sheet 2 from the measured values



the diagram

2. Complete the wiring in accordance with the circuit diagrams

3. Plug the completed PCB into the 31-pole connector of the electronics experimenter

4. Use the circuit via the variable resistor (voltage divider)5. Perform the measurements in accordance with the measurement circuits and

measurement sheet 1 and enter the values in the table

6. Draw the curves on measurement sheet 2 from the measured values

MCEIndustrietechnik

LE 21106-01DiodesExercise 1-3 10



32



tongs, conductor disconnector, side Clippers, 1.3 mm twist drill, holder for 1.3 mm twist

drill, desoldering device, steel rule.

Accident prevention measures

Be careful when handling the hot soldering iron (risk of burns and danger of fire).


injuries).


When the operating voltage is raised in a series circuit, the series resistor limits the diode

current.

The voltage drop at the germanium diode V1 rises sharply as the current increases. The

voltage drop at the silicon diode V2 remains almost constant.

The voltage drop at the LEDs B1 and B2 likewise remains almost constant.

The voltage drop at the Zener diodes B3 and B4 is in the range of their standard value.

Outside of the breakpoint zone, the voltage drop increases slightly.

MCEIndustrietechnik

LE 21106-01DiodesExercise 1-3 11



33

Trade: ____________________________Name:________________________________

Measurement sheet 1

Recording of characteristic curves:

In the series circuit made up of resistor and diode, adjust the currents desired indirectly

via the variable operating voltage.

In each instance, measure the voltage drop at the diode.

I 1 mA 2 mA 3 mA 5 mA 10 mA 20 mA

UV1

UV2

UB1

UB2

UV3

UV4

The value pairs produce measured points in the families of characteristic curves that

follow. For each diode, connect the measured points to form a characteristic curve. For

V1, V2, B1 and B2, the forward characteristics are obtained (measurement sheet 2, top).

For V3 and V4, the breakdown characteristics are obtained (measurement sheet 2,

bottom).

MCEIndustrietechnik

LE 21106-01DiodesExercise 1 - 3 12



35

Measurement sheet 1


In the series circuits made up of resistor and diode, adjust the currents desired indirectly

via the variable operating volage.

In each instance, measure the voltage drop at the diode.

I 1 mA 2 mA 3 mA 5 mA 10 mA 20 mA

UV1 0.37 V 0.5 V 0.6 V 0.77 V 1.14 V 1.76 V

UV2 0.56 V 0.6 V 0.62 V 0.64 V 0.68 V 0.71 V

UB1 1.51 V 1.54 V 1.55 V 1.57 V 1.6 V 1.62 V

UB2 1.81 V 1.84 V 1.85 V 1.88 V 1.94 V 2.02 V

UV3 5.4 V 5.5 V 5.55 V 5.6 V 5.7 V 5.75 V

UV4 8.0 V 8.05 V 8.05 V 8.1 V 8.2 V 8.25 V

The value pairs produce measured points in the families of characteristic curves that

follow. For each diode, connect the measured points to form a characteristic curve. For

V1, V2, B1 and B2, the forward characteristics are obtained (measurement sheet 2, top).

For V3 and V4, the breakdown characteristics are obtained (measurement sheet 2,

bottom).


LE 21106-01DiodesSolution Exercise 1 - 3 14



36

Measurement sheet 2

Germanium diode, silicon diode, LEDs

Zener diodes


LE 21106-01DiodesSolution 1 - 3 15



37


3 LE 21107-01 - Fundamentals of Electronic Components - Capacitors

Summary of contents

Contents: Capacitors

Skills to be acquired: Use of capacitors, storage of electrical energy

RC module

CR module

Dynamic inputProjects:

Projects: Capacitor circuits with DC and square-wave voltages




experimenter



This Learning Unit contains solution sheets TE 21107-01.






38

The use of capacitors

is absolutely necessary for the short-term storage of electrical energy.

If a DC voltage is applied to two metal plates that face one another with electrical

insulation between them, a difference develops in the electron population on the plates.

An electrical field is created in the space between the plates. The plates are isolated from

one another by means of the insulating material, referred to as the dielectric. The

capacitance of a capacitor depends on the size of the plates, the distance between the

plates and the dielectrics.

The plates facing one another can store a charge, i.e. a quantity of electricity determined

from current x time.

The more slowly the voltage drop at the plates rises on a capacitor during charging, thegreater its capacitance.

The farad is a very large unit. The following smaller units exist:

MCEIndustrietechnik

LE 21107-01 Capacitors 1



39

When a resistor is connected before or after a capacitor, an RC circuit or a CR circuit is

formed respectively.

The time required to charge or discharge the capacitor depends upon the effective

resistance R in the circuit and on the value of the capacitance C. The product R x C is an

indicator of the charging time. This product is called the time constant, = Tau:

A capacitor can be charged by 63 % and can be discharged by 37 % within the time

constant ( ). The charge or discharge is viewed as completed after 5 time constants.

Charging voltage level UC with a 10 V operating voltage with R = 50 k and C = 100 µF.

Because an uncharged capacitor does not offer any resistance to the current, the first

charging current peak is limited only by the resistor. The charging current peak produces

a voltage peak at the resistor. The increasing capacitor voltage opposes the operating

voltage and reduces the charging current. When the capacitor has been charged, the

capacitor voltage is exactly the same as the operating voltage.

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When charged, the capacitor itself has an almost infinitely high resistance. It blocks the

DC current. A discharge of the capacitor in the opposite direction produces a discharging

current peak at the resistor in the direction opposite to that during charging.

An RC circuit is frequently called an integrator circuit or a low pass. A CR circuit is called a

differentiation circuit or a high pass. Such circuits take a rectangular voltage and convert it

into a different voltage pattern.

With an RC circuit, the output voltage is tapped at the capacitor.

With a CR circuit, the output voltage is tapped at the resistor.

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Positive and negative needle pulses can bee tapped at the resistor of the CR circuit.

Either all positive or all negative needle pulses can be blocked for the circuit that follows,

or possibly for a load resistor, by using a diode. The CR circuit with an additional diode is

used in a logic circuit as a so-called dynamic input.

The dynamic input allows a quick change in voltage with a given switching edge to pass

through to the load resistor as a needle pulse.

Dynamic input for the positive switching edge

Dynamic input for the negative switching edge

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Capacitor types

Electrolytic capacitors

Electrolytic capacitors have a thin coat of oxides as a dielectric. This coat is formed by

applying the positive pole of a DC voltage to a metal foil within a fluid, the electrolyte. In

order to maintain the oxide coating, electrolytic capacitors may be operated only at the DC

voltage which is printed an them.

The most commonly used type is the aluminium electrolytic capacitor in sizes from 1 to

100,000 µF.

Metal-plastic capacitors

In metal-plastic capacitors, very thin coatings of metal are vaporized onto plastic foils. The

foils are rolled into round or flat coils. The different plastics used as the dielectric affect the

electrical properties of the capacitors. Typical capacitances are from 1 to 100,00 µF.

Ceramic capacitors

The development of ceramic masses with a very high insulating resistance has made

possible the simple and compact construction of capacitors with a ceramic dielectric. The

ceramic mass is simultaneously the basic body and the housing. Such capacitors are

made in versions from 1 pF to 1 µF.

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Circuit diagram

Structure of module

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LE 21107-01CapacitorsExercise 1 6



44


1 1 Fixed resistor R1 47 k


3 1 Fixed resistor R3 18 k 4 1 Fixed resistor R4 120 k



7 1 Electrolytic C1 100 µF

capacitor

8 1 Metal-plastic C2 100 nF

capacitor

9 1 Ceramic capacitor C3 47 nF


11 1 Diode V1 1N4148

12 1 Diode V2 1N4148


The components in


contained in the

accessories for

"Fundamentals of

Electronic

Components" 2GA5101-8F


MCEIndustrietechnik




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Circuit diagram


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capacitor

8 1 Metal-plastic capacitor C2 100 nF



11 1 Diode V1 1N4148

12 1 Diode V2 1N4148


The components in


contained in the



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Circuit diagram


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capacitor

8 1 Metal-plastic C2 100 nF

capacitor



11 1 Diode V1 1N4148

12 1 Diode V2 1N4148

13 1 Bare wire 0.5 mm, plated



16 1 Electronics 2GA5101-2A

experimenter

17 1 Function generator 5 V square-wave


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The use of capacitors in the storing of electric energy in an RC module, a CR module and

with a dynamic input



diagram

2. Complete the circuit in accordance with the circuit diagrams

3. Perform the measurements in accordance with the measurement sheets and enter the

missing curves





pieces of bare wire

2. Complete the circuit in accordance with the circuit diagram


missing curves



the diagram




missing curves


PCB holder, 1.0 mm tin-lead solder, 30 W soldering iron, flat-nosed pliers, solderingtongs, conductor interrupter, side Clippers, 1.3 mm twist drill, holder for 1.3 mm twist drill,

desoldering device, steel rule

MCEIndustrietechnik

LE 21107-01CapacitorsExercises 1-3 12



50

Measuring and testing equipment

1 oscilloscope

Accident prevention measure

Be careful when handling the hot soldering iron (risk of burns and danger of fire). Be

careful when cutting components to length; pieces of wire flying around (risk of eye

injuries). Be careful when handling the capacitors; there is a danger of explosion if the

opering voltage is too high or if the poles are incorrectly connected (risk of eye injuries).

Functional description RC circuit R1; C1

If 5 V DC is applied to the circuit, the charging of C1 across R1 begins. Initially, the

capacitor acts as if it is short-circuited, i.e. it offers no resistance to the charging current.

The height of the charging current is limited only by R1. The first charging current is large

but it then drops off quickly. After a time of one , the capacitor chare has reached 3 V.

There are 2 V left at R1, since the charging voltage opposes the operating voltage.

Further charging occurs significantly more slowly because the difference in voltage

between the operating voltage and the charging voltage continually decreases. After 5 ,

the capacitor is fully charged to 5 V. There is no longer any voltage differential at R1, and

therefore no further charging current flows there.

After the operating voltage is unplugged, the discharge is initiated by bridging over the

entire RC circuit. This corresponds to a parallel connection of R1 and C1. At first, the

capacitor voltage drives a high current across R1. The discharge current decreases

sharply during the first time of 1 , with the capacitor voltage dropping by 3 V. At C1, only

2 V are left.

Further discharge takes place significantly more slowly and ends after 5 with 0 V at Cl.

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Description of Operation RC circuit R2; C2

The circuit is operated using a DC voltage that is switched on and off alternately for 5 ms

each. This is called a 100 Hz square-wave voltage. The voltage levels are to be + 5 V and

0 V.

With each positive switching edge, C2 begins charging across R2. The capacitor voltage

rises logarithmically, at first quickly and then more and more slowly. With each negative

switching edge, C2 starts discharging across R2. The capacitor voltage drops

logarithmically, at first quickly and then more and more slowly. The time constant 1

selected was small enough to allow the capacitor to be fully charged and fully discharged

100 times per second.

Voltage characteristic at the capacitor

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Dynamic input with C3; R3; V1 and load resistor R4.

The circuit is operated using a 100 Hz square-wave voltage. The voltage levels are to be

+5 V and 0 V.

With each positive switching edge, the capacitor charging current peak produces a

positive voltage peak at R3 and R4. At R4, it is smaller than at R3 by the threshold voltage

of the diode V1.

With each negative switching edge, the capacitor discharging current peak produces a

negative voltage peak at R3. A capacitor discharge current cannot flow across the load

resistor R4 because V1 is blocking. Compared to R3, R4 is high-ohmic in order to ensure

that the charging time for the capacitor would not be noticeably shorter than the

discharging time.

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Dynamic input with C4; R5; V2 and load resistor R6.

The circuit is operated using a 100 Hz square-wave voltage. The voltage levels are to be

+5 V and 0 V.

With every positive switching edge, the capacitor loading current peak causes a positive

voltage peak at R5. Load resistor R6 is not affected by this because diode V2 is blocking.

With every negative switching edge, the capacitor discharging current peak causes a

negative voltage peak at R5 and R6. At R6 it is smaller than at R5 by the threshold

voltage of V2. Compared to R5, R6 is low-ohmic so that the discharge time of the

capacitor is clearly shorter than the charging time. The negative voltage peak R6 thus

becomes a narrow needle pulse.

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Trade: ____________________________Name:________________________________

Measurement sheet 1

Apply 5 V to the series circuit made up of R1 and C1. The charging of the capacitor may

be observed using an oscilloscope connected in parallel to C1. Draw the oscillogram for

the fully-charged C1. Adjust the zero line (0) with automatic triggering.

The scale on the x-axis is to be 5 ms per cm.

The scale on the y-axis is to be 1 V per cm.

The discharging of C1 can be observed when the series circuit of R1 and C1 is short-

circuited. The time constant (Tau R x C) can be determined by means of repeated

charging and discharging. The time constant corresponds to the time required for charging

or discharging C1 by 63 %, or by approx. 3 V.

t =

Calculations for the time constant are:

= R x C =

After 5 time constants (5 x R x C), the capacitor voltage is approximately the same as the

operating voltage.

UB =

UC1 =

MCEIndustrietechnik

LE 21107-01CapacitorsExercise 1 - 3 17



55

Trade: ____________________________Name:________________________________

Measurement sheet 2

The RC circuit made up of R2 and C2 is operated using a 100 Hz square-wave voltage

with +5 V and 0 V. A single oscillogram showing both the Input voltage at the RC module

and the output voltage at C2 is to be recorded. Select the positive trigger edge ( ).

Adjust the zero line (0) with automatic triggering. With manual triggering, set the triggering

level to just above zero.



Determine the time it takes to charge C2 by 63 % from the oscillogram.

t =

Calculate the time mathematically from the time constant.

= R x C =

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56

Trade: ____________________________Name:________________________________

Measurement sheet 3

The dynamic input consisting of C3, R3 and V1 is operated using a 100 Hz square wave

voltage with +5 and 0 V.

A single oscillogram showing the input voltage of the RC module and the output voltage at

the load resistor R4 is to be recorded. In an additional measurement, the voltage loss at

R3 is to be determined.

Select the positive trigger edge ( ). Adjust the zero line (0) with automatic triggering.

With manual triggering, set the trigger level to barely above zero.



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57

Trade: ____________________________Name:________________________________

Measurement sheet 4

The dynamic input consisting of C4, R4 and V2 is operated using a 100 Hz square-wave

voltage with +5 V and 0 V.

A single oscillogram showing the input voltage at the RC module and the output voltage at







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58

Measurement sheet 1

Apply 5 V to the series circuit made up of R1 and C1. The charging of the capacitor may

be observed using an oscilloscope connected in parallel to C1. Draw the oscillogram for

the fully-charged C1. Adjust the zero line (0) with automatic triggering.



The discharging of C1 can be observed when the series circuit of R1 and C1 is short-

circuited. The time constant (Tau R x C) can be determined by means of repeated

charging and discharging. The time constant corresponds to the time required for charging

or discharging C1 by 63 %, or by approx. 3 V.

t = 6 s

Calculations for the time constant are:

= R x C = 56 k x 100 µF = 5.6 s

After 5 time constants (5 x R x C), the capacitor voltage is approximately the same as the

operating voltage.UB = 5 V

UC1 = 5 V


LE 21107-01CapacitorsSolution Exercise 1 - 3 21



59

Measurement sheet 2

The RC circuit made up of R2 and C2 is operated using a 100 Hz square-wave voltage

with +5 V and 0 V. A single oscillogram showing both the Input voltage at the RC module

and the output voltage at C2 is to be recorded. Select the positive trigger edge ( ).

Adjust the zero line (0) with automatic triggering. With manual triggering, set the triggering

level to just above zero.



Determine the time it takes to charge C2 by 63 % from the oscillogram.

t = 0.5 cm x 2 ms / cm = 1 ms

Calculate the time mathematically from the time constant.

= R x C = 10 k x 0,1 µF = 1ms





60

Measurement sheet 3

The dynamic input consisting of C3, R3 and V1 is operated using a 100 Hz square wave

voltage with +5 and 0 V.

A single oscillogram showing the input voltage of the RC module and the output voltage at











61

Measurement sheet 4

The dynamic input consisting of C4, R4 and V2 is operated using a 100 Hz square-wave

voltage with +5 V and 0 V.

A single oscillogram showing the input voltage at the RC module and the output voltage at











62


4 LE 21108-01 - Fundamentals of Electronic Components - Transistor

Characteristics

Summary of contents

Contents: Transistor characteristics

Skills to be acquired: Use of transistors

Determining the input characteristic

Determining the output characteristic

Determining the current control characteristic

Projects: Recording of transistor characteristics

Emitter connection with negative current feedback




experimenter









63

The use of transistors

makes possible contactless controls.

A transistor can be compared to a relay switch.

On-off switch open On-off switch open

Relay switch open Transistor blocks

Signal light does Signal light does

not come on not come on

On-off switch closed On-off switch closed

Relay switch closed Transistor is conducting

Signal light comes on Signal light comes on

MCEIndustrietechnik

LE 21108-01 Transistor characteristics 1



64

When a small base current flows through the base-emitter circuit, the transmitter becomes

conductive in the collector-emitter circuit and a large collector current flows.

The base current and collector current are added to form the emitter current

IB + IC = IE

The ratio between the base current and the collector current is referred to as the DC gain

factor B. In the last diagram, this current gain was 100 x.

B = IC In the example: 100 mA = 100 x

IB 1 mA

Approximately 0.7 V is required to achieve conductance in the base-emitter circuit. Under

favourable conditions, conductance can be achieved in the collector-emitter circuit with

only

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65

Operation of the transistor

Corresponding to the flow of electrons, there is a simultaneous flow of holes in the

opposite direction because each electron that travels upward (car) leaves behind it a hole

that apparently travels downward (parking space). The technical direction of flow is that of

the holes. The base current controls a collector current. The voltage at the base is positive

with respect to the emitter. Electrons are conducted from the emitter into the base.

Because the base offers only a few spaces for electrons and the electrons in the base are

attracted by the positive voltage at the collector, they break through the PN-junction in the

direction of the collector.

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Transistors are usually made out of silicon. A NPN transistor made up of two diodes is

conceivable.

Structure:

Symbol and structure:

The circle (housing) Seen from below

can be omitted

Designations for semiconductors

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67

For every circuit which is to be dimensioned, a suitable transistor must be selected. Data

books can be used to help in doing this. The transistor data are classified into

characteristic data and limit data. The characteristic data are typical operating data for

transistors. By combining many characteristic data, it is possible, for example, to

determine input characteristics, output characteristics and current control characteristics.

The limit data indicate absolute limit values. If these are exceeded, you must expect the

transistor to be destroyed. It is possible to work out transistor data yourself from

measurements and experiments.

The input characteristic corresponds to a diode characteristic. The base current and the

base emitter voltage are measured an a transistor.

The higher the base current is set, the greater the voltage drop at the base-emitter track.

Measurement circuit

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68

Input characteristic

The Input characteristic can be used to determine the base current with a given base-

emitter voltage.

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69

An output characteristic indicates the relationship between the collector current and the

collector-emitter voltage for a constant base current.

A family of output characteristics is obtained by grouping together several characteristic

curves. The constant base current is the parameter, i.e. the constant quantity. The

collector-emitter voltage and the collector current are measured.

When the collector-emitter voltage rises, the collector current at first increases sharply,

and then remains approximately at the same level once the current gain B of the transistor

has been fully utilized.

IC = IB x B

Measuring circuit

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70

Output characteristic

In the family of output characteristics, the collector current can be determined for a given

collector-emitter voltage.

If the collector current is divided by the base current (parameter), the current amplification

B can be found.

In the overdriving range, the base current (parameter) is higher than would be necessary

for the collector current flowing at the moment. The full current gain B is not yet being

utilized here.

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71

The current control characteristic identifies the transistor current amplification.

Here, the base current and the collector current are measured: the higher the base

current, the higher the collector current.

Measuring circuit

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72

Current control characteristic

The current control characteristic can be used to find the collector current for a given base

current.

If the collector current is divided by the base current, the current gain B is obtained.

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73

Circuit diagram

Structure of module

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74


1 1 Fixed resistor R1 1 kΩ 1 W


3 1 Fixed resistor R3 120 kΩ 1 W4 1 Fixed resistor R4 820 kΩ 1 W

5 1 Transistor V1 BC 140-16


7 1 Transistor V3 BC 237-A

8 1 LED 81 CQV 20-5 red

9 1 Rheostat 100Ω 10 W


The components in

Item Nos. 1 - 9 are

contained in the

accessories for

"Fundamentals of

Electronic



MCEIndustrietechnik

LE 21108-01Transistor characteristicsExercise 1 12



75

Circuit diagram

Set-up of experimenter

MCEIndustrietechnik




76




3 1 Fixed resistor R3 120 kΩ 1 W4 1 Fixed resistor R4 820 Ω 1 W




8 1 LED B1 CQV 20-5 red

9 1 Bare wire 0.5 mm silver-

plated

10 2 Rheostat 100 Ω 10 W


The components in


contained in the

accessories kit. . 2GA5101-8A


MCEIndustrietechnik




77

Circuit diagram

Circuit board arrangement

MCE Industrietechnik LE 21108-01Transistor characteristicsExercise 3 15



78




3 1 Fixed resistor R3 120 kΩ 1 W4 1 Fixed resistor R4 820 Ω 1 W



7 1 Transistor V3 BC 237-A

8 1 LED B1 CQV 20-5 red

9 1 Bare wire 0.5 mm, silver-

plated



12 2 Rheostat 100 Ω 10 W


experimenter


MCEIndustrietechnik




79

Skills to be acquired - Transistor measurements: determining the input, output and control

current characteristics.



diagram.

2. Complete the circuit in accordance with the circuit diagrams.

3. Use the circuit via variable resistors (voltage dividers).


measurement sheets and enter the missing values in the table.









measurement sheet and enter the values in the table.

5. Form the curves from the measured values.



the diagram.




5. Perform the measurements in accordance with the measurement circuit and

measurement sheets and enter the values in the table.

6. Form the curve from the measured values.

Aids, tools and equipment - PCB holder, 1.0 mm tin-lead solder, 30 W soldering iron, flat-

nosed pliers, soldering tongs, conductor interrupter, side clippers, 1.3 mm twist drill,holder for 1.3 mm twist drill, desoldering device, steel rule.

Accident prevention measures - Be careful when handling the hot soldering iron (risk of

burns and danger of fire). Be careful when cutting components to length; pieces of wire

flying around (risk of eye injuries).

MCEIndustrietechnik




80

Trade: ____________________________Name:________________________________


Determine the input characteristic at transistor V1.

As the base voltage across R1 rises, the base current of V1 increases. It is limited by the

base series resistor R1. Each value for the base current causes a given drop in the base-

emitter voltage. The collector-emitter circuit of V1 must remain open because the high

base current (due to the low-ohmic series resistor R1) would allow too great a collector

current to pass.

Determine the output characteristics at transistor V2.

The base current of V2 can be adjusted using the operating voltage across R2. It is

permissible to connect a second source of operating voltage directly to the collector-

emitter circuit because the small base current (due to the high-ohmic series resistor R2)

will also allow only a small collector current to pass.

An increase in collector-emitter voltage has a very strong effect an the collector current in

the overdriving range, but only a very slight effect after that. The maximum collector

current depends upon the base current that has been set.

Determine the current control characteristic at transistor V3.

With increasing operating voltage across R3, the base current of V3 rises. A second

source of operating voltage allows a medium-large collector current depends upon the

base current. In the collector-emitter circuit, the voltage drop decreases as the base

current increases.

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Trade: ____________________________Name:________________________________

Measurement sheet 1

Recording of characteristic curves

Adjust the Base currents desired indirectly in the series circuit made up of series resistor R1

and the base-emitter circuit of transistor V1 using the variable operating voltage. In each

case, measure the voltage drop in the base-emitter circuit.

IB 1 mA 2 mA 3 mA 5 mA 10 mA 20 mA

UBE V1

The value pairs UBE, IB produce measured points for a

Transistor input characteristic

MCEIndustrietechnik

LE 21108-01Transistor characteristicsExercise 1 - 3 19



82

Trade: ____________________________Name:________________________________

Measurement sheet 2


Adjust the desired base current indirectly in the series circuit made up of series resistor R2 and

the base-emitter circuit of transistor V2 using the variable operating voltage. Using a second

source of operating voltage, change the collector-emitter voltage for each base current value.

The collector current IC is to be measured in each instance.

UCE 0 V 0.25 V 0.5 V 1.0 V 2.5 V 5.0 V

IC at IB = 50 µA

IC at IB = 40 µA

IC at IB = 30 µA

IC at IB = 20 µA

The value pairs UCE, IC produce measured points for four output characteristics for plotting an

measurement sheet 3.

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Trade: ____________________________Name:________________________________

Measurement sheet 3

Transistor output characteristics

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84

Trade: ____________________________Name:________________________________

Measurement sheet 4


Set the desired base current values for V3 with a variable operating voltage. Connect the

collector circuit to a second operating voltage source of 24 V.

In each case, measure the collector current.

IB 10 µA 20 µA 30 µA 50 µA 100 µA

IC

The value pairs IB, IC produce measured points for a

Transistor current control characteristic.

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85

Measurement sheet 1


Adjust the Base currents desired indirectly in the series circuit made up of series resistor R1

and the base-emitter circuit of transistor V1 using the variable operating voltage. In each

case, measure the voltage drop in the base-emitter circuit.

IB 1 mA 2 mA 3 mA 5 mA 10 mA 20 mA

UBE V1 0.64 V 0.67 V 0.69 V 0.72 V 0.76 V 0.81 V

The value pairs UBE, IB produce measured points for a

Transistor input characteristic


LE 21108-01Transistor characteristicsExercise 1 - 3 - Solution 23



86

Measurement sheet 2


Adjust the desired base current indirectly in the series circuit made up of series resistor R2 and

the base-emitter circuit of transistor V2 using the variable operating voltage. Using a second

source of operating voltage, change the collector-emitter voltage for each base current value.

The collector current IC is to be measured in each instance.

UCE 0 V 0.25 V 0.5 V 1.0 V 2.5 V 5.0 V

IC at IB = 50 µA 0 mA 7.20 mA 8.27 mA 8.33 mA 8.47 mA 8.62 mA




The value pairs UCE, IC produce measured points for four output characteristics for plotting an

measurement sheet 3.





87

Measurement sheet 3

Transistor output characteristics





88

Measurement sheet 4


Set the desired base current values for V3 with a variable operating voltage. Connect the

collector circuit to a second operating voltage source of 24 V.

In each case, measure the collector current.

IB 10 µA 20 µA 30 µA 50 µA 100 µA

IC 2.15 mA 4.33 mA 6.62 mA 11.18 mA 20.94 mA

The value pairs IB, IC produce measured points for a

Transistor current control characteristic.





89


5 LE 21109-01 - Fundamentals of Electronic Components - Setting the

Transistor Operating Point

Summary of contents

Contents: Setting the transistor operating points

Skills to be acquired: Setting the operating point in a common collector

connection.

Setting the operating point in a common emitter

connection

Projects: Collector connection

Emitter connection with negative current feedback




experimenter









90

Adjustment of the transistor operating point

is necessary to operate the transistor as an amplifier.

lt is usually done in such a way that the voltage drop in the C-E circuit is half the operating

voltage.

There is then a medium current flowing in the load current circuit. This adjustment

corresponds to an average operating point.

However, it is also possible to select a different operating point.

Collector circuit

One basic circuit is the collector circuit.

The adjustment of the desired operating point is made at the input side, using a voltage

divider, with simultaneous measurement of the output voltage.

The circuit input is located between base and ground. The circuit output is located

between emitter and ground.

If the input voltage UE is increased, the base current IB and with it the collector current I is

also increased.

The output voltage U A follows the input voltage at the base-emitter threshold voltage

interval UBE.

The operating point is stabilized by means of the full negative current feedback at the load

resistor RL.

MCEIndustrietechnik

LE 21109-01 Adjustment of transistor operatingpoint 1



91

Procedure for stabilizing the operating point: Collector circuit

As soon as the transistor warms up, there is an undesirable increase in its DC

amplification factor B:

IRL rises,

URL rises,

UBE drops because UE remains constant,

IB drops and

IRL drops approximately to the original value.

Since the stabilization is controlled by the current this is called negative current feedback.

MCEIndustrietechnik




92

Emitter circuit

Another basic circuit is the emitter circuit with negative current feedback. In addition to the

load resistor RL, it contains the emitter resistor RE for stabilization of the operating point.

The desired operating point is adjusted at the input side using a voltage divider, with

simultaneous measurement of the output voltage.

The circuit input is located between base and ground.

The circuit output is located between collector and ground, if the input voltage UE is

raised, the base current IB and with it the collector current IC is also increased, but the

output voltage U A drops because the voltage drop URL at load resistor RL increases.

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93

Procedure for stabilizing the operating point: Emitter circuit

As soon as the transistor warms up, there is an undesirable increase in its DC

amplification factor B.

IRE rises,

URE rises,

UBE drops because UE remains constant,

IB drops and

IRE drops to approximately the original value.

Since the stabilization is controlled by the current this is called negative current feedback.

MCEIndustrietechnik

LE 21109-01 Adjustment of transistor operating pointExercise 1 4



94

Circuit diagram

Module configuration

MCEIndustrietechnik




95


1 1 Fixed resistor R1 8.2 kΩ

2 1 Trimmer, vertical R2 47 kΩ

3 1 Fixed resistor R3 1 kΩ 4 1 Fixed resistor R4 6.8 kΩ


6 1 Fixed resistor R6 390 kΩ





11 1 Power pack 24 V DC

The components in

Item Nos. 1 - 9 are

contained in the

accessories for

"Fundamentals of

Electronic



MCEIndustrietechnik




96

Circuit diagram

Configuration of electronics experimenter

MCEIndustrietechnik




97












experimenter

The components in


contained in the



MCEIndustrietechnik




98

Circuit diagram


MCEIndustrietechnik




99













13 1 Electronics

experimenter


MCEIndustrietechnik




100


Use of transistor

Setting the operating point in a common collector connection

Setting the operating point in a common emitter connection



diagram.



missing values.




Component connection which are too short or too wide are prepared by soldering on


2. Complete the circuit in accordance with the circuit diagram.


missing values.



the diagram.




missing values.


PCB holder, 1.0 mm tin-lead solder, 30 W soldering iron, flat-no-sed pliers, soldering

tongs, conductor interrupter, side clippers, 1.3 mm twist drill, holder for 1.3 mm twist drill,

desoldering

Measuring and testing equipment1 multimeter

Accident prevention measuresBe careful when handling the hot soldering iron (risk of burns and danger of fire).Be careful when cutting components to length; pieces of wire flying around (risk of eyeinjuries).

MCEIndustrietechnik




101


Collector circuit

Various operating points can be adjusted using trimmer R2. At an average operating

point, an average collector current Ic flows. There is a voltage drop of 12 V in the C-E

circuit. The remaining 12 V drop occurs at load resistor R3. For this output level, an input

level of 12.7 V is to be set if UBE is assumed to be equal to 0.7 V.

Family of output characteristics

MCEIndustrietechnik

LE 21109-01 Adjustment of transistor operating pointExercise 1-3 12



103

Trade: ____________________________Name:________________________________

Measurement sheet 1

Collector circuit

1. Adjust the output voltage as desired (emitter to ground) by continuously changeling

the resistance of R2.

Calculate the emitter currents IE = IR3 = UR3 / R3

Work out the measurements line-by-line.

U A 16V 14V 12V 10V 8V

UCE

UE

UBE

IC

2. The value pairs UCE, IC produce operating points in the family of output

characteristics. The resistance straight line (operating straight line) of R3 can be

drawn in by connecting all operating points.


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104

Measurement sheet 2

Emitter circuit with negative current feedback

1. The Input voltage UE drops off at R5. By continuously changing R5, adjust the desired

output voltage values

U A = UCE + UR7

Calculate the collector current values Ic = IR6 = UR6 / R6

Work out the measured values line-by-line.

U A 8 V 10 V 12 V 14 V 16 V 18 V 20 V

UCE

UR6

UR7

UE

IC

2. The value pairs UCE, IC produce operating points on the resistance straight line

(operating straight line).


MCEIndustrietechnik




105

Collector circuit

1. Adjust the output voltage as desired (emitter to ground) by continuously changeling

the resistance of R2.

Calculate the emitter currents IE = IR3 = UR3 / R3

Work out the measurements line-by-line.

U A 16V 14V 12V 10V 8V

UCE 8.0 V 10.0 V 12.0 V 14.0 V 16.0 V

UE 16.6 V 14.6 V 12.6 V 10.6 V 8.6 V

UBE 0.63 V 0.62 V 0.62 V 0.61 V 0.60 V

IC 16 mA 14 mA 12 mA 10 mA 8 mA

2. The value pairs UCE, IC produce operating points in the family of output

characteristics. The resistance straight line (operating straight line) of R3 can be

drawn in by connecting all operating points.



LE 21109-01 Adjustment of transistor operating pointSolution Exercise 1-3 16



106

Measurement sheet 2

Emitter circuit with negative current feedback

1. The Input voltage UE drops off at R5. By continuously changing R5, adjust the desired

output voltage values U A = UCE + UR7

Calculate the collector current values Ic = IR6 = UR6 / R6

Work out the measured values line-by-line.

U A 8 V 10 V 12 V 14 V 16 V 18 V 20 V

UCE 3.8 V 6.4 V 8.7 V 11.3 V 13.8 V 16.5 V 19.1 V

UR6 16.1 V 14.0 V 12.0 V 9.9 V 7.9 V 6.0 V 3.8 V

UR7 4.2 V 3.6 V 3.1 V 2.6 V 2.0 V 1.5 V 1.0 V

UE 4.9 V 4.3 V 3.7 V 3.2 V 2.6 V 2.1 V 1.6 V

IC 41.3 mA 35.9 mA 30.8 25.4 20.3mA 15.4 mA 9.7 mA

2. The value pairs UCE, IC produce operating points on the resistance straight line

(operating straight line).



LE 21109-01 Adjustment of transistor operating pointSolution Exercise 1-3 17



107


6 LE 21110-01 - Fundamentals of Electronic Components - Transistors as

Amplif iers

Summary of contents

Contents: Transistors as amplifiers

Skills to be acquired: Use of the transistor as an amplifier

Amplification of a sinusoidal voltage

Projects. Emitter connection with negative current feedback

Collector connection




experimenter









109

Collector circuit

Due to its high input resistance together with its low output resistance, the collector circuit

is frequently used as an impedance converter to adjust the resistance of a low-ohmic

consuming device to a high-ohmic signal source. Impedance is AC resistance.

With respect to AC current, U and ground are short-circuited across the low-ohmic internal

resistance of the DC voltage source.

Equivalent AC circuit diagram for the collector circuit

MCEIndustrietechnik

LE 21110-01 Transistors as amplifiers 2



110

The phase angle of the AC output voltage relative to the AC input voltage is 0°.

The input resistance re is large.

r e Tr = approx. ß x (RL / / Ra) ( r e TR r e from the base of the transistor)

r e Scha = approx. R1 / / R2 / / r e Tr (r e Scha r e of the total circuit)

Example

r e Tr = 100 x (15 kΩ / / 1.2 kΩ ) = approx. 111 kΩ

r e Scha = 120 kΩ / / 120 kΩ / / 111 kΩ = approx. 40 kΩ

The output resistance r a is small

r e Scha = approx. Ri of the Signal source

ß

Example

r e Scha = 10 kΩ = 100 Ω

100

The max. voltage amplification Vu is 1 x.

Vu Scha = Ua

Ue

Example

Vu Scha = 20mV = 1

20 mV

The current amplification Vi

Vi Tr = approx. ß

Vi Sch = approx. ß x r e Scha

r e Tr

Example

Vi Tr = 100Vi Scha = 100 x 40 kΩ = 30

120 kΩ

ß = approx. DC amplification factor B

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112

The output voltage is 180° out-of-phase with the AC input voltage. The input resistance re

is small.

r e Tr = approx. rBE (re Tr r e of the base of the transistor)

r e Scha = approx. R1 / / R2 / / r e Tr (r e Scha re of the total circuit)

Example

r e Tr = approx. 1 kΩ

r e Scha = 22 kΩ / / 3.3 kΩ / / 1 kΩ = 740 Ω

The output resistance r a is, as a rule, small whenever RL / / Ra << r CE

r a Scha = approx. RL / / Ra

Example

r a Scha = 1 kΩ / / 1 kΩ = 0.5 kΩ

The voltage amplification Vu

Vu Scha = approx. ß x RL / / Ra

r e Tr

Example

Vu Scha = 100 x 1 kΩ / / 1 kΩ = 50

1 kQ

omitting Ra

Vu Scha 100 x 1 kΩ = 100

1 kΩ

The current amplification Vi

Vi Tr = approx. ß

Vi Scha = approx. ß x r e Scha x r a Scha

r e Tr RL

Example

Vi TR = 100

Vi Scha = 100 x 740 Ω x 0.5 kΩ 1 kΩ 1 kΩ

In order that the negative current feedback RE does not reduce the AC amplification, RE is

bridged with respect to AC current using the large capacitor CE. For that reason, RE has

no longer been drawn in the equivalent AC circuit diagram.

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113

Voltage frequency response

Transistor amplification depends an frequency. The frequency working range of an

amplifier is called the bandwidth. The bandwidth ends at the top limit frequency f go and the

bottom limit frequency f gu. At these points, the amplification Vu has dropped to 0.7 times

the rated value VuN (at the rated frequency f N = 1 kHz). Expressed in decibels, that means

a drop in amplification of -3 dB.

Vu = Ua VuN = Vu at f N

Ue

Vu [dB] = 20 x log Amplification at the selected frequency

VuN Amplification at the rated 1 kHz

Example

f N = 1 kHz Ua = 5.4 V Ue = 20 mV

VuN = 5.4 V = 270 Vu = 270 = 1 Vu = 20 x log 1.0 = 0 dB

20 mV VuN 270 VuN

f N = 100 Hz Ua = 3.7 V Ue = 20 mV

VuN = 3.7 V = 185 Vu = 185 = 0.68 Vu = 20 x log 0.68 = -3,3 dB

20 mV VuN 270 VuN

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114

Circuit diagram

Module configuration

MCEIndustrietechnik




115


1 1 Fixed resistor R1, 120 kΩ


3 1 Fixed resistor R3, 15 kΩ 4 1 Fixed resistor R4, 22 kΩ

5 1 Fixed resistor R5, 3.3 kΩ


7 1 Fixed resistor R7, 220 Ω

8 1 Transistor V1, BC 140-16


10 1 Electrolytic C1, 4.7 µF

capacitor

11 1 Electrolytic C2, 47 µF

capacitor


capacitor


capacitor


15 1 Power pack 24 V

The components in


contained in the

accessories for

"Fundamentals of

Electronic

Components" 2GA5101-8F '


MCEIndustrietechnik

LE 21110-01Transistors as amplifiersExercise 1 8



116

Circuit diagram


MCEIndustrietechnik




117




3 1 Fixed resistor R3, 15 kΩ 4 1 Fixed resistor R4, 22 kΩ






10 1 Electrolytic C1, 4.7 µF

capacitor


capacitor


capacitor


capacitor


plated


experimenter

The components in


contained in the

accessories kit 2GA5101-8A


MCEIndustrietechnik




118

Circuit diagram


MCEIndustrietechnik




120

Skills to be acquired - Use of the transistor as an amplifier



diagram


3. Perform the measurements in accordance with the measurement circuit and

measurement sheets 1 and 2

4. Form the curves from the measured values (measurement sheet 3)





pieces of bare wire







the diagram


3. Plug the complete PCB into the 31-pole connector of the electronics experimenter








Measuring and testing equipment - 2 multimeters, 1 oscilloscope, 1 sinus-wave generator Accident prevention measures



injuries).

MCEIndustrietechnik

LE 21110-01Transistors as amplifiersExercise 1 - 3 13



121


Collector circuit

An AC current is feed in across capacitor C1. The capacitive resistance XC is significantly

smaller than the circuit input resistance r e Scha. The current divides across R1, R2 and the

series circuit made up of V1BE and R3. The base current causes a current across R3

amplified by the factor ß. The output voltage drops across R3. The AC voltage at the

output is exactly as large as at the input but it lies lower by UBE V1 = 0.7 V. To decouple AC

voltage signal at R3, capacitor C2 is connected at the output. The capacitive resistance

XC2 depends an the consuming device that follows. If, for example, the following

consuming device R A has a resistance of 1.2 kΩ, V1 automatically conducts more current

so that the output voltage does not drop. The current amplification of the transistor = ß

remains constant. The base current, which likewise rises, adapts to the load current. The

voltage amplification factor remains 1x.

MCEIndustrietechnik




122

Collector circuit

Example with Ra = 1.2 kΩ, ß = 100

r e Tr = approx. ß x R3 / / Ra

= 100 x 15 kΩ / / 1.2 kΩ = 111 kΩ

r e Scha = approx. R1 / / R2/r e Tr

= 120 kΩ / / 120 kΩ / / 111 kΩ = approx. 40 kΩ

∆IB = Ue Vu Scha = Ua

r e Tr Ue

= 20 mV = 0.18 µA = 20 mV = 1

111 kΩ 20 mV

Ie = Ue Vi Scha = ∆Ia

r e Scha ∆Ie

= 20 mV = 0.5 µA = 16.6 µA = 33

40 kΩ 0.5 µA

Ic = approx. ∆IB x ß Vi Tr = Ic

= 0.18 µA x 100 = 18 µA IB

= 18 µA = 100

0.18 µA

R3/ /Ra = approx. 15 kΩ / / 1.2 kΩ = 1.1 kΩ

Ua = approx. ∆IC x R3 / / Ra

= 18 µA x 1.1 kΩ = 20 mV

Ia = Ua

r a

= 20 mV = 16.6 µA

1.2 kΩ

IR3 = Ua

r3

= 20 mV = 1.33 µA

15 kΩ

MCEIndustrietechnik




124

Emitter circuit

Example with Ra = 1 kΩ, ß = 100, r BE = approx. 1 kΩ

r e Tr = approx. r BE = approx. 1 kΩ Vu Scha = Ua = 1 V = 50

Ue 20 mV

r e Scha = approx. R4 / / R5 / / r e Tr

= 22 kΩ / / 3.3 kΩ / / 1 kΩ = 740 Ω

∆UBE = approx. Ue Vi Scha = Ia = 1 mA = 37

= 20 mV Ie 27 µA

∆IB = Ue = 20 mV = 20 µA

r e Tr 1 kΩ

Ie = Ue = 20 mV = 27 µA Vi Tr = ∆IC = 2 mA = 100

r e Scha 740 ∆IB 20 µA

∆IC = ∆IB x ß = 20 µA x 100 = 2 mA

r a = approx. R6 / / Ra = 1 kΩ / / 1 kΩ = 0.5 kΩ

Ua = approx. ∆IC x r a = 2 mA x 0.5 kΩ = 1 V

Ia = Ua = 1 V = 1 mA

Ra 1 kΩ

IR6 = Ua = 1 V = mA

R6 1 kΩ

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127

Trade: ____________________________Name:________________________________

Measurement sheet 3

Join connections a and b, c and d. Couple the input of the emitter circuit to the output of

the collector circuit across C2. From a sine-wave generator, feed in a sinusoidal voltage of

USS = 20 mV at the input of the collector circuit. Change the frequency step-by-step.

Determine the voltage values from the the input and output oscillograms.

Calculate Vu, Vu and Vu [dB]

VuN VN

VuN = Vu at 1 kHz

Enter the values for Vu and Vu [dB] into the diagram that follows

VuN VuN

Join up the marked points to form a transmission characteristic.

MCEIndustrietechnik




128

Trade: ____________________________Name:________________________________

Diagram:

MCEIndustrietechnik




129

Measurement sheet 1

Join connections a and b. Use connection c for decoupling. From a sine-wave generator,

feed in a sinusoidal voltage Uss = 20 mV with a frequency of 1 kHz at the input of the

collector circuit. The input signal and the amplified output signal can be recorded using the

oscilloscope. For Ue, select the positive trigger edge.

The scale on the x-axis is to be 0.2 ms per division.

The scale on the y-axis is to be 10 mV per cm for Ue.

The scale on the y-axis is to be 10 mV per cm for Ua.

Calculate the voltage amplification Vu from Ue and Ua.

Vu = Ua = 20 mV = 1

Ue 20 mV





130

Measurement sheet 2

Join connections c and d. Feed in across connection b. From a sine-wave generator, feed

in a sinusoidal voltage of USS = 20 mV/1 kHz to the input of the emitter circuit. The input

signal and the amplified output signal can be recorded using the oscilloscope. For Ue

select the positive trigger edge.

The scale on the x-axis is to be 0.2 ms per division.

The scale on the y-axis is to be 10 mV per cm for Ue.

The scale on the y-axis is to be 10 mV per cm for Ua.

Calculate the voltage amplification Vu from Ue and Ua.

Vu = Ua = 4.4 mV = 220

Ue 20 mV





132

Diagram:





133


7 LE 21111-01 - Fundamentals of Electronic Components - Transistors as

Switches

Summary of contents

Contents: Transistors as switches

Skills to be acquired: Use of the transistor as switches

Projects. Two-stage inverter

Switching amplifer




experimenter









134

Transistors used as switches

are found mainly in signal processing. There, innumerable transistor functions (transistors

as switches) are used in the form of integrated circuits (IC). Relatively few individual

transistors switch the end consuming devices an and off.

The actual switching element is the collector-emitter circuit. It can only work in a binary

manner, i.e. it must have two clearly defined switching conditions:

a) The transistor blocks - almost absolutely.

b) The transistor conducts - almost without losses.

The switching condition of a transistor passes a signal along to the transistors or signal

transmitters that follow. Machines can be controlled. Signals are often reversed (inverted)

and amplified. An inverter (e.g. a transistor in an emitter circuit) does the signal inversion.

A high level at the input brings about a low level at the output, and a low level at the input

brings about a high level at the output. A signal is amplified and may also be inverted by a

switching amplifier. One transistor output can provide the control current for several

following transistor inputs.

Fundamentals typical of the circuit

Transistors used as switches are overdriven, i.e. they receive a relatively high base

current in order to keep the transmission losses low. If the base current IB = IC / B

is multiplied by the overdriving factor m = 3, the collector-emitter voltage drops to a

saturation value of approx. UCEsat = 0.1 V.

The AC amplification factor B which is assumed is the value that can be expected in the

worst possible case. If the transistor used actually has better amplification than expected,

the overdriving is greater than 3x, which, as a rule, can only be advantageous.

MCEIndustrietechnik

LE 21111-01 Transistor as switches 1



135

Calculations for a transistor used as a switch

Problem: UB = 15 V, UBE = 0.7 V, UCESat = 0.1 V

RL = 330 Ω, B = 150, m = 3, RV = ?

Solution: IRL = URL = UB - UCESat = 5 V - 0.1 V = 45 mA

RL RL 330 Ω

IB = IC x m = IRL x m = 45 mA x 3 = 0.9 mA

B B 150

RV = URV = UB - UBE = 15 V - 0.7 V = 15.9 kΩ

IRV IB 0.9 mA

OP1 = Operating point 1 = Transistor is conducting

OP2 = Operating point 2 = Transistor is blocking

MCEIndustrietechnik

LE 21111-01 Transistor as switches 2



137




3 1 Fixed resistor R3, 330 Ω 4 1 Fixed resistor R4, 15 kΩ




8 1 Fixed resistor R8 560 Ω

9 1 Transistor V1 BC 237 B



12 1 LED B1 CQV 24-5

YELLOW


The components in


contained in the

accessories for

"Fundamentals of

Electronic



MCEIndustrietechnik

LE 21111-01Transistor as switchesExercise 1 4



138

Circuit diagram

Configuaration of electronics experimenter

MCEIndustrietechnik




140

Circuit diagram


MCEIndustrietechnik




142


Use of the transistor as a switch



diagram


3. Perform the measurements n accordance with the measurement sheets




Component connection which are too short or too wide are prepared by soldering on

pieces of bare wire


3. Perform the measurments in accordance with the measurement sheet

Job plan (pinted circuit board arrangement)


the diagram



4. Perform the measurements in accordance with the measurement sheet


PCB holer, 1.0 mm tin-lead solder, 30 W soldering iron, flat-nosed pliers, soldering tongs,

conductor interrupter, side clippers, 1.3 mm twist drill, holder for 1.3 mm twist drill,




injuries).

MCEIndustrietechnik

LE 21111-01Transistor as switchesExercises 1 - 3 9



143


When operating voltage has been connected and the transistor inputs of V1 and V3

remain open or are connected to negative, V1 and V3 are blocked and V2 conducts. If the

transistor inputs of V1 and V3 are connected to positive, V1 and V3 are conducting and

V2 is blocked. If the input of V3 is connected to the output of V2, V3 depends upon the

switching condition of V2 and V2 in turn depends an the switching conditions of V1. V1

and V2 are inverters. V3 is used as a switching amplifier. When V3 is conducting, B1

lights up. Load resistors R3, R5 and R8 limit the collector current. Series resistors R1, R4

and R6 limit the base current. R2 and R7 are base leakage resistors which guarantee that

V1 and V3 will block if the transistor inputs are opened. R8 is the series resistor before B1

to limit the LED current.

MCEIndustrietechnik

LE 21111-01Transistor as switchesExercises 1 - 3 10



144

Trade: ____________________________Name:________________________________

Measurement sheet 1

Voltage measurements

1. If transistors V1 and V3 are not energized, the following values can be measured:

2. If the positive operating voltage is appiied before R1 and V3 is energized across the

connection from V2 shown in a broken line, B1 lights. The switching conditions are:

V3 conducts V3 blocks



3. If negative is now connected before Rl instead of positive, the voltage levels can be

measured.

UCEV3 = Low UCEV3 = High



UCEV1

UR3

UR4

UBEV2

UR5

UCEV2

MCEIndustrietechnik

LE 21111-01Transistor as switchesExercise 1 - 3 11



145

Trade: ____________________________Name:________________________________

Measurement sheet 2

Measurements of voltage

Measure the voltages when V3 depends on V2 and V2 on V1.

If V1 is correctly energized, B1 is to light up.

UBEV1

UBEV2

UBEV3

UCEV1

UCEV2

UCEV3

UR1

UR4

UR6

UR3

UR5

UR2

UR7

MCEIndustrietechnik

LE 21111-01Transistor as switchesExercise 1 - 3 12



146

Measurement sheet 1

Voltage measurements

4. If transistors V1 and V3 are not energized, the following values can be measured:

5. If the positive operating voltage is appiied before R1 and V3 is energized across the

connection from V2 shown in a broken line, B1 lights. The switching conditions are:

X V3 conducts V3 blocks

V2 conducts X V2 blocks

X V1 conducts V1 blocks

6. If negative is now connected before Rl instead of positive, the voltage levels can be

measured.

UCEV3 = Low X UCEV3 = High

X UCEV2 = Low UCEV2 = High

UCEV1 = Low X UCEV1 = High

UCEV1 14.69 V

UR3 0.31 V

UR4 13.91 V

UBEV2 0.77 V

UR5 14.85 V

UCEV2 0.15 V


LE 21111-01Transistor as switchesSolution - Exercise 1 - 3 13



148


8 LE 21112-01 - Fundamentals of Electronic Components - Constant

Current Sources

Summary of contents

Contents: Constant current sources

Skills to be acquired: Use of a constant current source

- with NPN transistor

- with PNP transistor

- with field effect transistor (FET)

Projects. Constant current source

- with fixed resistor as load

- with linear capacitor charge

- with variable load resistor




experimenter




The electronics trainer, electronics experimenter and accessories contained in the lists ofequipment under Designation and Item No. can be found in the EA catalogue (Electrical




149

The use of constant current sources

is necessary to improve electronic switching functions.

It is frequently found in instrumentation and control systems. With constant current

sources, the internal resistance Ri is very high so the load current remains almost non-

dependent an the value of the load resistance. However, only small load currents are

possible.

High-ohmic constant current source with load resistor

MCEIndustrietechnik

LE 21112-01 Use of constant current sources 1



150

In order to keep larger currents constant, transistors are used. The constant voltage at the

emitter resistor RE keeps the emitter current constant.

The base current (IB) is not taken into consideration.

IC = approx. IE

A capacitor can be loaded with a constant current in such a way that its charging voltage

rises as a linear function.

MCEIndustrietechnik




151

As a rule, the circuits are constructed using an NPN or PNP tran-sistor. Because the field

effect transistor (FET) can also be used, its Operation is explained below.

D = Drain = Collector S = Source = Emitter G = Gate = Base

The field effect transistor selected is conducting in the N-channel. As the negative voltage

at the gate increases with respect to source (-UGS = UR Source) the drain-source cicuit

blocks. (The N-channel becomes high-ohmic). The transistor control is accomplished by

means of the electrical field between the N and P layers. No control current flows.

MCEIndustrietechnik




152

Compared with a transistor controlled by current, a negative gate source voltage (-UGS)

corresponds to a smaller base current (IB).

Comparison of output characteristics

MCEIndustrietechnik




153

Comparison of structures

BC 237 B

NPN low-frequency transistor

LF operating range

up to 50 kHz

BC 308 B

PNP low-frequency transistor

LF operating range

up to 50 kHz

BF 245 B

High-frequency field effect transistor

HF operating range

above 50 kHz

MCEIndustrietechnik




155


1 1 Fixed resistor R1 470 Ω 1 W


3 1 Fixed resistor R3 220 Ω 1 W4 1 Fixed resistor R4 560 Ω 1 W



7 1 Trimmer R7 1kΩlin 0.2 W

8 1 LED B1 CQV 13-6

YELLOW

9 1 Zener diode V1 BZX 83 C4 V7



12 1 Field effect V4 BF 245 B

transistor

13 1 Electrolytic C1 22 µF/40 V

capacitor


15 1 Power pack 12 V

The components in


contained in the

accessories for

"Fundamentals of

Electronic



MCEIndustrietechnik

LE 21112-01Constant current sourcesExercise 1 7



156

Circuit diagram


MCEIndustrietechnik

LE 21112-01Use of constant current sourcesExecise 2 8



157








8 1 LED B1 CQV 13-6

YELLOW





transistor


capacitor


plated


experimenter

The components in


contained in the



MCEIndustrietechnik




158

Circuit diagram


MCEIndustrietechnik

LE 21112-01Use of constant current sourcesExecise 3 10



159








8 1 LED B1 CQV 13-6

YELLOW





transistor


capacitor


plated



17 1 Electronic 2GA5101-2A

experimenter


MCEIndustrietechnik




161

Accident prevention measures


Be careful whenn cutting components to length; pieces of wire flying around (risk of eye

injuries).

Be careful when handling capacitors; there is a danger of explosion if the operating

voltage is too high or the poles are incorrectly connected (risk of eye injuries).


Constant current source with NPN transistor

R1 and V1 form a Zener diode voltage stabilization. The B-E circuit of transistor V2 and

the fixed resistor R3 are located parallel to the Zener diode. As a result, R3 is at a

constant voltage 0.7 V lower than the Zener voltage. This constant voltage causes a

constant current across R3. The collector current is almost the same size as the emitter

current. Any desired resistance can be therefore used in the collector circuit up to a given

maximum value. The constant current causes a voltage drop at load resistor R2.

Consideration: Assume load resistor R2 is male low-ohmic. The base voltage relative to

ground is kept constant by the Zener diode. The collector current (load current) and the

emitter current rise. There is a greater voltage drop at R3. The base-emitter voltage

decreases (UBE = UZ -UR3). The base current and the collector current drop.

This control process approximately restores the original load current. The principle of

Operation is negative current feedback.

Constant current source with a PNP transistor

When B1 and R4 are connected in series, there is a constant voltage drop at the LED B1.

R5 and the B-E circuit of V3 are parallel to B1. As a result, a constant voltage drop occurs

at R5 of 0.7 V less than the voltage drop at the LED. The current across R5 is likewise

constant. It fiows further via the emitter-collector circuit of V3 and charges capacitor C1 in

accordance with a linear function.The charging starts when the operating voltage is switched on. After the operating voltage

is taken away, C1 discharges according to a logarithmic function across the emitter-base

circuit of V3 and across R4, first quickly, and then more and more slowly.

MCEIndustrietechnik

LE 21112-01Constant current sourcesExercises 1 - 3 13



162

Constant current source with field effect transistor

When there is no voltage at its gate connection, field effect transistor V4 is conducting an

its drain-source circuit. The current causes a voltage drop at source resistor R6. After R6,

the voltage level is more negative than before R6, and at the same time more negative

than at the source. By means of the Feedback of this negative voltage level to the gate

(compared to source), the drainsource circuit blocks somewhat. At a given ratio between

the size of the drain-source current and the negative gate voltage, a constant transistor

operating condition sets in. A constant current flows across the transistor regardless of

how the variable load resistor R7 that follows is set.

Consideration: becomes larger, becomes smaller

As a result of this control process, the load current IR7 once again assumes approximately

its original value.

MCEIndustrietechnik


(no Tonger so negative)



163

Trade: ____________________________Name:________________________________

Measurement sheet 1

Measure voltages in the circuit with the NPN transistor V2.

After the first series of measurements, bridge R2 with a measuring line. This corresponds

to a load resistor R2 of 0 Ω.

Calculations of current show that the collector current (load current) remains

approximately constant.

IC before R2 is bridgedIC = approx. IR3 (IB is not considered)

IR3 = UR3 =

R3

IC when R2 is bridged

IC = approx. IR3 (IB is not considered)

IR3 = UR3 =

R3

MCEIndustrietechnik


R3 220 Ω

R2 220 Ω

UR1

UV1

UR3

UV2CE

UV2BE

UR2

R3 220 Ω

R2 0 Ω

UR1

UV1

UR3

UV2CE

UV2BE

UR2



165

Trade: ____________________________Name:________________________________

Measurement sheet 3

Measure voltages in the circuit with field effect transistor V4.

Calculate currents ID = IR6. Set the resistance values desired at R7.

The UDS ID value pairs "a" and "b" produce operating points an the load curves "a" and "b".

By connecting the operating points, the -UGS paramet can be obtained and ID can be read

off.

MCEIndustrietechnik


R6 220 Ω

R7 1 kΩ

UR6

UR7

UDS

Value pair a

ID

R6 220 Ω

R7 0 Ω

UR6

UR7

UDS

Value pair b

ID



169


9 LE 21113-01 - Fundamentals of Electronic Components - Constant

Voltage Sources

Summary of contents

Contents: Constant voltage sources

Skills to be acquired: Use of a constant voltage source

with transistors

Use of a constant voltage source

with a fixed voltage controller

Projects: Constant voltage source with Zener diode and

Darlington connection Constant voltage source with

fixed voltage controller




experimenter





equipment under Designation and Item No. can be found in the EA catalogue (ElectricalEngineering Training Equipment).



170

The use of constant voltage sources

makes it possible to provide the operating voltage for every active electronic instrument

economically. Without constant voltage sources, batteries or accumulators would be

necessary for that purpose.

The Zener diode voltage stabilization has been the basic circuit most frequently used in

the past.

With an additional transistor in a collector circuit, a constant voltage source for higher load

currents (RL = low-ohmic) is formed.

MCEIndustrietechnik

LE 21113-01 Constant voltage sources 1



171

Frequently a combination circuit using two transistors is used. This is called a Darlington

circuit.

The current amplification of the two transistors is multiplied. The base currents are not

taken into consideration.

Example: BV2 x BV3 = Btotal; 100 x 40 = 4000

A large change in load current requires only an insignificant change in base current. The

Darlington circuit is connected to a Zener diode voltage stabilization like the collector

circuit. It should be borne in mind that the base-emitter voltage drops must be added

together. The output voltage is lower than the Zener voltage by the amount of the

threshold voltage.

Example: Uthreshold = UBEV2 UBEV3

= 0.7 V + 0.7 V

= 1.4 V

U A = UZ Uthreshold

= 6 V - 1.4 V

= 4.6 V

MCEIndustrietechnik

LE 21113-01 Constant voltage sources 2



174




3 1 Fixed resistor R3 150 Ω 1 W4 1 Zener diode V1 BZY 97 C6 V2

5 1 Transistor V2 BC 107 A


7 1 Tantalum electrolytic

capacitor C1 1 µF 40 V

8 1 Tantalum electrolytic

capacitor C2 1 µF 40 V



11 1 Fixed voltage N1 TDB 7805 T

regulator


13 1 Power supply

The components in

Items Nos. 1-11 are

contained in the

accessories for

"Fundamentals of

Electronic



MCEIndustrietechnik

LE 21113-01Constant voltage sourcesExercise 1 5



175

Circuit diagram


MCEIndustrietechnik




177

Circuit diagram


MCEIndustrietechnik




179


Use of a constant voltage source with transistors.

Use of a constant voltage source with a fixed voltage controller.



diagram



mising values



diagram.

2. The component connections which are too short or too wide are prepared by soldering

on pieces of bare wire


4. Perform the measurements in accordance with the measurement sheets an enter the

missing values



the diagram




missing values





Measuring and test equipment - 1 Multimeter



injuries). Be careful when handling capacitors; there is a danger of explosion if the

operating voltage is too high or if the poles are incorrectly connected (eye injuries).

MCEIndustrietechnik

LE 21113-01Constant voltage sourcesExercise 1 - 3 10



184

Measurement sheet 1

Block the transistors with an external bridge so that R2 and B1 lie directly parallel to the

Zener diode.

Circuit

Measurements

UR1 5.68 VUZ 6.33 V

UR2 4.68 V

Calculations

IR1 = UR1 = 5.68 V = 56.8 mA

R1 100 Ω

IR2 = UR2 = 4.68 V = 31.2 mA

R2 150 Ω

IZ = IR1 - IR2 = 56.8 mA - 31.2 mA = 25.6 mA


LE 21113-01Constant voltage sourcesSolution - Exercise 1 - 3 15



186

Measurement sheet 3

Measure the Input voltage and the output voltage in the constant voltage source with a

fixed voltage regulator.

UE 12.0 V

U A 4.92 V


LE 21113-01Constant voltage sourcesSolution - Exercise 1 - 3 17



187


10 LE 21118-01 - Fundamentals of Electronic Components - High, Low, and

Band-pass Filters

Summary of contents

Contents: Voltage response of RC elements

Skills to be acquired: Designing low, high, and band-pass circuits

Projects: Integrating circuit Exercises 1a, b

Low-pass Exercises 2a, b

Differential element Exercises 3a, b

High-pass Exercises 4a, b

Band-pass Exercises 5a, b





189

Example of switching off:

The voltage across the capacitor (Uc) has fallen to 37 % of the operating voltage (UE).

Switching-off

MCEIndustrietechnik

LE 21118-01 Voltage response of RC elements 1



190

Integrating element

If the output voltage of the RC element is tapped at the capacitor, we call this an

integrating circuit.

When a voltage UE is applied, capacitor C is charged via resistor R. The time behavior of

the charging current depends on the time constant of the RC circuit. No current flows after

the charging process has been completed. There is a certain constant voltage across the

capacitor. If no voltage is now applied to the input, the capacitor C is discharged via the

consumer R A.

The discharge voltage and current are determined by the load resistor R A.

If square-wave pulses are applied to the circuit input, the capacitor charging and

discharging processes alternate periodically. The gradient of the rising and falling edges

may be modified by changing R or R A

The duration of the pulses supplied may be extended using an integrating circuit.

Compared with the pulse duration of UE, the time constant may be so large that several

rapidly successive pulses may be combined (= integrated) to form one Jong pulse.

MCEIndustrietechnik




191

Low-pass filter

A low-pass filter is a circuit which allows "low" frequencies to "pass" and which cuts out

high frequencies.

An integrating circuit is the simplest form of low-pass filter.

If a frequency mixture (music, speech) is supplied to its input, the low frequencies are

given preference.

The circuit practically acts as an AC voltage divider.

Since the capacitive reactance

1

XC = 1 where w = 2 x . f

w . C

is frequency dependent, the relation of XC to R also changes with frequency f. Because XC

is reduced with increasing frequency, there is a greater voltage drop across the resistor.

Therefore, the higher the frequency of the supplied AC voltage, the lower U A.

In order to make a clear statement an the effect of frequency, the concept of cut-off

frequency has been introduced. The cut-off frequency fg of an RC circuit is the frequency

at which XC = R (reactance is equal to active resistance). if this is true, then 70.7 % of the

AC input voltage is available at the circuit output (U A = 0.7 x UE).

MCEIndustrietechnik




193

High-pass filter

A high-pass filter is a circuit which allows "high" frequencies to "pass" and which cuts out

low frequencies.

A differential circuit operates like a high-pass filter.

High frequencies are given preference when a frequency mixture is transmitted. Thiscircuit is also an AC voltage divider.

Here, the resistance is parallel to the output.

The higher the frequency of the supplied AC voltage, the lower the value of XC and the

component voltages are displaced in favor of U A (UR).

Consequently, the higher the frequency of the supplied AC voltage the greater the value

for U A.

The following diagram shows that the output voltage rises steeply at first below the cut-off

frequency fg, but rises only slightly above fg.

MCEIndustrietechnik




194

Band-pass filter

In practice, it is often necessary to filter out only those frequencies lying within a certain

frequency range from a mixture of voltage at different frequencies.

The band-pass filter is an important instrument in this task.

In the simplest case, band-pass filters are equipped with two RC elements, one of which

is a low-pass filter and the other is a high-pass filter. It is important that the cut-off

frequency of the low-pass filter be higher than that of the'high-pass filter.

The cut-off frequency is the frequency at which the reactance XC is equal to the value of

the ohmic resistance R, whereby the outpuf voltage falls to 70.7 % of its maximum value.

The cut-off frequency of the high-pass filter is the lower cut-off frequency f u:

f u = 1

2 • n • R1 • C1

The cut-off frequency of the low-pass filter determines the upper cut-off frequency f o:

f o = 1

2 • n • R2 • C2

The difference between these two frequencies is the bandwidth

b = f o - f u

These equations apply only if both RC elements do not influence each other.

For this reason, it is sometimes necessary to decouple the two filter elements.

MCEIndustrietechnik




195

Circuit diagram:

Exercise 1a: set-up with electronics trainer - Items 1 - 2.

Exercise 1b: set-up an PC board - Items 3 - 14.

Solder R and C onto the soldering tag.

14 4 Soldering tag

13 1 Bare wire 0.5 mm,silvered

12 1 Capacitor C 1 µF/40 V

11 1 Capacitor C 470 nF/40 V



8 1 Capacitor C 2.2 nF/40 V

7 1 Resistor R 0.5 W 10 kΩ

6 1 Resistor R 0.5 W 2.2kΩ

5 1 Resistor R 0.5 W 1.2kΩ

4 1 Terminal strip DIN 41617 31 poles, SM2.54

3 1 PC board 100x160mm,SM2.54

2 1 Accessory set 2GA5101-8G Fund. of analog./

1 1 Electronics trainer 2GA5101-3B el. engineering

ItemNo.

Qty. Designation Detail No. Remarks

MCEIndustrietechnik

LE 21118-01Voltage response of RC elementsExercises 1a, b 8



196

Task

Designing an integrating element and observing and recording the deformation of a

square-wave signal.


1. Arrange all the required component modules on the assembly board.

2. Wire the circuit in accordance with the circuit diagram.

3. Carry out the measurements in accordance with the measurement sheet.

4. Enter the missing measurement values and curves on the measurement sheet.

Job plan (PC board set-up)

1. Fit and wire all the required components an the PC board in accordance with the

circuit diagram.


3. Enter the missing measurement values and curves an the measurement sheet

Auxiliaries, tools and working materials

PC board holder, tin-lead solder 1.0 mm, soldering iron 30 W, flat pliers, pointed soldering

pliers, diagonal cutter, desoldering device, steel rule.

Measuring and checkinq instruments

Multimeter, oscilloscope, frequency generator

Accident and damage prevention

Exercise care when handling the hot soldering iron (risk of burns and fire).

Exercise care when cutting off components because wire ends may fly off leading to eye

injuries.

MCEIndustrietechnik

LE 21118-01Voltage response of RC elementsExercises 1a, b 9



198

Trade: ____________________________Name:________________________________

Measurement sheet 2

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LE 21118-01Integrating circuitExercise 1a, b 11



199

Circuit diagram:

Exercise 2a: set-up with electronics trainer - Items 1 - 2.Exercise 2b: set-up an PC board - Items 3 - 7

7 1 Bare wire 0.5 mm, silvered


5 1 Resistor R 0.5 W 5.6 kΩ

4 1 Terminal strip DIN 41617 31 poles SM2.54

3 1 PC board 100x160mm, SM2.54

2 1 Accessory set 2GA5101 -8G Fund. of analog./

1 1 Electronics trainer 2GA5101 -3B el. engineering

ItemNo.


MCEIndustrietechnik

LE 21118-01Low-pass filterExercises 2a, b 12



200

Task

Design a low-pass filter. Study its effect by changing the Input frequency. Enter the

measured results in the table and draw the low-pass filter curve on the graph.







1. Fit and wire all the required components on the PC board in accordance with the

circuit diagram.


3. Enter the missing measurement values and curves on the measurement sheet


PC board holder, tin-lead solder 1.0 mm, soldering iron 30 W, fiat pliers, pointed soldering


Measuring and checking instruments





injuries.

MCEIndustrietechnik

LE 21118-01Low-pass filterExercise 2a, b 13



201

Trade: ____________________________Name:________________________________

Measurement sheet 3

Apply sinusoidal voltage of different frequencies to the input of the low-pass filter.

Measure the voltage values at the output and enter them in the table. Then transfer these

measured values to the graph and draw a curve.

Set a sinusoidal AC voltage UESS = 10 V an the frequency generator. Successively set the

frequencies given in the table.

Determine the voltage U ASS for each frequency.

Calculate the cut-off frequency:

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202

Trade: ____________________________Name:________________________________

Measurement sheet 4

Calculate the range for the coordinates

Insert the actual value of the cut-off frequency into the diagram

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203

Circuit diagram:


Exercise 3b: set-up an PC board - Items 3 - 14.

Solder R and C onto the soldering tag.

14 4 Soldering tag

13 1 Bare wire 0.5 mm, siivered

12 1 Capacitor C 1 F/40V

11 1 Capacitor C 470 nF/40V



8 1 Capacitor C 2.2 nF/40V

7 1 Resistor R 0.5 W 10 k

6 1 Resistor R 0.5 W 2.2 k

5 1 Resistor R 0.5 W 1.2 k

4 1 Terminal strip DIN 41617 31 poles, SM2.54

3 1 PC board 100x160 mm, SM2.5

2 1 Accessory set 2GA5101-8G Fund. of analog./

1 1 Electronics trainer 2GA5101-3B el. engineering

ItemNo.


MCEIndustrietechnik

LE 21118-01Differential circuitExercises 3a, b 16



204

Task

Design a differential element and observe and record the deformation of a square-wave

signal.








circuit diagram.








Accident and damalte prevention



injuries.

MCEIndustrietechnik

LE 21118-01Differential elementExercise 3a, b 17



205

Trade: ____________________________Name:________________________________

Measurement sheet 5

Using the frequency generator, supply a square-wave voltage to the input of the

differential element. Measure the deformation of the square-wave at the output using the

oscilloscope and draw the curve shape on the measurement sheets.

Set the frequency generator to 5 kHz and UESS = 4 V

Adjust the square-wave signal so that the botton line is on the zero line.

Oscilloscope: time base 50 gs/DIV, channel 1 is the trigger sources,

channels 1 and 2 = 2 V/DIV, input switch set to DC

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Trade: ____________________________Name:________________________________

Measurement sheet 6

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207

Circuit diagram:


Exercise 4b: set-up an PC board - Items 3 - 7



5 1 Resistor R 0.5 W 5.6 kΩ





ItemNo.


MCEIndustrietechnik

LE 21118-01High-pass filterExercises 4a, b 20



208

Task

Design a high-pass filter. Observe its effects by changing the Input frequency. Enter the

measured resuits in the table and draw the high- pass filter curve on the graph.








circuit diagram.



Auxiliaries, tools and workinq materials





Accident and damaqe prevention


Exercise care when cutting off components because wire ends may fly of leading to eye

injuries.

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209

Trade: ____________________________Name:________________________________

Measurement sheet 7

Apply sinusoidal voltage of different frequencies to the input of the high-pass filter.







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210

Trade: ____________________________Name:________________________________

Measurement sheet 8

Determine the scale for the ccordinate system.

Enter the actual value of the cut-off frequency on the graph.

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211

Circuit diagram:




7 1 Capacitor C2 1 nF/40 V


5 2 Resistor R1, R2, 0.5 W 1.5 kΩ





ItemNo.


MCEIndustrietechnik

LE 21118-01Band-pass filterExercises 5a, b 24



212

Task

Design a band-pass filter. Study its effects by changing the input frequency. Enter the

measured results in the table and draw the band- pass filter curve on the graph.

Job plan (electronic trainer)







circuit diagram.











injuries.

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213

Trade: ____________________________Name:________________________________

Measurement sheet 9

Apply sinusoidal voltage of different frequencies to the input of the band-pass filter.


measured values to the graph and draw a curve.Set a sinusoidal AC voltage UESS = 10 V an the frequency generator.

Successively set the frequencies given in the table.


Calculate the cut-off frequency and the band with:

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214

Trade: ____________________________Name:________________________________

Measurement sheet 10

Determine the scale for the coordinate system.

Enter the actual value of the cut-off frequency on the graph and determine the band width.

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215

Circuit diagram:






5 2 Resistor R1, R2, 0.5 W 1.5 kΩ





ItemNo.


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216

Measurement sheet 1


integrating element. Measure the deformation of the squarewave at the output using the

oscilloscope and draw the curve shape an the measurement sheets.

Set the frequency generator to 5 kHz and UEss = 4 V

Adjust the square-wave Signal so that the botton line is an the zero line (K2).




LE 21118-01Integrating circuitSolution - Exercise 1a, b 28



217

Measurement sheet 2


LE 21118-01Integrating circuitSolution - Exercise 1a, b 29



218

Measurement sheet 3

Apply sinusoidal voltage of different frequencies to the input of the low-pass filter.







In case of f g: XC = R f g = 1

2 x π x R x C

R = 1 = 1 f g = 6047 Hz

w x C 2 x π x f x C


LE 21118-01Low-pass filterSolution - Exercise 2a, b 30



219

Measurement sheet 4

Calculate the range for the coordinates

Insert the actual value of the cut-off frequency into the diagram


LE 21118-01Low-pass filterSolution - Exercise 2a, b 31



220

Measurement sheet 5


differential element. Measure the deformation of the square-wave at the output using the

oscilloscope and draw the curve shape on the measurement sheets.

Set the frequency generator to 5 kHz and UESS = 4 V

Adjust the square-wave signal so that the botton line is on the zero line.




LE 21118-01Differential elementSolution - Exercise 3a, b 32



221

Measurement sheet 6





222

Measurement sheet 7

Apply sinusoidal voltage of different frequencies to the input of the high-pass filter.







f g = 1 = 6047 Hz

2 x π x R x C


LE 21118-01High-pass filterSolution - Exercises 4a, b 34



223

Measurement sheet 8

Determine the scale for the ccordinate system.

Enter the actual value of the cut-off frequency on the graph.


LE 21118-01High-pass filterSolution - Exercises 4a, b 35



224

Measurement sheet 9

Apply sinusoidal voltage of different frequencies to the input of the band-pass filter.



Set a sinusoidal AC voltage UESS = 10 V an the frequency generator.

Successively set the frequencies given in the table.


Calculate the cut-off frequency and the band with:

f u = 1 = 1061 Hz

2 x π x R1 x C1

f o = 1 = 106.1 kHz

2 x π x R2 x C2

b = f o - f o = 105 Hz


LE 21118-01Band-pass filterSolution - Exercises 5a, b 36



225


Determine the scale for the coordinate system.

Enter the actual value of the cut-off frequency on the graph and determine the band width.


LE 21118-01Band-pass filterSolution - Exercises 5a, b 37



226


11 LE 21119-01 - Basic Electronics Circuits - Rectification

Summary of contents

Contents: One-pulse center-tap connection

Two-pulse center-tap connection

Two-pulse bridge connection

Rectification, demonstration model

Skills to be acquired: Setting up of rectifier circuits

Projects: One-pulse center-tap Exercises 1a, b

Two-pulse center-tap Exercises 2a, b

Two-pulse bridge

connection Exercises 3a, b





227

With ohmic load

The diode is forward-biased during the positive half-wave of the secondary voltage U 2.

The anode of diode V1 is negative in respect of the cathode during the negative half-wave

and the diode is reverse-biased.

The diagrams show the curves of the secondary voltage U2 and the voltage URL across

the load resistor RL.

The sympol u (lower case) expresses the instantaneous voltage value, the pulsating DC

voltage measured with the oscilloscope.

U (upper case) denotes the DC voltage measured using a moving-coil instrument (mean

time value).

The ideal no-load direct voltage Udi may be calculated in accordance with the following

formula:

Udi = 0.45 x U2

MCEIndustrietechnik

LE 21119-01 One-pulse center-tap connection 1



228

With ohmic load and capacitor

A pulsating DC voltage is not usually suitable as a power supply. For this reason, a

capacitor is parallel-connected to RL. The capacitor is called the charging capacitor CL.

The capacitor is charged by the positive half-wave of the secondary voltage (U 2) as Jong

as the "charging voltage" (U2) is larger than the capacitor voltage (u).If u2 falls below the capacitor voltage u, the capacitor is discharged via the load resistor

RL:

a) No discharge b) Slow c) Rapid

MCEIndustrietechnik

LE 21119-01 One-pulse center-tap connection 2



229

With ohmic load

The two-way or center-tap connection is basically the combination of two one-way

connections.

V1 is forward-biased during the positive half-wave, and V2 is forward-biased during the

negative half-wave.

In both cases, the current flows in the same direction through RL. This connection is also

called full-wave rectifier.

The output voltage URL has the curve shown below.

The voltage URL measured across RL using the measuring instrument is approximately

double that which was measured in the case of the one-way connection.

Udi ≈ 0.45 x (U2 + U3) since U2 = U3 it follows that:

Udi ≈ 0.9 x U2

MCEIndustrietechnik

LE 21119-01 Two-pulse center-tap connection 3



230


The charging capacitor CL causes "smoothing" of the output voltage in this circuit also. Forthis reason, it is also called a smoothing capacitor. The degree of smoothing depends an

the size of the capacitor.

High capacitor value - good smoothing (low ripple voltage).

Low capacitor value - poor smoothing (high ripple voltage).

Formula for Udi with C: Udi ≈ √ 2 x U2 (since U2 = U3)

MCEIndustrietechnik

LE 21119-01 Two-pulse center-tap connection 4



231

Ripple voltage and frequency

The voltage at RL, which is not fully smooth is considered as a DC voltage Udi with a

superimposed AC voltage Uü. This AC voltage is called a ripple voltage.

The ripple voltage Uü may be determined using an oscilloscope.

It is measured using the peak-to-peak values.

(Uüss: peak-to-peak superimposed AC voltage)

The frequencies for one-way and center-tap connections are different. One-way

connection f = 50 Hz, center-tap connection f = 100 Hz.

With ohmic load

The bridge connection is used most often. In the contrast to the one-way connection, both

half-waves are used.

In contrast to two-way connection, a transformer with center tap is not required.

MCEIndustrietechnik

LE 21119-01 Two-pulse bridge connection 5



232

The current I flows through the diodes V1 and V3 during the positive half-wave; V2 and V4

are reverse-biased. The current flows through the diodes V2 and V4 during the negative

half-wave; V1 and V3 are reversed-biased. The current therefore flows through R L in the

same direction during the positive and negative half-waves. A pulsating DC voltage is

available at RL.

Calculation formula: Udi ≈ 4.9 x U2


If a charging capacitor CL is connected in parallel with the load resistor RL, the voltage u is

smoothed.

The diagram shows the effect of the smoothing (charging) capacitor “CL”:

Calculation formula for Udi: Udi = √ 2 x U2

CL is charged during the period between t11 and t12. CL is discharged via RL between t12

and t21, and CL is recharged between t21 and t22 etc.

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233

AC component (ripple)

The concept of ripple is defined in accordance with DIN 41755. It

is understood as the ratio of the actual value of the superimposed AC voltage to the ideal

noload direct voltage.

W = US

Udi

Other ways of depicting a bridge connection

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234

Circuit diagram Suggestion for assembly diagram

Wiring examplesOne-pulse center-tap Two-pulse

bridge connection connection

Set the signal generator to a low frequency: 1Hz - 10Hz. Signal amplitude: 6.3 V.

7 1 Signal generator 0.5 mm, silvered

6 1 Bare wire 4 mm

5 10 Socket contacts R1-R4, RL100 1W

4 5 Resistor H1 - H4 red

3 4 LED 80 x 20 x 5 mm

2 1 Support blocks 70 x 60 mm

1 1 PC Board

ItemNo.


MCEIndustrietechnik

LE 21119-01RectificationDemonstration model 8



235

Circuit diagram

Soldering capacitor C1 on supports.


Exercise 2b: set-up on PC board - Items 3 - 8


7 1 Capacitor C2 elco 470 µF/25 V

6 1 Resistor RL 0.5 W 470 Ω

5 1 Diode V1, 1N4004



2 1 Accessory set 2GA5101-8G Fund. of analog.-

electr.Engineering

1 1 Electronics trainer 2GA5101-3B

ItemNo.


MCEIndustrietechnik

LE 21119-01One-pulse center-tap connectionExercises 1a, b 9



237

Trade: ____________________________Name:________________________________

Measurement sheet 1

Measure the voltage URL using the multimeter

Measured Calculated

Without charging capacitor V V

With charging capacitor V V

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238

Trade: ____________________________Name:________________________________

Measurement sheet 2

Record the oscillograph of the output voltage URL, at the load resistor.

Rectification without CL

Rectification with CL

How high is the ripple voltage? Uüss =

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240

Task

Study and measure the effect of rectification with and without a charging capacitor. Enter

the values in the table and draw the oscillographs.


1. Arrange all the required component modules on the assembly board






circuit diagram.

2. Connect the finished PC board to the power supply. (e.g. mounting rack LE 2195/96)

3. Carry out the measurements in accordance with the measurements heet.



PC board holder, tin-lead solder 1.0 mm, soldering fron 30 W, flat pliers, pointed soldering



Multimeter, oscilloscope.


Exercise care when handling the hot soldering fron (risk of bums and fire).


injuries.

Exercise care when handling capacitors. There is a danger of explosion if the operatingvoltage is too high or if the polarity is incorrect (eye injuries).

MCEIndustrietechnik




241

Trade: ____________________________Name:________________________________

Measurement sheet 3


Measured Calculated



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242

Trade: ____________________________Name:________________________________

Measurement sheet 4





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243

Circuit diagram

Solder capacitor onto the soldering tag!




7 1 Capacitor C2 elco 470 F/25 V

6 1 Resistor RL 0.5 W 470

5 4 Diode V1-V4 1N4004



2 1 Accessory set 2GA5101-8G Fund. of analog. -

electr.Engineering

1 1 Electronics trainer 2GA5101-3B

ItemNo.


MCEIndustrietechnik




244

Task

Study and measure the effect of rectification with and without a charging capacitor. Enter

the values in the table and draw the oscillographs.


1. Arrange all the required component modules on the assembly board

2. Wire the circuit in accordance with the circuit diagram





circuit diagram.

2. Connect the finished PC board to the Power supply. (e.g. mounting rack LE 2195/96)

3. Carry out the measurements in accordance with the measurements heet.



PC board holder, tin-lead solder 1.0 mm, soldering fron 30 W, flat pliers, pointed soldering



Multimeter, oscilloscope.


Exercise care when handling the hot soldering fron (risk of bums and fire).


injuries.

Exercise care when handling capacitors. There is a danger of explosion if the operatingvoltage is too high or if the polarity is incorrect (eye injuries).

MCEIndustrietechnik

LE 21119-01Two-pulse bridge connectionExercises 3a, b 18



245

Trade: ____________________________Name:________________________________

Measurement sheet 5


Measured Calculated



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246

Trade: ____________________________Name:________________________________

Measurement sheet 6





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247

Trade: ____________________________Name:________________________________

Measurement Sheet 7

Measurement success check

1. What is the maximum possible voltage URL across the output terminals of the center-

tap connection?

2. Which diagram shows a bridge connection?

3. A rectifier circuit (see circuit diagram below) is connected to 220 V AC. Which voltage

is available at the unloaded output?

MCEIndustrietechnik

LE 21119-01 Exercises 1 - 3 21



250

Trade: ____________________________Name:________________________________


6. The depicted voltage is measured using a moving-coil instrument set to the DC

voltage range. Which value is indicated ?

7. A frequency of 50 Hz is measured in a center-tap circuit with charging capacitor CL.

Which statement is correct?

The circuit is operating correctly

A diode is disconnected

The capacitance of the charging capacitor is too low

The capacitance of the charging capacitor is too high

The charging capacitor is missing

MCEIndustrietechnik

LE 21119-01 Exercises 1 - 3 24



251

Measurement sheet 1


Measured Calculated

Without charging capacitor 2.75 V 2.83 V

With charging capacitor 8.5 V 8.9 V


LE 21119-01One-pulse center-tap connectionSolution - Exercises 1a, b 25



252

Measurement sheet 2


How high is the ripple voltage? Uüss = 0.4 V





253

Measurement sheet 3


Measured Calculated







254

Measurement sheet 4




How high is the ripple voltage? Uüss = 0.3 V





255

Measurement sheet 5


Measured Calculated



At these relatively low voltages, the voltage drop across the diodes is extremely

pronounced; that is why the calculated values are between 10 % and 20 % above the

measured values.


LE 21119-01Two-pulse bridge connectionSolution - Exercises 3a, b 29



256

Measurement sheet 6



How high is the ripple voltage? Uüss = 290 mV


LE 21119-01Two-pulse bridge connectionSolution - Exercises 3a, b 30



257

Measurement Sheet 7

Measurement success check

1. What is the maximum possible voltage URL across the output terminals of the center-

tap connection?

2. Which diagram shows a bridge connection?

3. A rectifier circuit (see circuit diagram below) is connected to 220 V AC. Which voltage

is available at the unloaded output?


LE 21119-01 Solution - Exercises 1 - 3 31



258

Measurement sheet 8

4. The half-wave or one-way rectifier circuit is supplied with a sinusoidal AC voltalte from

the transformer.

Which oscillograph shows the curve of the voltalte URL?





260

Measurement sheet 8

6. The depicted voltage is measured using a moving-coil instrument set to the DC

voltage range. Which value is indicated ?

7. A frequency of 50 Hz is measured in a center-tap circuit with charging capacitor CL.

Which statement is correct?

The circuit is operating correctly

A diode is disconnected

The capacitance of the charging capacitor is too low

The capacitance of the charging capacitor is too high

The charging capacitor is missing



ee049 electronic components exercises pr inst

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