Electrical Engineering Department Electronics Lab Manual

Electrical Engineering Department

Electronics Lab Manual

2013

Revised by:Haneen Al-'Out

Nasim zayed

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An-Najah National University

Faculty of Engineering

Issue number: AD3-3

Department Name : Course Name:Electronic circuit lab Course # :63314

Report Grading Sheet

Instructor Name: Experiment name: Academic Year: Experiment #:

Students:1- 2-3- 4-

Report’s OutcomesILO A,B=( 50 ) % ILO D =( 20 ) % ILO A =( 20 ) % ILO K =( 10 ) %

Evaluation Criterion Grade PointsIntroductionSufficient,Clear and complete statement of objectives. Apparatus/ ProcedureApparatus sufficiently described to enable another experimenter to identify the equipment needed to conduct the experiment. Procedure sufficiently described.Experimental Results and CalculationsResults analyzed correctly. Experimental findings adequately and specifically summarized, in graphical, tabular, and/or written form.Conclusions Conclusions summarize the major findings from the experimental results with adequate specificity. ReferencesComplete and consistent bibliographic information that would enable the reader to find the reference of interest.Total

Total

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1Key 1: Report structure: Abstract, Introduction, results ….etc 2: References: Complete and consistent bibliographic information that would enable the reader to find the reference of interest. 3: Time keeping: Sticking to the allowed time for each student 4: Presentation skills: Personality, attention to audience, body language and gestures.



Issue number : AD4-1

Department Name:

Course name : Electronic Circuit Course # :63314

Mini- Project Evaluation Form

Project name : Semester : Instructor name : Project ILOs (with percentages) :

Final grade

Presentation evaluation Report evaluation

Students names Grade Answered questions

Presentation skills 4

Time keeping 3

GradeReferences

2 Addressing problems

Data analysis

Report structure 1

Points Points Points Points Points Points Points Points

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2Key 1: Report structure: Abstract, Introduction, results ….etc 2: References: Complete and consistent bibliographic information that would enable the reader to find the reference of interest. 3: Time keeping: Sticking to the allowed time for each student 4: Presentation skills: Personality, attention to audience, body language and gestures.



Issue number : AD4-1

Final grade

Presentation evaluation Report evaluation

Students names Grade Answered questions

Presentation skills 4

Time keeping 3

GradeReferences

2 Addressing problems

Data analysis

Report structure 1

Points Points Points Points Points Points Points Points

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63314 Electronic Circuits Lab - Department: Electrical Engineering

Semester Teaching Methods Credits

4Lecture Rec. Lab.

Project Work

HR Other Total

0 0 36 28 36 0 1001

Language English

Compulsory / Elective

major compulsory

Prerequisites

P1 : Electronic Circuits I (63260) OR Electronic Circuits I (63214)

Instructors

Eng Haneen Al-Out, [email protected]

Course Contents

Electronics lab has been prepared to equip the students with the necessary practical and theoretical knowledge of electronic principles. During the lab the students become very familiar with , Types of Diodes, Rectifier diode, Half wave rectifier, Bridge rectifier, On state and off state characteristic of zener diode, Testing the layering and rectifying of bipolar transistor, Characteristic of the transistor, Depletion layer Fets, Characteristic of the Fets, Multistage amplifier, Differential amplifier, Push pull output amplifier, Operational amplifier

Learning Outcomes and Competences

1 Basic Knowledge of Principles of Electrical circuits and analysis Knowledge and also to take measurements deferent type of Electronic circuits

A B 50 %

2 2) An ability To function and work the experiments as team D 20 %

3 An ability to identify, formulate, and solve electronics circuits problems A 20 %

4 An ability ORCAD methods to solve electronics circuits engineering analyses and designee

K 10 %

Textbook and / or Refrences

Electronic circuits lab

Assessment Criteria

Percent (%)

Projects 10 %

Laboratory Work 60 %

Final Exam 30 %

Week Subject

2 EXPERIMENT # 2: Junction diodes and applications

3 EXPERIMENT # 3: Zener diodes and applications

4 EXPERIMENT # 4 Bipolar Junction transistors

5 EXPERIMENT # 5: Soldering and Desoldering skills

6 EXPERIMENT # 6: Junction field effect Transistors

7 EXPERIMENT # 7: Amplifiers

8 EXPERIMENT # 8: Multistage amplifiers

9 EXPERIMENT # 9: DIFFERENTIAL AMPLIFIER

10 EXPERIMENT # 10: [10] Push Pull amplifiers

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1 Introduction -Pspice (ORCAD ) 1 I

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13 EXPERIMENT # 13: Dynamic Behavior of an Op-Amp

11 EXPERIMENT # 11 Operational amplifiers study

12 EXPERIMENT # 12: Op-amp circuits NON-INVERTING AMPLIFIER

13 EXPERIMENT # 13: Dynamic Behavior of Op-amps

14 EXPERIMENT # 14: Mini Project Discussion

15 Practical Exam

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16 Theoretical Exam

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An-Najah National University Electrical Engineering Department

Electronics Laboratory Safety Rules

Acquaint yourself with the location of the following safety items

within the laboratory:

a. Fire extinguisher

b. First aid kit

c. Telephone and emergency numbers:

The following Regulations and Safety Rules must be observed:

1. Avoid bulky, loose or trailing clothes. Avoid long loose hair. Remove metal bracelets,

rings or watchstraps when working in the laboratories.

2. Smoking, food, beverages and other substances are strictly prohibited in the laboratory at

all times. Avoid working with wet hands and clothing.

3. You may enter the laboratory only when authorized to do so and only during authorized

hours of operation.

4. Do not open, remove the cover, or attempt to repair any broken equipment or defective

parts.

5. When the lab exercise is over, all instruments must be turned off.

6. Concerning experiments, electrical equipments, materials and devices:

o Before equipment is energized ensure:

a. circuit connections and layout have been checked by a Teaching

Assistant (TA) and

b. all colleagues in your group give their assent.

o Experiments left unattended should be isolated from the power supplies. If for a special reason, it must be left on, a barrier and a warning notice are required.

o Make sure the power is off when wiring or making changes to circuit.

o Make sure instrumentation is set on proper range for the desired type of

measurement, BEFORE ENERGIZING CIRCUIT.

o Make sure components used are of a rating that will withstand applied current

and voltage.

7. It is the duty of all concerned (who use electronics laboratory and visitors) to take all

reasonable steps to safeguard the HEALTH and SAFETY of themselves.

Note: If any safety questions arise, consult the lab instructor or staff for guidance and

instructions. 000007000007

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[14] Mini Project Discussion

[15] Practical Exam

[16] Theoretical Exam

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[1] Introduction -Pspice (ORCAD )

[5] Soldering and Desoldering skills

[13] Dynamic Behavior of Op-amps 104

[7] Amplifiers 71

[10] Push Pull amplifiers 84

Schedule Lab: page

[2] Junction diodes and applications 16

[3] Zener diodes and applications 26

[9] Differential amplifiers 80

[8] Multistage amplifiers 76

[11] Operational amplifiers 88

[12] Op-amp circuits 95

[4] Bipolar Junction transistors 40

[6] Junction field effect Transistors 58

General Report Guidelines: The lab report reinforces the material that was learned in the lab and helps you develop effective technical communication skills. Development of both oral and written technical communication skills is one of the most important things you can learn as an undergraduate student. A technical report is expected to contain the following items or subsections:

• Cover page,

• Introduction,

• Experimental procedure and methodology

• experimental results,

• Conclusion or summary,

• References. Introduction:

It should contain a brief statement in which you state the objectives, or goals of the experiment.

The Procedure:

measurements were made.

Results/Questions:

This section of the report should be used to answer any

questions presented in the lab handout. Any tables and/or

circuit diagrams representing results of the experiment should

A brief conclusion summarizing the work done, theory

applied, and the results of the completed work should be

included here. Data and analyses are not appropriate for the

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be referred to and discussed/explained with detail.

It describes the experimental setup and how the

Conclusion:

conclusion.

References:Complete and consistent bibliographic information that wouldenable the reader to find the reference of interest.

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1) Opening PSpice

1. Find PSpice. Open Schematics and then click on the schematic icon .

2. You will see the window as shown in Figure 1.

Figure 1

2) Drawing the circuit

I. Getting the Parts

1. Clicking on the 'get new parts' button , or Pressing "Control+G".

2. Select a part that you want in your circuit. This can be done by typing in the name (part name) or scrolling down the list until you find it.

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Experiment # 1 Introduction –Pspice tutorial

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Figure 2

3. Upon selecting your part, click on the place button then click where you want it placed , if you need multiple instances of this part click again

4. If you want to take a part and close then you just select the part and click onplace& close.

II. Placing the Parts

1. Just select the part (It will become Red) and drag it where you want it.2. To rotate parts so that they will fit in you circuit nicely, click on the part and

press3. "Ctrl+R" (or Edit "Rotate"). To flip them, press "Ctrl+F" (or Edit

"Flip").If you have any parts left over, just select them and press "Delete".

III. Connecting the Circuit

1. You'll have to attach them with wires,go up to the tool bar andselect "Draw

Wire" or"Ctrl+W" .

2. With the pencil looking pointer, click on one end of a part, when you moveyour mouse around, you should see dotted lines appear. Attach the other end of your wire to the next part in the circuit.

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3. Repeat this until your circuit is completely wired.

IV. Changing the Name of the Part

1. To change the name, double click on the present name (C1, or R1 orwhatever your part is), t he top window, you can type in the name you want the part to have.

Figure 3

V. Changing the Value of the Part

• If you only want to change the value of the part, you can double click on the present value and type in the new value and press OK.

Figure 4

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VI. Making Sure You Have a GND

This is very important. You cannot do any simulation on the circuit if you don't have a ground. If you aren't sure where to put it, place it near the negative side of your voltage source.

VII. Voltage and Current Bubbles

These are important if you want to measure the voltage at a point or the current going through that point.To add voltage or current bubbles, go to the right side

of the top tool bar and select "Voltage/Level Marker" (Ctrl+M) or

"Current Marker" . To get either of these, go to "Markers" and either"Voltage/Level Marker" or "Current Marker".

3) Analysis Menu

Figure 5

To open the analysis menu click on the button.

A. DC Sweep

The DC sweep allows you to do various different sweeps of your circuit to see how it responds to various conditions.

• You need to specify a start value, an end value, and the number of points you wish to calculate.

• Another excellent feature of the DC sweep in PSpice, is the ability to do a nested sweep.A nested sweep allows you to run two simultaneous sweeps to see how changes in two different DC sources will affect your circuit.Once you've filled inthe main sweep menu, click on the nested sweep button and choose the second type of source to sweep and name it, also specifying the start and end values.(Note: In some versions of PSpice you need to click on enable nested sweep).

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• This is a simple, but incredibly useful sweep. It will not launch Probe and so give you nothing to plot. But by clicking on enable bias current display or enable bias voltage display, this will indicate the voltage and current at certain points within the circuit.

C. Transient

The transient analysis is probably the most important analysis you can run in PSpice, and it computes various values of your circuit over time

• Choose Analysis…Setup from the menu bar, or click on the Setup Analysis

button in the toolbar. The Analysis Setup dialog box opens.

• Click on the Transient button in the Analysis Setup dialog box. The Transient dialog box opens.

• Two very important parameters in the transient analysis are (see Figure6):

print step final time.

Figure 6

• The ratio of final time: print step (Keep print step atleast 1/100th of the final time) determines how many calculations PSpice must make to plot a wave form. However play with step time to see what works best for your circuit.

• You can set a step ceiling which will limit the size of each interval, thus increasing calculation speed.

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4) Graphing:

If you don't have any errors, you should get a window with a black background to pop up.

5) Finding Points:

There are Cursor buttons that allow you to find the maximum or minimum or just a point on the line. These are located on the toolbar (to the right).Select which curve

you want to look at and then select "Toggle Cursor" .Then you can

find the max, min, the slope, or the relative max or min ( is find relative max).

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1 Junction Diodes and applications 1.1 Effects of the p-n-junction of diodes 1.1.1 Basics

Diodes are two poles semiconductors. The semiconductor diode consists

of a semiconductor crystal (silicon, germanium), one side is n-doped, the

other one is p-doped. At the p-n-junction there is a barrier layer. The different

doped layers are equipped with ohm contacts and could be connected to

other components.

When voltage is applied, the barrier effect could be increased or nullified.

The semiconductor diode allows current flow in one direction and blocks the

other one.

The diode has following the circuit symbol in pic. 1.1.

1.1.2 Tests Test

Examine the effect of the rectifier diode’s p-n-junction on the current flow in relation to the applied

voltage and its polarity.

~-

+

Polarität1

Polarität2

RV=680Ω

U0 = 0...30 V V1 V1

IF

UF

V1 = Si-Diode 1A

pic. 1.2 Test proceeding:

• One by one apply following voltages UF according to chart 1.1 to the diode’s polarity 1 (pic. 1.2)

and measure the respective conducting current IF. Fill the values in chart 1.1. Use the current

error circuit.

pic. 1.1

P N

Electrical Engineering Department Electronics Lab Manual

1

Experiment #2

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• Then change the polarity of the diode (polarity 2) and repeat same test with the voltage

values of chart 1.2., remove the series resistor. (680 Ω) and set the voltage directly at the

power supply. Use the voltage error circuit for this measurement. An exact measurement of

the reverse current IR requires a highly sensible multimeter (100 nA).

• For construction of the diode characteristic curve use the values of both charts and fill in the

diagram pic. 1.3. chart 1.1 chart 1.2 chart 1.3 Test: What is the name of the voltage, when the diode gets conductive? Answer:

UF [V] 0 0,1 0,2 0,3 0,4 0,5 0,6 0,65 0,7 0,75

IF [mA]

UR [V] 0 2,5 5 10 15 20 25 30

IR [nA]

IF [mA]

25

20

15

10

5

30

U R [V]

20 10 0

20

0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8

U F [V]

40

60

I R [nA]

80


2

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1.2 Characteristics of Diodes with Various Semiconducting Materials 1.2.1 Basics

If AC voltage, with periodically changing instantaneous value between zero and the maximum value, is

applied to a diode, the diode characteristic could be presented completely at the display of the

oscilloscope. Therefore the diode voltage is applied to the X-amplifier. The diode current could be

measured at a series resistor as a proportional voltage. That voltage is applied to the inverted Y-

amplifier. 1.2.2 Tests Test

Store the characteristics of a silicon diode, a geranium diode and a gallium-arsenide diode (LED)

with your oscilloscope. Test proceeding:

• Set-up the circuit in pic. 1.4 and one by one present the voltage curve for the silicon diode, the

germanium diode and the gallium-arsenide diode (LED red) at the oscilloscope (X/Y

presentation).

G

R=330ΩY

X

uss = 6V Sinusf = 50 Hz

V1

0 V

pic.1.2.2.1

Note:

Due to the reversed polarity of the voltages, the Y-amplifier has to be inverted.

A mirror image is presented, when using an oscilloscope without inverting function.

Caution:

To avoid short-circuit of the measuring voltage across mutual earth, the generator output and the

oscilloscope inputs must be potential-free.

• Draw the curves of all three diodes into the diagram (pic. 1.4).


3

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pic. 1.4 Test 1: Which characteristics of the diodes vary?

Answer:

Threshold voltage:

Silicon diode:________ Geranium diode:________ Gallium arsenide Diode (LED red):________. Test 2: What is the difference of the internal resistors from each diode when current is changed

from 1,5 ... 4,5 mA? Answer:

Silicon diode:

Rdiff = differential internal resistance in Ω ∆IF = conducting current in A

∆UF = bias voltage in V Germanium diode: Gallium arsenide-diode (LED red):

=∆∆

=F

F

diff

I

UR =

∆∆

=F

F

diff

I

UR

Test 3: State advantages of dynamic curves by comparison to static ones?

Answer:

I F [mA] 4,5

3,0

1,5

-2,0 UR [V]

-1,5 -1,0 -0,5 0 0,5 1,0 1,5 2,0

U F [V]

=∆∆

=F

Fdiff

I

UR


4

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1.3 Half Wave Rectifier 1.3.1 Basics

If a semiconductor diode is integrated to a

circuit, current flow is only possible at one polarity of the applied voltage (bias direction).

When polarity is reversed the blocking function

is effective and no current will flow. If AC voltage is applied to such a circuit, it is only

conductive at the half wave with the respective

polarity. The other half wave is blocked. Current within the circuit will flow only in one

direction.

pic. 1.5 M1 single pulse center tapped circuit 1.3.2 Tests Test

Examine the rectifying effect of a semiconductor diode within a single pulse center tapped circuit M1

with a multimeter and an oscilloscope.

Test proceeding

• Set-up the circuit in pic. 1.6 without smoothing capacitor.

• Measure the input voltage UE and the DC voltage Ud with the multimeter and calculate the ratio

d E

G~

+1 2

Si 1AUeff = 5Vf = 50 Hz

Sinus

UERL=

10kΩ CG

Ud

pic. 1.6


5

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Y Y

between U and U .

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• Measure the input voltage UE and the DC voltage Ud with the oscilloscope and draw the curve in pic. 1.7.

Settings: Y1 = 5 V / part Y2 = 5 V / part X = 5 ms / part Determine by help of the oscilloscope picture in pic. 18.7 the peak-peak-value and the frequency of the ripple UBr . Note: Ripple is the AC portion of a pulsating DC voltage Ud. .

pic. 1.7 Then connect the smoothing - capacitor CG according to chart 1.3 in parallel to load resistance and repeat your measurements.

Please regard the polarity when

connecting the electrolytic capacitors! Fill

all determined values in chart 1.3.

chart 1.3

Settings:

Y1 = 5 V / part

Y2 = 5 V / part

X = 5 ms / part Draw the curve for the input voltage UE and the DC voltage Ud, with smoothing capacitor 10 µF into the chart in pic. 1.8.

pic. 1.8


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• Reverse the diode within the circuit in pic. 1.4, measure the voltages UE and Ud without smoothing capacitor.

• Draw the curves for both voltages into the chart 1.9

Settings: Y1 = 5 V / part Y2 = 5 V / part X = 5 ms / part

pic. 1.9 Test 1: What is the frequency of the ripple UBr? Answer: Test 2: What happens if the diode in circuit pic.1.6 is reversed in polarity? Answer: Test 3: At which connector of the diode is the plus pole of the obtained DC voltage Ud? Answer: Test 4: State the value of blocking voltage affecting the diode with smoothing capacitor CG? Answer: Test 5: What influence does the smoothing capacitor have on the peak-peak-value of the

ripple? Answer:

- 0 (Y2 ) = U d


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1.4 Full Wave Refctifier 1.4.1 Basics

At the single pulse center-tapped circuit M1 only a half-wave of the AC voltage is used. Disadvantage is the small portion of DC voltage with high ripple.

This disadvantage is avoided in the two-pulse bridge B2, as polarity of the opposed half-wave is changed and added to the DC voltage.

pic. 1.10 B2 Two-pulse Bridge Circuit 1.4.2 Tests Test Examine by help of the oscilloscope and multimeter the characteristics of two-pulse bridge B2.

V1 V2

a

G~

UE

RL

10 k

+CG

Ud

Y

X

b

V3 V4

V1 - V4 = Si Diode 1A

Ueff = 5Vf = 50 Hz

Sinus

pic. 1.11 Test proceeding:

• Set-up the circuit in pic. 1.11 without smoothing capacitor. Measure input voltage UE and DC voltage Ud with your multimeter and determine the ratio Ud to UE.

• Measure with the oscilloscope all the input voltages UE and DC voltages Ud and draw the curve

into pic. 1.12.

UE

UE

t

Ud

Ud

t


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• Determine by help of the oscilloscope curve in pic. 1.12 the peak-peak-value and the frequency of ripple voltage UBr. Note: Ripple voltage is the portion of AC voltage within pulsating DC voltage Ud.

Setting:

Y = 5 V / part

X = 5 ms / part - 0 (Y1 ) = U E

- 0 (Y2 ) = U d

pic. 1.12

• Then one by one connect the smoothing capacitors CG in parallel to

the load resistor RL according to chart

1.5 and repeat your measurements. Please regard the polarity when

connecting the electrolytic capacitor! • Fill the determined values in chart 1.5.

chart 1.5

Settings:

Y = 5 V / Teil

X = 5 ms / Teil

- 0 (Y1 ) = U E

Draw the curve for the input voltage UE

and the DC voltage Ud with smoothing

capacitor 10 µF in chart 1.13.

- 0 (Y2 ) = U d

pic. 1.13


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Test 1: What is the ratio between DC voltage Ud and the applied effective input voltage UE

(without smoothing capacitor)? Answer: Test 2: What is the frequency of the ripple voltage UBr? Answer:


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2 Zener Diodes and appliactions 2.1.1 Basics Z-Diodes are silicon-semiconductor diodes, with conduction curve similar to rectifier diodes. In

comparison to rectifier diodes, the Z-diode has a lower breakdown voltage within stop – or reverse band

(Z-Voltage). When exceeding the breakdown voltage, current increases steeply in reversed direction (Zener effect). In opposite to the rectifier diodes, Z-diodes are utilized in reversed direction.

To protect from overload, a resistor is connected in series to the Z-diode, its value is calculated as follows:

Ub = applied operating voltage UZ = Zener-voltage of the utilized diode-type IZ = average acceptable Z-current IL = current affecting the load resistance RL in parallel to the Z-diode

Circuit Symbol pic. 2.1 Due to its characteristics the Z-diodes are used for voltage stabilization and voltage limitation. 2.1.2 Tests Test

Present the curve of a Z-diode with your oscilloscope and determine the Z-voltage. Test proceeding:

• Set-up the circuit in pic. 2.2. Switch the oscilloscope to X/Y presentation.

Note: As both voltages are reversed poled to the reference point, the Y-amplifier has to invert.

Oscilloscopes without inverting function show a mirror like presentation.

G~

-Y

X

Zener-Diode 10 V / 40 mA

RV 1kΩ

Ueff = 24 Vf = 50 Hz

Sinus

pic. 2.2

LZ

ZbV

II

UUR

+−

=


11

Experiment # 3

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Note: The output of the power supply or the inputs of the oscilloscope have to be potential-free to avoid

short-circuit due to mutual earth. Draw the presented oscilloscope display into the chart 2.3.

Settings: Y = 10 V / part X = 2 V / part

- 0 (Y)

0 (X)

pic. 2.3 Test 1: Determine the Z-voltage UZ. Answer: UZ = Test 2: What is the value of the maximum current IZ? Answer:

R

UIz

R

=max

Test 3: What is the value of the threshold voltage US? Answer: US =


12

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2.2 DC Voltage regulation using Z-Diodes 2.2.1 Basics Because of the steep increasing current in the reverse band of a Z-diode, it is used for regulating DC

voltage.

Therefore a resistor, at which the difference between the unstable input voltage and the limited output voltage drops, is connected in series to the Z-diode.

Limited output voltage is equal to the Z-voltage and is depending on the Z-diode type. 2.2.2 Tests Test 1 Examine the relationship between output voltage and input voltage at a limiter circuit with Z-diodes. Test proceeding:

• Set-up the circuit in pic. 2.4. One by one set to DC voltages UE according to chart 2.1.auf.

Measure the respective output voltages with the multimeter and fill in chart 2.1.

• Graphically present the relation between output voltage and input voltage within the diagram

2.5.

~-

10 V40 mA

RV 330Ω

U=0...15V UE UA

pic. 2.4


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chart 2.1 pic. 2.5

UE [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

UA [V]


14

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Test 2

Examine the relationship between Zener current IZ and input voltage

Test proceeding: • Set-up the circuit in pic. 2.6. One by one set the DC voltages UE according to chart 2.6. • Measure the respective Z-currents IZ with the multimeter and fill the current values in chart 2.2.

• Graphically present the relation of Z-current IZ and input voltage UE in the diagram pic. 2.7.

~-

RV 330 Ω

10 V40 mA

Ugl = 0...15V

UE UA

IZ

pic. 2.6


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chart 2.2 IL [mA] pic. 2.7

UE [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

IZ [V]


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Test 3 Relationship between zener current Iz and load current I load Test proceeding:

• Set-up the circuit in pic. 2.8. Adjust the potentiometer to the load currents IL according to chart 2.3.

Note:

Increase the resistor R1 to 1 kΩ, 2,2 kΩ, 4,7 kΩ and 10 kΩ , use low load current.

~-

U = 15 V

RV 330 Ω R1 680 Ω

R 1 kΩ

ILIZ

UE 10 V40 mA

pic. 2.8

• Graphically present the relation between Z-current IZ and load current IL within the diagram pic. 2.9.


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chart 2.3 pic. 2.9 Test 1: What condition is needed that the output voltage remains constant in a limiter circuit

with Z-diode? Answer: Test 2: When does Z-current IZ flow? Answer: Test 3: Under which condition is the limiting effect maintained although under load? Answer:

IL [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

IZ [V]


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2.4 AC Voltage Limiting and Overvoltage Protection with Z-Diodes 2.4.1 Basics

If the voltage applied to a Z-diode is less than Z voltage, no current is flowing. Current only flows if the

value of Z voltage is reached or exceeded. Therefore this component could be used to protect other

parts (e. g. MOS-components) from over voltage.

To avoid a rectifying effect due to low threshold voltage, two Z-diodes are connected in opposite,

thus a polarity independent voltage limitation is received.

2.4.2 Tests

Test 1

Statically record the curve UA = f (UE) of two in opposite directed Z-diodes.

Test proceeding:

• Set-up the circuit in pic. 2.11. One by one set to DC voltages UE according to chart 2.6. Measure the respective output voltage UA with your multimeter and fill in chart 2.6.

• Then reverse the poles of input voltage UE and go on with your measurements. Note the respective output voltages UA in chart 2.7.

V1 = Z-Diode, 10 V / 40 mA V2 = Z-Diode 3,3 V / 130 mA

pic. 2.11


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• Graphically present the relation between output voltage UA and input voltage UE in chart 2.7. chart 2.6 chart 2.7 pic. 2.12

UE [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

UA [V]

-UE [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

-UA [V]

UA [V] 10

8

6

4

2

-16 -UE [V]

-14 -12 -10 -8 -6 -4 -2 -2

-4

2 4 6 8 10 12 14 16

UE [V]

-6

-8

-UA [V] -10


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Test 2 Examine with your oscilloscope the voltage limiting effect of two opposite directed Z-diodes. Test proceeding:

• Apply a sine-wave AC voltage Ueff = 10 V; f = 50 Hz to the circuit in pic. 2.14 and set the input voltage to UE eff = 2 V at the potentiometer.

V1=Z-Diode 10V/40mA V2=Z-Diode 3,3 V/130mA

pic. 2.14

• Measure the input voltage UE and output voltage UA with your oscilloscope and draw the curve into diagram pic. 2.15.

• Then increase input voltage to UE eff = 10 V.

• Repeat your measurements and fill the values into pic. 2.16.

Notes: Y1 = input voltage UE

Y2 = output voltage UA

- 0 (Y1 )

Settings: X = 5 ms / part

Y1 = 2 V / part Y2 = 2 V / part

- 0 (Y2 ) pic. 2.15


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Notes: Y1 = Input voltage UE

Y2 = Output voltage UA

- 0 (Y1 )

Settings: X = 5 ms / part Y1 = 10 V / part Y2 = 5 V / part

- 0 (Y2 ) pic. 2.16 Test 1: Do you know the application possibilities for a Z-diode? Answer: Test 2: State the advantage of two opposed directed Z-diodes? Answer:


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2.5 Voltage stablization circuit using Z-Diodes 2.5.1 Basics Z-diodes are not only used for stabilizing long-period fluctuations, but also for short-period fluctuations like the ripple of a rectified and pre-soothed AC voltage. 2.5.2 Tests Test

Examine the stabilizing effect of a Z-diode to a high ripple DC voltage.

Therefore rectify AC voltage with a rectifier diode and pre-smooth with a load capacitor. Test proceeding:

• Set-up the circuit in pic. 2.17. Measure both voltages UE and UA with the oscilloscope and draw

the cure into the chart 2.18. Therefore set the inputs of the oscilloscope to DC.

Note: Please see the polarity of electrolytic capacitors!

G~

+

R 470Ω

Ueff=24Vf=50HzSinus

CG 1µF UE

Si 1A RV=330Ω

10 V40 mA

UA

Y2 Y1

pic. 2.17

• Set the inputs of the oscilloscope to AC (Y2 = 0,1 V / part). This will suppress the DC portion of the voltage and only the ripple is indicated. Draw the curve for voltages UE and UA into the chart 2.19.


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- 0 (Y2) pic. 2.18 pic. 2.19 Settings: Notes:

X = 5 ms / part Y1 = voltage

UE Y1 = 5 V / part Y2 = voltage

UA Y2 = 5 V / part

Set inputs to DC

Settings: Notes: X = 5 ms / part Y1 = voltage UE

Y1 = 5 V / part without DC portion

Y2 = 0,1 V / part Y2 = voltage UA

without DC portion set inputs to DC

Test 1: Measure in pic. 2.18 ripple voltage ∆UE at the smoothing capacitor CG. Answer:

∆UE = ∆UE = peak-peak-value input voltage UE

Test 2: Measure in pic. 2.19 ripple voltage ∆UA at the Z-diode. Answer:

∆UA = ∆UA = -peak-value output voltage UA

Test 3: Determine the smoothing factor G (absolute stabilizing factor)? Answer:

=∆∆

=A

E

U

UG

Test 4: Determine the relative stabilizing factor S? Answer:

=•=•∆•∆

=E

A

EA

AE

U

UG

UU

UUS

S = relative stabilizing factor

Measure voltages UE and UA with a multimeter.

- 0 (Y1)

- 0 (Y1)

- 0 (Y2)


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4 Bipolar Junction Transistors

4.1 The Rectifying Function of Bipolar Transistors 4.1.1 Basics

Transistors are 3-layers semiconductor

components with two p-n-junctions each. The n-p-n transistors have a thin p-conductive layer

between the two n-conductive layers. The p-n-p

transistors have a thin n-conductive layer embedded between two p-conductive layers.

The p-n-junction between the mid-layer (base) and the two outer layers (emitter and

collector) have rectifying effect, this could be

examined likewise to any other rectifier diode.

4.1.2 Tests

Examine the effect of the p-n-junctions of a n-p-n transistor on the conductivity in dependence of the

supply voltage and its polarity. Repeat same the test with a p-n-p transistor. Note the basic differences

to the n-p-n transistor.

Test proceeding

• Set-up the circuit in pic. 4.2. One by one adjust the potentiometer with help of a multimeter

to the voltages UF in chart 4.1. Measure each respective current IF and fill in chart 4.1.

Graphically present the relation between current IF and voltage UF within the diagram

• pic. 4.2.

Note: Make sure to apply correct type of

measurement: voltage error or current error

circuit

pic. 4.2

pic. 4.1


30

Experiment #4

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• Set-up the circuit in pic. 4.2 (picture 2). One by one adjust to the voltages UF according to chart

4.3. Measure the respective current IF and fill the values in chart 4.3. Graphically represent the

relation between current IF and voltage UF within the diagram pic.4.4.

• For following measurements according to pic. 4.2 remove your multimeter and adjust the

voltages directly at the power supply. For protective measures do not remove the series resistor RV.

• Set-up the circuit in pic. 4.2 (picture 3). One by one adjust to the voltages UR according to chart 4.2. Measure the respective current IR and fill the values in chart 4.3. Graphically represent the

relation between current IF and voltage UF within the diagram pic. 4.3.

• Set-up the circuit in pic. 21.2 (picture 4). One by one adjust to the voltages UR according to

chart 4.5. Measure the respective current IR and fill the values in chart 4.5 (multimeter with 0,1-

µA measuring range required). Graphically present the relation between current IF and voltage UF

within the diagram pic.4.4.

UF [V] 0 0,1 0,2 0,3 0,4 0,5 0,6 0,65 0,7 0,75 0,76

IF [mA chart 4.1 pic.1 (base/ emitter junction)

* Due to tolerances within the capacitors, the voltage UF for the threshold needs to be adjusted. UR [V] 0 2 4 6 8 8,1 8,2 8,3

IR [mA chart 4.2 pic. 3 (base/ emitter junction) pic. 4.3

I F [mA]

2,0

1,5

1,0

0,5

14 U R [V]

12 10 8 6 4 20

10,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8

UF [V]

2

3

I R [mA] 4


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UF [V] 0 0,1 0,2 0,3 0,4 0,5 0,6 0,65 0,7 0,75 0,8

IF [mA chart 4.3 pic. 2 (collector/ base junction)

* Due to tolerances within the capacitors, the voltage UF for the threshold needs to be adjusted. UR [V] 0 5 10 15 20 25 30

IR [mA chart 4.5 pic. 4 (collector/ base junction) pic. 4.4

I F [mA]

2,0

1,5

1,0

0,5

35 U R [V]

30 25 20 15 10 50

20

0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8

U F [V]

40

60

I R [nA]

80


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• Then repeat the measurements with a p-n-p transistor (Art no. 503.130.004) and give your statement for the base/ emitter path and the collector/base path at what polarity the junctions of p-n-p transistors and n-p-n transistors are conductive or blocked. Fill your results in chart 4.6.

Polarity NPN-Type PNP-Type

base/ emitter junction (conductive or blocked)

base + / emitter -

base - / emitter +

collector/ base junction conductive or blocked)

collector - / base +

collector + / base -

chart 4.6 Test 1: Name the basic characteristics of the two p-n-junctions of a transistor. Answer: Test 2: What characteristic of the p-n-junction from base and emitter is different from base and

collector? Answer: Test 3: What needs to be regarded when changing a circuit from n-p-n transistors to p-n-p

transistors?

Answer:


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4.2 Current Distribution within a Transistor and Control Function of Base Current

4.2.1 Basics Due to the conductive p-n-junction load carriers from the emitter will be accelerated into the extreme thin basis layer, they will break down the opposing blocked p-n-junction between collector and base, and are biased towards the collector. Base current IB is less than emitter current IE.by the value of this collector current IC. Base current influences the value of the collector current. The relation between those both currents is called the current amplifier ratio B.

CE IIB −= B

C

I

IB =

Low signal current amplification β is: B

C

I

I

∆∆

=β

The terminal connection of the transistor with negative or positive polarity is given by its layering. With n-p-n transistors base and collector are positive in reference to the emitter, with the p-n-p transistor it is negative. 4.2.2 Tests Test 1

Examine statically the influence of the collector current in reference to the base current. Repeat same test for the n-p-n transistor. Test proceeding

• Apply DC voltage of Ugl = 20 V on the circuit in pic. 4.5. Measure the base current IB with

interrupted collector conductance (remove potentiometer) and fill the value in chart 4.7.

• Add potentiometer to the circuit and adjust the collector currents according to chart 4.7. Add

the values for the base current into chart 4.7.

~-

U = 20V

0V

R14,7kΩ

R21kΩ

RV

R 1kΩ IC

IB

R310Ω

NPN, 20 V100 mA

Basis links

pic. 4.5


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• Graphically represent the relation between base current IB and collector current IC (with constant base/emitter current) within the diagram pic. 4.6.

IB = (IC), UBE constant chart 4.7 pic. 4.6

RV [Ω] ∞ 1000 680 470 470 330 220

IC [mA] 0 20 25 30 40 50 60

IB [mA]

I B [mA]

4 3,5

3

2,5

2

1,5

1

0,5

0

0 10 20 30 40 50 60

I C [mA]


35

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Test 2

Examine statically the influence of base current to the collector current. Use n-p-n transistors for this

test. Test proceeding:

• Set-up the circuit in pic. 4.7. Adjust the base current according to the values in chart 4.8.

Measure the respective collector current IC and fill the values in chart 4.8.

~-

U = 20V

R2 100Ω

R14,7 kΩ

R 1 kΩR3

10 Ω

0 V

IB

IC

NPN 20 V100 mA

Basis links

pic. 4.7

• Graphically present the relation between collector current and base current within the diagram pic. 4.8.

IC = f (IB)


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chart 4.8 pic. 4.8 Test 1: What do we learn from the curve in pic. 4.8? Answer: Test 2: What is the current amplification factor B at IC = 55 mA (see pic. 4.8)?

Answer: ==B

C

I

IB

Test 3: What is the small signal current amplification β (see pic. 4.8)?

Answer: small signal current amplification at ∆IC = 40 mA - 20 mA =∆∆

=B

C

I

Iβ

small signal current amplification at ∆IC = 80 mA - 70 mA =∆∆

=B

C

I

Iβ

IB [mA] 0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 0,4 0,45 0,5

IC [mA]

IC [mA]

110

100

90

80

70

60

50

40

30

20

10

0

0 0,1 0,2 0,3 0,4 0,5 0,6

IB [mA]


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4.3 Characteristics of a BJT Transistor 4.3.1 Basics Characteristics of the transistor could be presented in 4 curves. Input characteristic

The input characteristic show the relation between base current IB and base/ emitter voltage UBE

(at short-circuit output). Output characteristic

The output characteristic shows the relation between collector current IC and collector/ emitter

voltage UCE with different constant base currents. Control characteristic

The control characteristic shows the relation between collector current IC and base current IB. Reverse voltage transfer characteristic

Reverse voltage transfer characteristic show the relation between different constant base currents UBE

and respective base/ emitter voltages UCE. 4.3.2 Tests Test

Measure the electric characteristics of a transistor and draw the 4 curves of the transistor (4 quadrants

presentation).

Test proceeding: • Set-up the circuit in pic. 4.9. One by one adjust the potentiometer to the base currents IB

according to chart 4.9. Measure the respective collector currents IC and fill in chart 4.9.

Graphically present the relation between collector current and base currents IC = f (IB) in the 2nd

quadrant of the diagram pic. 4.13.

~-

U = 10V

0 V

R14,7 kΩ

R 1kΩ

IB

IC

R210Ω

NPN20 V 100 mA Basis links

pic. 4.9


38

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Note: At this and following measurements there are high tolerances in the measured values due to component heating. Temperature stabilizing measures are not practicable, as they would basically

falsify the characteristic curves. Therefore we recommend reducing the current to zero for about 30 s

after each measurement and short action for reading the measured values when current is applied. When drawing the curve the variations within the measurements have to be compensated by interpolating. Modify the circuit according to pic. 4.10, then one by one adjust the potentiometer to base/ emitter UBE

currents according to chart 4.10.

Measure the respective base currents IB and fill the values in chart 4.1. Graphically present the relation

between the base current and the base/ emitter voltage IB = f (UBE) in the 3rd quadrant of the diagram

4.13.

~-

U = 10 V

R14,7 kΩ

R 1 kΩR2 10 Ω

UBE

IB NPN 20 V 100 mA

Basis links

0 V pic. 4.10

• Then modify the circuit according to pic. 4.11, then measure the collector current IC for different base currents IB and collector/ emitter voltages UC according to chart 4.11 and fill the values

into chart 4.11. Graphically present the relation between collector current and collector/ emitter current IC = f (UCE) for different base currents in the 1st quadrant of diagram 4.13.

~+

~

+

U = 15 V

U = 0...30 V

R14,7 kΩ

R 1 kΩ

R3 10 kΩ

R210 Ω

IC

UCE

NPN 20 V 100 mA

Basis links

pic. 4.11


39

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• Modify the circuit according to pic. 4.12, then measure the base/ emitter voltage UBE for

different base currents IB and different collector/ emitter voltages UCE according to chart 4.12

and fill the values in chart 4..12. Graphically present the relation between base/ emitter voltage and collector/ emitter voltage UBE = f (UCE) for different base currents in the 4th

quadrant of the diagram 4.13.

• Last examine the influence of the resistor R2. Therefore adjust the circuit according to pic. 4.9 to

a collector current of IC = 3 mA bridge the resistor for about 2-3 s and regard the behavior of both

ammeters.

~+

~

+

U = 15 V

U = 0...20 V

R14,7 kΩ

R1 kΩ

R3 10 kΩ

R210 Ω

UCE

UBE

IBNPN

20 V 100 mABasis links

pic. 4.12 Test 1: What is the duty of resistor R2 within the emitter line? Answer: Test 2: Which parameter can control the collector current IC? Answer:


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chart 4.9 chart 4.10 chart 4.11 chart 4.12

UCE = 10 V IB [∝A] 0 20 40 60 80 100 120 140 160

IC [mA]

UCE = 10 V

UBE [V] 0 0,5 0,6 0,65 0,7 ca. 0,71 ca. 0,71 ca. 0,71 ca. 0,71 ca. 0,71 IB [µA]

UCE [V] 0 0,2 0,5 2 4 6 8 10 12 14

IC [mA] at IB = 20 µA




UCE [V] 2 4 6 8 10 12 14

UBE [V] at IB = 20 µA




41

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pic. 4.13


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4.4 Effect of the Working Resistor on Transistor Characteristics 4.4.1 Basics A transistors variation of collector current IC due to base current IB is converted into voltage variation UCE by a series connected working resistor Ra.

Variation in base current IB is caused by base/ emitter voltage UBE.

Voltage amplification of a transistor is given by the ratio of both voltage variations. As the working resistor Ra affects the collector/ emitter voltage variations, it also controls the voltage amplification.

BE

CE

U

U

Uv

∆∆

=

4.4.2 Tests Test

By doing tests, examine the effect of the working resistor on voltage amplification and on frequency of a

transistor amplifier. Test proceeding:

• Set-up the circuit in pic. 4.14. f = 1 kHz / Ra = 100 Ω. Utilize the transistor for an AC voltage amplifier. The capacitors C1 and C2 will avoid DC voltage portions at input and output side.

G

~+U = 10V

U = 0...1Vf = 10Hz...50kHz

R3 10 kΩ

R41kΩ

R14,7 kΩ

Ra

Y1Y2

C20,22µF

R210Ω

R 1kΩ

C11µF

∆UBEUE

∆UCEUA

NPN links20V

100mA

A

pic. 4.14 Note: Before measurement, adjust the quiescent collector current with a potentiometer, so the display of

the oscilloscope shows the biggest possible sine-wave amplitude.

(operating point adjustment).

• Apply to the input (point A) AC voltages UE (f = 1 kHz) according to chart 4.13 with different

working resistors Ra and measure with the oscilloscope the respective output voltages UA.


43

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• Calculate the amplification ratio with bellow equation:

• Fill the calculated values in chart 21.13.

• Draw the curve for input voltage UE and output voltage UA (at Ra = 4,7 kΩ) in pic. 4.15.

• Graphically present the relation between amplification ratio and working resistor in diagram

4.16.

E

A

U

U

Uv =

- 0 (Y1)

- 0 (Y2) pic. 4.15 Settings: Notes:

X = 0,1 ms / part Y1 = input voltage UE

Y1 = 20 mV / part Y2 = output voltage UA

Y2 = 2 V / part


44

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chart 4.13 * depending on the transformer type voltage values have to be decreased pic. 4.16

Ra [kΩ] 0,1 0,47 1 4,7 10 22

UE ss [mV]* 400 200 100 40 20 20

UA ss [V]

vU

v U

300 200

100

20

10

2

1

0,1 0,2 1 2 10 20 R [ kΩ]

30


45

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chart 4.14 * for the test you will need a function generator with frequency 100 kHz ... 5 MHz pic. 4.17

• In another test measure the influence of the working resistor on the limiting frequency. Therefore

adjust the output voltage at f = 1 kHz to a fixed value, which is considered 100%. Now increase

frequency until output voltage is 70,7% of its original value. The respective frequency is called the upper limiting frequency. Determine the upper limiting frequency for each working resistor in chart 4.14 and fill the values in chart 4.14. Please see the adjusted input voltage UE does not overload the

transistor.

• Graphically present the relation between upper limiting frequency and working resistor in diagram

4.17.

Ra [kΩ] 0,1 0,47 1 4,7 10 22

f0 [kHz]*

f [MHz]

10

5

2

1

0,5

0,2

0,1 0,2 1 2 10 20 Ra [ kΩ]

30


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Test 1: State the influence of the working resistor on the amplification factor. Answer: Test 2: State the influence of the working resistor on the upper limiting frequency. Answer: Test 3: What is the phase shift angle between input and output voltage? Answer:


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5 Unipolar Transistors (Junction Field-Effect Transistor) 5.1 Layering Check and Rectifying Characteristics of FETs 5.1.1 Basics

At field-effect transistors (FET), the current

carriers do not pass the PN junction between

the different conductive layers, but flow within the same conductive channel. Therefore we

call them unipolar transistors. There are

electrodes (gates) to control the current carriers, which are isolated from the channel

either by p-n-junction (barrier FET) or by quartz

layer (MOS-FET).

5.1.2 Tests Test Examine the characteristics of the p-n-junction between gate-electrode and the main electrodes (source

and drain) of a n-channel-FET.

Measure the relation between current and applied voltage with your multimeter. Repeat this test for a p-channel-FET. Test proceeding

• Set-up the circuit in pic. 5.2 auf (pic. 1) and determine with your multimeter if the p-n-junction is conductive or blocked. Repeat your measurements for pic. 2,3 and 4. Fill your results (conductive/

blocked) into chart 5.1.

~-

AB

BA

B

A

A

B

Typ N-Kanal Art. 503.130.015

Bild 1 Bild 2 Bild 3 Bild 4

U = 2V

0 V

R 1 kΩ

R 1kΩ

IF(IR)

A

B

UF(UR)

Gate/Source-Strecke

Gate/Source-Strecke

Drain/Gate-Strecke

Drain/Gate-Strecke

pic. 5.2

N-channel-FET P-channel-FET

D D

N P P

P N N

G G

S S

Circuit Symbol Circuit Symbol

pic. 5.1


48

Experiment #6

000059000059

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Picture 1 2 3 4

n-channel-type

p-channel-type chart 5.1

• Substitude the n-channel-FET with a p-channel-FET (pic. 5.3). Determine the states of the p-n-

junction with current measurements (pic. 1...4) and fill the results in chart 5.1.

~-

A

BA

B

B

AB

A

U = 2 V

0 V

R 1kΩ

R 1kΩ

IF(IR)

UF(UR)

A

B

Typ P-Kanal 503.130.016

Bild 1 Bild 2 Bild 3 Bild 4

Gate/Source-Strecke

Gate/Source-Strecke

Drain/Gate-Strecke

Drain/Gate-Strecke

pic. 5.3 Test 1: When is the p-n-junction of a n-channel-FET blocked? Answer: Test 2: When is the p-n-junction of a p-channel-FET blocked? Answer:


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5.2 Current Curve of the PN-Junction of the FETs-Gates 5.2.1 Basics The junction-FET has a rectifying effect between gate and channel. Although practically it is not relevant, you have to know its current characteristic to understand the control behaviour of the FET. 5.2.2 Tests Test 1 By measurement examine by the current curve of the p-n-junction between gate and channel electrodes of a junction field-effect transistor.

Do this test only with the n-channel-FET. These results are also valid for p-channel-types except of the polarity. Test proceeding: • Set-up the circuit in pic. 5.4 (with gate/ source path). One by one adjust the voltage UF according to

chart 5.2. Measure the respective currents IF with your multimeter and fill in chart 5.2.

• Repeat your measurements with the drain/ gate path and fill the current values IF in chart 5.3. • Graphically present the current curve of the p-n-junction IF = f (UF) in diagram 5.5.

~-

A

B

B

A

U = 0...30V

O V

R1 kΩ

R1 kΩ

IF

UF

Art. 503.130.015

Gate/Source-Strecke

Drain/Gate-Strecke

pic. 5.4


50

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Gate/Source Path

UF [V 0 0,2 0,4 0,6 0,7 0,75 0,8 0,85 0,9 1,0

IF [mA] chart 5.2

Gate/Drain Path

UF [V 0 0,2 0,4 0,6 0,7 0,75 0,8 0,85 0,9 1,0

IF [mA] chart 5.3 pic. 5.5 Test: What is the meaning of the distortion between both current curves? Answer:

IF [mA]

20 18

16

14

12

10

8

6

4

2

0

0 0,2 0,4 0,6 0,8 1

UF [V]


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5.3 Control Function of the N-Channel-FET 5.3.1 Basics Current flow through the channel of a field-effect transistor (Source/ Drain) is controlled by the gate

potential. Unlike bipolar transistors, no voltage is required, as long s the p-n-junction between gate and

channel is blocked. Input characteristic or control characteristic of a FET shows the relation between gate/ source voltage UGS and drain current ID.

With the characteristic curve steepness S could be determined, which shows the exact voltage amplification:

S = Steepness in mA/V ∆ID = Change of Drain Current in mA

∆UGS = Change of Gate/Source Voltage in mA

GS

D

U

IS

∆∆

=

5.3.2 Tests Test

Examine the influence of the gate/ source voltage on the gate current and drain current. Draw the

control curve: IG = f (UGS); ID = f (UGS) Test proceeding:

• Set-up the circuit in pic. 5.6 and change the gate/ source voltage UGS in steps shown in chart

5.4.Measure each gate current IG and drain current ID with your multimeter and fill the values in

chart. 5.4 and 5.5.

~+

~

+

U = + 15V

U = - 15V

R1 1kΩ

R 1kΩ

R2 1kΩ

ID

IG

UGS

503.130.015 25 V10 mA

N-Kanal Gate links

pic. 5.6


52

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• Graphically present the relation between gate IG current and gate/ source voltage UGS in diagram 5.7.

IG = f (UGS) chart 5.4 UDS = 15 V pic. 5.7

UGS [V] -4 -3 -2 -1 0 +0,5 +0,6 +0,7 +0,75

IG [mA]

IG [mA]

0,8

0,6

0,4

0,2

-4 -3 -2 -1 0 1 2 UGS [V]


53

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• Graphically present the relation between drain current ID and gate/ source voltage UGS in diagram 5.8.

ID = f (UGS) chart 5.5 UDS = 15 V pic. 5.8 Test 1: What is the steepness S of a FETs when change of gate/ source voltage is

∆UGS = 1,0 V and a respective change of drain current ∆ID = 3,8 mA? Answer: Test 2: When is the field-effect-transistor controlled without voltage? Answer:

UGS [V] -4 -3 -2 -1 0 +0,5 +0,6 +0,7 +0,75

ID [mA]

ID [mA] 22

20

18

16

14

12

10

8

6

4

2

-4 -3 -2 -1 0 1 2 UGS [V]


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5.4 Output Characteristics of FETs 5.4.1 Basics Semiconductor-manufacturers not only state control characteristics of field-effect-transistors, but also output characteristics, showing the relation between drain current and a variety of constant drain/ source voltages. Output characteristics are stated without working resistance (static values). The practically used working resistance is drawn into the characteristic curve, the graph shows the voltage amplification. 5.4.2 Tests Test 1

Statically measure the relation between drain current and drain/ source voltage for different gate/

source voltages. ID = f (UDS) Test proceeding:

• Set-up the circuit in pic. 5.9, then measure with your multimeter the drain current ID for each

gate/ source voltage UGS and drain/ source voltage UDS according to chart 5.6.

Fill the values for drain current ID into chart 5.6.

~+

~+

~

+

U = 0...20 V

U = +15V

U = -15V

R2,2 kΩ

R2,2 kΩ

R1 kΩ

ID

UDS

UGS

503.130.01525 V 10 mA

N-Kanal, Gate links

pic. 5.9


55

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chart 5.6

• Graphically present the relation between drain current ID and drain/ source voltage UDS for different gate/ source voltages UGS.

UDS [V] 0 0,5 1 1,5 2 3 4 6 8 10 12 14 16 18 20

ID [mA] at UGS = -1,5 V

ID [mA] at UGS = -0,5 V

ID [mA] at UGS = 0 V

ID [mA] at UGS = 0,5 V


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16 14 12 10 8 6 4 2 0 ID [mA] pic. 5.10

20

18

16

14

12

10

8

6

4

2

0

UDS

[V]


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Test 2

Examine the influence of working resistance to the field-effect-transistor. Test proceeding:

• Set-up the circuit in pic. 5.11 and measure with your multimeter the output voltage UA for all

working resistances Ra and input voltages UE of chart 5.7.

• Fill your determined values into chart 5.7.

• Calculate the amplification v for different working resistances Ra with following equation:

∆UA = UA1 – UA2

E

A

U

Uv

∆∆

=

∆UE = UE1 – UE2

~

+

~+

U = +15V

U = -15V

R2 1kΩR 1kΩ

Ra

ID

UGS = UE

UDS = UA

503.130.01525 V 10 mA

N-Kanal Gate links

pic.5.11


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chart 5.7

• Graphically present in chart 5.12 the relation between voltage amplification v and working

resistance Ra. in chart 5.12

• Draw the graph for the working resistance with Ra = 2 kΩ into the diagram 5.10. pic. 5.12

Ra [kΩ] 1 2 kΩΩΩΩ 4,7

-UE 1 [V] 0,5 1,0 1,5

-UE 2 [V] 1,0 1,5 2,0

UA 1 [V]

UA 2 [V]

∆UE [V]

∆UA [V] v = ∆UA / ∆UE

20

v 18

16

14

12

10

8

6

4

2

0

0 2 4 6 8 10

Ra [kΩ]


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Test 1: How does the amplification factor v react, when working resistance Ra increases? Answer: Test 2: What is the amplification factor v at UE = -0,5 V - (-1,5 V)? Diagram in pic. 5.7 may help

to find the answer. Answer: Test 3: What do the negative results for UA indicate? Answer:


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6 Amplifiers 6.1 Basic Amplifier Circuits using BJT 6.1.1 Introduction

Transistors are operated in different basic amplifier circuits. Those circuits are named according to the

electrode, which is the common reference point for in- and output signal voltage. Most amplifier circuits

are operated by NPN-transistors. Operation with PNP-transistors is also possible, but polarity of the supply voltage has to be reversed.

According to that we distinguish between:

1. emitter circuit 2. collector circuit

3. base circuit 6.1.2 Tests

Determine by measurement the electric characteristics of the three basic amplifier circuits: - Voltage amplification factor vU - input resistance RE

- upper limiting frequency fo - output resistance RA

- phase shift angle ϕ

Operate the basic amplifier circuit as pure AC voltage amplifier, for isolation of supply voltage from signal voltage use capacitor C1, C2 and additionally for the base circuit C3. Test proceeding:

• Apply sine-wave voltage to the input A of emitter circuit in pic. 6.1.: f = 1 kHz, uss = 30 mV

Note: For more convenient measurement of the low input voltage, select a voltage divider (RT 1 / RT 2)

at the generator output. Thus, input voltage UE‘ at the measuring point B is 11 x UE = 0,33 V.

Adjust the operation point of the thyristor with the potentiometer, until undistorted voltage is applied to

measuring point C.

• Measure input – and output voltage with the oscilloscope and draw the curve into diagram pic. 6.4 .Determine the phase shifting between UE and UA and calculate the voltage amplification

factor with following equation:

vU = voltage amplification factor UA = output voltage UE = input voltage

==E

A

U

U

Uv


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G

~+

U = 10 V

f = 1 kHz

RT1 1kΩ

UE` =11x UE

RT2100Ω

UEUSS

30mV

R110 kΩ

Ra4,7 kΩ

R1 kΩ

R2470Ω

R310Ω

C20,22µF

UA

Y1Y2NPN20V

100 mAB A

C

pi. 6.1 Emitter circuit

• Determine the upper limiting frequency. It is the frequency, when output voltage drops to

70,7% of its maximum value if frequency is increased.

• Determine the input resistance RE. According to pic. 6.1 connect resistor RV with 470 Ω in series to the input (A). This will affect, that output voltage of value U1 drops to value U2. From both

voltages and the familiar series resistor, you can calculate the input resistance RE:

==

ss

ss

V

E

U

U

RR

2

1

RE = input resistance in Ω RV = series resistance in Ω U1ss = output voltage without series resistance U2 ss = output voltage with series resistance

• Then determine the output resistance RA. Therefore connect the load resistor RL in parallel to

output (C to earth), thus output voltage with value U1 drops to value of U2. From both voltages

and the familiar load resistor, you can calculate the output resistance with following equation:

• =

−•= 1

2

1

U

URR LA

RA = output resistance in Ω RL = load resistance in Ω U1 = output voltage without load resistance U2 = output voltage with load resistance

• Note the determined values in chart 61. • Then set-up the collector circuit in pic. 6.2. Repeat your measurements in the same way. Draw

the curve into diagram 6.4 and note your values in chart 6.1.

Note: For determination of the upper limiting frequency the function generator should have

a frequency range of 5 MHz.


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G

~+

U = 10 V

f = 1kHz

R110 kΩ

C10,22µF

UEUSS= 1V RT2

100Ω

C20,22µF

UA

Y2Y1

NPN20 V

100 mA

pic. 6.2 collector circuit

• Set-up the base circuit in pic. 6.3 and repeat your measurements. Draw the curves in diagram 6.6. and note your values in chart 6.1.

Note: Use the measuring frequency of 10 kHz, as otherwise the reactance of the input coupling

capacitor is not low enough in respect to input resistance.

G

~+

U = 10 VR1

10 kΩRa

4,7 kΩ

RT11 kΩ

R 1kΩ

f =10kHz

U´E=11xUE

RT2100Ω

UEUSS=

30mV

R3470Ω

R2680Ω

C20,22µF

UA

Y2 Y1

C30,47µF

C11µF

B A C

NPN20V

100mA

pic. 6.3 Base circuit


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chart 6.1

Settings: Y1 = 0,1 V / part Y2 = 1 V / part

X = 0,1 ms / part pic.6.4

Settings: Y1 = 0,5 V / part Y2 = 0,5 V / part X = 0,1 ms / part

pic. 6.5

Emitter Circuit Collector Circuit Base Circuit

UE 30 mV / 1 kHz 1 V / 1 kHz 30 mV / 10 kHz

UA

vU

fo ϕ

RE (RV = 470 Ω) (RV = 4,7 kΩ) (RV = 220 Ω) RA (RL = 4,7 kΩ) (RL = 1kΩ) (RL = 4,7 kΩ)

- 0 (Y1)

= U E’

- 0 (Y2)

= U A

- 0 (Y1)

= U E’

- 0 (Y2)

= U A


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Settings: Y1 = 0,1 V / part Y2 = 0,2 V / part X = 10 µs / part

pic. 6.6 Test 1: Which one of the three circuits does have inverting effects? Answer: Test 2: For what task do we use the collector circuit due to its characteristics? Answer: Test 3: What are the both significant differences between base circuit and emitter circuit? Answer:

- 0 (Y1)

= U ’E

- 0 (Y2)

= U A


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7 Multistage Amplifiers 7.1 IntroductionIf the amplification factor of one transistor is not enough, then more could be connected, thus amplification is multiplied. An exception is the distributed amplifier, used for ultra-high-frequency, here all partial amplification factors are added.

When connecting the single transistors, they have to be coupled in a way that the operation values (collector – and basic-closed-currents, base -/ emitter-pre-load) will not influence one another. For AC voltage amplifiers it is affected by coupling across RC-elements or transformers. At composite amplifiers (for amplification of DC- and AC voltages) all stages are potential-matched and then coupled together. 7.2 Two-Stage AC Voltage Amplifier 7.2.1 The Basics Voltage gain of both transistors in an emitter circuit is fully utilizable. Resistor R4 and R6 are for stabilizing the operation point. As input- and output stages are coupled by RC elements, no DC voltage could be transferred. Only AC voltages over a certain frequency will be transferred. 7.2.2 Tests Measure the curves for input – and output voltages, the partial voltage amplification factors, total voltage amplification factor and the limiting frequencies for a two-stage AC voltage amplifier Test proceeding:

• Apply to the input A of the circuit in Fig. 7.1 a sine-wave voltage with a frequency of f = 1

kHz. Set the voltage to 1 mV.

Note: For better adjustment of the low input voltage with the oscilloscope, select a voltage divider (RT 1 / RT 2) at the generator output. Input voltage UE‘ at measuring point B is 1001 x UE = 1 V.

• Adjust the operating point of the transistors at the potentiometers 1 kΩ and 10 kΩ, so the maximum and undistorted output voltage is given.


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• Measure with your oscilloscope the curve for the output voltage at the collector of transistor V1

(UA 1) and the curve for the output voltage at the collector of transistor V2 (UA 2).

• Trigger the oscilloscope with input voltage UE‘. • Then measure the amplification factor v1 and v2, as well as the total amplification factor vges of

both transistors.

• At least measure with constant input voltage the output voltage for different frequencies and note the values in chart 7.4.

• In pic. 7.15 graphically present the dependency of amplification factor vges from frequency.

Maximum amplification factor vges = 100 %.

• Determine then the upper and lower limiting frequency. It is the frequency, when output voltage drops to 70,7% of its maximum value if frequency is increased or decreased.

• Note: Multistage amplifiers are susceptible for noise from the 50-Hz-power supply (so called

ripple voltage) due to the high amplification factor. For keeping this a minimum, all reference

potentials of the circuit have to be connected to a single point ( see point E pic. 7.13) (negative pole of the operating voltage, reference point of the function generator and the oscilloscope, base point of RT 2, potentiometer 1 kΩ and 10 kΩ, R4, and R6).

If necessary (ripple of power supply > 1 mV), please increase the output voltage of the power supply to 20V and smooth with R7 and Z-diode ZPD 10.

G

~+

20V mit Z-Diode

R7 330Ω

U = 10V R24,7kΩ

R310kΩ

10Hz...5MHz

RT1 10kΩ

B

UE´=1001*UE

RT210Ω

UEUSS=1mV

C11µF

R1kΩ

R122kΩ

R422Ω

C21µF

UA1

R522kΩR

10kΩ R622Ω

C30,47µF

Y2 Y1

UA2

NPN, links20V, 100mA

NPN, links40V, 1A

Punkt E

Fig 7.1


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Settings:

Y1 = 0,5 V / part

Y2 = 20 mV / part Y2 = 5 V / part

X = 0,2 ms / part pic.7.14 chart 7.3 chart 7.4

- 0 (Y1)

= U E

- 0 (Y2)

= U A

v1 v2 vges

f [kHz] 0,01 0,02 0,05 0,1 0,2 0,5 1 2 5 10 20 50

UE [mV] 1 1 1 1 1 1 1 1 1 1 1 1

UA [V]

v = U A

U E

vges [%]


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pic.7.15 Test 1: What is the phase shift between UE and UA 1? Answer: Test 2: What is the phase shift between UE and UA 2? Answer: Test 3: What is the interrelation between the single amplification factor and the total

amplification factor? Answer: Test 4: What is the lower limiting frequency? Answer: Test 5: What is the upper limiting frequency? Answer:

Vges [%] 100

90

80

70

60

50

40

30

20

10

0 10 20 50 100 200 500 Hz 1 2 5 10 20 50 100 kHz

f


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9. Differential Amplifiers 9.1 The Basics

Differential amplifiers increase the difference between two input voltages. Same voltages at both inputs

cause no output voltage. Basically the differential amplifier is similar to an emitter-coupled phase inverter. Instead of an emitter resistor, another transistor is utilized for constant voltage supply. It

provides that the sum of the both emitter currents within the emitter-coupled circuit remains constant.

Thus, emitter current of one transistor drops the same rate than emitter current at the other increases. than increases at the other. 9.2 Tests Test

Examine the characteristics of a differential amplifier first for DC voltage and second for superimposed

AC voltage. Test proceeding:

• Apply to circuit 9.26 DC voltage of Ugl = 20 V. Adjust the voltage UE2 by potentiometer (10 kΩ)

to 5,5 V and keep this constant.

• Then step by step change the voltage UE 1. Note the respective values of the output voltage in

chart 9.8.

~+

U = 20 V

R12,2 kΩ

R1 kΩ

R21 kΩ

R31 kΩ

R62,2 kΩ

R710kΩ

UA

R4 100Ω R5 100ΩR8

10 kΩ

R10 kΩ

UE2

R10470 Ω

R91 kΩ

20 V

UE1

NPN, links40 V1 A

NPN, rechts40 V1 A

NPV, links20 V

100 mA

Z-Diode3,3 V

pic. 9.26


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• Calculate the input voltage differentials UE Diff = UE 1 - UE 2 and note the values in chart 9.8.

• Then set the voltage UE 1 to 5,5 V and keep it constant. Step by step change the voltage UE 2 and

note the respective output voltages in chart 9.9.

• Again calculate the input voltage differentials and note the results in chart 9.9.

• Graphically present in diagram 9.27 the dependency of output voltage from the difference of the input voltages, first for constant voltage UE 2, then for constant voltage UE 1.

chart 9.8 UE 2 = 5,5 V (constant) chart 9.9 UE 1 = 5,5 V (constant) pic. 9.27

UE 1 [V] 4,6 4,8 5 5,2 5,5 5,8 6 6,2

UA [V]

UE Diff [V]

UE 2 [V] 4,6 4,8 5 5,2 5,5 5,8 6 6,2

UA [V]

UE Diff [V]

UA [V]

8 6

4

2

-1,0 -0,8 -0,6 -0,4 -0,2 0,2 0,4 0,6 0,8 1,0

-2 U Diff [V]

-4

-6

-8


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G

~+

U = 20 V

R12,2 kΩ

C11µF

R1 kΩ

UE1

R21 kΩ

R31 kΩ

R62,2 kΩ

R710 kΩ

UA1

C30,47 µF

UA

UA2

C40,47 µF

R4 100Ω R5 100Ω

R810 kΩ

C21 µF

R10 kΩ

Y2 Y1

UE2C

R91 kΩ

R10470 Ω

Z-Diode3,3 V

130 mA

NPN, links40 V1 A

NPN, rechts40 V1 A

NPN, links20 V

100 mA

20 V

B A

pic. 9.28

• In another test examine the dynamic characteristics of the circuit in pic. 9.28. Therefore adjust both DC input voltages UE 1 and UE 2 to 5,5 V (UA = 0 V). First superpose input voltage UE 1 with

AC voltage UE ss = 0,6 V; f = 1 kHz. Draw the respective curves for output voltages UA 1 and UA 2

in diagram 9.29.

Note: To allocate the input voltages, the oscilloscope has to be triggered with the input

voltage UE at the external trigger input. Set the trigger source to “extern“.

Settings:

Y1 = 1 V / part

Y2 = 1 V / part

X = 0,2 ms / part UE 1 = trigger signal

pic. 9.29

- 0 (Y1)

= U A 1

- U E 1

- 0 (Y2)

= U A 2


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• Then superimpose the AC voltage of the output voltage UE 2 (C) and draw the curves of the

output voltages UA 1 and UA 2 in diagram 9.30.

• Then superpose the AC voltage of both input voltages (UE 1 and UE 2) (B and C) and draw the curve of the output voltages UA 1 and UA 2 into diagram 9.31.

Settings: Y1 = 1 V / part Y2 = 1 V / part

X = 0,2 ms / part UE 2 = trigger signal

pic.9.30

Settings: Y1 = 1 V / part Y2 = 1 V / part

X = 0,2 ms / part

UE 1 = UE 2 = trigger signal pic. 9.31 Test 1: What is the output voltage when both inputs are controlled the same? Answer: Test 2: In which integrated circuit is the differential amplifier the base circuit? Answer:

- 0 (Y1)

= U A 1

- U E 2

- 0 (Y2)

= U A 2

- 0 (Y1)

- 0 (Y2)

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10. Push-Pull-Amplifier 10.1 Basics

Amplifiers have the task to apply a certain signal to a load resistor. For load resistors there are actuators, which transfer

electrical signals into meachnical power (i.e. loud speakers,

actuator motors, valves). In todays amplifier technology mostly complementary collector stages are used. In pic. 10.37 you see a its circuit diagram. During the positive half-wave V1 is conductive

and V2 is blocked, at the negative half-wave it is vice versa. Therefore one after the other half-wave is applied to the load RL.

Without signal both transistors are blocked, its closed-circuit current is IC = 0.

10.2 Tests Test Examine by measurement the characteristics of a push-pull-amplifier with complementary transistors. Test proceeding:

• Apply a sine-wave voltage of Uss = 2 V; f = 4 kHz to the input (A) of the push-pull-amplifier

according to pic. 10.38 and display the dependency of output voltage to input voltage on the oscilloscope (X/Y-presentation). Adjust the operating point of the transistor V1 at the

potentiometer, so undistorted output voltage is at measuring point C.

+

G

~+

U = 30 V

R11 kΩ

R1 kΩ

R222 kΩ

R31 kΩ

R4100 Ω

C10,47 µF

UE

C2470 µF

UA

RL22 Ω

Y1Y2

NPN, links20 V

100 mA

NPN, links40 V1 A

PNP, links- 40 V- 1 A

pic. 10.38

pic. 10.37

G

+Ub

- Ub

RL

V1

V2

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Caution: Regard the polarity of the electrolytic capacitor!

Note: This open circuit set-up holds the possibility that the circuit energizes itself and starts with high-

frequent oscillation. If this occurs you have to twist the input signal wire with the ground wire or use an

isolated cable. Do not double-ground your measuring devices.

Settings:

Y = 2 V / part

X = 0,2 V / part

• Draw the screen shot into diagram 10.39.

• Calculate by help of the measured voltage the voltage applied to the load resistor RL. Note those values

together with the values for the effective load current in chart 10.12.

pic. 10.39 chart 10.12

Settings:

Y = 0,5 V / part

X = 50 mV / part

• Calculate the efficiency and note the values in chart 10.12.

• Reduce step by step the input voltage

to Uss = 0,5 V and draw the curve (X / • Y-presentation) in pic. 10.40.

pic. 10.40

- 0

0

uA ss UA eff P Ieff ϕ RA η

0

- 0


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• Select the two-channel mode with time dependant presentation for the oscilloscope, draw the curve for input- and output voltage in diagram10.41.

Settings:

Y1 = 0,2 V / part

Y2 = 1 V / part

X = 50 ms / part pic. 10.41

• Then increase the input voltage to Uss = 2 V and draw the Oscilloscope-presented curve in

diagram 10.42.

Settings:

Y1 = 1 V / part

Y2 = 5 V / part

X = 50 ms / part pic. 10.42

• Measure the open circuit voltage UA and draw into diagram 10.43.

• Determine the phase shift between UA and UE. Note the values in chart 10.12. • Use following equation for the calculation of the internal resistance at the output side.

=

−•= 1

2

1

U

URR LA

RA = output resistance in Ω RL = load resistor in Ω U1 = output voltage without load resistor U2 = output voltage with load resistor

- 0 (Y1)

- 0 (Y2)

- 0 (Y1)

- 0 (Y2)


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Settings:

Y1 = 1 V / part

Y2 = 10 V / part

X = 50 ms / part pic. 10.43 Test 1: How big is in this circuit the effective output voltage and output power? Answer: Ueff = P = Test 2: Determine the effective load current across the load resistor. Answer: Ieff = Test 3: On which parameter depends the operating current that the amplifier consumes from the

power supply? Answer: Test 4: Determine the input power of the circuit at hand. What is the efficiency of this? Answer: PV = Test 5: What is the internal resistance at output side? Answer:

==Pzu

Pabη

Test 6: What is the difference in the curve for low control in respect to high control? Answer:

=

−•= 1

2

1

U

URR LA

Test 7: How is the curve for open circuit output voltage? Answer:

- 0 (Y1)

- 0 (Y2)


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1. Operational Amplifiers Introduction

Operational Amplifiers (op-amp) are ideal DC voltage amplifiers with high ohmic differential input and very high amplification factor. Op-amps are offered as integrated standard components (e.g. A 741). In principal all circuit set-ups with single transistors could also be done with op-amps. Following tests are for examination of the basic characteristics of this kind

of components.

Circuit Symbol

pic. 15.1

15.1 Op-Amp serving as inverting Amplifier 15.1.1 Basics The inverting amplifier is a circuit that inverts the input voltage at the output. With AC voltage there is a phase shifting of 180° between in- and output voltage. With an inverting amplifier we can either get a gain or loss at the input signal, depending on the resistance of RR and RE.

Amplification factor:

or

pic.11.2 Circuit of the op-amp as inverting amplifier

Calculation could be simplified as follows: - Within the dynamic range is the difference of both inputs 0. - Input current IE within the amplifier is 0. Amplification factor is v = -1 (at RR = RE), means the amplitude at input and output are the same.

v = − U A

U E v = −

RR

RE

R R

R E -E

U E +E

U A

8-

+


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Experiment #11

000089000089

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11.1.2 Tests Test

Graphically present the relation between output voltage and input voltage and examine in a further test

the influence of different load resistors on the output voltage. Test proceeding:

• Set-up the circuit in pic. 11.3 and measure output voltage UA for different feedback

resistors RR and input voltages UE according to chart 11.1 with your multimeter.

• Fill your determined values in chart 11.1. Then graphically present the relation of output

voltage UA to input voltage UE and feedback resistors RR.

pic. 11.3

~

-

U gl = 15 V

R

R 1 1 k Ω

RE

10 kΩ -E

R R

10 kΩ

R2 1 k Ω

UE+E

U A R L

~- Ugl = -15 V


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chart 11.1 *To get a linear curve, you have to insert an additional measuring point UE, at the

limiting point of the op-amp.

pic. 11.4

UE [V]* -10 -8 -6 -4 -2 0 2 4 6 8 10 UA [V] at RR = 10 kΩ UA [V] at RR = 22 kΩ

UA [V] 14

12

10

8

6

4

2

-10

-U E [V]

-8 -6 -4 -2

-2

-4

2 4 6 8 10

U E [V]

-6

-8

-10

-12

-UA [V] -14


87

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/ 108

• For another test adjust input voltage UE to -5 V. RR and RE = 10 kΩ. Connect different loads according to chart 11.2. Now measure the respective output voltages UA.

• Fill the determined values in chart 11.2. Then graphically present the relation of output voltage UA to load resistance RL in diagram 11.5.

chart 11.2 pic. 11.5 Test 1: What is the polarity of the input voltage UE in reference to the output voltage UA at an

inverting amplifier? Answer: Test 2: What components specify the amplification factor v at the inverting amplifier? Answer: Test 3: What is the amplification factor at RR = 10 kΩ and RE = 1 kΩ? Answer: Test 4: What statement could be given to the characteristic curve pic. 11.5? Answer:

RL [Ω] 1000 680 470 330 220 100

UA [V]

UA [V]

7

6

5

4

3

2

1

0

0 200 400 600 800 1000

R L

1200 [Ω ]


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1.2 Op-Amp serving as non-inverting Amplifier 1.2.1 Basics At the non-inverting amplifier input – and output voltages are of same polarity, with AC voltages phase shifting is 0°. Like shown in pic. 11.6 input voltage at the not inverted amplifier is applied to input +E. As within the dynamic range the difference between both inputs is 0, also the input –E is voltage supplied. Use following equation for the amplification factor :

pic. 11.6

11.2.2 Tests Test Graphically present the relation of output voltage to input voltage with different feedback resistors. Test proceeding

• Set-up the circuit in pic. 11.7 and measure with your multimeter the output voltages UA for different feedback resistors RR and input voltages UE according to chart 11.3.

• Fill the determined values in chart 11.3, then graphically present the relation between output voltage UA and input voltage UE as well as the feedback resistors RR within the diagram 1.8

pic. 11.7

R E

∆U = 0

-E +E

R R

U A

U E U E

v = 1 + RR

RE

v = U A

U E

~

-

U 15 V

R 3

1 k Ω

R 10 k Ω

R E

10 kΩ

R 1

10 kΩ

U E

-E +E

R R

U A

~ -

U -15 V R 4

1 k Ω


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chart 11.3 * To get a linear curve, you have to insert an additional measuring point UE, at the limiting point of the op-amp. pic. 11.8

UE [V]* -10 -8 -6 -4 -2 0 2 4 6 8 10 UA [V] at RR = 10 kΩ UA [V] at RR = 22 kΩ

UA [V] 14

12

10

8

6

4

2

-10

-U E [V]

-8 -6 -4 -2

-2

-4

2 4 6 8 10

U E [V]

-6

-8

-10

-12

-UA [V] -14


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/ 108

Test 1: What components define the amplification factor? Answer: Test 2: What is the output voltage UA when RR = 47 Ω, RE = 10 kΩ, UE = 2 V? Answer: Test 3: What is the polarity of input voltage UE with respect to output voltage UA? Answer:


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12. Op- amp circuits 12.1 The Adder

With an adder multiple analog voltages could be summed up with respect to their algebraic sign. Basically adder also could be set-up with resistors. However, a disadvantage would be that the input resistors could not be decoupled. Pic. 12.9 shows a circuit that could sum up two voltages UE 1 and UE 2. If only input E1 is connected to the input voltage, we will have an inverter gain.

pic. 12.9

If input E2 is connected to earth, nothing will change at the characteristic of the inverter. As –E has

ground potential, the voltage drop at the input resistor is 0. Hence, input current is I E 2 = 0. At the

voltages E1 = 1 V and E2 = 3 V, for each input resistor is the voltage respective to its input voltage.

Hence, is IE 1 = 0,1 mA IE 2 = 0,3mA.

Both input currents have to flow across the feedback resistor R R. Hence, voltage at the output has to

be UA =- (IE 1 + IE 2) . RR = - (0,1 mA + 0,3 mA) . 10 k_ = - 4 V.

Due to the sum of the input currents, also the input voltages are summed-up. Please consider the

change of the algebraic sign due to the inverting effects of the amplifier.

This can be corrected by a downstream connected inverter.

If one of the input voltages is negative, the input current reverses its direction. One input current is

subtracted from the other one. Hence, only the difference between the input currents flows across the

feedback resistor, and at the output there is also only the difference of the input voltages.

Multiple inputs could be added to the circuit and you will receive the sum of all input voltages at the

output. Following is true for output voltages:

at RE 1 = RE 2 = ... RE n = RE is true: ( )EnEE

E

R

A UUUR

RU +++−= ...21

at RR = RE is true: UA = - ( UE 1 + UE 2 + ... + UE n )

+++−=

En

En

E

E

E

E

RA

R

U

R

U

R

URU ......

2

2

1

1

E1 I E1

I E2

E2

U E1

U E2

R E1

10 k Ω R E2

10 k Ω

I E

I E1 + I E2

-E

+E

R R 10 k Ω

UA

v = − RR

RE1

= − 1


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12.2 Tests Test

Examine an adder by measurement.

Test proceeding

• Set-up the circuit in pic. 12.10 and adjust the input voltage to UE 2 = 2 V the input voltage UE 1

is according to chart 12.4 then measure with the multimeter each output voltage UA.

Repeat those measurements with input voltage of UE = -2 V.

• Fill the determined values into chart 12.4 and graphically present the relation of UA and input voltage UE 1 [UA = f (UE 1)] in diagram 12.11.

• Then repeat all measurements for an input resistance of RE 1 = 4,7 kΩ, fill the values for the output kΩ, voltages UA in chart 12.5, then draw the curve into the diagram 12.12.

pic. 12.10

~ ~

-

-

Ugl = 15 V Ugl = 0 ... 10 V

U E1

R1

1 kΩ

R 1 k Ω

R2

1 k Ω

RE1

10 kΩ

RE2

10 k Ω U E2

-E +E

RR

10 k Ω

UA

~-U -15 V


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chart 12.4 pic. 12.11

RE 2 = 10 kΩ

UE 1 [V] 0 2 4 6 8 10

UA [V] at UE 2 = 2 V

UA [V] at UE 2 = -2 V

UA [V] 4

2

0 -2

2 4 6 8 10

UE1 [V]

-4

-6

-8

-10

-12

-14


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chart 12.5 * To get a linear curve, you have to insert an additional measuring point UE, at the limiting point of the . op-amp. pic. 12.12

RE 1 = 4,7 kΩ

UE 1 [V]* 0 2 4 6 8 10


UA [V] at UE 2 = -2 V

UA [V] 4

2

0 -2

2 4 6 8 10

UE1 [V]

-4

-6

-8

-10

-12

-14


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Test 1: Input currents IE 1 and IE 2 have to flow across which resistor? Answer: Test 2: What kind of circuit is obtained if only one input is connected? Answer: Test 3: What is the output voltage if one input current is positive and the other one negative? Answer:


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12.3 Op-Amp serving as Differential Amplifier 12.3 Basics The differential amplifier is considered to be a subtraction element with amplification. Both input voltages gain with the same amplification factor. For a good common mode rejection the amplification factors for both inputs have to be equal. Following equation is for the common mode rejection G: At the non-connected amplifier it is µA 741 approx. 90 dB. With subtraction element is RE 1 = RE 2 = RE and R3 = RR output voltage UA is:

E

R

E

E

RE

RE

R

EA

R

RU

R

RR

RR

RUU •−

+•

+•= 12 ( )

E

R

EEA

R

RUUU •−= 12

If resistance RR and both input resistances RE are totally equal, then only the difference between the two inputs gain by amplification ratio RR to RE. Op-amps in differential amplifier circuits are e.g. used for measuring amplifiers. pic. 12.13

G = 20 log v

vcommon mode

/ dB

RR

RE1 -E

UE1

UE2

RE2 +E RR

UA


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12.3.1 Tests Test

Examine the differential amplifier for common mode rejection by measurement. Test proceeding

• Set-up the circuit in pic. 12.14, then measure for different input voltages UE 1 and UE 2

according to chart 12.6 the output voltage UA. Fill the values for voltage in chart 12.6 and graphically present the relation of output voltage UA to input voltages UE 1 and UE 2

in diagram. 12.15. pic. 12.14

~

- Ugl = 15 V

R 1

1 k Ω R 3

470 Ω R 5

10 k Ω -E

R 7

10 k Ω

R

U E1

R R 6

22 kΩ +E

~- U -15 V

10 k Ω

R 2

1 k Ω

1 k Ω R 4

470 Ω U E2 R8

22 k Ω

U A


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UE 1 [V]* -10 -6 -2 +2 +6 +10

UA [V] at UE 2 = -4 V



chart 12.6 * To get a linear curve, you have to insert an additional measuring point UE, at the limiting

point of the op-amp. pic.12.15

UA [V] 14

12

10

8

6

4

2

-16 -14 -UE1 [V]

-12 -10 -8 -6 -4 -2 -2 -4

2 4 6 8 10 12 14 16

UE1 [V]

-6

-8

-10

-12

-UA [V] -14


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Test 1: When do we obtain a common mode rejection? Answer: Test 2: What circuit is considered to be a differential amplifier? Answer: Test 3: What output voltage is obtained when both inputs are supplied? Answer:


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13. Dynamic Behavior of an Op-Amp 13.1 Basics Op-amps are broadband DC voltage amplifiers, not only amplifying DC voltages but also AC voltages in some frequency ranges. Negative feedback caused by external circuitry does not only define the amplification factor but also the bandwidth and the input resistance. 13.2 Tests Test

Examine my measuring the behavior of an op-amp when applying sinus voltage.

Test proceeding: • Apply sinus voltage Uss = 1 V, f = 1 kHz to the input of circuit in pic. 13.16. Measure

with the oscilloscope the input voltage UE and the output voltage UA, and draw the respective curve into pic. 13.18.

• Determine the amplification factor v by following equation:

E

A

U

U

Uv =

pic. 13.16

~

-Ugl = 15 V

R 2

1 k Ω

R 1

10 k Ω

G~ UE

+

UA

~-

U -15 V


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• Then determine the upper limiting frequency. It is the frequency, when output voltage drops to 70,7% of its maximum value, if frequency is increases.

• Determine the input resistance RE. Therefore connect a resistor RV in series to the input. This will affect a receding of the output voltage from value U1 to value U2.. Calculate the input resistance RE with both voltage values and the given series resistor with following equation:

−=

12

1

U

U

RR

V

E

RE = input resistance in Ω RV = series resistor in Ω U1 = output voltage with no series resistor U2 = output voltage with series resistor

• Then determine the output resistance RA, by connecting load resistance RL to earth at the

output. This affects a drop of output voltage from value U1to value U2. • Calculate the output resistance with both voltage values and the given load resistance with

following equation:

−= 12

1

U

URR LA

RA = output resistance in Ω RL = series resistance in Ω U1 = output voltage with no resistor U2 = output voltage with resistor

• Fill the determined values in chart 13.7.

• Then set-up the circuit in pic. 13.17, apply sinus voltage of uss = 1 V; f = 1 kHz to the input UE.

Repeat all measurements and calculations, draw the curves of input- and output voltages into diagram 13.19 and fill your values in chart 13.7.

pic.13.17

~

-

Ugl = 15 V

A

R 2

R 1

G

B

10 k Ω

- R2 1 kΩ

+

~ U E R 4 R 3 UA 1 k Ω 10 k Ω

~-

U -15 V


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Inverting Input

(R1 = 10 kΩ, R2 = 1 kΩ) Non-inverting Input

(R1 = R3 =10 kΩ, R2 = R4= 1 kΩ) UE 1 V 1 V

UA

vU

fo

RE (RV = 100 Ω) (RV = 10 kΩ)

RA (RL = 100 Ω) (RL = 100 Ω)

Tab. 13.7

Settings: Y1 = 5 V / part Y2 = 0,5 V / part X = 0,2 ms / part

pic. 13.18

Settings: Y1 = 5 V / part Y2 = 0,5 V / part X = 0,2 ms / part

pic. 13.19

• Now disconnect point A in the circuit of pic. 13.17 from reference potential () and connect to point B. Hence, both inputs are fed by same voltage. Measure the output voltage with your oscilloscope.

- 0 (Y1)

= U A

- 0 (Y2)

= U E

- 0 (Y1)

= U A

- 0 (Y2)

= U E


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Test 1: What is the phase shift angle between input and output voltage? Answer:

a) When inverting input is used:

b) When non-inverting input is used: Test 2: The input resistance RE is similar to what resistance value at the inverting input? Answer: Test 3: The input resistance RE is similar to what resistance value at the non-inverting input? Answer: Test 4: Does output resistance RA is in any relation to the feeding input RE? Answer: Test 5: What is the value of the output voltage RA if both inputs of the op-amp are fed by

same voltage?

Answer:


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Required voltage supply and measuring instruments Function generator

functions: Sinus

- reactive voltage: Uss = 20 mV ... 7 V

- frequency: f = 10 Hz ... 5 MHz

- impedance: Ri = 50 Ω

Voltage supply

- mains supply: 230 V AC; 50 ... 60 Hz

- DC voltage: 0 ... 30 V; 0,3 A (adjustable); +/-15 V; 0,3 A (fix) - AC voltage (Sinus): 0 ... 24 V (effective); 50 Hz; 0,1 A (adjustable) AC Generator

- Line voltage: 7 V

- Conductor voltage: 12 V - Conductor Current: 50 mA Multimeter

- DC voltage

range: 0,1 V ... 30 V input

resistance: RE > 100 kΩ - AC voltage (Sinus)

range: 1 V ... 15 V (effective) input

resistance: RE > 100 kΩ frequency:

f < 50 Hz - DC voltage*

range: 0,1 µA ... 300 mA

voltage drop : < 250 mV - tolerances +/- 3%

* For some test with lower currents you will need a measuring range of < 1µA. Oscilloscope (with X/Y-operation) - channels: 2 (one of it invert)

- range: f ε 10 MHz - distortion coefficient Y: 0,02 - 0,5 - 1 - 2 - 5 - 10 - 20 V / part

X: 0,1 - 1 - 2 - 5 - 10 ms / part

X: 0,1 - 1 - 2 - 5 - 10 ms / part Those technical data are minimum requirements to do all the tests within this manual.


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Electrical Engineering Department Electronics Lab Manual

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