Laboratory Guide for Electronics 1: Basic Components

Laboratory Guide for

Electronics I

Basic Components

Prepared by

Francisco Glover, S.J.

Version 1.1

Electronic and Communication Engineering Ateneo de Davao University

June, 2001

Ateneo de Davao University

i

Preface This present text is a laboratory guide for introductory electronics,

and is designed to accompany eleven laboratory modules. The 27 experiments contained here are loosely connected with Chapters 1 through 9 of the 7

th edition of Electronic Devices and Circuit Theory by

Boylestad & Nashelsky, although the sequence of the chapters has been modified Three different preferences for the conduct of a laboratory course in electronics may be noted: 1] The students are given the components, a breadboard and hook-up wire and are to construct the circuit starting from scratch. 2] The entire laboratory is done by computer simulation, using programs such as Electronic Workbench, PSpice or Tina. 3] Each circuit is a hard-wired visual circuit with all components accessible. Each of these methods has its advantages and disadvantages, and perhaps a mixture of all three methods might be the best choice. The present manual is written for use with a specific set of modules, developed at the Ateneo de Davao University. This manual and its associated modules are a part of a larger project to cover most of the advanced courses required for a degree in Electronic and Communication Engineering . The original vision for the project stems from Engr. R. U. Espina, Coordinator of the Ateneo ECE program. He has provided continuing assistance and consultation. The project would not have been possible without the gracious support of Mrs. Perla Funa, College Dean, and Dra. Nenita Maluluan, Chair of the Engineering Division. Acknowledgement is also due to Rafael Gaid, Nelson Pelotan and Vincent Pera, ECE students at the Ateneo, for their invaluable assistance in design, construction and evaluation of this material. From the Physics Department staff design and construction work was provided by Romulo Pepeña, and logistic support by Judith Perias. June, 2001 Ateneo de Davao University

ii

Electronics I

1: Oscilloscope Measurements

2: Ohm’s Law EL1-A

3: Diode Characteristics EL1-B

4: Diode Load Line EL1-B

5: Half- and Full-wave Rectifiers EL1-C

6: Bridge Rectifiers EL1-C

7: Zener Voltage Regulator EL1-D

8: Clippers EL1-E

9: Clampers EL1-E

10: Voltage Multiplier EL1-F

11: BJT Charactistics EL1-G

12: BJT Biasing: Fixed Bias EL1-H

13: BJT Biasing: Voltage Divider EL1-H

14: BJT Biasing: Collector Feedback EL1-H

15: BJT Biasing: Common Base EL1-H

16: BJT Biasing: Emitter Follower EL1-H

17: N-channel JFet Characteristics EL1-I

18: FET Biasing: Self Bias EL1-J

19: FET Biasing: Voltage Divider EL1-J

20: BJT Small Signal: AC, ro and re EL1-H

21: BJT Small Signal: Fixed Bias EL1-H

22: BJT Small Signal: Voltage Divider Bias EL1-H

23: BJT Small Signal: Collector Feedback Bias EL1-H

24: BJT Small Signal: Emitter Follower Bias EL1-H

25: FET Small Signal: gm and rd EL1-J

26: FET Small Signal: Self Bias EL1-J

27: FET Small Signal: Source Follower EL1-J


1

Electronics I Experiment #1 : Oscilloscope Measurements Materials: Oscilloscope, multimeter, function generator, power supply

Basically an oscilloscope is a measuring instrument that displays a voltage vs. time graph of an input signal. Many different models are commercially available, from basic to extremely advanced. Although the controls differ from model to model, all models have certain controls in common, which are briefly described below. As with any graph, we are free to select the origin, and the range along the vertical and horizontal axes; the following controls let us do this:

Vertical axis: Voltage is displayed in the vertical direction:

Vertical Position: Moves the entire display up or down Vertical Gain: Discrete: Sets vertical amplifier gain to fixed values in volts/cm Variable: Adjusts the vertical amplifier gain in a continuous manner; Set to maximum or CAL when making measurements Input: AC Ground DC

AC: Input passes through a capacitor, removing any constant voltage Ground: Input signal disconnected from vertical amplifier DC: Input signal connected directly to vertical amplifier

Dual Channel: If the oscilloscope is a dual channel model, two separate traces may be displayed. In this case there are two sets of vertical controls, for channel A and B, along with a selector for A, B, Both, A+B, A-B

Horizontal axis: Time is displayed along the horizontal axis.

Horizontal Position: Moves the entire display left or right. Horizontal Time Base:

Discrete: Sets horizontal sweep speed in mSec/cm or Sec/cm Variable: Adjusts sweep speed in a continuous manner; Set to maximum or CAL when making measurements

Triggering: Determines when the horizontal sweep begins

Mode: Auto: Trace visible even if no input signal. Normal: Trace begins, depending on Trigger Source Trigger source: Channel A, Channel B, Line voltage, External source


2

Trigger character:

Signal Level: + or - : Sweep starts if input greater than or less than

some selected value

Signal Slope: + or - : Sweep starts if signal is increasing or

decreasing

Trace controls: Brightness: sets intensity level of trace Focus: makes trace sharp or fuzzy Rotation: makes trace horizontal (earth’s magnetic field can tip trace)

Test point: Reference signal, 1000 Hz square wave, 1.0 volt, peak-to-peak

Procedure: 1: With the oscilloscope model you are using, identify and adjust each of the

controls noted above. 2: DC Voltage Measurement: Set the vertical gain to 1.0 volt/cm, and the input

to DC. Move the trace to the bottom line of the screen. Connect the power supply to the input and vary this input from 2.0 to 10.0 volts, in 1.0 volt steps, as measured from the oscilloscope screen. Also with a digital multimeter measure and record the minimum and maximum input voltages that correspond to the same voltage level as viewed on the oscilloscope screen. This is an indication of the precision of the oscilloscope measurement. (Current may be measured simply by passing it through a known resistor, measuring the voltage drop, and applying Ohm’s law.)

3: AC Voltage Measurement: Set the vertical gain to 2.0 volt/cm, and the input

to AC. Set the function to 500 Hz, sine wave and connect this, in parallel with the multimeter, to the oscilloscope input. Vary the function generator output from 2.0 to 16.0 volts, peak-to-peak, in 2.0 volt steps. Record the oscilloscope peak-to-peak readings and the AC multimeter RMS readings and calculate Vp-p (2

-3/2) and %-difference 200% x |A-B|/(A+B)

AC multimeters normally have poor high-frequency response. Repeat

the above measurements for 5,000 Hz and 50,000 Hz and note any differences.


3

4: Time base Measurements: Set the function generator to square wave and

the time base variable to CAL. Use the following settings to compare accuracy of the sweep speed:

5: Triggering: Set the oscilloscope time base to 1 mSec / cm and CAL. Set the

function generator to 500 Hz, sine wave, and adjust amplitude display to 8.0 Vp-p. The display should appear as shown. Adjust the Trigger Level and Slope as shown, and record the screen display:

Sweep speed Function Generator

10 mSec / cm 50.0 Hz

1 mSec / cm 500 Hz

100 μSec / cm 5,000 Hz

10 μSec / cm 50.0 kHz

Level Slope

0 V –

+2.0V +

–2.0V +

–2.0V –


4

Oscilloscope Setting

Multimeter Minimum

Multimeter Maximum

2.0

3.0 4.0

5.0 6.0

7.0 8.0

9.0 10.0

Data Sheet: Electronics I Experiment # 1 Oscilloscope Measurements

Name:______________________________ Date:______ DC Voltage Measurement:

AC Voltage Measuremen 500 Hz

5.00 kHz 50.0 kHz

Vp-p VRMS Vp-p (2-3/2) %-Diff

2.0

4.0 6.0

8.0 10.0

12.0 14.0

16.0


2.0 4.0

6.0 8.0

10.0 12.0

14.0 16.0


2.0 4.0

6.0 8.0

10.0 12.0

14.0 16.0


5

Time Base Measurements:

Triggering: Sketch oscilloscope display

Level = 0 Slope = negative Level = +2.0 Slope = positive

Level = -2.0 Slope = positive Level = -2.0 Slope = negative

Sweep speed Function Generator

Over / Under

10 mSec / cm 50.0 Hz

1 mSec / cm 500 Hz

100 μSec / cm 5.0 kHz

10 μSec / cm 50.0 kHz 5,000 Hz

50.0 kHz

6 2 Ohm’s Law

Electronics I Experiment #2: Ohm’s Law Materials: Module EL-1-A, power supply, two multimeters

The ratio of the potential difference or voltage, V, across a circuit element to the current, I, through it is defined as the element’s resistance. The ratio of the

change in potential difference, V, across a circuit

element to the change in current, I, through it is defined as the element’s dynamic resistance. If a graph of current vs. voltage shows a straight line passing through the origin, the resistance and dynamic resistance are constant and equal. In all other cases they may not be equal and depend on the selected point on the graph line. In this experiment you are to investigate the current-voltage relationship for several circuit elements.

Procedure: 1: Use a full-size sheet of graph paper to make a current vs. voltage graph for R1. Plot as many points as are needed to draw a smooth graph line. Limit the voltage (horizontal ) range to 10.0 volts. Select the current (vertical) scale to give optimum resolution. For 3.0, 6.0 and 9.0 volts determine the resistance and dynamic resistance of the R1. 2: Repeat step 1 for each of the other elements on the module.

7 2 Ohm’s Law

Data Sheet: Electronics I Experiment # 2 Ohm’s Law

Name:______________________________ Date:______ For R1: (include full-size current-voltage graph )

For R2: (include full-size current-voltage graph )



Voltage Current Resistance Dynamic Resistance

3.00

6.00

9.00


3.00

6.00

9.00


3.00

6.00

9.00


3.00

6.00

9.00

3: Diode characteristics 8

Electronics I Experiment #3: Diode Characteristics Materials: Module EL-1-B, power supply, two multimeters

In this experiment, you are to look at the behavior of four different diodes. You apply a known voltage across the diode and measure the resulting current, which depends both on the size of the voltage and its polarity. Too large a voltage may destroy a diode. For this reason three of the

diodes to be used, ZD1, ZD2, and LED have a resistor in series to limit the current. The current through the resistor and diode is the same; however measure the voltage across the diode alone. Also, for some diodes the voltage range of interest is 2 volts and below. Since the regulated power supply used may not extend this low, it can be attached to the voltage divider (potentiometer). Notice that for the zener diodes the voltage is applied in the reverse direction.

Procedure: 1: For each diode, draw a graph of current (vertical axis) against voltage (horizontal axis). Do not use current in excess of 30 milliamperes (only 20 mA for the LED). The number of pairs of values you use depends on the shape of the graph; where the slope is changing, measure more data pairs. Individual data points are not to be connected by straight lines; rather a smooth curve is to be drawn, representing the best fit of the data presented. For each graph select a horizontal scale that is most appropriate to the voltage range. 2: For the diode, D, compute the DC (static) and AC (dynamic) resistance at current levels of 2, 10, 20, 30 mA. 3: For the LED compute the power at current levels of 2, 10, and 20 mA.

3: Diode characteristics 9

Data Sheet: Electronics I Experiment # 3

Diode Characteristics

1: Attach full-size graphs of current vs. voltage for D, ZD1, ZD2, and LED.

2: Static and dynamic diode resistance

I (mA) V (mV) R = V / I I (mA) V (mV) r = V / I 1.90

0.2

2.10

9.50 1.00

10.50

19.50 1.00

20.50

29.50 1.00

30.50

3: LED Power consumption:

I (mA) V (V) Power (mW)

2.00

10.00 20.00

4: Diode Load Line 10

Electronics I Experiment #4: Diode Load Line Materials: Module EL-1-B, two multimeters

Prior experiments considered current-voltage graphs for various resistors and diodes. Here we connect to a fixed voltage source a series combination of a resistor and diode. Since in series, each element carries the same current while the sum of their voltage drops just equals the fixed voltage.

It is convenient to superimpose on the diode graph an inverted copy of the resistor graph. The line of the inverted resistor graph is the load line, and the point of intersection is the Q-point. At the Q-point the resistor and diode currents are the

same; the diode voltage is VQ, the resistor voltage

is VS-VQ so their total is VS as expected. This

load line technique for a series combinations of two elements across a fixed voltage will occur again in the analysis of transistors.

Procedure: 1: For diode D construct a current-voltage graph. Limit the current to 10 mA and use a full sheet of graph paper for an enlarged scale Set

resistor R to 1000 and Vs to 10.00 volts and connect the diode and resistor, R, in series (if necessary by-pass the input potentiometer). Measure and record VQ and IQ. From a load line on your graph compare these values of VQ and IQ with

the experimental results. Repeat for R = 2000

and R = 4000 2: Repeat step 1 for zener diode ZD2 . Bypass the series resistor connected to this zener diode. 3: Repeat step 1 for LED. Bypass the series resistor connected to this LED.

4: Diode Load Line 11


Diode Load Line Name:______________________________ Date:______ For diode D: (attach current-voltage graph)

For zener diode ZD2: (attach current-voltage graph)

For diode LED: (attach current-voltage graph)

Resistance VQ IQ measured IQ load line % Diff (for IQ) 1000

2000 4000

Resistance VQ IQ measured IQ load line % Diff (for IQ) 1000 2000

4000

5 Half and Full-Wave Rectifiers 12

Electronics I Experiment # 5 Half- and Full-Wave Rectifiers

Materials: Module EL1-C, AC power supply, oscilloscope, multimeter

Rectifier circuits are used to convert an alternating to a direct voltage. Three common types are the half-wave, full-wave and bridge. The half-wave circuit uses a single diode which conducts only during the positive half of each

cycle. Points A and C are alternately positive and negative to each other. Charge is stored in C1 during the conduction period, and released during the non-conduction period. The voltage appearing across the load may contain a periodic component, the ripple voltage, the peak-to-peak value of which may be measured with an oscilloscope (AC input ) . The direct component of the output is measured across the load with a multimeter (DC volts). With a greater load current (decrease in load resistance) the output voltage normally decreases. Increasing the capacitance and the load resistance normally decreases the ripple.

The full-wave circuit uses two diodes connected as shown which alternately conduct on either half cycle. This configuration provides a somewhat larger output voltage, with less ripple, than the half-wave.

Procedure:

Connect the module board, EL1-C, to the power supply module to provide 60 Hz power. Set up each of the two circuits, and make the measurements shown on the Data Sheets attached.



Half- and Full-wave Rectifiers

Name:______________________________ Date:______

Half-Wave: Peak-to-Peak Ripple measurement Load = 7.9 k

Half-Wave: DC output voltage R = 470

Which capacitor, C1 or C2, is more effective in maintaining a high output voltage

across the load? _________

Wave forms: R = 470 Load = 7.9 k

C1 = 0 C2 = 0 C1 = 10 C2 = 0 C1 = 10 C2 = 10

R C1= 0 C2= 0

C1= 10 C2= 0

C1= 10 C2= 10

C1= 10 C2= 100

C1= 100 C2= 0

C1= 100 C2= 10

C1=100 C2= 100

100

200

470

Load C1= 10 C2= 0

C1= 10 C2= 10

C1= 10 C2= 100

C1= 100 C2= 0

C1= 100 C2= 10

C1=100 C2= 100

7.9k

4.7k

3.2k

1.0k


Data Sheet: Electronics I Exp. # 5 Cont.

Half- and Full-wave Rectifiers

Full-Wave: Peak-to-Peak Ripple measurement Load = 7.9 k

Full-Wave: DC output voltage R = 470

Which capacitor, C1 or C2, is more effective in maintaining a high output voltage across the

load? _________

Wave forms: R = 470 Load = 7.9 k

C1 = 0 C2 = 0 C1 = 10 C2 = 0 C1 = 10 C2 = 10

R C1= 0 C2= 0

C1= 10 C2= 0

C1= 10 C2= 10

C1= 10 C2= 100

C1= 100 C2= 0

C1= 100 C2= 10

C1=100 C2= 100

100

200

470

C1= 10 C2= 0

C1= 10 C2= 10

C1= 10 C2= 100

C1= 100 C2= 0

C1= 100 C2= 10

C1=100 C2= 100

7.9k

4.7k

3.2k

1.0k

6 Bridge Rectifiers 16

Electronics I Experiment #6: Bridge Rectifiers Materials: Module EL1-C, oscilloscope, function generator, multimeter

The full-wave rectifier requires two diodes and a center tap on the

transformer secondary. The bridge rectifier considered here requires four diodes but does not require a center-tapped transformer. During each half cycle one pair of diodes is ON the other OFF. For real diodes this means 1.2 volt drop for the bridge rectifier, as opposed to 0.6 volt diode drop for the full-wave.

Ripple considerations are basically the same for both bridge and full-wave.

Twice each cycle the filtering capacitor is fully recharged. The rate of discharge depends of the effective time constant of the resistor-capacitor pair, namely, C x (R + Load); the larger the time constant, the slower the discharge rate. The duration of this discharge increases with the period of the signal (decreases with the frequency). The peak-to-peak ripple voltage depends on the amount of discharge during each half cycle, decreasing as frequency, capacitance and resistance increase.

Note: In Experiment # 5, Half- and Full-wave Rectifiers ,we used a 60 Hz power-

line input, while here we use a function generator to provide a variety of frequencies and wave-shapes. Either type of input may be used with any of the rectifier types considered.


Procedure:

1: With Module EL1-C, set up a bridge rectifier circuit, as shown, and connect the function generator between points A and B. Set f = 100 Hz , R = 0

, C1 = 0, C2 = 0, Load = 7.9 k . Use, in succession a sine wave, square wave and triangular wave. Connect the oscilloscope first across points A and B, and then across the Load. In each case, sketch the screen display. Use the oscilloscope DC input mode 2: Connect the oscilloscope across the load and set the other values as Sine wave, R = 0 Ω, C1 = 10 μf, C2 = 0 . Set the frequency successively to 50, 100, 200, 400 and 800 Hz and record the peak-to-peak ripple across a load

of 1.0 k and 7.9 k . In each case, sketch the screen display. Use the oscilloscope DC input mode

3: Set f = 50 Hz, R = 0 , C1 = 10 μf, C2 = 0 . Use, in succession, a sine, triangular and square wave input, and sketch the screen display. . Use the oscilloscope DC input mode.


Data Sheet: Electronics I Experiment # 6 Name:____________________________________ Date: _______

Bridge Rectifiers

Part 1: f = 100 Hz , R = 0 , C1 = 0, C2 = 0, Load = 7.9 k

Wave form across points A and B: (before rectification)

Sine wave Triangular wave Square wave

Wave form across Load: (after rectification)


Part 2: Peak-to-Peak ripple voltage:

Sine wave, R = 0 , C1 = 10 μf, C2 = 0,

Part 3: f = 50 Hz, R = 0 , C1 = 10 μf, C2 = 0


frequency Load = 1.0 k Load = 7.9 k

50 100

200 400

800

7:Zener voltage regulator 19

VZ IZ

4.05 18.4 4.10 22.4

4.15 27.6

4.20 34.7

4.25 47.6

Electronics I Experiment # 7: Zener Voltage Regulator Materials: Module EL-1-D, power supply, two multimeters

An ideal zener diode, with reverse bias, acts as an open circuit for voltage between 0 and VZ, and as a short circuit for voltages more negative than VZ. Real zener diodes only approximate this condition. In this experiment we explore the zener diode used as a voltage regulator.

As shown, the zener diode is placed in parallel with the load so for regulation we select a zener with VZ equal to the desired load voltage, VL. But to provide regulation, the zener current, IZ, must be greater than zero so I1 must be sufficient to turn “on” the zener and provide the desired voltage drop across the load. The current-voltage graph of an ideal zener indicates a vertical graph line at voltage –VZ, while for a real zener the graph line near –VL is steep but not really vertical. As a first step in this experiment we need accurate data for the ZD1, ZD2 and ZD3. The attached table shows a typical zener relation, at 0.05 volt intervals, for current near 30 mA.

A: Design a Zener Regulator

Design a regulator to give 4.15 volts across a 1,000Ω load, from a 8.00 volt source, using the zener data shown. Determine IL, I1, IZ and R1.

IL = VL/RL = 4.15/1000 = 0.00415 A = 4.15 mA From the table, IZ = 27.6 mA Therefore I1 = IL + IZ = 4.15 + 27.6 = 31.8 mA

R1 = (VIN – VZ) / I1 = (8.00 – 4.15) / 0.0318 = 121 It is possible to set up such a circuit, measure the currents and verify the results. B: With constant VIN , determine range of RL for VL = 4.15 ± 0.10 volts:


From the table, IZ = 18.4 mA for VZ = 4.15 – 0.10 volts V1 = VIN – VZ = 8.00 – 4.05 = 3.95 volts I1 = V1 / R1 = 3.95 / 121 = 0.0326 A = 32.6 mA IL = I1 – IZ = 32.6 – 18.4 = 14.2 mA

RL = VL / IL = 4.05 / 0.0142 = 285 = minimum load resistance

From the table, IZ = 47.6 mA for VZ = 4.15 + 0.10 volts V1 = VIN – VZ = 8.00 – 4.25 = 3.75 volts I1 = V1 / R1 = 3.75 / 121 = 0.0310 A = 31.0 mA Notice that even if IL = 0 (for infinite RL) IZ can only reach 31.0 mA corresponding

to a VZ value of about 4.18 volts. Therefore the upper limit load is RL = .

C: With constant RL, determine range of VIN for VL = 4.15 ± 0.10 volts:

IL = VL / RL = (4.15 – 0.10) / 1000 = 4.05 / 1000 = 0.00405 A = 4.05 mA From the table, IZ = 18.4 mA for VZ = 4.15 – 0.10 volts I1 = IL + IZ = 4.05 + 18.4 = 22.5 mA V1 = I1 R1 = (.0225)(121) = 2.72 volts VIN = V1 + VZ = 2.72 + 4.05 = 6.77 volts = minimum for VIN

IL = VL / RL = (4.15 + 0.10) / 1000 = 4.25 / 1000 = 0.00425 A = 4.25 mA From the table, IZ = 47.6 mA for VZ = 4.15 + 0.10 volts I1 = IL + IZ = 4.25 + 47.6 = 51.9 mA V1 = I1 R1 = (.0519)(121) = 6.28 volts VIN = V1 + VZ = 6.28 + 4.25 = 10.53 volts = maximum for VIN

Procedure: 1: For each zener, ZD1, ZD2 and ZD3, make a separate table of current-voltage in the neighborhood of 30 mA. Measure the current at “IR”, leave “IL” open, and measure the voltage directly across the diode.

2: For each zener: A: Design a Zener Regulator, with RL = 1000 Ω, VIN = 8.00 volts B: With constant VIN , determine range of RL for ± 0.10 volt change in load

voltage, and verify results by direct measurement. C: With constant RL determine range of VIN for ± 0.10 volt change in load

voltage, and verify results by direct measurement.

Note: In making any current measurement, you must take into account the input resistance of

the ammeter range used.


VZ IZ

Calculated Measured %-Diff IL I1

R1 RL-min

RL-max VIN-min

VIN-max

VZ IZ

Calculated Measured %-Diff IL I1

R1 RL-min

RL-max VIN-min

VIN-max

VZ IZ

Calculated Measured %-Diff

IL I1

R1 RL-min

RL-max VIN-min

VIN-max

Data Sheet: Electronics I Experiment # 7 Zener Regulator

Name:______________________________ Date:______

For ZD1: RL=1,000Ω, VIN=8.00 volts; Select for IZ near 30 mA: VZ = _____ IZ = ____

For ZD2: RL=1,000Ω, VIN=8.00 volts; Select for IZ near 30 mA: VZ = _____ IZ = _____

For ZD3: RL=1,000Ω, VIN=8.00 volts; Select for IZ near 30 mA: VZ = _____ IZ = ______

8 Clippers 22

Electronics I Experiment # 8 Clippers: Materials: Module EL1-E, Oscilloscope, Function generator

A clipper circuit, using a resistor and a diode, removes a portion of the input signal. In a series configuration (shown below) the diode is in series with the load. The diode must be turned ON (conducting) to obtain an output.

In a parallel configuration the diode is in parallel with the load. The diode must be turned OFF

(non-conducting) to obtain an output.

Either the positive or negative portion of the input is removed, depending on the orientation of the diode

In A the diode conducts

whenever the input is positive. In

B the battery is working against

the diode, which conducts only

when the input is greater than Vb.

In C the battery is helping the

diode, so there is conduction whenever the input is greater than

–Vb.

Procedure:

In the data sheets attached, various series and parallel clipper circuits are shown. Wire each of these circuits on the module board, and use a 1000 Hz sine wave as input. View two full cycles of output on an oscilloscope. Sketch the oscilloscope display and record the pertinent data.

8 Clippers 23

Data Sheet: Electronics I Experiment # 8 Clippers

Name:______________________________ Date:______ Serial Clippers:

Data Sheet: Electronics I Exp. # 8 continued

Clippers Name:______________________________ Date:______ Parallel Clippers:

8 Clippers 24

Data Sheet: Electronics I Exp. # 8 continued Clippers

Name:______________________________ Date:______ Parallel Clippers

9 Clampers 25

Electronics I Experiment # 9 Clampers: Materials: Module EL1-E, Oscilloscope, function generator

A clamper circuit shifts the DC level of a periodic waveform, using a capacitor, diode and optionally a constant voltage

source, Vb. The resistor shown in

the diagram represents an optional circuit resistor in parallel with the resistance of the load. The time constant of the circuit (product of the capacitance and the load resistance) must be long in comparison with the period of the input signal, so that the capacitor remains fully charged at all times. The diode, always in parallel with the load, effectively prevents the capacitor from discharging, making the capacitor appear as a battery. If the capacitor used is electrolytic, care must be taken to observe correct

polarity. The (ideal) diode acts like a simple switch, ON or OFF depending on a forward or reverse bias.

Procedure: In the data sheets attached, various clamper circuits are shown. Wire each of these circuits on the module board, and use a square wave as input with a period much shorter than the circuit time constant. View two full cycles of output on an oscilloscope. Sketch the oscilloscope display and record the pertinent data. Make sure you use the oscilloscope DC input rather than the AC.

9 Clampers 26

Data Sheet: Electronics I Experiment # 9 Clampers

Name:______________________________ Date:______

10 Voltage Multipliers 27

Electronics I Experiment #10: Voltage Multipliers Materials: Module EL1-F , function generator, oscilloscope, multimeter

Is it possible to obtain a 20 volt output from a 5 volt AC source without the use of a transformer? If only a small current is needed, a voltage multiplier can do it. Imagine charging 4 capacitors each to 5 volts; then connect them in series. The combined voltage would be 20 volts,

but once you start drawing a current, this voltage drops rapidly. Basically a voltage multiplier circuit is just such a series of capacitors, but with provision to continually re-charge them. The diagram below shows an exploded view of this module. Notice that units #1 and #3 are identical, while #2 and #4 are upside-down versions of #1. In each case the diode permits the capacitor to be charged but prevents it from discharging through the diode. The voltage reference point is the junction of the capacitor and resistor. Let’s see how this works…

Let Vm be the

maximum amplitude of the alternating source

voltage ( VRMS =

Vm/ 2 ) Now connect unit #1 to the source.

Diode D1 conducts

only when point B is positive with respect to A. At this moment we know from Kirchoff’s loop relation

that the voltage across C1 is Vm, (assuming the diode to be ideal), and V1-A = Vm.

Next, attach unit #2. Diode D2 conducts only when point A is positive with

respect to B. Kirchoff’s loop relation now tells us that the voltage across C2 is

2Vm, since it is charged through the diode by the combined voltage of the source

and C1 ; also V2-B = 2Vm.


Attach unit #3. Diode D3 conducts and charges C3 only when point B is

positive. Again by Kirchoff, the voltage across C3 is 2Vm, sum of the source plus

C2 minus C1, and therefore V3-A is 3Vm. Finally attach unit #4, and by similar

reasoning we find the voltage across C4 = 2Vm and V4-B = 4Vm. The DC

output voltage between B and point V4 is four times the peak value of the AC input voltage. If more diode-capacitor units are added, this output increases by

2Vm for each additional unit. Although the output may be very high, the voltage

across each capacitor never exceeds 2Vm and the reverse voltage across any

diode is limited to Vm . In the above analysis, the diodes are assumed to be

ideal, with zero voltage drop when conducting. The actual drop is in the neighborhood of 0.6 volts, so all capacitor voltages will be slightly decreased. The good news is that we can get a very high DC output from a low AC input, with low voltage ratings for the diodes and capacitors The bad news is that if the load resistance is not very high, a ripple appears in the output. Ripple voltages are normally measured in peak-to-peak values, the difference between the maximum and minimum. The capacitors are re-charged only during one-half cycle. The ripple can be decreased by using larger capacitors which discharge more slowly with the same load, and also by using an AC source of higher frequency, effectively shortening the time between recharging.

Procedure: 1: Set the AC source VRMS to 3.54 volts so that Vm = 5.00 volts. Set the input

frequency to 1,000 Hz, and include all units by closing S2, S3 and S4.

2: Set all capacitor switches to 1 f. Record the DC voltage across each diode,

capacitor, and the values V1-A , V2-B, V3-A, and V4-b . 3: Repeat step 2: for 10 f .

4: Set all capacitor switches to 1 f. Use input frequencies of 400, 1000 and 4000

Hz and place load resistors ( between points B and V4 ) of 1, 10 and 100 k , and with an oscilloscope measure the ripple across the load.

5: Repeat step 4: for capacitor values of 10 f .


Data Sheet: Electronics I Experiment # 10 Voltage Multipliers

Name:______________________________ Date:______

D C Voltages Cap D1 D2 D3 D4 C1 C2 C3 C4 V1-A V2-B V3-A V4-B

1 f

10 f

AC Ripple Voltage Capacitors set to 1 f Load \ Freq 400 Hz 1000 Hz 4000 Hz

1 k

10 k

100 k

AC Ripple Voltage Capacitors set to 10 f Load \ Freq 400 Hz 1000 Hz 4000 Hz

1 k

10 k

100 k

1: Is the voltage V1-B AC or DC ? Examine with an oscilloscope and explain.

2: Does the voltage V2-B change if S3 and S4 are open or closed? Explain.

3: Does the voltage V2-B change if C3 or C4 are set to 1 f or 10 f ? Explain.

11 BJT Characteristics 30

Electronics I Experiment #11: BJT Characteristics Materials: Module EL1-G , two multimeters, power supply

A transistor is a three-terminal device, the emitter, base and collector. For the transistor itself, apart from the external circuit elements to which it may be connected, six parameters may be

considered: the currents in each of the three terminals, IE, IB and

IC, and the difference in potential between any pair of terminals,

VBE, VCB, and VEC. These six parameters are not all independent

of each other. If two currents are known the third is determined by Kirchoff’s node method:

IE + IB + IC = 0. (a current leaving a node is considered as negative)

Likewise if two of the voltages are known the third is determined by Kirchoff’s mesh method:

VBE + VCB + VEC = 0.

For an NPN transistor, the direction of the base and collector

currents, IB and IC, are into the transistor; the emitter current, IE, is

directed out of the transistor. For a PNP the directions are just opposite. From the viewpoint of the engineer, not the transistor, any one of the three terminals may be considered as common, and either of the other two as input side or output side.

Normally, the base current is only a tiny fraction of the emitter current, while the emitter current is the sum of the base and

collector current . The symbols and define these relations:

IC = IE, IC = IB .

In practice is taken as 1, while will vary from transistor to transistor.

Transistor characteristics refer to a graphical presentation of the relationships between these variables. For each configuration we consider separately the input (input voltage and current) and output (output voltage and current) characteristics. However, since the output (emitter or collector) current is controlled by the level on the input (base) current, a third variable is involved with the output characteristics. A complete graphical presentation of three variable would require a three-dimensional graph which is a bit hard to sketch. As a compromise output current and voltage are plotted along the Y and X axes, and separate graph lines are shown for various values of base current. The most often used display is the common emitter. In this experiment you are to determine the input (base-emitter) and output (collector-emitter) characteristics of a NPN and PNP bipolar junction transistor (BJT) . The main selector switch, marked NPN/PNP switches one of the two transistors into the circuit and disconnects the other. It also automatically changes the polarity of the base and collector supply voltages.


INPUT Characteristics: A graphic input presentation displays the base current, IB , along the vertical axis and the

base-emitter voltage, VBE, along the horizontal, for zero collector-emitter voltages, VCE. The

resulting graph line is quite similar to that of a forward-biased diode. The ratio VBE / IB is the

static or DC input resistance; The ratio VBE / IB is the dynamic or AC input resistance

OUTPUT Characteristics A graphic output presentation displays the collector current, IC, along the vertical axis and the

collector-emitter voltage, VCE, along the horizontal, for a series of fixed base current values,

IB. The product of collector current, IC, and collector-emitter voltage, VCE, equals the power

dissipated as heat within the transistor. Frequently an additional curve is displayed showing the locus of points with a given power dissipation.

. The ratio VCE/IC is the static or DC output resistance; The ratio VCE/ IC is the

dynamic or AC output resistance. The static current gain, DC, is defined for any point as IC/IB

: The dynamic current gain, AC, is defined for any point as IC/ IB . These values may

easily be obtained from the output characteristics.

Procedure: 1: Make a plot of the input characteristics of the NPN transistor for VCE = 0 volts (IB along

vertical axis, VBE along horizontal) Do not let the base current, IB exceed 250 A . Place the


ammeter at the indicated position, A, on the module to measure IB, and the voltmeter between the base and emitter leads. Compute the dynamic input resistance for a base current of 50

A and also 200 μA. Compare these values with the textbook expression: ri = (26mV)/IB

where IB is expressed in milliamperes. Also compute the static input resistance at the same two points 2: Make plots of the output characteristics of the NPN transistor for values of base current, IB = 50, 100 and 150 μA. For each curve determine at VCE = 5.00 V the dynamic output

resistance, ro = VCE / IC and also the static output resistance. Repeat these resistance

measurements at VCE = 8.0 volts.

For convenience place the voltmeter at the indicated position, V, on the module. With the

small switch to the left you can read directly VCE ; with the switch to the right the collector current, IC, in amperes is given by the meter reading divided by 100 (shift the decimal point two places left for amperes, shift 1 place right for milliamperes). 3: Add a curve to the graphs of step 2, showing power dissipation of 500 milliwatts. 4: Repeat steps 1 to 3 for the PNP transistor.


Data Sheet: Electronics I Experiment # 11 BJT Characteristics

Name:______________________________ Date:______

NPN transistor: INPUT resistance:

Static: @ 50 A _______ : @ 200 A _______ Dynamic (measured): @ 50 A _______ : @ 200 A _______

Dynamic (theoretical): @ 50 A _______ : @ 200 A _______

OUTPUT resistance:

Static: @ 50 A _____ @ 100 A _____ @ 150 A _____ Dynamic : @ 50 A _____ @ 100 A _____ @ 150 A _____ CURRENT GAIN

DC ______ AC ______ [ Attach graphs of input and output characteristics]

PNP transistor: INPUT resistance:

Static: @ 50 A _______ : @ 200 A _______ Dynamic (measured): @ 50 A _______ : @ 200 A _______

Dynamic (theoretical): @ 50 A _______ : @ 200 A _______

OUTPUT resistance:

Static: @ 50 A ______ @ 100 A ______ @ 150 A ______ Dynamic : @ 50 A ______ @ 100 A _____ @ 150 A ______ CURRENT GAIN

DC ______ AC ______ [ Attach graphs of input and output characteristics]

12 BJT Biasing – Fixed Bias 34

Electronics I Experiment #12 : BJT Biasing: Fixed Bias Materials: Module EL-1-H, power supply, two multimeters

For any transistor, the “common emitter characteristics” reflect the relation between the base, emitter and collector currents and voltages at the time of manufacture, although the actual values do vary from unit to unit. In use, the transistor is incorporated into an external circuit, involving fixed DC voltages and currents, as well as time-varying AC voltages and currents. The Principle of Superposition allows us to treat separately the DC and the AC areas. Biasing refers to the DC area

The DC voltages are used to bias the transistor, that is, maintain the quiescent values of currents and voltages with no AC or signal components present. The Q-point on the

common emitter output characteristics represents a set of values for VCE, IC, and IB. The same transistor may be used with different external circuits (different bias configurations). The challenge to the electronic engineer is to select the correct external components for each configuration to maintain the desired Q-point.

Basic Fixed Bias

For each bias configuration to be studied, there is given a value for the supply voltage,

VCC , and also the Q-point, VCE, and IC . The usual analysis tool is Kirchoff’s mesh method,

for writing separate equations for the input and output sides of the circuit. For the basic fixed bias configuration, shown above, these are:

Input side: VCC = IB RB + VBE Output side: VCC = IC RC + VCE


For the output side

the only unknown is RC

which may be determined at once. For the input side

both RB and IB are

unknown. If the full common-emitter output characteristics are available, we can determine

IB by noting which IB=constant curve passes through the Q-point. Otherwise we can

determine directly the IB

current that gives the desired Q-point values of

VCE, and IC. You can use

Module EL-1-H in the configuration shown here to

determine the desired IB.

From this value you can also determine for this Q-point : DC = IC / IB .

The fact that differs from transistor to transistor poses a problem for the design

engineer. For a fixed VCC and RB the value of IB does not change; it is independent of .

However, if the original transistor is replaced by one with a different value, both IC and VCE

will be different, since DC = IC / IB . The Q-point has shifted! The emitter stabilized fixed bias

provides a partial solution to this difficulty.

Emitter Stabilized Fixed Bias

The emitter-base junction of any transistor is essentially just a diode. When conducting, a small change in voltage causes a large change in current. This is clearly seen in the common-

emitter input characteristics. So, in the transistor, a slight change in VBE causes a significant

change in IB. In the emitter-stabilized configuration, the voltage of the emitter with respect to

ground, VE, equals IERE ICRE. If IC were to increase, so also VE would increase, which

would decrease VBE. This decrease would in turn lessen slightly IB and so decrease IC,

tending to push it back to it’s original value. For the emitter-stabilized configuration Kirchoff’s equations now become:

Input side: VCC = IB(RB+βRE)+VBE Output side: VCC = IC(RC+RE)+VCE

Here we have two equations but three unknowns: RE, RC and RE. We need an additional

condition. The output equation above states that the voltage across the collector resistor,


emitter resistor and the transistor itself equals the supply voltage. Since VCE and the supply

voltage, VCC are given, we may still specify a given voltage across either of the two resistors,

or alternately, the voltage at the emitter ( with respect to ground, the negative terminal of the

supply voltage ), VE, or VC, the voltage at the collector with respect to ground. Changes in

RE change VE and VC, while the emitter-collector voltage, VCE is unchanged as long as (RC

+ RE) remains constant.

Procedure:

Basic Fixed bias:

1: Determine the value of the transistor in your module.

2: Given VCC = 12.00 volts, IC = 20.0 mA, VCE = 6.00 . Determine RB and RC. Then

configure Module EL1-H with these values of VCC, RB and RC, and measure IC and VCE and

compare these with the desired Q-point values. For RB use either R1 or R2 or both in series.

In measuring IC with the multimeter, remember that on the milliampere range the meter has a

certain input resistance which must be considered when setting the value for RC.

3: Repeat step #2 for: VCC = 12.00 volts, IC = 25.0 mA, VCE = 5.00.


Emitter Stabilized Fixed bias:

4: Given VCC = 12.00 volts, IC = 20.0 mA, VCE = 4.00 , VE = 2.00 V Determine RB , RE and RC. Then configure Module EL1-H with these values of VCC, RB, RE and RC, and

measure IC and VCE and compare these with the desired Q-point values.

5: Given VCC = 12.00 volts, IC = 25.0 mA, VCE = 5.00 , VE = 2.00 V Determine RB , RE and RC. Then configure Module EL1-H with these values of VCC, RB, RE and RC, and

measure IC and VCE and compare these with the desired Q-point values.

NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here.


Data Sheet: Electronics I Experiment # 12 BJT Biasing: Fixed Bias

Name:______________________________ Date:______

1: Value of ________

2, 3: Fixed Bias, VCC = 12.0 volts

4, 5: Emitter Stabilized Bias, VCC = 12.0 V, VE = 2.00 V

Calculated Measured Given Q-point RB RC IC VCE IC VCE

20.0 mA 6.00 V

25.0 mA 5.00 V

Calculated Measured Given Q-point RB RC RE IC VCE IC VCE

20.0 mA 6.00 V

25.0 mA 5.00 V

13: BJT Biasing: Voltage Divider 39

Electronics I Experiment #13: BJT Biasing: Voltage Divider Materials: Module EL1-H , two multimeters, power supply The voltage divider bias configuration, like the emitter-stabalized fixed bias, also

attempts to minimize the differences in from

transistor to transistor. We are given values of

the supply voltage, VCC, and the Q-point (IC and VCE) and the emitter voltage, VE and are

required to find suitable values for the four external resistors.

Recall the definition of as IC/IB and from Kirchoff’s current law: IC + IB = IE . So

IE = IC(1 + 1/ ) = IC( + 1)/

Now apply Ohm’s law to the emitter resistor, RE

RE = VE / IE = (VE/IC) / ( /( +1))

The voltage across the collector resistor, RC, is

VCC – VCE – VE and so, again by Ohm,

RC = (VCC – VCE – VE)/ IC

The voltage at the midpoint of the divider

formed by R1 and R2 is VB which is determined not by R1 or R2 individually but by

their ratio. Therefore we may assign any convenient value to R1 and solve for the

corresponding value of R2.

For the upper and lower portions of the divider we have :

Upper half: (I2 + IC/ ) R1 = VCC – VCE – VE

Lower half: I2 R2 = VBE + VE

For the upper, assign any convenient value to R1 and solve for I2. Then use this value

of I2 in the lower to obtain R2.

Notice that the smaller the value you assign to R1, the larger is the value of I2 . In fact

if I2 (with a small R1) is large in comparison with IC/ , you may safely assume that IE = IC . However, such a large I2 can put an undesirable load on the system power

supply. It is a trade-off to be made by the design engineer in selecting R1.


Procedure:

1: Measure and record the value of the transistor in your module.

2: Given the following: VCC = 12.0 V, IC = 5.00 mA, VCE = 4.00 V, VE = 2.00 V Assume VBE = 0.52 V Set R1 = 25.0 k

Calculate values for RE, RC, R2, and I2 . Set these R values in the module and

measure I2, IC, VCE and VE . Suggestion: Connect the 1000 ohm resistor in

series with R2 and include its value with R2. The voltage across it, in millivolts, equals

the value of I2, in microamperes.

3: Repeat step 2 but change R1 to 8.0 k

4: Given the following: VCC = 12.0 V, IC = 10.00 mA, VCE = 4.00 V, VE = 4.00 V Assume VBE = 0.52 V Set R1 = 25.0 k

Calculate values for RE, RC, R2, and I2 . Set these R values in the module and

measure I2, IC, VCE and VE .

5: Repeat step 4 but change R1 to 8.0 k


Data Sheet: Electronics I Experiment # 13 BJT Voltage Divider Bias

Name:______________________________ Date:______

1: Value of ________

2, 3:

VCC = 12.0 V, IC = 5.00 mA, VCE = 4.00 V, VE = 2.00 V , VBE = 0.52 V

Calculated Measured R1 R2 RC RE I2 mA I2 mA IC mA VCE V VE V

25.0 k

8.0 k

4, 5:

VCC = 12.0 V, IC = 10.00 mA, VCE = 4.00 V, VE = 4.00 V , VBE = 0.52 V

Calculated Measured R1 R2 RC RE I2 mA I2 mA IC mA VCE V VE V

25.0 k

8.0 k

NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here

14: BJT Biasing Colletor Feedback 42

Electronics I Experiment #14: BJT Biasing: Collector Feedback Materials: Module EL1-H , two multimeters, power supply

For the input loop we obtain:

IB RB = VCB or ICRB/ = VCE - VBE

and for the output side we obtain:

VCC = VCE + IC (RC + RE) .

If IC were to increase, due to a transistor with a

different , the voltage across both the emitter

and collector resistors would increase,

decreasing VCE. This in turn would lessen the

voltage across RB, reducing the bias current, IB

and IC as well. Initially we have to know the

value in order to determine RB. Use the method discussed in Experiment #12 to

determine this value. In computing the proper

value for RB add the 1 k series, as shown in the diagram. The voltage, measured in

millivolts equals IB in microamps.

The supply voltage, VCC, equals the sum of the voltage drops across each element:


VCC = ICRC + VCE + VE or

RC = (VCC – VCE – VE) / IC.

Next apply Ohm’s law to the emitter resistor: VE = IE RE or

RE = VE / (IC ( /( +1))

And for the feedback loop, IB RB = VCB = (VCE – VBE) or

RB = (VCE – VBE) / IC .

Procedure:

1: Measure the value of for the transistor in your module.

2: Given the following values:

VCC = 10.00 V, VE = 2.00 V, VCE = 4.00 V, IC = 10.0 mA, VBE = 0.52 V

Calculate appropriate values for RC, RE, RB, and IB. Use these resistance values

in setting up the module. Measure and record IB, VE , VCE and IC. 3: Repeat step #2 with the following values:

VCC = 12.00 V, VE = 3.00 V, VCE = 6.00 V, IC = 15.0 mA, VBE = 0.52 V


Data Sheet: Electronics I Experiment # 14 BJT Biasing: Collector Feedback

Name:______________________________ Date:______

1: Value of ________

2: VCC = 10.00 V, VE = 2.00 V, VCE = 4.00 V, IC = 10.0 mA, VBE = 0.52 V

Calculated Measured

RB RC RE IB mA IB mA IC mA VCE V VE V

3: VCC = 12.00 V, VE = 3.00 V, VCE = 6.00 V, IC = 15.0 mA, VBE = 0.52 V

Calculated Measured

RB RC RE IB mA IB mA IC mA VCE V VE V


15: BJT Bias: Common Base 45

Electronics I Experiment # 15 : BJT Biasing: Common Base Materials: Module EL-1-H, two multimeters, power supply, 3-volt source

At first the common base configuration may appear strange. The signal is applied to the emitter rather that directly to the base. Notice the dual power supply: VEE provides the forward bias for the emitter-base junction, VCC provides the reverse bias for the base-collector junction.

Given a Q-point and the power supply voltages, you are to find RE and RC . Kirchoff’s loop method is useful here:

Emitter loop: VEE = VBE + IE RE Collector loop: VCC = ICRC + VCB

VCB = VCE – VBE and assume IC ≈ IE

It is possible to solve these equations for RE and RC in terms of known given quantities. When using Module EL-1-H, VCC is provided by the external power source while VEE is provided by a pair of pen-light batteries, which have appreciable internal resistance. Initially measure this VEE with no load; a correction for internal resistance can be made later;

Procedure: 1: Assume VBE = 0.50 V, set VCC to 10.00 V, measure the un-corrected VEE and select the Q-

point as IC = 5.00 mA , VCE = 5.00 V. Solve the above equations for RE and RC.

2: Use these values with Module EL-1-H, and measure the actual values of VEE and VBE in the circuit. With these corrected values again solve the equations for RE and RC. 3: Use these new values with Module EL-1-H. Now measure and record IC and VCE and compare these measured values with the desired Q-point.

4: Repeat steps 1 to 3, using the corrected values of VEE and VBE , but with VCC = 12.00 V

and a new Q-point as IC = 10.00 mA , VCE = 6.00 V.

NOTE: In this analysis we are not concerned with values of IB.


Q-point values Measured values % Difference

IC = 5.00 mA VCE = 5.00 volts

Q-point values Measured values % Difference IC = 10.00 mA

VCE = 6.00 volts


BJT Biasing: Common Base

Name:______________________________ Date:______ Part 1: Q-point: IC = 5.00 mA , VCE = 5.00 V.

Supply : VCC = 10.00 V VEE (no load) = _______ Assume VBE = 0.50 V,

Compute: RE = ________ RC = _______ (configure module with these values)

Re-Measure: VEE (with load) = _______ VBE = _________

Re-Compute: RE = ______ RC = _______ (re-configure module with these values)

Part 2: Q-point: IC = 10.00 mA , VCE = 6.00 V.

Supply : VCC = 12.00 V VEE (no load) = _______ Assume VBE = 0.50 V,

Compute: RE = ________ RC = _______ (configure module with these values)

Re-Measure: VEE (with load) = _______ VBE = _________

Re-Compute: RE = ______ RC = _______ (re-configure module with these values)

(Include all calculations with this report) NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here

16: BJT Bias: Emitter Follower 48

Electronics I Experiment # 16 : BJT Biasing: Emitter follower Materials: Module EL-1-H, power supply, two multimeters

The basic emitter follower is shown. Given the supply voltage, VCC, and a definite Q-point ( VCE and IC

); you are to calculate the corresponding values for RB and RE. The governing equations are: .

Input side: VCC = IB RB + VBE + IC RE

Output side: VCC = VCE + IC RE

Note there are three unknowns, IB, RB and RE.

Since is not given, use the method outlined in

Experiment #12 to evaluate it. At the Q-point IB = IC /

.

For this circuit, VB = VBE + VE . Since VBE is

normally quite small ( ≈ 0.5 volts ) the output emitter

voltage, VE, closely “follows” the input base voltage, VB. This is the reason for the name “emitter follower”

This circuit is also known as a common collector. The collector may not seem to be common. However when viewed from the signal or AC viewpoint, the power supply and the ground are at the same potential, since they are connected by the comparatively large filtering capacitors. In this AC view the collector is common to the input and output loops,


Procedure: 1: Set VCC = 10.0 volt. Select the Q-point as : IC = 10.00 mA, VCE = 4.00 volts and assume VBE = 0.45 volts.

2: Determine IB for this Q-point, as explained above.

3: Calculate values for RB and RE and configure the module using these values.

4: Measure the actual value of VEB and re-calculate for RB and RE. Configure the module again with these values, and compare measured values with the desired Q-point.

5: Repeat steps 1 to 4 for VCC = 12.00 V, IC = 20.00 mA, VCE = 6.00 volts.


Q-point Measured value % difference

IC = 10.00 mA,

VCE = 4.00 volts

Q-point Measured value % difference

IC = 20.00 mA,

VCE = 6.00 volts

Data Sheet: Electronics I Experiment # 16 BJT Biasing: Emitter follower

Name:______________________________ Date:______

Part A:

Set VCC = 10.0 volt. Assume VBE = 0.45 volts.

Select the Q-point as : IC = 10.00 mA, VCE = 4.00 volts

IB = _________ DC = _________ Corrected VEB = ____________

Part B:

Set VCC = 12.0 volt. Assume VBE = 0.45 volts.

Select the Q-point as : IC = 20.00 mA, VCE = 6.00 volts

IB = _________ DC = _________ Corrected VEB = ____________


17 N-channel JFet Characteristics 51

Electronics I Experiment # 17 : N-channel JFet Characteristics Materials: Module EL-1-I, power supply, 3-volt source, two multimeters The bipolar junction transistor, BJT, is a current-controlled device; the collector-emitter

current is directly proportional to the base-emitter current (the proportionality constant is ).

The field-effect transistor, FET, is a voltage-controlled device; the source-drain current is controlled by the gate-source voltage (the proportionality factor is not linear). The source-drain

current, IDS, can be expressed in terms of the variable gate-source voltage, VGS, and two

constants: IDSS, the source-drain current with VGS=0, and VP, the minimum value of VGS that

sets IDS=0. Notice that on a graph of source-drain current ( vertical ) against gate-source

voltage (horizontal ), each of these constants lies on a separate coordinate axis. The graph line joining these two points is not a straight line, but a segment of a parabola. The Shockley equation expresses this relation:

Notice the polarity of the battery in the input circuit. The gate-source junction is reverse biased so ther is no gate current. As with the BJT, we specify an operating, or Q, point by assigning definite values to VDS and IDS. The drain-source voltage term, VDS, does not appear in the Shockley equation for IDS, suggesting that the drain-source current is independent of the drain-source voltage! Over a wide voltage range this is approximately true.

IDS = IDSS ( 1 – VGS/VP )2 .


The circuit shown here is used to measure the input and output characteristics of a given N-channel FET. In the diagram the conventional current enters the drain terminal and leaves at the source, which at first may seem a backwards way of naming terminals. Recall that we are dealing with an N-channel device in which the

charge carriers are the negative electrons. Thus the source terminal is where the carriers enter, the drain is where the carriers leave. A similar convention is followed for P-channel FETs .

Procedure:

1: Draw the input characteristics. Maintain VDS at 8.0 volts. Attach the external 3.0 volt supply

where indicated. Connect a voltmeter between the gate and source terminals, for VGS, and

place a second voltmeter where shown on the diagram. The small switch lets you use this

meter to read either VDS or the source drain current, ID.

When measuring IDSS, use alligator clips to short-circuit the gate and source. When

measuring VP, use the lowest possible meter range to measure current. Display these points

on your graph: [0, IDSS], [0.3 VP, IDSS/2], [0.5 VP, IDSS/4], [VP, 0].

2: Draw the output characteristics, at 0.50 volt intervals for VGS.



N-channel JFet Characteristics

Name:______________________________ Date:______

Attach graphs of the input and output JFet characteristics

53 18: FET Biasing: Self Bias

Electronics I Experiment # 18 : FET Biasing: Self Bias Materials: Module EL-1-J, power supply, two multimeters

As with the Bipolar Junction Transistors (BJT), biasing implies selecting external circuit components so that the DC (no signal) voltage across the transistor and the current through the transistor have definite values (Q-point). For the BJT, this means a definite base current determined by the transistor characteristics; for the FET the gate-source voltage, VGS must have a definite value, again determined by the transistor characteristics. In measuring the FET characteristics in

the previous experiment, two power supplies were used. For practical applications a

single power supply is preferable. In the self bias diagram above, the voltage rise, VDD across the power supply equals the voltage drop across the transistor and the two

resistors, RS and RD. Since there is no current in the gate circuit, there is no voltage

drop across RG . Therefore VGS just equals the voltage drop across RS. (RG could be

omitted for biasing, but it is needed when a signal is applied to the gate).

Given an operating or Q-point, ID ( or IDS ) , VDS, and the supply voltage, VDD, find values for RS, RD and RG. From the input characteristic curve, or the Shockley

equation, IDS = IDSS ( 1 – VGS/VP )2 , we can find the required VGS . From this we

find the value of the source resistor, RS = VGS/IDS . Kirchoff’s mesh equation for the

output loop is

VDD = VDS + IDS (RS + RD)

where RD is the only unknown. And for biasing purposes RG may have any value, even

zero.

Procedure:

1: For the FET in your module EL1-J, measure VP and IDSS. You may have measured these values in Experiment #17, using module EL1-I, but each FET has its own characteristics. The diagram below suggests how you can use module EL1-J, with

its single voltage source, to measure VP and IDSS. For IDSS , place a jumper between the gate and source terminals, so VGS =

zero; the value of ID is then IDSS.


For VP, remove the jumper and place a voltmeter between the gate

and source terminals. At pinch-off, the gate must be sufficiently negative

with respect to the source to reduce to zero the drain current, ID . To do

this connect the gate to ground. Next join together in series RD and RS to

form a voltage divider between the supply voltage and ground; connect the

junction of RD and RS to the source.. Now adjust RD and RS, until the drain

current, ID , reads zero on the most sensitive current range of the

multimeter. The value of VP is the gate-source voltage, VGS

2: Use the circuit configuration shown below for the self-bias configuration. Set VDD = 12.00 V. Take Q-point as: ID = 2.00 mA, VDS = 5.00 V; Calculate values for RS and RD . Set the module to these values, and record the corresponding Q-point. 3: Repeat step #2 for Set VDD = 10.00 V. Take Q-point as: ID = 1.50 mA, VDS = 6.00 V;


Data Sheet: Electronics I Experiment # 18 FET Biasing: Self Bias

Name:______________________________ Date:______

1: VP = ____________ IDSS = ______________

2: VDD = 12.00 V, ID = 2.00 mA, VDS = 5.00 V; RS = ___________ RD = ___________

Desired Measured % Difference

ID 2.00 mA

VDS 5.00 V

3: VDD = 10.00 V, ID = 1.50 mA, VSD = 6.00 V; RS = ___________ RD = ___________


ID 1.50 mA

VDS 6.00 V NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here

19: FET Biasing: Voltage Divider 60

Electronics I Experiment # 19 : FET Biasing: Voltage Divider Materials: Module EL-1-J, power supply, two multimeters In any FET,the drain current, ID (or IDS), depends on the gate-source voltage, VGS. In the self bias configuration we have considered, VGS is provided by the voltage drop, IDRS, across the source resistor, RS. VS, the voltage at the source terminal with

respect to ground ( the negative side of the supply voltage, VDD) is determined by RS, while VD, the drain terminal voltage, is determined by RD. Normally any input signal would be applied to the gate, and the output taken from either source or drain. With voltage divider biasing, introduced here, for a given source-drain current, ID, we may set to any value not only the drain terminal voltage, VD, but also the source terminal voltage, VS. The total current provided by the supply. VDD, equals the two separate currents, ID through the transistor, and I12 through the two series resistors, R1

and R2. Since there is no gate current, this resistor pair acts as a voltage divider. Thus the voltage drop across R2 is VDD R2 /(R1 + R2). From the diagram it is easy to see that the gate-source voltage, VGS, is the difference between the source-ground voltage and the gate-ground voltage:

VGS = ID RS - VDD R2 / (R1 + R2). [A]

Here RS can have any value greater than or equal to VGS/ID The pair of values, R1 and R2, are selected so that [A] is satisfied. For the loop including the supply voltage and the transistor we have

VDD = VDS + ID(RD + RS) [B]

VD = VDD – ID RD , VS = ID RS [C]

Given a value for VDD and the Q-point ( ID and VDS ), and values for VS and VD, how do we determine R1, R2, RS and RD ?

(1) Determine VGS from the Shockley equation or the input characteristics. If the given VS is less than this VGS then no value of RS may be found to satisfy [A], and the problem cannot solved.

(2) From [C] determine RS and RD.


(3) From [A] determine a pair of values for R1 and R2

Procedure: 1: For the FET in your module EL1-J, measure VP and IDSS.

2: Given VDD=12.00 V, VDS=5.00 V, ID=1.50 mA, VS=3.00 V, VD=8.00 V

Find suitable values for R1, R2, RS and RD.


Find suitable values for R1, R2, RS and RD.


Data Sheet: Electronics I Experiment # 19 FET Biasing: Voltage Divider

Name:______________________________ Date:______

1: VP = ____________ IDSS = ______________


RS = ______ RD = _______ R1 = ______ R2 = _______


ID 1.50 mA

VDS 5.00 V

VS 3.00 V

VD 8.00 V


RS = ______ RD = _______ R1 = ______ R2 = _______


ID 0.75 mA

VDS 5.00 V

VS 1.00 V

VD 6.00 V


20: BJT Small Signal: bAC, r0, re

63

Electronics I Experiment #20: BJT Small Signal: AC, ro, and re Materials: Module EL1-H , two multimeters, function generator, power supply

A transistor may be viewed as a current valve, where the handle is the base-

emitter current for a BJT and the gate-source voltage for a FET. We have already considered three important, and inter-related parameters for each type:

BJT: emitter-base current, IB; collector current, IC; collector-emitter voltage, VCE FET: gate-source voltage, VGS; drain current, ID; drain-source voltage, VDS These are not the only parameters that could have been selected (for a three terminal device, there are three terminal currents, and three voltages between different pairs of terminals) but they are the most common. The input and output characteristics displays the relationships between these quantities. From these characteristic curves, three additional values are derived for AC or signal values, which are quite important for transistor circuit analysis. For the BJT these are:

re = emitter-base

resistance;

ro = emitter-collector

resistance;

AC = output to

input current ratio

These three parameters are calculated for a given Q-point ( VCE, IC). In each case we are concerned with the ratio of changes about a given value for particular quantities.

To explain the method, let us select the following: VCC=12.00 V, VCE=6.00 V, IC= 20.0 mA.

Evaluate βAC: = IC / IB Place jumpers on Module EL1-H as shown in the diagram below, which is similar

to the Fixed Bias configuration considered in a prior experiment. Here RB is the series

combination of R1, R2 and the 1000 ohm reference resistor. This 1000 ohm reference

resistor is added to simplify measuring the base current, IB, since the voltage across it,


64

measured in millivolts, is numerically equal the current through it, measured in

microamperes. Adjust R1, R2 or both so that the collector current, IC, equals the

prescribed 20.0 mA.

Notice the voltage across the transistor, VCE plus the voltage drop across the

collector resistor, IC RC equals the supply voltage, VCC; therefore adjust the value of

RC so VCE equals the prescribed 6.00 V. Next adjust RB so that the collector current, IC

is slightly above an again slightly below the Q-point value, and record both IB and IC, as shown in the sample data below. Notice that IC and IB are measured on different meter ranges. Use the range that gives the most number of significant digits.

Above Below Difference AC = IC / IB

IC 22.0 mA 18.0 mA 4.0 mA 89

IB 0.262 mA 0.127 ma 0.045 ma


65

Evaluate ro: = VCE / IC

Use the same set-up as above. First adjust RB and RC to get back to the initial Q-point values. Next, without changing RB, adjust RC to obtain slightly above and below values for VCE :

Evaluate re: = VBE / IE

First restore the Q-point values. Measure the base-emitter voltage, VBE, which came to

0.605 V in the sample data used here. Next vary RB to attain values of VBE slightly

above and below this value, and record the corresponding emitter current, IE. Notice in the diagram that the ammeter is positioned to measure the collector current, IC. Recall

that IE = IC( +1/ ), or IE IC

Procedure:

1: For VCC=12.00 V, and a Q-point of VCE=6.00 V, IC= 20.0 mA, determine

the value of AC, re and ro . Compare the measured value of re with the expression:

26 mV / (IE mA)

2: Repeat step 1 for VCC=10.00 V, and a Q-point of VCE=4.00 V, IC= 25.0 mA. NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here

Above Below Difference ro = VCE / IC

VCE 6.99 V 5.02 V 1.97 V

21.9 k IC 20.31 mA 20.22 mA 0.09x10

-3 A

Above Below Difference re = VBE / IE

VBE 0.610 0.600 0.010

1.5 IE 21.70 mA 15.10 mA 6.60x10

-3 A


66

Data Sheet: Electronics I Experiment # 20 BJT Small Signal: AC, ro and re

Name:______________________________ Date:______

1: VCC=12.00 V, VCE=6.00 V, IC= 20.0 mA

2: VCC=10.00 V, VCE=4.00 V, IC= 25.0 mA.

Above Below Difference βAC = IC / IB

IC

IB


VCE

IC


VBE

IE

Above Below Difference βAC = IC / IB

IC

IB


VCE

IC


VBE

IE

22: BJT Small Signal: Fixed Bias

67

Electronics I Experiment #21: BJT Small Signal: Fixed Bias

Materials: Module EL1-H , two multimeters, function generator, power supply, oscilloscope.

In Experiment #12 we considered the fixed bias configuration: the diagram there is reproduced below:

In many practical circuits a capacitor ( 50 f … 470 f) is placed in parallel with the emitter resistor, RE. This does not affect the DC biasing, but for the small signal analysis this bypass capacitor effectively short-circuits the emitter resistor, so we may

take RE 0.

Recall that the three parameters, AC, ro, and re, are properties of the

transistor itself, independent of the bias configuration. Recall that re is typically in the

neighborhood of 25 while ro is some 1000 times greater. Recall, too, that for a large

and small resistor, when in series we may ignore the small resistor, when in parallel we may ignore the large resistor. In some of the expressions that follow, we may ignore ro,

and also take +1. It our small signal analysis we are interested in four additional parameters that

depend on both the transistor properties and also on the particular bias configuration. In this and the following experiments, we will set-up the module for a given Q-point, measure these four values experimentally, and compare the results with the predicted values.


68

Z i : input impedance = Vi / Ii Z o : output impedance = Vo

* / Io

*

A v : voltage gain = - Vo / Vi

A i : current gain = Io / Ii

Here Vi , Ii , Vo and Io represent the input and

output signal voltages and currents. Recall the definition of Zo : Apply zero AC signal at

the input terminals; apply a voltage, Vo* to the output terminals, and measure the

resulting current as Io*; then Zo = Vo

* / Io

*.

If there is no emitter resistor, or if it is by-passed by a

large capacitor, set RE to zero in the expressions shown. We

have assumed that ro is in parallel with smaller resistors and

make be neglected. The derivation of these expressions may be found in any standard electronics textbook.

Set up the module as shown below. RB is represented

by R1 on the module. Three variations may be had by

shorting out RE, bypassing it with a capacitor, or letting it

stand alone.

Procedure:

Use the following values: VCC=12.00 V, VE=2.00 V, VCE=6.00 V, IC=20.0 mA Input Signal ≈ 2 kHz.

1: Measure and record the values of AC, ro, and re, using the methods of

experiment #20.

2: Compute the appropriate values of RB, RC and RE. Then use these values to set up the module board, EL-1-H for the Common Emitter, emitter stabilized configuration, as shown in the diagram above. Next fine-tune these three resistors to meet the Q-point

values as closely as possible. Measure and record these final values for RB, RC and

RE.

Common Emitter

bias stabilized

Z i = RB (RE + re)

Z0 = RC AV = – RC / (RE + re)

A i = Z i / (RE + re)


69

3: Measure and record the values of Vi, Vo, Ii and Io. Measure VI and Vo as the voltage between ground ( the negative terminal of the input power ) and Signal In or

Signal Out. First adjust Vi so that Vo = 2.00 VAC. Check the form of the output signal

with the oscilloscope to make sure there is no distortion.

Find Ii and Io by measuring the voltage drop across the 1kΩ resistor or RC, and

divide by the corresponding resistance value ( Ohm’s law!). Remember to make the

current and voltage measurements in the AC meter mode. 4: From these values of Vi, Vo, Ii and Io calculate the values of ZI, Av and AI .

5: Connect the IN side of CIN to ground. Apply an AC signal of approximately one volt

( Vo* ) between the OUT side of Cout and ground, but passing through the 1kΩ resistor,

as shown in the diagram below. Use Ohm’s law to calculate the current ( Io* ) through

this resistor. From these values calculate Zo. It is interesting to note that the ratios Vo / Io and Vo

* / Io

* are approximately equal for this configuration. This will not always be

the case (for example, the emitter follower configuration considered in Experiment #24).


70

6: Compute the values of ZI, Zo, Av and AI using the measured values of RB, RC , RE,

AC and re. Compare these with the results of step 4 above.

7: Connect the bypass capacitor, CE, in parallel with the emitter resistor, RE, and repeat steps 3, 4, 5 and 6. NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here


71

Data Sheet: Electronics I Experiment # 21 BJT Small Signal: Fixed Bias

Name:______________________________ Date:______

1: AC _______ , ro ________, re _________

2: RB ________, RC _______, RE ________ Emitter NOT bypassed:

Vi ____, Vo ____, Vo* ____, Ii ____, Io____ , Io

*____

Emitter bypassed :

Vi ____, Vo ____, Vo* ____, Ii ____, Io____ , Io

*____

From currents / voltages From resistances % Difference

ZI

Zo

Av

Ai


Zi

Zo

Av

Ai

22: BJT Small Signal: Voltage Divider Bias

72

Electronics I Experiment #22: BJT Small Signal: Voltage Divider Bias

Materials: Module EL1-H , two multimeters, function generator, power supply, oscilloscope

In Experiment #13 we considered the voltage divider bias configuration: the diagram there is reproduced below. For small signal analysis, this circuit is similar to the emitter

stabilized circuit of Experiment 21; however RB is taken as the parallel combination of

R1 and R2. The emitter may or may not be bypassed by a capacitor.

Procedure

Use the following values: VCC=12.00 V, VE=2.00 V, VCE=6.00 V, IC=20.0 mA Input Signal ≈ 2 kHz.

1: Measure and record the values of AC, ro, and re, using the methods of

Experiment #20.

2: Compute the appropriate values of R1, R2, RC and RE. Then use these values to set up the module board, EL-1-H for the Common Emitter, voltage divider configuration, as shown in the diagram below. Then fine-tune these values to satisfy the Q-point.

Common Emitter

Voltage Divider Bias

RB = R1R2 / (R1 + R2)

Z i = RB (RE + re) Z0 = RC AV = – RC / (RE + re)

A i = Z i / (RE + re)


73

3: Use a 1 kHz sine wave as input signal, and set its value so that the output voltage is approximately 2.0 volts (use an oscilloscope, if available, to check for distortion).

Measure and record the values of Vi, Vo, Vo*, Ii , Io and Io

*. Measure Vi and Vo as

the voltage between ground ( the negative terminal of the input power ) and Signal In or

Signal Out. Find Ii and Io by measuring the voltage drop across the 1kΩ resistor or RC,

and divide by the corresponding resistance value ( Ohm’s law!). Measure Vo* and Io

*

using the method explained Experiment #20. Remember to make the current and

voltage measurements in the AC meter mode. 4: From these values of Vi, Vo, Vo

*, II, Io and Io

* calculate the values of ZI, Zo, Av and

AI .

5: Compute the values of ZI, Zo, Av and AI using the measured values of RB, RC , RE,


6: Connect the bypass capacitor, CE, in parallel with the emitter resistor, RE, and repeat steps 3, 4 and 5. NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here


74

Data Sheet: Electronics I Experiment # 22 BJT Small Signal: Voltage Divider Bias

Name:______________________________ Date:______

1: AC _______ , ro ________, re _________

2: R1 ______, R2 _________, RC _______, RE ________ Emitter NOT bypassed:

Vi ____, Vo ____, Vo* ____, Ii ____, Io____ , Io

*____

Emitter bypassed :

Vi ____, Vo ____, Vo* ____, Ii ____, Io____ , Io

*____


ZI

Zo

Av

AI


Zi

Zo

Av

Ai

75 23: BJT Small Signal: Collector Feedback Bias

Electronics I Experiment #23: BJT Small Signal: Collector Feedback Bias

Materials: Module EL1-H , two multimeters, function generator, power supply In Experiment 14 we considered collector feedback biasing. We now make a small signal analysis of the same configuration. We may optionally bypass the emitter resistor with a large capacitor;

Notice that K 1 for a very large value of the feedback resistor, RF (or RB),

and so Zi = (RE + re). This is also the value of Zi for the fixed and voltage divider

bias configurations when bias resistor, RB, is very large.

Procedure

Use the following values: VCC=12.00 V, VE=2.00 V, VCE=6.00 V, IC=20.0 mA Input Signal 2 kHz. Use the series combination of R1 and R2 as RF.

1: Measure and record the values of AC, ro, and re, using the methods of the

experiment #20.

Common Emitter

Collector Feedback Bias

K = 1 + b(RE + re + RC)/RF

Z i = RB (RE + re)

Z0 = RC AV = – RC / (RE + re)

A i = Z i / (RE + re)


2: Compute the appropriate values of RF, RC and RE. Then use these values to set up

the module board, EL-1-H for the Common Emitter, collector feedback configuration, as shown in the diagram above.

3: Measure and record the values of Vi, Vo, Vo*, Ii, Io and Io

*. Measure VI and Vo

as the voltage between ground ( the negative terminal of the input power ) and Signal In

or Signal Out. Find Ii and Io by measuring the voltage drop across the 1k resistor or

RC, and divide by the corresponding resistance value ( Ohm’s law!). Measure Vo* and

Io* using the method explained Experiment #20. Remember to make the current and

voltage measurements in the AC meter mode. 4: From these values of Vi, Vo, Ii and Io calculate the values of ZI, Zo, Av and AI .

5: Compute the values of ZI, Zo, Av and AI using the measured values of RF, RC , RE,


6: Connect the bypass capacitor, CE, in parallel with the emitter resistor, RE, and repeat steps 3, 4 and 5. NOTE The stand-alone computer program, TRNSTR-1.EXE can be useful here


Data Sheet: Electronics I Experiment # 23 BJT Small Signal: Collector Feedback Bias

Name:______________________________ Date:______

1: AC _______ , ro ________, re _________

2: RF _________, RC _______, RE ________ Emitter NOT bypassed:

Vi ____, Vo ____, Vo

* ____, Ii ____, Io____ , Io

*____

Emitter bypassed :

Vi ____, Vo ____, Vo* ____, Ii ____, Io____ , Io

*____


Zi

Zo

Av

Ai


Zi

Zo

Av

Ai

78 24: BJT Small Signal: Emitter Follower Bias

Electronics I Experiment #24: BJT Small Signal: Emitter Follower Bias Materials: Module EL1-H , two multimeters, function generator, power supply In Experiment 16 we considered emitter follower biasing. We now make a small signal analysis of the same configuration. In this configuration we do NOT bypass the emitter resistor; otherwise the output signal would be lost.

Common Emitter

Emitter Follower

Z i = RB (RE + re) Z0 = RE + re

AV = RE / (RE + re)

A i = – Z i / (RE + re)


Procedure

Use the following values: VCC=12.00 V, VCE=6.00 V, IC=20.0 mA Input Signal ≈ 2 kHz. Take R1 as RB.

1: Measure and record the values of AC, ro, and re, using the methods of the

experiment #20.

2: Compute the appropriate values of RB and RE. Then use these values to set up the module board, EL-1-H for the Common Emitter, emitter follower configuration, as shown in the diagram above.

3: Measure and record the values of Vi, Vo, Ii and Io. Measure VI and Vo as the

voltage between ground ( the negative terminal of the input power ) and Signal In or

Signal Out. Find Ii and Io by measuring the voltage drop across the 1k resistor or RE,

and divide by the corresponding resistance value ( Ohm’s law!). Measure Vo* and Io

*

using the method explained Experiment #20. Note that the ratios Vo / Io and Vo* / Io

*

are quite different for the emitter follower configuration.

Remember to make the current and voltage measurements in the AC meter mode. 4: From these values of Vi, Vo, Ii and Io calculate the values of ZI, Zo, Av and AI .

5: Compute the values of ZI, Zo, Av and AI using the measured values of RB , RE, AC and re. Compare these with the results of step 4 above.



Data Sheet: Electronics I Experiment # 24 BJT Small Signal: Emitter Follower Bias

Name:______________________________ Date:______

1: AC _______ , ro ________, re _________

2: RB _________, RE ________ Vi ____, Vo ____, Vo

* ____, Ii ____, Io____ , Io

*____


ZI

Zo

Av

Ai

81 25 FET Small Signal – gm and rd

`Electronics I

Experiment # 25 : FET Small Signal: gm and rd

Materials: Module EL-1-J, power supply, function generator, two multimeters

We considered the characteristics of the BJT in Experiment #11, and those of the FET in Experiment #17. In Experiment #20 we considered the BJT small signal

parameters, AC, ro and re. In the present experiment we consider the FET small

signal parameters, gm and rd . The figure below displays the similarity and the

differences for the two transistor types.

For the BJT, is the ratio of the change in the output current to the change in the

controlling input current ; is dimensionless, the ratio of two currents. For the FET the

counterpart of is gm, the ratio of the change in output current to the change in the

controlling input voltage; gm has the dimensions of reciprocal ohms or siemens, and is called transconductance.

The FET drain resistance, rd is similar to the BJT output resistance, ro,. There is

no FET counterpart to the BJT input resistance, re, since the FET has zero input or gate current. Notice that the values of all these parameters depend on the Q-point, the selected point on the output characteristics. They may be evaluated by the graphical methods presented in Experiment #20.

Recall there exists an algebraic relation between VGS and IDS given by the

Shockley equation, presented in Experiment #17:

IDS = IDSS ( 1 – VGS/VP )2

which is just the equation of the FET input characteristics, shown below, in the range of

VGS from VP to zero. It has the form of a segment of a parabola.


Therefore we may obtain an algebraic expression for gm by differentiating the

Shockley equation with respect to VGS.

Graphical determination of gm and rd

IDSS is the drain current, IDS, for VGS = 0, and VP is the value of VGS for IDS = 0; with these two values and the above equation we can find gm for any value of VGS. A graph of gm against VGS is a straight line in the interval from VP to zero, with a

positive slope of 2 IDSS / VP2 .

We may use a method similar to that used in Experiment #20 to determine

graphically the value of gm and rd at a particular Q-point. We measure the values a little

above and a little below the Q-point and form the ratio of their differences. Consider sample data to illustrate the method;

Given: VDD = 10.00 V, ID = 1.00 mA, VDS = 6.00 V ; Find gm and rd

Use the circuit configuration shown below. 1: Set the supply voltage to VDD = 10.00 V

2: Adjust RS to obtain the desired ID = 1.00mA

3: Adjust RD to obtain the desired VDS = 6.00 V

4: Measure and record the value of VGS 5: For gm, hold VDS constant and adjust RS so ID is slightly above and slightly

below 1.00 mA , and record the corresponding values of VGS

gm = dIDS / dVGS = 2 ( IDSS / |VP| )( 1- VGS / VP )


VDS maintained constant

6: For rd, hold VGS constant and adjust RD so VDS is slightly above and slightly

below 6.00 V , and record the corresponding values of ID

VGS maintained constant

Above Below Difference gm = ID / VGS

ID 1.10 mA 0.90 mA 0.20 ma

1.8X10-3

S VGS 0.71 V 0.82 V 0.11 V

Above Below Difference rd = VDS / ID

VDS 7.0 V 5.0V 2.0V

20 kΩ ID 1.10 mA 1.00 mA 0.10 mA


Procedure:

1: For the FET in Module EL-1-J calculate the value of VP and IDSS . In computing VP use the most sensitive current range to determine the exact value of VGS that makes ID

= 0.

2: . Set VDD = 12.00 V. Take Q-point as: ID = 2.00 mA, VDS = 5.00 V. Use the

graphical method to determine rd.

3: With these same values use the graphical method to determine gm.

4: Compute gm using the expression from the derivative of the Shockley equation, and

compare this with the value from the graphical method.

5: Repeat steps 2 to 5 for : VDD = 10.00 V, ID = 1.50 mA, VDS = 6.00 V;


Data Sheet: Electronics I Experiment # 25 FET Small Signal: gm and rd

Name:______________________________ Date:______

1: VP = ____________ IDSS = ______________

2, 3: VDD = 12.00 V, ID = 2.00 mA, VDS = 5.00 V; 4:

5: VDD = 10.00 V, ID = 1.50 mA, VDS = 6.00 V;


VSD

ID


ID

VGS

gm measured gm computed % Difference


VSD

ID


ID

VGS

gm measured gm computed % Difference

86 27: FET Small Signal – Self Bias

Electronics I Experiment # 26 : FET Small Signal: Self Bias Materials: Module EL-1-J, power supply, function generator, two multimeters In Experiments #18 and #19 we considered separately FET self and voltage divider bias. The DC circuits for these are repeated here.

In either case the source resistor, RS,

may be bypassed by a capacitor. However for small signal analysis the AC equivalent circuit for both configurations may

be combined, as shown. For

self bias, RG’ represents RG;

for voltage divider bias RG’ represents the parallel

combination of R1 with R2. If

the source resistor, RS, is

bypassed by a large

capacitor, RS is replaced by a short circuit. Because the input and output sides are totally separated,

the input impedance, Zi, is

RG’. By definition the output impedance, Zo, is the impedance “looking back” into the

output terminals, with the input terminals shorted. In this case the shorted input makes

VGS = 0, so the current generator, gm VGS, is effectively an open circuit. In the previous

experiment it was found that rd is usually quite large, so Zo RD. The voltage gain is

given by:

Av = - gm ( RD / (1 + gm RS))


Procedure:

Set VDD = 12.00 V. Take Q-point as: ID = 2.00 mA, VDS = 5.00 V. Use

the methods of Experiment #25 to determine gm and rd.

Self Bias:

1: Using the methods of Experiment # 18, determine the values for RS, RD and R2 to satisfy the Q-point values. Configure Module EL1-J using these values.

2: Use a sinusoidal input signal, Vi, of approximately 2 kH, and adjust its amplitude so

that the output signal, Vo, equals 2.00 VAC. If necessary use R1 in series with the

function generator to reduce its amplitude. Calculate the ratio Vo / Vi and compare this

with the formula for Av given above.

3: Use CS to bypass the source resistor, RS, and repeat step 2 above.


Voltage Divider

1: Using the methods of Experiment # 18, determine the values for RS, RD, R1 and R2 to satisfy the Q-point values. Configure Module EL1-J using these values, and adjust them, if necessary, to maintain the desired Q-point..

2: Use a sinusoidal input signal, Vi, of approximately 2 kH, and adjust its amplitude so

that the output signal, Vo, equals 2.00 VAC. Calculate the ratio Vo / Vi and compare

this with the formula for Av given above.

3: Use CS to bypass the source resistor, RS, and repeat step 2 above.


Data Sheet: Electronics I Experiment # 26 FET Small Signal: Self Bias

Name:______________________________ Date:______

VDD = 12.00 V, ID = 2.00 mA, VDS = 5.00 V.

gm _______ , rd ______

Self Bias: RS ______, RD ______, R2 _________

Voltage Divider: RS ______, RD ______, R1 _________, R2 _________

Av (by formula) V0/Vi % Difference NO Bypass

Bypass

Av (by formula) V0/Vi % Difference NO Bypass

Bypass

27: FET Small Signal: Source Follower 90

Electronics I Experiment # 27 : FET Small Signal: Source Follower Materials: Module EL-1-J, power supply, function generator, multimeter

The FET source follower configuration is somewhat similar to self bias. However,

RD is set to zero, RS is never bypassed, and the signal is taken from the source terminal

rather than from the drain. However the voltage gain, Av, and output impedance, Zo, are

quite different.

Av = gm RS / ( 1 + gm RS) Zo = RS / ( 1 + gm RS)

Notice that the denominator in the expression for Av is one greater than the numerator,

indicating that the voltage gain is always less that one. The output impedance, Zo, is

always less ( and sometimes much less ) than the source resistor, RS.


Procedure:

1: Set VDD = 12.00 V. Take Q-point as: ID = 2.00 mA. Use the

methods of Experiment #25 to determine gm.

2: Determine the value for RS to satisfy the Q-point values. Configure

Module EL1-J using this value, or adjust, if necessary, to maintain desired Q-point.

3: Use a sinusoidal input signal, Vi, of approximately 2 kHz, and adjust its

amplitude so that the output signal, Vo, equals 0.20 VAC. Check the wave

shape with an oscilloscope, if available If necessary use R1 in series with

the function generator to reduce its amplitude. Calculate the ratio Vo / Vi and compare this with the formula for Av given above.

4: Connect the in side of Cin to ground. Apply a sinusoidal 2 KHz , 0.20 VAC, Vo

*, to the out side of Cout through resistor RD (set to any

convenient value) and measure the current through it, Io* ( use Ohms law).

From these values compute Zo = Vo* / Io

*.

5: Set VDD = 10.00 V. Take Q-point as: ID = 1.50 mA. Use the

methods of Experiment #25 to determine gm. Repeat steps 2 to 4 for these

values.



FET Small Signal: Source Follower

Name:______________________________ Date:______

VDD = 12.00 V, ID = 2.00 mA,

gm _______________ RS _______________

Vo ________ Vi _________ Vo* ______ Io

* _______

VDD = 10.00 V, ID = 1.50 mA,

gm _______________ RS _______________

Vo ________ Vi _________ Vo* ______ Io

* _______

Av (by formula) Vo / Vi % Difference

Zo (by formula) Vo* / Io

* % Difference

Av (by formula) Vo / Vi % Difference

Zo (by formula) Vo* / Io

* % Difference

93

Ateneo de Davao University

Electronic Communication Series

Electric Circuits I: Direct Current Electric Circuits II: Alternating Current Electronics I: Basic Components Electronics II: Amplifiers and Oscillators Electronics III: Operational Amplifiers Communications I: AM and FM Communications II: Digital Communications Digital Logic Circuits, with Verilog HDL Industrial Electronics LOGO! PLC: Learning a Programmable Logic Controller

We are a university in a Third World country, the Philippines. We believe that more than chalk and whiteboard pens are needed to train a communication engineer for today’s world. “Hands on” is a must for every student. Excellent student laboratory equipment is readily available on the world markets. Yet the funding necessary for us to purchase such equipment, and in the quantity we desired, was completely unavailable. Our only viable option was to design and fabricate locally the materials of which before we only dreamt. For each item of laboratory equipment student instructional material had to be prepared, as shown in the above listing. With a view to share with other institutions the fruit of our own endeavors, we are making these student manuals freely available. Permission is given to copy this material, and to suitably modify it to the needs of a particular institution.

Laboratory Guide for Electronics 1: Basic Components

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Transcript of Laboratory Guide for Electronics 1: Basic Components