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INTRODUCTIONComponent Content

Chapter 3

BASIC ELECTRICITY AND ELECTRONICS Assignments using

STUDENTS MANUALWorkboard 12-200-B

Chapter 3

BASIC ELECTRICITY AND ELECTRONICS Assignments using

STUDENTS MANUALWorkboard 12-200-B

3 Assignments using Workboard 12-200-B

3.1 The Semiconductor Diode Assignment

3.1.1 Objectives

To recognise diodes in various physical forms.

To determine the diode polarity and to understand the need for correct connection.

To obtain knowledge of the forward voltage/current characteristic and the conduction voltage for a silicon diode.

3.1.2 Prerequisite Assignments

Resistance

Resistor Networks

Resistors in Series and Parallel

Superposition Theorem

Thvenin's Theorem

Power

3.1.3 Knowledge Level

Before working this assignment you should:

Know the operation of series dc circuits.

3.1.4 Equipment Required

Qty

Apparatus

1

Basic Electricity and Electronics Module 12-200-B

1

Power Supply Unit, 0 to 20 V variable dc, regulated(eg, Feedback Teknikit Console 92-300).

2

Multimeters

OR

Feedback Virtual Instrumentation may be used in place of one of the multimeters

1

Function Generator, 100 Hz 5 kHz 20 V pk-pk sine(eg, Feedback FG601)

Background

A Semiconductor Junction Diode (or just Diode) is made from a piece of P-type and a piece of N-type semiconductor joined together. See fig 1.

Fig 1 Junction Diode

If a voltage (potential difference) is applied across the two terminals, the Diode will conduct electricity.

The amount of current that flows depends upon the magnitude and polarity of the applied voltage.

A diode will conduct electricity if connected one way round in a circuit and will not conduct if connected the other way round.

The Diode is represented in circuits by the symbol shown in fig 2.

Fig 2 Diode Symbol

3.1.4.1 Forward/Reverse Biased Connections

When a diode is connected so as to conduct it is said to be Forward Biased.

When a diode is connected so as NOT to conduct, it is Reverse Biased.

Fig 3 shows the two methods of connecting diodes.

Fig 3 Diode Bias

A knowledge of the conduction characteristic when a diode is forward biased is very important .

Fig 4 Typical Silicon Diode - Forward Characteristic

3.1.5 Practical 1

Firstly in this Practical you will identify two physical forms of semiconductor diode that are provided in the equipment.

Then you will investigate the circuit shown in fig 5, below.

Fig 5

Initially you will measure the voltage across the circuit and the current that flows through the diode. You will then reverse the connections of the diode and measure the new voltage and current values.

From these measurements you will be able to understand the basic operation of a diode.

Connect up the circuit as shown in the Patching Diagram for this Practical.

Practical 1 Patching Diagram

3.1.5.1 Perform Practical

The Diode is represented in circuits by the symbol shown in fig 6.

Fig 6 Diode Symbol

Two types of diode are shown in fig 7.

Fig 7 Two Types of Diode

The power diode can handle larger currents and power than the 1N4007 type. This is because of its larger size and metal case that can be bolted to a heat-sink to dissipate higher power.

Identify the two forms of diode provided. You will be using the 1N4007 type in this Assignment.

Determining Diode Polarity

Ensure that you have connected up the circuit as shown in the Patching Diagram and that it corresponds with the circuit diagram of fig 8.

Note that the resistor limits the current to a safe value.

Fig 8 Diode Test Circuit

Switch on the power supply.

Set the power supply control to give 10 V on the meter.

Go to the Results Table section of this Assignment and copy fig 9 to tabulate your results.

Record the current measurement in the first row of the table.

Now, switch off the power supply and reverse the 1N4007 diode to give the circuit of fig 10

Fig 10 Diode reversed

Switch on the power supply and readjust the voltage to 10 V.

Read the new value of diode current and record it in the second row of your table.

Questions

1. Which side of a diode should be connected to the positive voltage supply to make it conduct current?

2. When the diode was connected the opposite way round was the current

a) slightly smaller?b) much smaller?or c) too small to measure?

3.1.6 Practical 2

In this Practical you will examine in more detail the current that flows with the forward biased connection of the diode and you will plot its forward characteristic.

The circuit that you will be using is shown in figure 11.

The current that flows through the diode also flows through the 100 resistor and the voltage across the resistor will be directly proportional to that current. As the resistor value is low, it will have negligible effect on the overall working of the circuit.

Fig 11 Test Circuit




The Characteristics of Forward Biased Diodes

Ensure that you have connected up the circuit as shown in the Patching Diagram and that it corresponds with the circuit diagram of fig 12.

The 2.2 k potentiometer will provide fine control over the applied voltage.

Fig 12 Test Circuit

NOTE: Vd = Vs - Vr and Vr = If x 100

mA

r

10V

f

I

1000mA

A

100

r

V

A

100

r

V

f

I

=

\

=

=

\

Go to the Results Table section of this Assignment and copy fig 13 to tabulate your results.

Turn the potentiometer to zero; fully anti-clockwise.

Switch on the power supply and adjust it to supply 20 V.

Adjust the potentiometer to give a voltage of 1 V on the voltmeter showing Vs.

Now use the power supply variable control to set Vs to:

0, 0.1 V, 0.2 V, etc, up to 1.0 V.

Note Vr for each setting and enter it in your table.

Now, with the power supply variable control set to supply 20 V, use the potentiometer to set Vs to:

1.5 V, 2.0 V, 2.5 V and 3.0 V.

Again enter the values of Vr in your table.

Calculate Vd and If as shown in fig 13 and enter these also in the table.

Use the spreadsheet or prepare a graph like Fig 14 on which to plot your results.

Plot Vd against If on the axes of the graph.

Fig 14 Diode Forward Characteristics

Questions

1. At what approximate value of Vd does the current If begin to rise noticeably?

2. Does Vd rise much above this value for larger values of If?

3.1.7 Results Required

When you have performed this Assignment you should have:

identified the diodes provided,

determined how diode conduction is dependent on applied voltage polarity,

measured and plotted the forward characteristic of a diode,

determined the typical conduction voltage for a silicon diode.

Your report should contain:

the circuits that you investigated,

the results that you achieved,

the graph plotted,

conclusions on your findings in the Assignment.

To produce your report you should use a word processing package.

To achieve the calculated values you could use a spread sheet package.

3.1.8 Practical Considerations and Applications

Both the 1N4007 and the power diode are made of Silicon and the forward conduction voltage of about 0.6 V is typical of silicon junctions.

Also typical of silicon diodes is the very small reverse current.

The power diode passes a greater reverse current than 1N4007. This is because it is designed for much larger forward currents - up to 6 A average. At the low voltage used in this experiment the reverse amount will still be very small.

Diodes can withstand high reverse voltage but will eventually break down at some voltage and may be irreparably damaged.

Diodes have very many applications at many different power, voltage and current levels. A very important application is the production of direct voltage from alternating voltage and this is dealt with in the Assignments which cover Rectification.

Results Tables

Circuit

Current (mA)

Fig 8

Fig 10

Fig 9

Vs(V)

Vr(V)

Vd = Vs Vr(V)

If = 10 Vr(mA)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.5

2.0

2.5

3.0

Fig 13

3.1.9 Further Work

Construct the circuit of fig 15 and apply the method used earlier in this Assignment to find Vd and If.

Fig 15 Diode Power Dissipation

Calculate the power dissipation of the diode and check to see if it becomes warm to the touch after a few minutes operation.

Half-Wave Rectification Assignment

3.1.10 Objectives

To learn to recognise a half-wave rectified sinusoidal voltage.

To understand the term 'mean value' as applied to a rectified waveform.

To understand the effect of a reservoir capacitor upon the rectified waveform and its mean value.


The Semiconductor Diode



Know the operation of transformers.

Know the operation of series and parallel ac circuits.

Know how to use an oscilloscope.


Qty

Apparatus

1


1

Power Supply Unit, ac. supply; 12 V rms; 50 or 60 Hz (isolated from other supplies).(eg, Feedback Teknikit Console 92-300).

1

Multimeter

OR

Feedback Virtual Instrumentation may be used in place of the multimeter

1

Oscilloscope

Background

In The Semiconductor Diode Assignment you found that a diode conducts current in one direction (from anode to cathode) but not in the reverse direction.

A widely used application of this feature is the conversion of alternating voltages to direct voltages (fig 1). This assignment studies the simplest circuits for achieving this conversion, which is called Rectification or, in some cases, Detection.

Fig 1

3.1.14 Practical 1

In this Practical you will see the typical waveform of a half-wave rectified sinusoidal voltage.

You will measure the period and the peak voltage of the rectified waveform and you will calculate the mean value of the voltage.

You will reverse the rectifying diode and observe its effect.

The circuit that you will be using is shown in fig 2.

Fig 2




Ensure that you have connected up the circuit as shown in the Patching Diagram and that it corresponds with the circuit shown in fig 3.

Fig 3 Half-wave Rectification

Switch on the oscilloscope and the sinusoidal supply.

With the oscilloscope dc coupled adjust the timebase and the Y amplifier sensitivity to obtain a steady trace of about 4 cm vertical and 5 ms/cm horizontal.

You should see a waveform as in fig 4.

Fig 4 Half-wave Rectified Waveform

Measure and record time T and peak voltage Vpk.

Sketch the waveform and label it to show the periods when the diode is conducting and those when it is not. Time T depends upon the frequency of your power supply. For a 50Hz supply it should be 20 ms and for 60 Hz it should be 17 ms.

Confirm this. Vpk should be very nearly equal to the peak voltage of the alternating supply.

The waveform of fig 4 goes only positive relative to zero volts. If you connect a moving coil dc voltmeter across the output as in fig 3, the mechanical inertia of the meter will not allow the needle to respond to the rapid voltage changes. Instead, it indicates the MEAN voltage of the waveform.

The mean value of a half-sinusoid can be shown by geometry to be:

p

pk

2V

But at every half-cycle the voltage is zero. The mean value of the waveform, therefore, is:

p

p

pk

pk

V

2

T

0

2

T

2V

T

1

=

+

Note the mean voltage indicated by the voltmeter, and compare it with 0.32 Vpk.

The mean voltage you obtained is positive relative to zero.

Rearrange the circuit to give a negative voltage. Confirm this by experiment.

Questions

1. Why is Vpk not be exactly equal to the peak voltage of the alternating supply?

2. By how much does it differ?

3. The mean voltage you obtain from this Practical is positive relative to zero. How could you obtain a negative voltage?

3.1.15 Practical 2

In this Practical you will see the effect of adding a capacitor across the output of a half-wave rectifier circuit. This is shown in fig 5, below.

Fig 5 Half-wave Rectifier with Reservoir Capacitor

You will see how this capacitor smoothes the output voltage to give a more nearly constant dc voltage. However, you will also see that there are still variations (ripple) on this output voltage.

You will change the capacitor to one with a larger value and observe any changes in output voltage waveform.




Very often when rectifying an alternating voltage, you wish to produce a steady direct voltage free from variations of the sort observed in fig 5. One way of doing this is to connect a capacitor in parallel with the load resistor as in fig 6.

Fig 6 Half-wave Rectifier with Reservoir Capacitor

The effect of a capacitor can be seen with reference to fig 7.

Fig 7 The Effect of a Reservoir Capacitor

The capacitor C (usually called the reservoir capacitor) becomes charged-up by the current through the diode during the positive half-cycle. Then, when the supply voltage starts to reduce again, the capacitor keeps the output voltage high and the diode cuts off. Capacitor C then discharges through R until the next positive half-cycle occurs.

Ensure that you have connected up the circuit as shown in the Patching Diagram and that it corresponds with the circuit shown in fig 6. Initially, the 2.2 F capacitor should be connected in circuit as capacitor C.

Observe the output waveform on the oscilloscope and note the value of the peak-to-peak variations in voltage. Note also the new mean voltage on the voltmeter.

Now replace the 2.2 F capacitor by a much larger value of 47 F, making sure to connect the + side of the capacitor to the diode cathode (the capacitor is electrolytic and MUST be connected in the correct polarity) and answer Questions 2 and 3 for this Practical.

Questions

1. Is the mean voltage with the 2.2 F capacitor added greater or less than it was before?

2. The variations on the rectified waveform are called RIPPLE. Is the ripple with the 47 F less than or more than it was with the lower value capacitor?

3. Is the mean rectified voltage now greater or less?



seen how a diode circuit can convert an alternating voltage into a direct voltage,

measured the peak and calculated the mean voltage values for a half-wave rectifier circuit,

observed the effect of adding a reservoir capacitor to the basic half-wave rectifier and the ripple voltage resulting,

observed the effect of the capacitor value on the degree of smoothing and the amplitude of the ripple voltage.




the waveforms drawn,





When rectification is used to provide a direct voltage power supply from an alternating source, the ripple is an undesirable feature. For a given capacitor value, a greater load current (smaller load resistor) discharges the capacitor more and so increases the ripple obtained.

Fig 8 shows this.

Fig 8 The Effect of Load Current

Several methods are available to reduce ripple:

1. Larger capacitors, the uses of which are limited due to cost and size, and also because large capacitors can require very large charging currents to be supplied through the diode.

2. Electronic stabilisation This reduces ripple as well as keeping the output voltage steady when the load or input voltage changes.

3. Full-wave rectification. With this, the ripple is much reduced as every half-cycle of the input, instead of every other half-cycle, contributes to the rectified output.

In fig 9 it can be seen that capacitor charging occurs at half the previous interval and the amount of discharge for a given load current is therefore less.

Fig 9 Full-wave Rectification

The Full-Wave Rectification Assignment deals with methods of achieving full-wave rectification.

When diodes are used for detection purposes in the reception of modulated radio signals, quite different considerations apply. These are not discussed in detail here.

3.1.18 Further Work

A half-wave rectifier, as in fig 10, produces a certain amplitude (peak-to-peak) of ripple.

Fig 10 Rectifier Circuit

If the load resistor is reduced to half of its original value, what increase in capacitor value will restore the ripple to the same value as before?

Confirm your answer by practical experiment, starting with:

R = 10 k and C = 47 F.

Full-Wave Rectification Assignment

3.1.19 Objectives

To learn to recognise a full-wave rectified sinusoidal voltage waveform.

To understand the working of a diode bridge circuit as a full-wave rectifier and its advantage over half-wave rectification.

To understand the effect of a reservoir capacitor upon the rectified waveform and its mean value.

To have an awareness of the two diode form of full-wave rectifier.


Half-wave Rectification



Know the operation of transformers.




Qty

Apparatus

1


1

Power Supply Unit, ac supply; 12 V rms; 50 or 60 Hz. (isolated from other supplies).(eg, Feedback Teknikit Console 92-300).

1

Multimeter

OR


1

Oscilloscope

Background

At the end of the Half-wave Rectification Assignment ways of reducing the ripple or voltage variation on a rectified direct voltage were discussed. One of these was to use every half-cycle of the input voltage instead of every other half-cycle.

A circuit which allows this is shown in fig 1, and is known as the diode bridge.

Fig 1 A Diode Bridge Rectifier

During the positive half-cycle of the supply 'A' is more positive than 'B'. Diodes D1 and D2 therefore conduct while diodes D3 and D4 are reverse-biased. The current flows as shown in fig 2.

Fig 2 Positive Half-cycle

Fig 3 Negative Half-cycle

During the negative half-cycle the current flow is as represented by fig 3.

In each case the current in the load is in the same direction.

A typical bridge rectifier component is shown in fig 4a.

Note that, on your equipment, the bridge rectifier component may be a small cube, as shown in fig 4a, or may be cylindrical in construction.

Fig 4b shows the circuit symbol and how the rectifier terminals are labelled.

Fig 4 Bridge Rectifier

The terminals labelled + and - are so called because these are the polarities that will exist across the load.

3.1.23 Practical 1

In this Practical you will see the typical waveform of a full-wave rectified sinusoidal voltage.

You will measure the period and the peak voltage of the rectified waveform and you will calculate the mean value of the voltage.


Fig 5

You will add a reservoir capacitor and observe its effect.




A Bridge Rectifier with Resistive Load


Note:

Prior to connecting an ac power supply to the board, ensure that the supply is isolated from the ground.

Fig 6 Test Circuit

With the oscilloscope dc coupled, adjust the controls to obtain a steady trace of about 4 cm vertical and 5 ms/cm horizontal. You should observe a waveform as in fig 7. Time 'T' will be 10 ms for 50 Hz supply, and 8.5 ms for 60 Hz.

Fig 7 Full-wave Rectified Waveform

Note the value of Vpk and also the mean value of output voltage indicated on the voltmeter. Compare these figures with those obtained in Half Wave Rectification.

p

pk

2V

As the mean value of a half-cycle of a sine wave is, and every half-cycle is present, this should be the mean value measured.

Confirm this from your readings.

The Effect of a Reservoir Capacitor

Add a 2.2 F capacitor in parallel with the load resistor and note the new mean value and the peak-to-peak ripple amplitude of the rectified waveform. Compare these figures with those obtained in the Half-wave Rectifier Assignment for the same load and capacitor values.

3.1.24 Questions

1. Should Vpk be the same as it was for a half-wave rectifier? Does your observation confirm your answer?

2. How does the mean value compare with that found for half wave rectification?

3. Fig 9 shows the discharge curve for a reservoir capacitor in half-wave and full-wave rectification, for the same load and capacitor values.

Fig 9

A - Start of dischargeB - End of discharge; full waveC - End of discharge; half wave

A capacitor discharges into a resistor in an exponential fashion, that is with a rate of discharge that reduces as the discharge progresses.

With this in mind, would you expect the peak-to-peak ripple in full-wave to be:

(a) half that in half-wave(b) less than half(c) more than half

Explain your answer and confirm it by reference to measurements made in the Half-wave Rectification Assignment and this Assignment for similar load conditions.



seen how a bridge rectifier circuit can convert an alternating voltage into a direct voltage,

measured the peak and calculated the mean voltage values for a bridge rectifier circuit,

observed the effect of adding a reservoir capacitor to the circuit.




the waveforms drawn,

a comparison of the performance of half-wave and bridge rectifier circuits,





The alternating input voltage to a rectifier is usually obtained from the main supply through a transformer, for two reasons:

1. 1.To obtain the desired voltage by choice of the transformer ratio.

2. To provide isolation from the main supply for safety reasons.

Fig 9 shows such an arrangement with a bridge rectifier.

Fig 9 Transformer-fed Bridge Rectifier

In this figure, although the load current is always in one direction, the current in the transformer secondary is alternating.

Fig 10 shows another method of full-wave rectification, using a centre-tapped transformer winding and two diodes.

Fig 10 Full-wave Rectifier using Two Diodes

The arrows show how current flows on alternate half-cycles. The value of the output waveform is exactly the same as that for a bridge circuit provided each half of the transformer windings has the same rms voltage as the whole of the winding in fig 9.

The circuit saves two diodes, but increases the cost of the transformer. In fig 10 each half-secondary winding must have the same voltage rating as the single secondary of fig 9. Suppose the half-secondaries were wound with wire of half the cross-sectional area, so as to fit the two into the same space as the one secondary of fig 9, and use the same amount of copper. Each half-secondary would then have twice the resistance.

The current flows in each half-secondary only on alternative half-cycles, but would generate twice the I2R loss in the active cycle.

Each half-secondary would thus develop as much heat as the single secondary of fig 9, ie, twice as much for both. A larger transformer would therefore be required to avoid excessive heating. Its greater cost would usually outweigh the cost of the two diodes saved.

In full-wave rectification the basic repetition rate of the ripple is twice that of the supply (eg, 100 Hz for a 50 Hz supply). In half-wave the frequency is the same as the supply frequency. This is often useful as an indication that one half of a bridge or full-wave rectifier is faulty.

Notes

The Zener Diode Assignment

3.1.27 Objectives

To recognise Zener diodes in various physical forms and to distinguish them from rectifying diodes.

To understand the constant-voltage characteristic of a reverse-biased Zener diode.

To understand the use of a Zener diode in a simple voltage regulator circuit.





Know what is meant by internal resistance and the effect it has on terminal voltage.


Qty

Apparatus

1


1

Power Supply Unit, 0 to 20 V variable dc, regulated.(eg, Feedback Teknikit Console 92-300).

2

Multimeters

OR


Background

In the Semiconductor Diode Assignment you found that a reverse-biased diode passes negligible current. You also learnt that it will eventually suffer breakdown and damage if the reverse voltage is made too high. See fig 1.

Fig 1 Reverse Breakdown of a Diode

Zener diodes are specially constructed to break down at controllable voltages and to do so without damage to the device. As we shall see, this feature can be put to good use.

Two Zener diodes are contained in the 12-200. They are 10 V and 7.5 V types and are shown in fig 2 with the standard circuit symbol.

Fig 2 Zener Diodes and Symbol

Zener diodes look very similar to rectifier diodes and the terminal names and identification methods are the same. The larger types have greater power and current capacities.

The two types of diode can usually be distinguished only by their type numbers. For Zener diodes these often, but not always, contain the letter Z.

3.1.30.1 A Simple Zener Diode Voltage Regulator

The Zener diode has a region in its reverse characteristic of almost constant voltage regardless of the current through the diode. This can be used to regulate or stabilise a voltage source against supply or load variations.

Fig 3 shows an unregulated voltage source supplying current to a variable load.

Fig 3 An Unregulated Power Source

If either Vs or RL changes, so will the voltage across the load VL.

One way of keeping this voltage more constant is to connect across the load a Zener diode whose breakdown voltage is the desired constant voltage.

Fig 4 shows a Practical circuit of this kind.

Fig 4 A Simple Zener Diode Regulator

3.1.31 Practical 1

In this Practical you will plot the typical reverse characteristic of a zener diode.

You will measure the voltage across the diode and also that across a series resistor.

You will use Ohms Law to calculate the current through the diode.


Fig 5

You will plot the reverse characteristic and you will also calculate and plot the power dissipation against the diode voltage.




As shown in the Patching Diagram for this Practical, construct the circuit of fig 6.

Fig 6 Test Circuit

The method of obtaining the voltage-current characteristic is the similar to that of the Semiconductor Diode Assignment but notice that the Zener diode is reverse-biased.

Using the power supply variable control, set Vs to the following values:

0 V, 2 V, 4 V, 6 V, 6.5 V, 7.0 V, 7.5 V, 8.0 V, 8.5 V, 9.0 V, 10 V, 15 V and 20 V

For each value record Vr.

Go to the Results Tables section of this Assignment and copy fig 7 to tabulate your results.

Calculate:

Vd = Vs - Vr and Id = EQ \f(Vr,1000)= Vr mA

Prepare a graph like fig 8 and plot Id against Vd.

Fig 8 The Zener Diode Characteristic

Calculate the power dissipated in the diode for each value of Vd and Id, and enter it into the last column of your table.

Pd = Vd x Id mW

Plot Pd against Vd on your graph.

Questions

1. Describe the Zener diode characteristic in your own words.

2. The nominal voltage of the BZY55C7V5 is 7.5 V. Does your graph agree with this exactly? If not, can you suggest a reason for any difference?

3. Why is the series resistor in Fig 6 necessary?

4. The maximum allowable power dissipation of type this type of Zener diode is 400 mW. Does your maximum value of Pd approach this limit?



seen that, in its breakdown region a Zener diode has an almost constant voltage regardless of diode current,

understood how this feature can be used to stabilise a varying voltage,

understood the need to keep the diode power dissipation below the allowable maximum.


the circuit that you investigated,


the characteristic graph drawn,





Zener diodes are widely used as voltage stabilisers and voltage references.

They are manufactured with Zener voltage ratings of between about 2.7 volts and 200 volts, usually in 'preferred' voltage steps, eg, 2.7, 3.3, 3.6, 3.9, 4.3, 4.7, 5.1 etc, just as 1% tolerance resistors are manufactured.

The power dissipation rating of a Zener diode is an important parameter.

Zener diodes are manufactured with power ratings between 250 mW and in excess of 100 watts.

Although Zener voltages are fairly insensitive to changes in diode current, they are however sensitive to temperature changes. Normally the Zener voltage is specified at a temperature of 25C, but diodes will have a temperature coefficient. Typical figures for this range from -5.0 mV/deg C for a 2.7-volt device to +60 mV/deg C for a 75-volt device. The zero temperature coefficient is given around 5.6 volts Zener voltage.

Zener diodes have many uses other than for providing stable or reference voltage sources (eg, they can be used for clipping, thereby doing away with the need for a voltage source as the clipping reference).

Zener diodes are often used for over-voltage protection, being connected across the load. The Zener voltage is chosen such that under normal operating conditions the diode is reverse-biased below the Zener voltage, so the device acts as an ordinary diode (i.e. non-conducting).

If however the voltage rises above the Zener voltage, the diode will break down and pass a heavy current. The excess voltage may be dropped in a resistor, as in fig 9(a), or the fuse will blow, as in fig 9(b).

Fig 9 A Zener Diode used as Load Protection

3.1.34 Results Tables

Vs(V)

Vr(V)

Vd =Vs - Vr(V)

Id = Vr(mA)

Pd =Vd x Id(mW)

0

2

4

6

6.5

7.0

7.5

8.0

8.5

9.0

10.0

15.0

20.0

Fig 7

Notes

Transistor Familiarisation Assignment

3.1.35 Objectives

To recognise transistors in various physical forms and to identify their terminals.

To understand the basic construction of PNP and NPN transistors.

To understand junction biasing and the direction and magnitude of current flows.





Know the operation of parallel dc circuits.


Qty

Apparatus

1


1

Power Supply Unit, +5 V and -15 V variable dc, regulated.(eg, Feedback Teknikit Console 92-300).

3

Multimeters

OR


Theory

Transistors are three-terminal devices constructed in the form of two semiconductor junctions, rather like two junction diodes.

Fig 1 shows the two types, NPN and PNP, governed by the physical arrangement of the P- and N-type semiconductor materials.

Fig 1 Two Types of Transistor

Each of the PN junctions in this diagram behaves individually like the simple diode you studied in the Semiconductor Diode Assignment but, when joined together in this way, the behaviour is very different.

In normal use the emitter-base diode is forward-biased and behaves almost exactly like an independent diode. The collector-base diode, however, is reverse-biased and normally you would expect it to pass no current. But if the E-B diode is conducting forward current, this influences the reverse-biased C-B diode and causes it to pass almost as much reverse current.

Fig 2 shows this for PNP and NPN types. The small difference current flows in the base circuit.

Fig 2 Transistor Current Flow

The ratio

E

C

I

I

is usually called hfb.

Because IC is almost as big as IE, hfb is nearly 1.

(

E

C

I

I

= hfb= nearly 1 (eg, 0.99)

The ratio

B

C

I

I

is usually called hfe

IB + Ic = IE =

fb

C

h

I

ThusIC (

fb

h

1

1) = IB

andhfe =

B

C

I

I

=

=

ifhfb = 0.99,

hfe=

= 99.

It is this large ratio between IC and IB that makes the transistor a useful amplifying device when connected so that IB is derived from an input and IC provides an output.

In the Assignment you will first identify some actual transistors and then confirm the directions and magnitudes of currents, finding hfb and hfe in the process.

3.1.39 Practical 1

In this Practical you will investigate the currents that flow in a transistor.

The transistor that you will use is a general purpose NPN transistor with the type number BC107. It is of the type called a low power or small signal transistor to distinguish it from transistors that can operate at significantly higher power levels they are called power transistors.


Fig 3

You will measure the base current of the device and the resulting collector current. You will then change the base current and see how the collector current changes.

You will also measure the base-emitter voltage and compare it with that of a forward biased diode.

In the circuit of fig 3, the capacitor is provided to ensure that the circuit is stable and has no effect on your measurements.




Fig 4 illustrates a typical low power transistor.

Fig 4 Typical Low-power Transistor (as supplied)

Transistors are made in many other physical forms. Fig 5 shows some other types you could have to recognise.

Fig 5 Different Transistor Styles

Make sure you can accurately identify the terminals of the transistors in the kit.


Fig 6 Transistor Test Circuit

As stated, this uses the BC107-NPN transistor and the capacitor is provided to ensure that the circuit is stable and has no effect on your measurements.

Turn the potentiometer to zero (anti-clockwise) and switch on both power supplies.

Slowly increase VEB by turning the potentiometer clockwise until IC just begins to flow.

Connect the voltmeter temporarily between E and B on the transistor (3 V dc range) and note the value of VEB in the table.

Note: if you are using virtual instrumentation to measure this voltage you will need to measure between emitter and 0 V and then base and 0 V and subtract the readings to obtain VEB.


Remove the voltmeter and then continue increasing VEB until IC approximately equals 1 mA and record the values of IC and IB in the table.

Now increase VEB until IC approximately equals 10 mA; again record IC and IB.

Satisfy yourself that both IB and IC are flowing in the directions shown in fig 6

For each set of readings calculate hfe, IE and hfb as follows:

E

C

fb

B

C

E

B

C

fe

I

I

h

;

1000

I

I

I

1000;

I

I

h

=

+

=

=

Questions

1. Did your reading of VEB confirm that the forward-biased E - B junction is acting like a simple diode? Explain.

2. Do your results show that hfb and hfe increase, decrease or stay constant as Ic increases?

3. Fig 8 shows an incomplete circuit of PNP transistor in common-emitter connection.

Fig 8

Complete the circuit with a suitable collector bias voltage and show the direction and size of the collector current IC.

Also find hfb and hfe


The measurements in this Assignment are made with the BC107 transistor used in a circuit in which the E and C terminals are biased with voltages relative to the base B. For this reason this circuit is called a common base connection.

It is also possible to bias the junctions with voltages relative to the emitter or collector, giving common emitter and common collector connections as in fig 9.

Fig 9 Bias Arrangements for an NPN Transistor

In common-emitter, VCE must be larger than VBE to ensure that the C - B junction remains reverse-biased.

In common-collector, VEC must be larger than VBC to ensure that the E - B junction remains forward-biased.

These three connections have important differences in their responses to inputs.

The common-emitter and common-collector circuits are the most important connections since the common-base is used only in special circumstances.

As with diodes, transistors can be made from Germanium instead of Silicon, but Germanium devices are rarely used nowadays.



seen that transistors have two basic forms: the PNP and the NPN,

learnt that a transistor comprises two diode junctions, one forward and one reverse-biased,

seen that the base current is much smaller than either the emitter or collector current, which are themselves nearly equal,

learnt that there are three basic bias connections for a transistor.



the calculations that you performed,





Results Table

IC(mA)

VEB(V)

IB(mA)

B

C

fe

I

I

h

=

IE = IC + IB(mA)

E

C

fb

I

I

h

=

justmeasurable

-

-

-

-

0.99

-

10

-

Fig 7

Notes

The Common-Emitter Transistor Assignment

3.1.42 Objectives

To become familiar with the common-emitter output (collector) characteristics.

To provide an understanding of the meaning and importance of Operating Point and Load Line.



Transistor Familiarisation


Know the operation of a potential divider circuit.


Qty

Apparatus

1


1

Power Supply Unit, 0 to 20 V variable dc regulated and +5 V dc regulated (eg, Feedback Teknikit Console 92-300).

3

Multimeters

OR


Background

In the Transistor Familiarisation Assignment you learnt that the emitter and collector currents of a transistor are nearly equal. Also, you learnt that the base current is the difference between them and is therefore much smaller.

You learnt too that the transistor junctions can be biased in three different ways, called 'common-base', 'common-emitter' and 'common-collector'. This assignment looks more closely at the common-emitter connection, using an NPN type of transistor.

Fig 1 is to remind you of some of the results from The Transistor Familiarisation Assignment

Fig 1 NPN Transistor Common-emitter Connection

If IB is derived from an input signal and IC is used to generate an output, then the ratio IC/IB represents the gain of the transistor in terms of the currents.

That is:

hfe =

NT

INPUTCURRE

ENT

OUTPUTCURR

= CURRENT GAIN

What is required is a graph that will tell us exactly how IC, IB and VCE are related to one another.

The Collector Characteristic, which is found in this Assignment, is one such graph.

3.1.46 Practical 1

In this Practical you will investigate how the collector current IC is controlled by the base current IB and the collector-emitter voltage VCE.


Fig 2

In this circuit the input is applied to the base and the output is from the collector. The emitter is common to both input and output. This connection is therefore called the common emitter connection.

You will set a value of collector-emitter voltage and then vary the base current of the device and measure the resulting collector current. You will then change the collector-emitter voltage and see how the base current and collector current change.

You will then plot IC against VCE for each value of IB on your graph.





Fig 3 Test Circuit

Go to the Results Tables section of this Assignment and copy fig 4 to tabulate your results and prepare a graph as shown in fig 5, for your results.

Fig 5 BC107 Collector Characteristics

Set VCE to 0.5 V, then use the potentiometer to adjust IB to each value given in the table of fig 5.

At each setting record IC in the appropriate column. Then repeat for each other VCE value.

Plot IC against VCE for each value of IB on your graph.

The graphs you now have are the 'output' or 'collector characteristics'.

Questions

1. What happens to IC when VCE becomes less than 0.6 V?

What is the significance of this value?

2. What do you notice about the effect of VCE upon IC when VCE is greater than about 1.0 V?

Practical 2

In this Practical you will begin to investigate what happens when a load resistor is connected in the collector circuit.


Fig 6

This circuit is the same common emitter connection, as used in Practical 1, but has a load resistor R connected in series with the collector.

Now the supply voltage is designated VCC, to distinguish it from the collector-emitter voltage VCE.

You will construct a load line on the output characteristic plotted in Practical 1 corresponding to the load resistor value.

This Practical is a paper-based exercise and requires no patching.


You should have concluded from the first Practical that IC is very little affected by VCE . Instead however, it is almost entirely controlled by IB.

We say that the output circuit represents a constant current source.

If we wish to produce an output voltage, as we might in certain kinds of amplifier, this current may be passed into a resistor to generate a voltage.

Fig 7 shows a circuit in which the collector bias is applied through a load resistor.

Fig 7 Addition of a Load Resistor

VCC is the term now used for the supply voltage to distinguish it from VCE, because when a current IC is flowing these two will be different.

By Ohm's Law:

VCE = VCC - IC R.

Therefore,

when IC = 0, VCE = VCC

and

when VCE = 0 IC =

On your graph plot these two points VCE and IC for:

IC = 0, R = 2k, and VCC = +10 V.

Plot two or three more points for other values of IC such as 2 and 5 mA using the same values of R and VCC.

Join up the points by a line.

Your line should be straight because the equation (VCE = VCC - IC R) is a linear one; there being no square or cubic terms in it.

What you have now drawn is called a load line. It shows how VCE varies with IC and in turn with IB.

Construct a second load line for:

VCC = 8 V and

R =1k ohms.

3.1.47 Practical 3

In this Practical you will connect a load resistor in the collector circuit and set an operating point.

You will construct the circuit and investigate its operation.


Fig 8

This circuit is a practical implementation of the basic circuit of fig 7, but includes components to enable the supply voltage and the base current to be adjusted and meters to measure the resulting effects.





Go to the Results Tables section of this Assignment and copy fig 9 to tabulate your results and prepare a graph as shown in fig 5, for your results.

Fill in the gaps in your table of Practical 2 by examination of your graph.

Make estimates if an exact value cannot be obtained.

Now alter your constructed circuit to include a load resistor of 1k and set VCC = 8 V. The circuit now looks like fig 10. Also see the Patching Diagram.

Fig 10 Test Circuit

Check your answers to items No. 1 and 4 in the table, by setting the given parameter and measuring the other two.

Now change VCC and R to 10 V and 2k respectively and check your answers to items 2 and 3.

A circuit like fig 10 is widely used to amplify alternating signals. It is then necessary to set the initial value of VCE to allow the output to vary both up and down. This is called setting the operating point and is what you have just done.

Very often the operating point will be set to VCE which approximately equals

to allow the maximum possible swing in either direction.

Mark this point on each of the two loads lines on your graph (fig 6) and label the points as operating points.

Estimate the extreme values of VCE for each load line if IB varies by (10 A about the operating point value. Draw a thick line along the section of load line included within these limits. This represents a typical operating range.

Your graph should now look like fig 11.

Fig 11 Typical BC107 Characteristics

Questions

1. Use the value of hfe found in the Transistor Familiarisation Assignment for the BC107 to estimate the peak-to-peak variation of VCE in the circuit of fig 12 when IB varies by 2 A.

Fig 12

Also what mean level of IB must you use to establish the operating point correctly?(HINT: Variation of VCE = R x variation of IC)

Results Required


learnt that the collector or output characteristics for a common-emitter transistor can be used to predict IC, given VCE and IB,

constructed a load line can on the characteristic to show the effect of a resistor in the collector lead,

seen that the load line can be used to determine a suitable operating point, which can be set by adjustment of the base current,

learnt that the variation in base current determines the operating range.



the calculations that you performed,






Setting the operating point is an important matter in transistor amplifiers. The Practicals in this Assignment have effectively done this by applying a voltage to the base-emitter junction through a resistor, as in fig 13.

Fig 13 Simple Bias Circuit

Unfortunately this does not give a stable operating point in practice because temperature changes cause a considerable change in IC for a given IB. Also individual transistors of one type show some variation in characteristics.

Fig 14 shows a circuit which is much used to stabilise the operating point against such variations.

Fig 14 A Stabilised Bias Circuit

R1 and R2 set a voltage at the base, VB, of say 4 V and IE is made to pass through resistor RE. The emitter voltage, by Ohm's Law must be:

VE = IE x RE

and since VBE is about 0.6 V, then:

VB = VE + 0.6 = IE RE + 0.6

Thus, if

VB = 4 V, VE = 3.4 V.

Now if IE increases, this causes VBE to reduce and in turn reduces IB and hence IE.

Thus the original change is counteracted. If IE reduces the result is similar. The operating point has effectively been set by VB and RE, because:

IC ( IE =

(

)

R

0.6

V

E

B

-

Results Table

IB

IC (mA) for Vce = ......... ..(V)

(A

0.5

1

2

5

10

0

10

20

30

40

50

Fig 4

No.

VCC (V)

R ()

VCE (V)

IB (mA)

IC (mA)

1

8

1000

25

2

10

2000

4

3

10

2000

5

4

8

1000

5.5

Fig 9

The Silicon Controlled Rectifier Assignment

3.1.49 Objectives

To recognise SCR's in different physical forms.

To understand the two-transistor analogy and the different ways of triggering an SCR.

To appreciate the meaning of the terms 'Breakdown Voltage' and 'Holding Current'.

To appreciate the areas of application of SCR's.








Qty

Apparatus

1


1

Power Supply Unit, 0 to 20 V variable dc regulated and +15 V dc regulated.ac supply; 12 V rms 50 or 60 Hz (isolated from other supplies.(eg, Feedback Teknikit Console 92-300).

2

Multimeters

OR


Background

The Silicon Controlled Rectifier, or SCR, is one of several semiconductor devices which are capable of acting as fast switches for large currents. The general name for these devices is thyristor.

Fig 1 shows the SCR symbol and some of the physical forms in which it is found.

Fig 1 Types of SCR and Graphical Symbol.

The SCR resembles a rectifier diode but if the anode is held positive relative to the cathode no current flows until a positive current is injected into the gate. The diode then switches on and will not switch off until the anode-cathode voltage is removed. Hence the name controlled rectifier.

Fig 2 shows the SCR as a three-junction device.

Fig 2 The Two-transistor Analogy of an SCR.

This can be regarded as two inter-connected two-junction transistors, one PNP and the other NPN.

If A is made positive relative to K, and G is left unconnected, no current will flow because each transistor gets its base-emitter current from the other's collector emitter current. So, until one of the transistors is given some base current, nothing can happen.

If now a current is injected into the base of transistor '2', the resulting collector current flows in the base of '1'. This in turn causes a collector current in '1' which increases the base current of '2' and so on. Very rapidly the two transistors force each other to conduct to saturation; the current being limited only by resistance in the external circuit.

If the anode-cathode voltage is now reduced, the current also reduces until it goes below some critical value, and the transistors switch off again, just as rapidly.

If a reverse voltage is applied to the SCR (anode negative to cathode) it behaves very much like an ordinary diode. No current passes until at, some high voltage, it breaks down completely.

3.1.53 Practical 1

In this Practical you will measure the Trigger Current and the Saturation Voltage of a forward biased SCR.

You will also measure the Holding Current for the device.


Fig 3

You will set a value of supply voltage and then increase the gate current of the device until the lamp lights.

You will also measure the anode-cathode voltage when the SCR is on (conducting).

Finally, you will find the holding current for the device.





Fig 4 Test Circuit for Trigger Current.

Set the variable dc volts to 12 V. Turn the potentiometer to zero (clockwise) and switch on the supplies.

Slowly rotate the potentiometer, observing the gate current meter continuously, until the lamp suddenly lights. Record the gate current at which this occurs.

Switch off the supplies and return the potentiometer to zero.

Copy the results table as shown in fig 5, found in the Results Table section of this assignment, for your results.

Repeat the measurement several times. to ensure that you have the correct value. What you have found is the Trigger Current (IGT). Enter it into your table.

Now trigger the SCR on again and measure the voltage from anode to cathode.

This is the Saturation Voltage VAK (sat). Enter this in the table.

Finally connect the multimeter (on range 100 mA) in series with the lamp as in fig 7.

Fig 6 Test Circuit for Holding Current.

Trigger the SCR on. The meter reads the lamp current for a 12 V supply voltage.

Temporarily disconnect the gate connection.

Slowly reduce the supply voltage until the SCR current suddenly falls to zero.

Note the value of the current at which this occurs.

Repeat this procedure several times to ensure that you have the correct value.

What you have found is the Holding Current (IH). Enter this also into your table.

Questions

1. Look at fig 7.

Fig 7

Do you expect the saturation voltage to be greater or less than 0.6 V? Explain.

2. What do you think will happen in the circuit of fig 4 if you trigger the SCR on, and then reduce the gate current to zero again?

Confirm your answer by experiment.

3.1.54 Practical 2

In this Practical you will replace the variable dc anode supply with a 12 V ac supply and investigate the performance of the circuit under ac conditions.


Fig 8

Connect up the circuit as shown in the Patching Diagram for this Practical. Remember to use ac meters.



In Practical 1, you found that the anode current must be reduced to below IH to switch the SCR off. This is the only way of switching off. You CANNOT do it by reducing the gate current.

If the anode supply is an alternating voltage it will go negative every half-cycle, reducing the anode current to zero.


Fig 9 Test Circuit with an ac Supply

Repeatedly increase and decrease the gate current and observe what happens.

Observe the brightness of the lamp and compare this with that for the 20 V dc supply.

Questions

1. What do you observe when you repeatedly increase and decrease the gate current? Explain what you think is happening.

2. Why does the lamp burn less brightly with the ac supply than it did with the 12 V dc supply?

3. The SCR in fig 10 has a maximum steady anode current capability of 10 A. Given that VAK (sat) is 1.5 V at this current, find:

Fig 10

(a) The power dissipation in the SCR at maximum anode current.(b) The value of R.(c) The power dissipation in R.(d) The efficiency of the SCR as a switch for a 10 A anode current.

Results Required


seen that SCR's can be regarded as two interconnected transistors,

learnt that SCR's can be triggered on by a gate current but triggered off only by reducing the anode current,

investigated the operation of an SCR for both dc and ac anode supply voltages,

learnt that SCR's can be triggered unintentionally by high temperature, over voltage or a rapidly rising anode voltage.







SCR's can be triggered unintentionally in several ways and you have to be aware of these as they could be the cause of wrong operation. Fig 11 illustrates this.

Fig 11 False Triggering Mechanisms

In (a) a high temperature increases the leakage current of the two 'transistors' in the SCR. This is the small collector-emitter current that flows when there is zero base current. If it becomes too great it will be enough to initiate the trigger action.

If a very high forward voltage is applied, as in (b), the 'transistors' can break down and this too will initiate triggering.

Fig 11(c) shows a very rapidly rising anode-cathode voltage. Every transistor has some capacitance from collector to emitter as shown in fig 12. A fast-rising anode-cathode supply causes small currents in these capacitors and can act to cause triggering.

Fig 12 Stray Capacitance Positions.

Steps must always be taken in practice to avoid each of these possible false trigger mechanisms.

SCR's are available to carry currents from less than 1 A up to 1000 A or more.

They therefore find use in the switching of heavy electrical equipment, where they replace contactors. The following advantages should be obvious:

No moving parts

No contact arcing

No bad contacts due to corrosion or dirt

In addition to simply switching currents on and off, SCR's can be made to control the mean value of a load current without dissipating large amounts of power. In this application they can replace bulky high wattage rheostats and save electrical energy at the same time. A good example of this is the control of theatre lighting.

The introduction to the Trigger Devices Assignment explains how this can be done.

Results Table

TRIGGER CURRENT

(IGT)

mA

SATURATION VOLTAGE

(VAK (sat))

V

HOLDING CURRENT

(IH)

mA

Fig 5

The TRIAC Assignment

3.1.56 Objectives

To recognise a TRIAC device.

To understand the bidirectional nature of the TRIAC and its areas of application.

To appreciate the different behaviour of the device in the four operating quadrants.


The Silicon-Controlled Rectifier (SCR)



Know how to use Cartesian axes involving four quadrants.


Qty

Apparatus

1


1

Power Supply Unit, +5 V dc regulated and +15 V dc regulated,ac supply; 12 V rms 50 or 60 Hz isolated from other supplies(eg, Feedback Teknikit Console 92-300).

1

Multimeter

OR


3.1.60 Background

In the SCR Assignment you learnt that the SCR can be used to switch a unidirectional current but it will not conduct in reverse.

An alternating supply is often necessary to ensure that the SCR will switch off. It is rather inefficient, because it conducts only every other half-cycle (like a half-wave rectifier as described in the Half-wave Rectification Assignment).

Fig 1 shows the basic single-SCR circuit and also one way of using four SCR's in a bridge to achieve controlled full-wave rectification.

Fig 1 Half-wave and Bridge SCR Circuits

Note that, in a bridge circuit, gates G1/G2 are triggered together on one half-wave and G3/G4 on the next.

A simpler and less expensive way of obtaining bidirectional conduction is to use a TRIAC.

Fig 2 shows a typical device and its graphical symbol.

Fig 2 A TRIAC and its Graphical Symbol

A TRIAC, like an SCR, is a type of thyristor. Although it has the same basic four layers of semi-conductor materials, its detailed construction is too involved to be described here.

The behaviour of the TRIAC is very similar to that of the SCR but it can be triggered into conduction by gate current for either polarity of the voltage between terminals T1 and T2.

Fig 3 shows the four modes in which a TRIAC can be operated.

Fig 3 TRIAC Triggering Modes

With reference to Quadrant I, the TRIAC is usually triggered by a positive gate current (mode I +), but can be triggered by negative gate current, (mode I -).

Similarly in Quadrant III negative IG is usual (mode III-), but positive IG is possible (mode III+).

Modes I- and III+ are, however, less sensitive than the usual modes, I+ and III-.

3.1.61 Practical 1

In this Practical you will confirm that a TRIAC may be triggered in any of the four modes referred to in the Background section.


Fig 4

Initially, you will set the gate and supply voltages to +15 V and then increase the gate current of the device until the lamp lights and you will note the gate current required to achieve this.

You will then, in turn, set up the other three combinations of positive and negative gate and supply voltages to provide the four quadrants of operation of the device, again noting the gate currents required to obtain conduction.

Finally, you will find the holding current for the device.





Fig 5 Test Circuit for Trigger Current


Set the potentiometer to zero (anti-clockwise) and switch on the power supplies.

Slowly increase the gate current until the lamp lights, noting the value of IGT when this occurs. Record this in your table under IGT for mode I+.

Switch off and move link 1 to apply -15 V to the gate circuit instead of +15 V.

Repeat the measurement and record the value of IGT for mode I-, reversing the meter connections if necessary.

Now move link 2 to apply -15 V to the lamp.

Repeat the measurement of IGT for mode III-.

Finally restore the gate supply to +15 V and measure IGT for mode III+.

You should find that Modes I+ and III- have similar values of IGT but that modes I- and III+ require comparably greater gate currents to cause triggering.

TRIACs, like SCRs, require a minimum current, called the HOLDING CURRENT, to keep them in conduction. You can confirm this if you wish for mode I+ by the same method as used in the SCR Assignment.

3.1.62 Practical 2

In this Practical you will set up a simple TRIAC switch circuit and investigate its operation.


Fig 7





Fig.8 Simple TRIAC Switch Circuit.

Verify that:

If link A is not connected there is no gate current and the lamp does not light.

If A is connected to point B, gate current flows at every half-cycle and the lamp lights.

Questions

1. In fig 8 what do you expect to happen if A is connected to point C? Confirm by experiment and explain your answer.

2. Which mode or modes are in use for:

'A' connected to 'B','A' connected to 'C'?

Results Required


learnt that a TRIAC is a four-layer thyristor device similar to the SCR,

seen that a TRIAC can be triggered into conduction in either direction,

Learnt that there are four triggering modes, of which two are preferred.







TRIACs have the same breakdown voltage, temperature and rate of voltage rise limitations as SCR's. They are available in a wide range of voltage and current ratings.

Apart from the bidirectional current capability and the need to consider the triggering modes, TRIACs are in most other respects similar to SCRs in their range of applications.

Results Table

MODE

I+

I

III

III+

IGT(mA)

Fig 6

3.1.64 Further Work

Fig 9 shows a simple TRIAC switch circuit.

Fig 9 Simple TRIAC Switch Circuit.

Use your measured results for IGT and assume that the voltage from G to T1 will be about 1 V.

Suggest a value for R which, when the switch is closed, will give:

Half-wave operation of the lamp.

Full-wave operation of the lamp.

Trigger Devices The DIAC and UJT Assignment

3.1.65 Objectives

To be able to recognise DIAC and UJT devices and their symbols.

To understand the need for trigger devices when used with thyristors.

To appreciate the main features of the DIAC and UJT.


The Silicon-controlled Rectifier (SCR)

The TRIAC





Qty

Apparatus

1


1

Power Supply Unit, 0 to 20 V variable dc regulated+15 V dc regulated(eg, Feedback Teknikit Console 92-300).

1

2-channel oscilloscope

1

Multimeter

OR

Feedback Virtual Instrumentation may be used in place of the oscilloscope and multimeter

Background

In the SCR and TRIAC Assignments two semiconductor switches (or thyristors) were studied; the SCR and the TRIAC. In ON-OFF switching applications they could be triggered by simple circuits producing steady gate currents.

Fig 1 is an example of such a circuit using an SCR.

Switch open

- no gate current

- no conduction

Switch closed- Gate current flows

- SCR Conducts on positive half-cycles

Fig 1 An SCR Circuit.

REMEMBER: A thyristor will switch off only when its anode-cathode voltage falls to zero.

If it is required to control the mean value of a load current, rather than just switch it on and off, there is only one method available. This is illustrated in fig 2 for an SCR.

Fig 2 SCR Conduction for Different Delay Times.

A steady gate current would allow conduction over the full period of the positive half-cycle.

If instead, a short pulse of gate current is applied at the trigger points, conduction occurs over part of the half-cycle only. This reduces the mean current.

The mean current can be varied by changing the delay time T between the start of the cycle and the trigger. This is known as phase control.

Fig 3 explains why.

Fig 3 Phase Control of an SCR.

The supply wave A is delayed by the phase shift to give B. When B reaches a certain trigger level the trigger circuit generates a gate pulse C for the SCR.

To achieve phase control, then, two things are needed:

a. A variable phase shift circuit (usually passive components such as resistors and capacitors).

b. A trigger circuit that can produce a pulse when the delayed waveform reaches a certain level.

In this assignment you will look at two devices which serve as trigger generators, the DIAC (DIode AC switch) and the UJT (Uni-Junction Transistor).

3.1.69 Practical 1

In this Practical you will investigate the operation of a DIAC.


Fig 4

Initially, you will set the dc voltage to zero and then you will increase this voltage until the device conducts.

You will use the oscilloscope to observe and measure the voltages around the circuit and you will construct an approximate characteristic for the DIAC.




Figure 5 shows the symbol and a typical voltage-current characteristic. An alternative DIAC symbol is shown in fig 6.

Fig 5 The DIAC, its Graphical Symbol and Characteristic

Fig 6 An alternative DIAC symbol

Identify the DIAC on your 12-200-B workboard.

The DIAC is made like a transistor but has no base connection. When a voltage greater than VBR is applied, breakdown occurs. In an ordinary diode the voltage would then remain constant as the current increased. In the DIAC the transistor action causes the voltage to reduce as the current increases. This gives the characteristic a negative resistance, as shown in fig 5.

The DIAC is symmetrical and therefore has the same characteristic for negative voltages. It is the negative resistance that makes the DIAC suitable as a trigger for an SCR or TRIAC.

To test this, ensure that you have connected up the circuit as shown in the Patching Diagram and that it corresponds with the circuit shown in fig 7.

Fig 7 The DIAC Test Circuit

Set the variable dc supply to zero and switch on the supplies.

Slowly increase the variable dc voltage until the waveform at Y2 suddenly appears. That is, the DIAC 'switches on'.

Notice the very rapid rise of VR, produced by the negative resistance. See fig 8.

Fig 8 DIAC Waveforms

Measure VBR and VR from the oscilloscope.

VBR is the DIAC breakover voltage.

VR is the load voltage.

is the DIAC current immediately after switch-on.

VBR - VR is the voltage across the DIAC immediately after switch-on.

From these figures it is possible to construct an approximate characteristic for the DIAC as in fig 9.

Fig 9 The DIAC Characteristic and Load Line

Fig 9 represents the conditions just before switch-on, and Q the conditions just after. The values shown are not necessarily the correct ones.

Prepare a graph like fig 10. on which to plot your results and use your own measurements to draw a graph like fig 9.

Fig 10 The DIAC Characteristic

3.1.70 Practical 2

In this Practical you will investigate the operation of a UJT.


Fig 11

Again, you will set the variable dc voltage to zero and then you will increase this voltage until the device switches on.

You will construct a characteristic for the UJT.




Fig 12 shows the appearance of a Unijunction Transistor (UJT) (like an ordinary transistor), its graphical symbol, and gives an indication of its construction.

Fig 12 Unijunction Transistor Details

Identify the Unijunction Transistor (UJT) on your 12-200-B workboard.

Base B2 is biased positive relative to B1. This sets up a reverse bias at the diode PN junction. This bias is overcome when a sufficient positive voltage is applied to E. Then emitter current flows and the effect of the current already flowing from B2 to B1 is to give the E B1 diode a negative resistance, similar to that in the DIAC.


Fig 13 A UJT Test Circuit and Characteristic

Set the variable dc voltage to zero and switch on the power supply. Slowly increase the variable dc until the emitter current suddenly increases. That is the UJT 'switches on'.


Record the value of VEB1 just before switch-on (point P), also the values of VEB1 and IE just after switch-on (point Q).

Slowly reduce the variable dc voltage until the emitter current suddenly switches off again.

Record, in your table the values of IE and VEB1 just before switch-off (point R) and the value of VEB1 just after switch-off (point S).

You can now construct the characteristic for your UJT similar to the example shown in fig 15.

Fig 15 Graph Layout for UJT Characteristic.

Layout a graph like fig 16 on which to plot your results.

Fig 16 The UJT Characteristic

Enter points P, Q, R and S and the load lines on the axes of your graph of using your measured values from your table. Sketch in the UJT characteristic.

Questions

1. What do you think will happen if VEB1 is made negative?

2. Study your graphs for the DIAC and UJT and determine what are the main differences between the characteristics of the two trigger devices?



learnt that SCRs and TRIACs used to control the mean value of a load current need phase-delayed trigger pulses,

learnt that the DIAC and UJT are suitable devices to produce these pulses for the TRIAC and SCR respectively,

learnt that the DIAC and UJT both possess negative-resistance characteristics which allow them to switch on rapidly once a certain applied voltage level is reached.







The DIAC, because of its bidirectional operation, is primarily used to trigger TRIACS; the UJT, being unidirectional, is suitable only for use with SCR's.

Both devices are normally used in a circuit such as fig 17.

Fig 17 A Relaxation Oscillator used as a Trigger Generator

This circuit forms a Relaxation Oscillator and operates as follows:

R and C form a variable delaying circuit to obtain the necessary phase control as described in the introduction. When C is sufficiently charged (ie, up to VBR for the DIAC or VP for the UJT), the device triggers and C discharges into RL forming a short pulse. When C is discharged sufficiently (e.g. down to point R on the UJT characteristic) the device turns off and C starts to recharge.

All the time the supply voltage is large enough, the circuit goes on oscillating in this way.

Fig 18 shows a typical output for a UJT driven by a full-wave rectified input.

Fig 18 Waveforms of a Relaxation Oscillator.

A chain of trigger pulses is often better than just one pulse as it gives a greater certainty of triggering the SCR or TRIAC.

Results Table

Condition

IE (mA)

VEB1 (V)

POINT

Just before switch-on

0

P

Just after switch-on

Q

Just before switch-off

R

Just after switch-off

0

S

Fig 14 UJT Measurements

3.1.73 Further Work

Fig 19 shows a practical relaxation oscillator circuit using a UJT.

Fig 19 UJT Relaxation Oscillator Circuit

Construct the circuit and observe the output on an oscilloscope. Sketch what you see.

The Field Effect Transistor Assignment

3.1.74 Objectives

To understand the difference between bipolar and field-effect transistors

To distinguish between JFET and MOSFET types and between N and P channel construction.

To be able to recognise the basic characteristics of a JFET.

To know the principal advantages of FETs and some applications.



The Common-Emitter Transistor


The Common-Emitter Transistor



Qty

Apparatus

1


1

Power Supply Unit, 0 to 20 V variable dc regulated+15V dc regulated(eg, Feedback Teknikit Console 92-300).

1

2-channel oscilloscope

1

Function Generator, 2 V pk-pk at 1 kHz

3

Multimeters

OR

Feedback Virtual Instrumentation may be used in place of the oscilloscope and two multimeters

3.1.78 Background

Field-Effect Transistors (FETs) are made in various forms. One type, the Junction FET (JFET) has a construction quite similar to the UJT (Trigger devices Assignment) but works in a different way.

Fig 1 shows the construction, graphical symbol and physical appearance of a typical JFET.

Fig 1 JFETs - Construction, Symbols, and Appearance.

As can be seen, the JFET has two forms; the N-channel and P-channel which are analogous to PNP and NPN in ordinary transistors. You will look more closely at the N-channel type.

Fig 2 shows an N-channel JFET and its bias voltages.

Fig 2 The Bias Arrangement for an N-channel JFET.

The channel is a resistive path through which voltage VDS can drive a current ID.

A voltage gradient is thus formed down the length of the channel, the voltage becoming less positive as you go from drain to source. The PN junction thus has a high reverse bias at D and a lower reverse bias at S. This bias causes a 'DEPLETION LAYER', whose width increases with the bias.

Depletion means a reduction of available electrons to carry current. If VGS is made more negative, the depletion layer increases in width at all points. The values of VDS and VGS both influence the width of the depletion layer. This alters the effective channel resistance and hence ID. Fig 3 shows this.

Fig 3 The Depletion Effect.

As VGS increases negatively the channel is 'squeezed', reducing the current ID. But the GATE-CHANNEL junction is like a reverse-biased junction diode and thus carries only a very small current. ID is controlled by VGS through a 'field effect'.

Hence the name FET.

This Assignment investigates how VDS and VGS affect ID.

3.1.79 Practical 1

In this Practical you will investigate the bias requirements of an N-channel Junction FET.


Fig 4

You will measure the drain current and the gate-source voltage for a number of set drain-source voltages and plot the output (drain) characteristics for the FET.





Fig 5 Test Circuit for a Typical N-channel JFET

Set the potentiometer anti-clockwise and the variable dc voltage to zero. Switch on the power supply.


Now set VDS to the first value in the table and then read ID for each value of VGS.

Repeat for all the values of VDS in the table, recording the corresponding ID values.

Prepare a graph like fig 7.

Fig 7 JFET Drain Characteristics.

Plot the results from your table onto your graph, drawing one curve of ID against VDS for each value of VGS.

3.1.79.2 Questions

1. Study your graphs and answer the following questions:

a) Above which values of VDS is ID almost unaffected by VDS when VGS = 0?

b) For a given value of VDS, (say 10 V), do equal changes of VGS cause equal changes of ID?

3.1.80 Practical 2

In this Practical you will endeavour to measure the gate current of the JFET.


Fig 8




Ensure that you have connected up the circuit as shown in the Patching Diagram.

On your circuit set VDS to 10 V and VGS to -1.0 V. Then alter the circuit to place the ammeter in place of the link in the gate lead as in fig 9. and try to measure IG.

Fig 9 Measuring Gate Current.

Record if you can measure IG and, if so, its value.

3.1.80.2 Questions

1. Can you measure IG or is it too small?

3.1.81 Practical 3

In this Practical you will connect up the JFET in common-source connection as an amplifier.


Fig 10

You will apply a sinusoidal input voltage to the amplifier at 1 kHz and you will observe the output voltage on an oscilloscope.

You will calculate the voltage gain of the circuit.

You will also estimate the input resistance of the amplifier.




An FET can be used to amplify signals in a manner similar to a transistor in common-emitter connection. In this case it is called common-source.

To obtain an output voltage you insert a load resistance in the drain lead, the effects of this being represented on the characteristic by a load line.

Fig 11 shows a Practical amplifier circuit with a typical characteristic and load line.


Fig 11 Test Circuit and Characteristic

On your graph draw a load-line from VDD = 15 V at an angle suitable for the 1kohm load and select an operating point at about VDS = +10 V.

Switch on the supplies and adjust VGS to give this operating point.

Now apply an input of 2 V peak-to-peak at 1000 Hz from the generator and observe the output on the oscilloscope.

Measure the peak-to-peak output voltage and calculate the voltage gain

.

Now change the resistor RG, as indicated in the Patching Diagram. Find the one that makes the output signal about half its original size. This value is equal to the input resistance of the amplifier as fig 12 shows.

Fig 12 Input Resistance Measurement

3.1.81.2 Questions

1. Is the output from your amplifier a good sine-wave?

2. The input resistance of fig 11 can not be greater than the bias resistor RG. Is it , however, much less than this? If not, what does this indicate?

3. An important parameter of an FET used as an amplifier is its transconductance. This is defined by:

Transconductance (gs ) =

(

)

(

)

GS

d

V

I

in

change

in

change

mA/V (common source)

Study your graph, fig 11 and estimate the change in Id for a 0.5 V change in VGS when VDS = 10 V and VGS = 1.0 V.

Then find gs.

The voltage gain (A) for a load resistor R is given by:

A =

, where R is in ohms ()

Use this expression to verify the voltage gain

measured in the Assignment.


When you have performed this Assignment you should:

Understand the difference between bipolar and field-effect transistors,

Know how to bias a JFET for correct operation,

Have plotted and be able to recognise the basic characteristics of a JFET,

Know the principal advantages of FETs.


the circuits that you investigat

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