semiconductors and amplifiers circuits lab report

Name of student:

Chandan Mungra

Student ID:

008672

Module Name:

Electromechanical Systems (MM1EM1)

Title of work:

Semiconductors and Amplifiers Lab Report

Lecturer Name:

Dr Woo Ko Choong

Laboratory conducted on:

19th July 2011

Group: G1

PART 1

CALIBRATION EXPERIMENT

TABLE OF CONTENTS

SUMMARY.............................................................................................. 3

1.1 AIM OF EXPERIMENT................................................................... 4

1.2 BACKGROUNG INFORMATION.................................................. 4

1.3 EQUIPMENT AND MATERIAL..................................................... 6

1.4 EXPERIMENTAL PROCEDURES................................................. 7

1.5 DATA COLLECTION, CALCULATION AND ANALYSIS......... 7

1.6 DISCUSSION..................................................................................... 10

1.7 CONCLUSION.................................................................................. 11

1.0 DIODES

1.1 Diode I-V curve

The circuit was setup as shown below:

Figure 1: Showing the experimental setup

The 0-2 V DC knob was varied to give different nominal voltages across the diode. The corresponding actual voltage and current were then read from the multimeters. These results were then were tabulated.

Nominal Voltage (mV)

Actual Voltage (mV)

Current (μA)

100 99.2 0.00

200 199 0.02

300 300 1.50

400 404 11.9

500 500 86.6

550 552 260

600 599 660

650 651 1940

700 702 4980

750 750 10690

800 800 21000

850 850 35100

Table 1: Showing the actual voltage and corresponding current

Using the values from Table 1, a graph of current against voltage was then plotted.

0 100 200 300 400 500 600 700 800 9000

5000

10000

15000

20000

25000

30000

35000

40000

Voltage (mV)

Curr

ent

(μA)

Figure 2 Graph of current against voltage

Question 1:

At what voltage does the current start to flow significantly (specifically, at what voltage does it exceed 5mA)?

By using construction lines at I= 5000 μA (5mA), the voltage was found to be about 705mV.

Experiment 1.2: Half wave rectifier

The circuit board was set up as follows;

Figure 3: Showing the experimental set up

The voltage and timebase setting are as follows:

CH1 2 V/div

CH2 2 V/div

Time base 5 ms/ div

Table 2: Showing the voltage and timebase setting

The oscilloscope traces were sketched on the following graph.

Time/ms

Volta

ge/V

Figure 4: Showing the oscilloscope trace obtained for experiment 1.2

Peak voltage for CH1: 3.5 V x 2V/div = 7.0 V


Period for CH1=Period for CH2 = (4div) x 5 ms/ div

= 20ms

Frequency = 1 / 20ms

= 50Hz

Question 2

Identify the two main differences between the AC and rectified signals.

i. From Figure 4 it can be seen that for the AC signal is a full cycle

(Signal changes from 0 to +ve, down to 0 again, then fall down to -ve

and then raise up to 0 and repeat the same cycle.)

For the rectified signal only half of the original cycle was obtained;

signal changes from 0 to +ve, then fall down into 0 and stays zero

until the negative cycle of the original signal raise up to +ve ,and fall

back to 0 and repeat the same cycle.

ii. The amplitude variation for AC signal is fixed but for rectified signal,

amplitude depends on the way the diodes are connected. A forward-

biased diode conducts current and drops a small voltage across it.

Experiment 1.3: Smoothing circuit and reservoir capacitor

The circuit diagram for this experiment is as follows:



CH1 2 V/div

CH2 2 V/div

Time base 5 ms/ div


The oscilloscope traces were sketched on the following graph.

CH1CH2

Time/ms

Volta

ge/V

Figure 6: Showing the oscilloscope trace obtained for experiment 1.3




= 20ms


= 50Hz

2.0 Transistors and Amplification

For this part an NPN bipolar junction transistor was used. It has three connections: collector, base and emitter as shown in the figure below.

Figure 7

Its main function is amplification, particularly amplification of small currents and of small variations in voltage.

2.1 Collector output characteristics experiment

The circuit was setup as follows:

Figure 8

Values of the base current and collector current were then collected and tabulated.

Base Current Target Value (μA)

Base Current Actual Value (μA)

Collector Current for V c = 1V (mA)

10 10.3 2.20

20 20.5 4.82

30 30.6 7.22

40 40.4 9.58

50 50.0 11.49

Table 4: Showing the values of the base current and corresponding collector current

A graph of collector current against base current was then plotted using the tabulated values.

5 10 15 20 25 30 35 40 45 50 550

2

4

6

8

10

12

14

f(x) = 0.235164852993068 x − 0.0776049368695571R² = 0.99820486191681

Base Current (μA)

Co

llec

tor

curr

ent

(mA

)

Figure 9: Showing the graph of collector current Ic against base current Ib

Question 3: Rearrange equation 1 to show current gain in terms of the collector current, Ic and base current, Ib.

Hence calculate the current gain β (or hFE) of the transistor (using the last

row of your table of values)

Ic = βIb

β= I c

I b

β =11.49 x10−3

50 x10−6

β = 229.8

β is dimensionless (no units)

2.2 Voltage amplification: AC experiment

The circuit was setup as follows:



CH1 2 V/div

CH2 2 V/div

Time base 5 ms/ div


The oscilloscope traces were then sketched on the following graph.

CH1CH2

Time/ms

Volta

ge/V




= 20ms


= 50Hz

Question 4: Calculate the gain of the amplifier as:

AC voltage gain of amplifier=Amplitude of AC signal output (Channel2)Amplitude of AC signal input (Channel1)

= 2.2 x 2V

2.2 x 50 x 10−3 V

= 40 (dimensionless)

3.0 Operational Amplifiers

In this part the Digiac 3000 Op Amp Board was used.

3.1 Inverting Amplifier Experiment

The wires were connected as follows:


Channel 1 Setting

CH 1 Probe Setting

Channel 2 Setting

CH 2 Probe Setting

Time base Setting

1 V/div x1 2 V/div x1 0.2 ms/div1 V/div x1 5 V/div x1 0.2 ms/div

Table 1

Rf = VR1 Vin Vo

Is output signal inverted?

Measured Gain,

G=Vo/Vin

Theoretical Gain,

G= - Rf / Rin

10 Ω 2 V 7.2 V No 3.6 -140 Ω 2 V 11 V No 9 -4

Table 2


CH1 1V /div

CH2 2 V/div

Time base 0.2 ms/ div

Table 8

The oscilloscope trace was then sketched;

CH1CH2

Time/ms

Volta

ge/V

Figure 13: Showing the oscilloscope trace for experiment 3.1



Period for CH1=Period for CH2 = (4div) x 0.2 ms/ div

= 0.8 ms

Frequency = 1 / 0.8ms

= 1250Hz


CH1 1V /div

CH2 5 V/div

Time base 0.2 ms/ div

Table 9

The oscilloscope trace was then sketched;

CH1CH2

Time/ms

Volta

ge/V




= 0.8 ms


= 1250Hz

3.2 Operational Amplifiers in practice; non-inverting amplifier

The wires were connected as follows:


The values used and results were tabulated:

CH 1 Setting

CH1 Probe Setting

CH2 Setting

CH2 Probe Setting

Timebase Setting

1 V/div x 1 5 V/div x 1 0.2 ms/div

Table 10

Vin V0

Is output signal

inverted?

Measured Gain,

G=Vo/Vin

Theoretical Gain,

G=R9+R10

R10

1 V 7 V YES 7 4.191

The oscilloscope trace was then sketched.

CH1CH2

Time/ms

Volta

ge/V




= 1.2 ms


= 833.3Hz

QUESTIONS

5. Explain the purpose of a diode.

A diode is an electronic component made from semiconductor materials (Typically silicon). Diodes function as one-way valves when placed inside circuits, thereby allowing current to flow in one direction only.As shown by experiment 1.1, Diodes conduct when certain threshold voltages are reached, and they are essentially closed or open to current flow based on their position in the circuit and current direction. They are, therefore, used as switches. They are used as rectifiers in this way, converting alternating current (AC) to direct current (DC) by periodically reversing direction to DC.

Light-emitting diodes (LEDs) provide illumination, and are widely used in

many devices as indicator lamps. Also unlike light bulbs, LEDs rarely burn out

unless their current limit is passed.

Photodiodes are specially formulated to detect light, including infrared. This makes them useful in devices such as remote controls, robotics and burglar alarms.

Diodes have many other purposes, such as voltage regulation, current regulation, voltage multiplication and surge protection. They are used as gates in logic circuits, where they perform "OR" and "AND" operations. They are also used for Temperature Sensing, Signal demodulation, Radio receiver and Random noise generators.

6. Explain the purpose of rectificationThere are two types of rectification, namely half wave and full wave rectification;In half wave rectification, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one half of the input waveform reaches the output, it is very inefficient if used for power transfer. Half-wave rectification can be achieved with a single diode in a one-phase supply, or with three diodes in a three-phase supply.

Figure 17: Half wave rectification using a single diode

A full-wave rectifier converts the whole of the input waveform to one of constant polarity (positive or negative) at its output. Full-wave rectification converts both polarities of the input waveform to DC (direct current), and is more efficient. However, in a circuit with a non-center tapped transformer, four diodes are required instead of the one needed for half-wave rectification. Four diodes arranged this way are called a diode bridge or bridge rectifier.

Figure 18: Full-wave rectification using 4 diodes.

The primary application of rectification is to derive DC power from an AC supply. Virtually all electronic devices require DC, so rectifiers find uses inside the power supplies of virtually all electronic equipment.

Converting DC power from one voltage to another is much more complicated. One method of DC-to-DC conversion first converts power to AC (using a device called an inverter), then use a transformer to change the voltage, and finally rectifies power back to DC.

Rectification also finds a use in detection of amplitude modulated radio signals. The signal may be amplified before detection, but if un-amplified, a very low

http://en.wikipedia.org/wiki/Three-phase

http://en.wikipedia.org/wiki/File:Gratz.rectifier.en.svg

http://en.wikipedia.org/wiki/File:Gratz.rectifier.en.svg

voltage drop diode must be used. When using a rectifier for demodulation the capacitor and load resistance must be carefully matched. Too low a capacitance will result in the high frequency carrier passing to the output and too high will result in the capacitor just charging and staying charged.

Rectification is also used to supply polarised voltage for welding. In such circuits control of the output current is required and this is sometimes achieved by replacing some of the diodes in bridge rectifier with thyristors, whose voltage output can be regulated by means of phase fired controllers.

7. Explain the process of smoothing when applied to a rectifier circuit.

Figure 19

The output waveform in second figure above shows how the smoothing process works. During the first half of the voltage peaks from the rectifier, when the voltage increases, the capacitor charges up. Then, while the voltage decreases to zero in the second half of the peaks, the capacitor releases its stored energy to

keep the output voltage as constant as possible. Such a capacitor is called a 'smoothing' or 'reservoir' capacitor when it is used in this application.

The smoothing capacitor converts the full-wave rippled output of the rectifier into a smooth DC output voltage. Generally for DC power supply circuits the smoothing capacitor is an Aluminium Electrolytic type that has a capacitance value of 100uF or more with repeated DC voltage pulses from the rectifier charging up the capacitor to peak voltage. However, there are two important parameters to consider when choosing a suitable smoothing capacitor and these are its Working Voltage, which must be higher than the no-load output value of the rectifier and its Capacitance Value, which determines the amount of ripple that will appear superimposed on top of the DC voltage. Too low a value and the capacitor has little effect but if the smoothing capacitor is large enough (parallel capacitors can be used) and the load current is not too large, the output voltage will be almost as smooth as pure DC.

8. Describe how a transistor operates in terms of collector current and base current.

Figure 20

The base-emitter junction behaves like a diode. A base current IB flows only when the voltage VBE across the base-emitter

junction is 0.7V or more. The small base current IB controls the large collector current Ic. Ic = hFE × IB (unless the transistor is full on and saturated)

hFE is the current gain (strictly the DC current gain), a typical value for hFE is

100 (it has no units because it is a ratio) A value of 40 was obtained in experiment 2.1

The collector-emitter resistance RCE is controlled by the base current IB: o IB = 0 RCE = infinity transistor off o IB small RCE reduced transistor partly on o IB increased RCE = 0 transistor full on ('saturated')

A transistor that is full on (with RCE = 0) is said to be 'saturated'. When a transistor is saturated the collector-emitter voltage VCE is reduced to

almost 0V. When a transistor is saturated the collector current Ic is determined by the

supply voltage and the external resistance in the collector circuit, not by the transistor's current gain. As a result the ratio Ic/IB for a saturated transistor is less than the current gain hFE.

The emitter current IE = Ic + IB, but Ic is much larger than IB, so roughly IE =

Ic.

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semiconductors and amplifiers circuits lab report

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