I. OBJECTIVES a. Collect current-voltage ( I-V ) data for one resistor and several light emitting diodes b. Plot I versus V for the resistor and the diodes c. Compare the I-V plot of the resistor with those of the diodes
II. EQUIPMENT
Resistor, diodes, 10 volt DC power supply, voltmeter, ammeter, connecting wires
III. INTRODUCTION
An incandescent lamp consists of a metal filament made out of tungsten which is heated to a high temperature by an electric current that passes through it. The filament is enclosed in an evacuated glass envelope to avoid oxidation of the filament by the surrounding air. Most of the light emitted by an incandescent lamp is in the infrared and therefore is not visible. Incandescent lamps are inefficient light sources because only a small part of the electrical power they consume is converted into visible light. Light emitting diodes (LEDs) which are made out of semiconductor materials such as gallium arsenide (GaAs )are much more efficient light sources.
Below we describe briefly the operating principle of these diodes. Pure semiconductors (also called “intrinsic”) do not contact electricity at room temperature and are thus not useful for the fabrication of electric and electronic devices.
Semiconductors conduct electricity if we dope them with a small amount of impurity atoms. In the case of Silicon (Si) these would be any atom other than Si. Impurities are divided into two broad categories: Donors and Acceptors. A donor is an impurity atom that has more electrons in its outermost shell that is necessary to form chemical bonds with their neighboring atoms. Each Silicon atom has four electrons in its outermost shell; these electrons are involved in forming the chemical bonds with the neighboring Si atoms. An Arsenic (As) atom on the other hand has 5 electrons in its outermost shell. If we add it as an impurity silicon it will act as a donor because it has one more electron than the necessary 4 to form chemical bonds with its nearest Si atom neighbors. The extra electron on As can easily be removed from the As donor on which it was originally localized. If this happens, the electron can move freely in the semiconductor material and conduct electric current. This is shown in fig.1a. A semiconductor doped with donors is known as n-type material. An acceptor is an impurity atom that has in its outermost shell fewer electrons than needed for the formation of chemical bonds with its neighbors.
A boron (B) atom has three electrons in its outermost shell. If we add it as an impurity in Si it will act as an acceptor because it has one electron less than the necessary 4 to form chemical bonds with its nearest Si neighbors. The missing electron in the chemical bond is called a hole . It too can be easily removed from the vicinity of the B atom and move freely in the semiconductor material. This is illustrated in fig.1b. A semiconductor doped with acceptors is known as p-type material.
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A diode is formed if we place a slab of a p-type semiconductor in close contact with an n-type slab made of the same material. If we connect the p-type section with the positive terminal of a battery and the n-type section with the negative terminal we say that the diode is forward biased (see fig.2a). A current flows through a forward biased diode because the electrons in the n-type section are attracted by the positive battery terminal. In a similar fashion the holes in the p-type section are attracted by the negative battery terminal. The electrons and holes encounter each other at the junction of the ptype and n-type slabs. One electron and one hole destroy each other (the technical term is: annihilate ) in a process called recombination and at the same time a new particle known as the photon is created. The electron and hole energy is carried away by the newly created photon. Light in general can be thought of a stream of photons. These particles have the following exotic properties: a) they have zero mass b) they travel at the speed of light c
  8
3 10 m/s .
If we connect the p-type section of the diode with the negative battery terminal and the n-type section with the positive terminal as shown in fig.2b, no current flows through the diode and no light is emitted. The reason is that the negative battery terminal repels the electrons in the n-type section of the diode; at the same time the positive battery terminal does repels the holes in the p-type section. Thus no electrons or holes cross the junction (also known as p-n junction ). The diode is said to be reversed biased .
IV. EXPERIMENTAL METHOD
A schematic of the circuit used in this experiment is shown in fig. 3. A realistic picture of the circuit is shown in fig.4. Here A, B, and C represent the banana posts used to hook up the voltmeter, the ammeter, and the power supply using wires with banana terminals on both ends. A variable voltage DC power supply is connected between points A and C .
The voltmeter is connected between points A and B and measures the voltage V across the diode. The diode is also connected between points A and B. The ammeter is connected between points B and C and measures the current I that flows through the diode. The voltage V is set to the desired value using the voltage knob of the power supply. The I-V plot for a resistor is shown in fig.5a. The points lie on a straight line that passes through the origin. The resistance R is defined as the ratio R

V
I
. For the resistor the value of R is constant and does not depend on the value of the voltage or its polarity. Such conductors are known as Ohmic . The I-V plot for a diode is given in fig.5b. For voltages between 0 and a threshold value V th
the current is zero even though the diode is forward biased. Current starts to flow through the resistor when V exceeds
V th
. For V

V th
the current increases steeply with increasing voltage V . For V

0
(reverse bias) the current through the diode is practically zero. By looking at the diode I-
V plot of fig.5b one can see that the ratio R

V
I property is called non-Ohmic .
is not constant. A conductor with this
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Note: The diode voltage must be kept below a maximum value V max otherwise excessive current will pass through the diode and damage the device. The maximum voltage for each diode is given in table 1.
V. PROCEDURE.
Built the circuit shown in fig. 3. Do not turn the power on till the instructor checks your circuit for errors. You can take data after the instructor gives you the OK.
V1.
(Resistor). Use voltage values from 0 to 2.0 volts in steps of 0.2 volts.
( V = 0, 0.2, 0.4, 0.6 , …, 2.0 volts). The voltage V is given by the voltmeter and the current I by the ammeter in the circuit. Record the values for each pair of V and I in your data tables.
V2.
(Red LED). Use voltage values from 0 to 2.0 volts
From 0 to 1.7 volts use steps of 0.1 volts. ( V = 0, 0.1, 0.2, 0.3, …, 1.7 volts)
From 1.7 to 2.0 volts use steps of 0.05 volts ( V = 1.7 , 1.75, 1.80, 1.85, …, 2.0 volts)
Record the values for each pair of V and I in your data tables. Note the voltage V R e
  at which the red LED starts emitting light.
V3.
(Yellow LED). Use voltage values from 0 to 2.2 volts
From 0 to 1.8 volts use steps of 0.1 volts. ( V = 0, 0.1, 0.2, 0.3, …, 1.8 volts)
From 1.8 to 2.2 volts use steps of 0.05 volts ( V = 1.80 , 1.75, 1.80, 1.85, …, 2.2 volts)
Record the values for each pair of V and I in your data tables. Note the voltage V Y e
  at which the yellow LED starts emitting light.
V4.
(Green LED). Use voltage values from 0 to 2.3 volts
From 0 to 1.8 volts use steps of 0.1 volts. ( V = 0, 0.1, 0.2, 0.3, …, 1.8 volts)
From 1.8 to 2.3 volts use steps of 0.05 volts ( V = 1.80 , 1.75, 1.80, 1.85, …, 2.3 volts)
Record the values for each pair of V and I in your data tables. Note the voltage V G e
  at which the green LED starts emitting light.
V5.
(Blue LED). Use voltage values from 0 to 4.0 volts. Change the voltage V from 0 to 4 volts in steps of 0.1 volts ( V = 0, 0.1, 0.2, 0.3, … 4.0 volts)
Record the values for each pair of V and I in your data tables. Note the voltage V B e
  at which the blue LED starts emitting light.
VI. FOR THE REPORT
VI-1.
Tabulate the data from section V1. Plot I versus V for the resistor. Determine the resistance R V I for V = 0.5, 1.0, and 1.5 volts.
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VI-2.
Tabulate the data from section V2. Plot I versus V for the red LED. Determine the resistance for V = 1.0, 1.7 and 1.9 volts. Mark V R e
 
on the plot.
VI-3.
Tabulate the data from section V3. Plot I versus V for the yellow LED. Determine the resistance for V = 1.0, 1.9 and 2.1 volts. Mark V Y e
 
on the plot.
VI-4.
Tabulate the data from section V4. Plot I versus V for the green LED.
Determine the resistance for V = 1.5, 2.0 and 2.2 volts. Mark V G e
 
on the plot.
VI-5.
Tabulate the data from section V5. Plot I versus V for the blue LED. Determine the resistance for V = 2.0, 3.3 and 3.8 volts. Mark V B e
 
on the plot.
VII QUESTIONS
VII 1.
Are the LEDs Ohmic or non-Ohmic conductors? Explain
VII 2 . Instead of a constant (DC ) voltage assume that we use in the circuit of fig.3 the time-dependent voltage shown in fig.6a. Draw the current I through the blue LED in fig.6b.
V
+3V t
Fig.6a
-3V
I
Fig.6b t
VII 3.
The amount of energy consumed per second (also known as the power P ) is given by the equation: P

IV . Calculate the maximum power consumed by the resistor and each of the four LEDs. The electric energy consumed is converted into a different form of energy. Try to guess what is this form for the resistor and the LEDs.
Note: If the voltage V is measured in volts and the current I in amperes, the power units come out in Watts .
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Table I : Maximum allowed voltages for the LEDs used in this experiment
LED V max
(volts)
Red 2.0
Yellow 2.2
Green 2.3
Blue 4.0
Fig.1a: Silicon doped with Arsenic donors (from J.Singh’s book: “Semiconductor devices, McGraw Hill)
Fig.1b. Silicon doped with Boron acceptors (from J.Singh’s book “Semiconductor devices, McGraw Hill)
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p n
LED
V
+ p n
I
+ -
V
Fig.2a. Forward biased diode
I = 0
- +
V
Fig.2b. Reverse biased diode
V
+
+
A
I
Fig.3
A
Voltmeter
LED
B
Ammeter
DC power
Fig.4
C
6

I
V
I
V
V max
V th
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