Semiconductor Electronics: Materials, Devices & Simple Circuits, Study notes of Physics

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Chapter - 14 Semiconductor Electronics: Materials, Devices

and Simple Circuits

Introduction

Devices in which a controlled flow of electrons can be obtained are the basic building blocks of all the electronic circuits. Before

the discovery of transistor in 1948, such devices were mostly vacuum tubes (also called valves) like the vacuum diode which has

two electrodes, viz., anode (often called plate) and cathode; triode which has three electrodes – cathode, plate and grid; tetrode

and pentode (respectively with 4 and 5 electrodes).

The seed of the development of modern solid-state semiconductor electronics goes back to 1930’s when it was realised that some

solid-state semiconductors and their junctions offer the possibility of controlling the number and the direction of flow of charge

carriers through them. Simple excitations like light, heat or small applied voltage can change the number of mobile charges in a

semiconductor. Note that the supply and flow of charge carriers in the semiconductor devices are within the solid itself, while in

the earlier vacuum tubes/valves, the mobile electrons were obtained from a heated cathode and they were made to flow in an

evacuated space or vacuum. No external heating or large evacuated space is required by the semiconductor devices. They are

small in size, consume low power, operate at low voltages and have long life and high reliability. Even the Cathode Ray Tubes

(CRT) used in television and computer monitors which work on the principle of vacuum tubes are being replaced by Liquid Crystal

Display (LCD) monitors with supporting solid state electronics.

Classification of Metals, Conductors and Semiconductors

On the basis of conductivity

On the basis of the relative values of electrical conductivity (σ) or resistivity (ρ = 1 /σ), the solids are broadly classified as:

(i) Metals: They possess very low resistivity (or high conductivity).

ρ ∼ 10

− 2

− 10

8

S/m

σ ∼ 10

2

− 10

− 8

Ωm

(ii) Semiconductors: They have resistivity or conductivity intermediate to metals and insulators.

ρ ∼ 10

− 5

− 10

− 6

S/m

σ ∼ 10

5

− 10

6

Ωm

(iii) Insulators: They have high resistivity (or low conductivity).

ρ ∼ 10

− 11

− 10

− 19

S/m

σ ∼ 10

− 11

− 10

− 19

Ωm

The values of ρ and σ given above are indicative of magnitude and could well go outside the ranges as well. Relative values of the

resistivity are not the only criteria for distinguishing metals, insulators and semiconductors from each other. There are some

other differences, which will become clear as we go along in this chapter.

Our interest in this chapter is in the study of semiconductors which could be:

(i) Elemental semiconductors: Si and Ge

(ii) Compound semiconductors:

Examples are:

• Inorganic: CdS, GaAs, CdSe, InP, etc.
• Organic: anthracene, doped pthalocyanines, etc.
• Organic polymers: polypyrrole, polyaniline, polythiophene, etc.

Most of the currently available semiconductor devices are based on elemental semiconductors Si or Ge and compound inorganic

semiconductors. However, after 1990, a few semiconductor devices using organic semiconductors and semiconducting polymers

have been developed signaling the birth of futuristic technology of polymer electronics and molecular electronics. In this chapter,

we will restrict ourselves to the study of inorganic semiconductors, particularly elemental semiconductors Si and Ge. The general

concepts introduced here for discussing the elemental semiconductors, by-and-large, apply to most of the compound

semiconductors as well.

On the basis of energy bands

According to the Bohr atomic model, in an isolated atom the energy of any of its electrons is decided by the orbit in which it

revolves. But when the atoms come together to form a solid, they are close to each other. So, the outer orbits of electrons from

neighbouring atoms would come very close or could even overlap. This would make the nature of electron motion in a solid very

different from that in an isolated atom.

Inside the crystal each electron has a unique position, and no two electrons see exactly the same pattern of surrounding charges.

Because of this, each electron will have a different energy level. These different energy levels with continuous energy variation

form what are called energy bands. The energy band which includes the energy levels of the valence electrons is called the valence

band. The energy band above the valence band is called the conduction band. With no external energy, all the valence electrons

will reside in the valence band. If the lowest level in the conduction band happens to be lower than the highest level of the valence

band, the electrons from the valence band can easily move into the conduction band. Normally the conduction band is empty. But

when it overlaps on the valence band electrons can move freely into it. This is the case with metallic conductors.

If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no

free electrons are available in the conduction band. This makes the material an insulator. But some of the electrons from the

valence band may gain external energy to cross the gap between the conduction band and the valence band. Then these electrons

will move into the conduction band. At the same time, they will create vacant energy levels in the valence band where other

valence electrons can move. Thus the process creates the possibility of conduction due to electrons in conduction band as well as

due to vacancies in the valence band.

Let us consider what happens in the case of Si or Ge crystal containing N atoms. For Si, the outermost orbit is the third orbit (

n = 3 ), while for Ge it is the fourth orbit (n = 4 ). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence,

the total number of outer electrons in the crystal is 4N. The maximum possible number of electrons in the outer orbit is 8 ( 2s +

6p electrons). So, for the 4N valence electrons there are 8N available energy states. These 8N discrete energy levels can either

form a continuous band or they may be grouped in different bands depending upon the distance between the atoms in the crystal

(see box on Band Theory of Solids).

At the distance between the atoms in the crystal lattices of Si and Ge, the energy band of these 8N states is split apart into two

which are separated by an energy gap E g

(Fig. 14.1). The lower band which is completely occupied by the 4N valence electrons at

temperature of absolute zero is the valence band. The other band consisting of 4N energy states, called the conduction band, is

completely empty at absolute zero.

FIGURE 14.1 The energy band positions in a semiconductor at 0 K. The upper band, called the conduction band, consists of

infinitely large number of closely spaced energy states. The lower band, called the valence band. consists of closely spaced

completely filled energy states.

The lowest energy level in the conduction band is shown as E C

and highest energy level in the valence band is shown as E

V

. Above

E

C

and below E

V

there are a large number of closely spaced energy levels, as shown in Fig. 14.1. The gap between the top of the

valence band and bottom of the conduction band is called the energy band gap (Energy gap E g

). It may be large, small, or zero,

depending upon the material. These different situations, are depicted in Fig. 14.2 and discussed below:

Case I : This refers to a situation, as shown in Fig. 14.2(a). One can have a metal either when the conduction band is partially filled

and the balanced band is partially empty or when the conduction and valance bands overlap. When there is overlap electrons

from valence band can easily move into the conduction band. This situation makes a large number of electrons available for

electrical conduction. When the valence band is partially empty, electrons from its lower level can move to higher level making

conduction possible. Therefore, the resistance of such materials is low or the conductivity is high.

Fig. 14.2 Difference between energy bands of (a) metals. (b) insulators and (c) semiconductors.

Case II: In this case, as shown in Fig. 14. 2 (b), a large band gap E g

exists (E

g

3eV). There are no electrons in the conduction

band, and therefore no electrical conduction is possible. Note that the energy gap is so large that electrons cannot be excited from

the valence band to the conduction band by thermal excitation. This is the case of insulators.

Case III: This situation is shown in Fig. 14.2(c). Here a finite but small band gap (E g

< 3eV) exists. Because of the small band gap,

at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the

conduction band. These electrons (though small in numbers) can move in the conduction band. Hence, the resistance of

semiconductors is not as high as that of the insulators.

In this section we have made a broad classification of metals, conductors and semiconductors. In the section which follows you

will learn the conduction process in semiconductors.

Intrinsic Semiconductor

We shall take the most common case of Ge and Si whose lattice structure is shown in Fig. 14.3.

FIGURE 14.3 Three-dimensional diamond-like crystal structure for Carbon. Silicon or Germanium with respective lattice

spacing an equal to 3.56 ,5 .43 and 5.66 AA.

These structures are called the diamond-like structures. Each atom is surrounded by four nearest neighbours. We know that Si

and Ge have four valence electrons. In its crystalline structure, every Si or Ge atom tends to share one of its four valence electrons

with each of its four nearest neighbour atoms, and also to take share of one electron from each such neighbour. These shared

electron pairs are referred to as forming a covalent bond or simply a valence bond. The two shared electrons can be assumed to

shuttle back-and-forth between the associated atoms holding them together strongly.

FIGURE 14.4 Schematic two-dimensional representation of Si or Ge structure showing covalent bonds at low

temperature (all bonds intact). +4 symbol indicates inner cores of Si or Ge.

Figure 14.4 schematically shows the 2-dimensional representation of Si or Ge structure shown in Fig. 14.3 which overemphasises

the covalent bond. It shows an idealised picture in which no bonds are broken (all bonds are intact). Such a situation arises at low

temperatures. As the temperature increases, more thermal energy becomes available to these electrons and some of these

electrons may break-away (becoming free electrons contributing to conduction). The thermal energy effectively ionises only a

few atoms in the crystalline lattice and creates a vacancy in the bond as shown in Fig. 14.5(a). The neighbourhood, from which

the free electron (with charge −q ) has come out leaves a vacancy with an effective charge (+q). This vacancy with the effective

positive electronic charge is called a hole. The hole behaves as an apparent free particle with effective positive charge.

In intrinsic semiconductors, the number of free electrons, n e

is equal to the number of holes, n h

. That is

n e

= n

h

= n

i

where n i

is called intrinsic carrier concentration.

Semiconductors possess the unique property in which, apart from electrons, the holes also move. Suppose there is a hole at site

1 as shown in Fig. 14.5(a). The movement of holes can be visualised as shown in Fig. 14.5(b).

FIGURE 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron due to thermal energy at

moderate temperatures. (b) Simplified representation of possible thermal motion of a hole. The electron from the lower

left hand covalent bond (site 2) goes to the earlier hole site1. leaving a hole at its site indicating an apparent movement

of the hole from site 1 to site 2.

An electron from the covalent bond at site 2 may jump to the vacant site 1 (hole). Thus, after such a jump, the hole is at site 2 and

the site 1 has now an electron. Therefore, apparently, the hole has moved from site 1 to site 2. Note that the electron originally set

free [Fig. 14.5(a)] is not involved in this process of hole motion. The free electron moves completely independently as conduction

electron and gives rise to an electron current, I e

under an applied electric field. Remember that the motion of hole is only a

convenient way of describing the actual motion of bound electrons, whenever there is an empty bond anywhere in the crystal.

Under the action of an electric field, these holes move towards negative potential giving the hole current, I h

. The total current, I

is thus the sum of the electron current I e

and the hole current I

h

:

I = I

e

+ I

h

It may be noted that apart from the process of generation of conduction electrons and holes, a simultaneous process of

recombination occurs in which the electrons recombine with the holes. At equilibrium, the rate of generation is equal to the rate

of recombination of charge carriers. The recombination occurs due to an electron colliding with a hole.

An intrinsic semiconductor will behave like an insulator at T = 0 K as shown in Fig. 14.6(a).

FIGURE 14.6 (a) An intrinsic semiconductor at T=0 K behaves like an insulator. (b) At T>O K. four thermally generated

electron-hole pairs. The filled circles (•) represent electrons and empty circles (O) represent holes.

It is the thermal energy at higher temperatures (T > 0K), which excites some electrons from the valence band to the conduction

band. These thermally excited electrons at T > 0 K, partially occupy the conduction band. Therefore, the energy-band diagram of

an intrinsic semiconductor will be as shown in Fig. 14.6(b). Here, some electrons are shown in the conduction band. These have

come from the valence band leaving equal number of holes there.

Extrinsic Semiconductor

The conductivity of an intrinsic semiconductor depends on its temperature, but at room temperature its conductivity is very low.

As such, no important electronic devices can be developed using these semiconductors. Hence there is a necessity of improving

their conductivity. This can be done by making use of impurities.

When a small amount, say, a few parts per million (ppm), of a suitable impurity is added to the pure semiconductor, the

conductivity of the semiconductor is increased manifold. Such materials are known as extrinsic semiconductors or impurity

semiconductors. The deliberate addition of a desirable impurity is called doping and the impurity atoms are called dopants. Such

a material is also called a doped semiconductor. The dopant has to be such that it does not distort the original pure semiconductor

lattice. It occupies only a very few of the original semiconductor atom sites in the crystal. A necessary condition to attain this is

that the sizes of the dopant and the semiconductor atoms should be nearly the same.

There are two types of dopants used in doping the tetravalent Si or Ge:

(i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous (P), etc.

(ii) Trivalent (valency 3); like Indium (In), Boron (B), Aluminum (Al), etc.

We shall now discuss how the doping changes the number of charge carriers (and hence the conductivity) of semiconductors. Si

or Ge belongs to the fourth group in the Periodic table and, therefore, we choose the dopant element from nearby fifth or third

group, expecting and taking care that the size of the dopant atom is nearly the same as that of Si or Ge. Interestingly, the

pentavalent and trivalent dopants in Si or Ge give two entirely different types of semiconductors as discussed below.

(i) n-type semiconductor

Suppose we dope Si or Ge with a pentavalent element as shown in Fig. 14.7.

FIGURE 14.7 (a) Pentavalent donor atom (As, Sb. P. etc.) doped for tetravalent Si or Ge giving n type semiconductor, and

(b) Commonly used schematic representation of n-type material which shows only the fixed cores of the substituent

donors with one additional effective positive charge and its associated extra electron.

When an atom of +5 valency element occupies the position of an atom in the crystal lattice of Si, four of its electrons bond with

the four silicon neighbours while the fifth remains very weakly bound to its parent atom. This is because the four electrons

participating in bonding are seen as part of the effective core of the atom by the fifth electron. As a result, the ionisation energy

required to set this electron free is very small and even at room temperature it will be free to move in the lattice of the

semiconductor. For example, the energy required is ∼ 0 .01eV for germanium, and 0 .05eV for silicon, to separate this electron

from its atom. This is in contrast to the energy required to jump the forbidden band (about 0 .72eV for germanium and about

1 .1eV for silicon) at room temperature in the intrinsic semiconductor. Thus, the pentavalent dopant is donating one extra electron

for conduction and hence is known as donor impurity. The number of electrons made available for conduction by dopant atoms

depends strongly upon the doping level and is independent of any increase in ambient temperature. On the other hand, the

number of free electrons (with an equal number of holes) generated by Si atoms, increases weakly with temperature.

In a doped semiconductor the total number of conduction electrons n e

is due to the electrons contributed by donors and those

generated intrinsically, while the total number of holes n h

is only due to the holes from the intrinsic source. But the rate of

recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get

reduced further.

Thus, with proper level of doping the number of conduction electrons can be made much larger than the number of holes. Hence

in an extrinsic semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority

carriers. These semiconductors are, therefore, known as n-type semiconductors. For n-type semiconductors, we have, n e

≫ n

h

(ii) p-type semiconductor

This is obtained when Si or Ge is doped with a trivalent impurity like Al, B, In, etc. The dopant has one valence electron less than

Si or Ge and, therefore, this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to

offer to the fourth Si atom. So the bond between the fourth neighbour and the trivalent atom has a vacancy or hole as shown in

Fig. 14.8.

FIGURE 14.8 (a) Trivalent acceptor atom (In, Al, B etc.) doped in tetravalent Si or Ge lattice giving p-type semiconductor.

(b) Commonly used schematic representation of p-type material which shows only the fixed core of the substituent

acceptor with one effective additional negative charge and its associated hole.

Since the neighbouring Si atom in the lattice wants an electron in place of a hole, an electron in the outer orbit of an atom in the

neighbourhood may jump to fill this vacancy, leaving a vacancy or hole at its own site. Thus the hole is available for conduction.

Note that the trivalent foreign atom becomes effectively negatively charged when it shares fourth electron with neighbouring Si

atom. Therefore, the dopant atom of p-type material can be treated as core of one negative charge along with its associated hole

as shown in Fig. 14.8(b). It is obvious that one acceptor atom gives one hole. These holes are in addition to the intrinsically

generated holes while the source of conduction electrons is only intrinsic generation. Thus, for such a material, the holes are the

majority carriers and electrons are minority carriers. Therefore, extrinsic semiconductors doped with trivalent impurity are

called p-type semiconductors. For p-type semiconductors, the recombination process will reduce the number

(

n

i

)

of intrinsically

generated electrons to n e

. We have, for p-type semiconductors n

h

n

e

Note that the crystal maintains an overall charge neutrality as the charge of additional charge carriers is just equal and opposite

to that of the ionised cores in the lattice.

In extrinsic semiconductors, because of the abundance of majority current carriers, the minority carriers produced thermally

have more chance of meeting majority carriers and thus getting destroyed. Hence, the dopant, by adding a large number of current

carriers of one type, which become the majority carriers, indirectly helps to reduce the intrinsic concentration of minority

carriers.

The semiconductor's energy band structure is affected by doping. In the case of extrinsic semiconductors, additional energy states

due to donor impurities (E D

) and acceptor impurities (E

A

) also exist. In the energy band diagram of n-type Si semiconductor, the

donor energy level E D

is slightly below the bottom E

C

of the conduction band and electrons from this level move into the

conduction band with very small supply of energy. At room temperature, most of the donor atoms get ionised but very few

(∼ 10

12

) atoms of Si get ionised. So the conduction band will have most electrons coming from the donor impurities, as shown in

Fig. 14.9(a).

FIGURE 14.9 Energy bands of (a) n-type semiconductor at T>0 K, (b) p-type semiconductor at T>0 O.

Similarly, for p-type semiconductor, the acceptor energy level E A

is slightly above the top E

V

of the valence band as shown in Fig.

14.9(b). With very small supply of energy an electron from the valence band can jump to the level E A

and ionise the acceptor

negatively. (Alternately, we can also say that with very small supply of energy the hole from level E A

sinks down into the valence

band. Electrons rise up and holes fall down when they gain external energy.) At room temperature, most of the acceptor atoms

get ionised leaving holes in the valence band. Thus, at room temperature the density of holes in the valence band is predominantly

due to impurity in the extrinsic semiconductor. The electron and hole concentration in a semiconductor in thermal equilibrium

is given by

n e

n

h

= n

i

2

Though the above description is grossly approximate and hypothetical, it helps in understanding the difference between metals,

insulators and semiconductors (extrinsic and intrinsic) in a simple manner. The difference in the resistivity of C, Si and Ge depends

upon the energy gap between their conduction and valence bands. For C (diamond), Si and Ge, the energy gaps are 5 .4eV, 1 .1eV

and 0 .7eV, respectively. Sn also is a group IV element but it is a metal because the energy gap in its case is 0eV.

p-n Junction

A p-n junction is the basic building block of many semiconductor devices like diodes, transistor, etc. A clear understanding of the

junction behaviour is important to analyse the working of other semiconductor devices. We will now try to understand how a

junction is formed and how the junction behaves under the influence of external applied voltage (also called bias).

p-n junction formation

Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely, a small quantity of pentavalent impurity, part of

the p-Si wafer can be converted into n-Si. There are several processes by which a semiconductor can be formed. The wafer now

contains p-region and n-region and a metallurgical junction between p-, and n- region.

Two important processes occur during the formation of a p-n junction: diffusion and drift. We know that in an n-type

semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of

holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. During the

formation of p-n junction, and due to the concentration gradient across p-, and n - sides, holes diffuse from p-side to n-side (p →

n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carries gives rise to diffusion current across the

junction.

When an electron diffuses from n → p, it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is

immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n → p, a layer of positive charge (or

positive space-charge region) on n-side of the junction is developed.

Similarly, when a hole diffuses from p → n due to the concentration gradient, it leaves behind an ionised acceptor (negative

charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-

side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as

the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges (Fig. 14.10).

The thickness of depletion region is of the order of one-tenth of a micrometer. Due to the positive space-charge region on n-side

of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards

negative charge develops. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction

moves to pside. The motion of charge carriers due to the electric field is called drift. Thus, a drift current, which is opposite in

direction to the diffusion current (Fig. 14.10) starts.

FIGURE 14.10 p-n junction formation process.

Initially, diffusion current is large and drift current is small. As the diffusion process continues, the space-charge regions on either

side of the junction extend, thus increasing the electric field strength and hence drift current. This process continues until the

diffusion current equals the drift current. Thus, a p-n junction is formed. In a p-n junction under equilibrium there is no net

current.

The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction

of the two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium

exists.

FIGURE 14.11 (a) Diode under equilibrium ( 𝐕 = 𝟎 ). (b) Barrier potential under no bias.

Figure 14.11 shows the p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and p

material has acquired electrons. The n material is thus positive relative to the p material. Since this potential tends to prevent the

movement of electrons from the n region into the p region, it is often called a barrier potential.

Semiconductor Diode

A semiconductor diode [Fig. 14.12(a)] is basically a p-n junction with metallic contacts provided at the ends for the application

of an external voltage. It is a two-terminal device. A p-n junction diode is symbolically represented as shown in Fig. 14.12(b).

FIGURE 14.12 (a) Semiconductor diode, (b) Symbol for p-n junction diode.

The direction of arrow indicates the conventional direction of current (when the diode is under forward bias). The equilibrium

barrier potential can be altered by applying an external voltage V across the diode. The situation of p-n junction diode under

equilibrium (without bias) is shown in Fig. 14.11(a) and (b).

p-n junction diode under forward bias

When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the

battery and n-side to the negative terminal [Fig. 14.13(a)], it is said to be forward biased. The applied voltage mostly drops across

the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. (This is because the resistance

of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side and p-side.) The

direction of the applied voltage (V) is opposite to the built-in potential V 0

. As a result, the depletion layer width decreases and the

barrier height is reduced [Fig. 14.13(b)].

FIGURE 14.13 (a) p-n junction diode under forward bias, (b) Barrier potential (1) without battery, (2) Low battery

voltage, and (3) High voltage battery.

The effective barrier height under forward bias is (V 0

− V).

If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, and only a small

number of carriers in the material-those that happen to be in the uppermost energy levels-will possess enough energy to cross

the junction. So the current will be small. If we increase the applied voltage significantly, the barrier height will be reduced and

more number of carriers will have the required energy. Thus the current increases.

Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries).

Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under forward

bias is known as minority carrier injection. At the junction boundary, on each side, the minority carrier concentration increases

significantly compared to the locations far from the junction.

Due to this concentration gradient, the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-

side.

Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. 14.14).

Fig. 14.14 Forward bias minority carrier injection.

This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion

current and conventional current due to electron diffusion. The magnitude of this current is usually in mA.

p-n junction diode under reverse bias

When an external voltage (V) is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse

biased [Fig.14.15(a)]. The applied voltage mostly drops across the depletion region. The direction of applied voltage is same as

the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in

the electric field. The effective barrier height under reverse bias is (V 0

• V), [Fig. 14.15(b)].

Fig. 14.15 (a) Diode under reverse bias, (b) Barrier potential under reverse bias.

This suppresses the flow of electrons from n → p and holes from p → n. Thus, diffusion current, decreases enormously compared

to the diode under forward bias.

The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close

to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current. The drift current is of the order

of a few μA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the

junction. The drift current is also there under forward bias but it is negligible (μA) when compared with current due to injected

carriers which is usually in mA.

The diode reverse current is not very much dependent on the applied voltage. Even a small voltage is sufficient to sweep the

minority carriers from one side of the junction to the other side of the junction. The current is not limited by the magnitude of

the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. independent up

to a critical reverse bias voltage, known as breakdown voltage (V br

). When V = V

br

, the diode reverses current increases sharply.

Even a slight increase in the bias voltage causes large change in the current. If the reverse current is not limited by an external

circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed. Once it exceeds the rated value,

the diode gets destroyed due to overheating. This can happen even for the diode under forward bias, if the forward current

exceeds the rated value.

The circuit arrangement for studying the V - I characteristics of a diode, (i.e., the variation of current as a function of applied

voltage) are shown in Fig. 14.16(a) and (b). The battery is connected to the diode through a potentiometer (or reheostat) so that

the applied voltage to the diode can be changed. For different values of voltages, the value of the current is noted. A graph between

V and I is obtained as in Fig. 14.16(c).

FIGURE 14.16 Experimental circuit arrangement for studying V-I characteristics of a p-n junction diode (a) in forward

bias, (b) in reverse bias. (c) Typical V- 1 characteristics of a silicon diode.

Note that in forward bias measurement, we use a milliammeter since the expected current is large (as explained in the earlier

section) while a micrometer is used in reverse bias to measure the current. You can see in Fig. 14.16(c) that in forward bias, the

current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After the

characteristic voltage, the diode current increases significantly (exponentially), even for a very small increase in the diode bias

voltage. This voltage is called the threshold voltage or cut-in voltage (∼ 0. 2 V for germanium diode and ∼ 0. 7 V for silicon diode).

For the diode in reverse bias, the current is very small (∼ μA) and almost remains constant with change in bias. It is called reverse

saturation current. However, for special cases, at very high reverse bias (break down voltage), the current suddenly increases.

This special action of the diode is discussed later in Section 14.8. The general-purpose diode is not used beyond the reverse

saturation current region.

The above discussion shows that the p-n junction diode primarily allows the flow of current only in one direction (forward bias).

The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages

as discussed in the next section. For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage

ΔV to a small change in current ΔI :

r d

=

ΔV

ΔI

Application of Junction Diode as a Rectifier

From the V-I characteristic of a junction diode we see that it allows current to pass only when it is forward biased. So if an

alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This

property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. If an alternating voltage is

applied across a diode in series with a load, a pulsating voltage will appear across the load only during the half cycles of the ac

input during which the diode is forward biased. Such rectifier circuit, as shown in Fig. 14.18, is called a half-wave rectifier. The

secondary of a transformer supplies the desired ac voltage across terminals A and B.

FIGURE 14.18 (a) Half-wave rectifier circuit, (b) Input ac voltage and output voltage waveforms from the rectifier circuit.

When the voltage at A is positive, the diode is forward biased and it conducts. When A is negative, the diode is reverse-biased and

it does not conduct. The reverse saturation current of a diode is negligible and can be considered equal to zero for practical

purposes. (The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of

the transformer to protect the diode from reverse breakdown.)

Therefore, in the positive half-cycle of ac there is a current through the load resistor R L

and we get an output voltage, as shown

in Fig. 14.18(b), whereas there is no current in the negative halfcycle. In the next positive half-cycle, again we get the output

voltage. Thus, the output voltage, though still varying, is restricted to only one direction and is said to be rectified. Since the

rectified output of this circuit is only for half of the input ac wave it is called as half-wave rectifier.

The circuit using two diodes, shown in Fig. 14.19(a), gives output rectified voltage corresponding to both the positive as well as

negative half of the ac cycle. Hence, it is known as full-wave rectifier. Here the p-side of the two diodes are connected to the ends

of the secondary of the transformer. The n-side of the diodes are connected together and the output is taken between this common

point of diodes and the midpoint of the secondary of the transformer. So, for a full-wave rectifier the secondary of the transformer

is provided with a centre tapping and so it is called centre-tap transformer. As can be seen from Fig.14.19(c) the voltage rectified

by each diode is only half the total secondary voltage. Each diode rectifies only for half the cycle, but the two do so for alternate

cycles. Thus, the output between their common terminals and the centre-tap of the transformer becomes a full-wave rectifier

output. (Note that there is another circuit of full wave rectifier which does not need a centre-tap transformer but needs four

diodes.) Suppose the input voltage to A with respect to the centre tap at any instant is positive. It is clear that, at that instant,

voltage at B being out of phase will be negative as shown in Fig. 14.19(b). So, diode D 1

gets forward biased and conducts (while

D

2

being reverse biased is not conducting). Hence, during this positive half cycle we get an output current (and a output voltage

across the load resistor R L

) as shown in Fig. 14.19(c).

FIGURE 14.19 (a) A Full-wave rectifier circuit; (b) Input wave forms given to the diode 𝐃 𝟏

at 𝐀 and to the diode 𝐃 𝟐

at 𝐁 :

(c) Output waveform across the load 𝐑 𝐋

connected in the full-wave rectifier circui t.

In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive.

In this part of the cycle diode D 1

would not conduct but diode D

2

would, giving an output current and output voltage (across R

L

) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as the negative half of

the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the halfwave rectifier.

The rectified voltage is in the form of pulses of the shape of half sinusoids. Though it is unidirectional it does not have a steady

value. To get steady dc output from the pulsating voltage normally a capacitor is connected across the output terminals (parallel

to the load R L

). One can also use an inductor in series with R

L

for the same purpose. Since these additional circuits appear to

filter out the ac ripple and give a pure dc voltage, so they are called filters.

Now we shall discuss the role of rectifier circuit. capacitor in filtering. When the voltage across the capacitor is rising, it gets

charged. If there is no external load, it remains charged to the peak voltage of the rectified output. When there is a load, it gets

discharged through the load and the voltage across it begins to fall. In the next half-cycle of rectified output, it again gets charged

to the peak value (Fig. 14.20).

FIGURE 14.20 (a) A full-wave rectifier with capacitor filter, (b) Input and output voltage of rectifier in (a).

The rate of fall of the voltage across the capacitor depends inversely upon the product of capacitance C and the effective resistance

R

L

used in the circuit and is called the time constant. To make the time constant large value of C should be large. So, capacitor

input filters use large capacitors. The output voltage obtained by using capacitor input filter is nearer to the peak voltage of the

rectified voltage. This type of filter is most widely used in power supplies.

Multiple Choice Question

1. What bonds are present in a semiconductor?

A. Monovalent B. Bivalent

C. Trivalent D. Covalent

Answer: D

Explanation:

Covalent bonds are present in a semiconductor.

2. The number of electrons in the valence shell of a semiconductor is

A. 1 B. 2

C. 3 D. 4

Answer: D

Explanation:

The number of electrons in the valence shell of a semiconductor is 4.

3. What happens to the forbidden energy gap of a semiconductor with the fall of temperature?

A. Decreases B. Increases

C. Unchanged D. Sometimes decreases and sometimes increases

Answer: B

Explanation:

The forbidden energy gap of a semiconductor increases with the fall of temperature.

4. In a p-type semiconductor, the current conduction is due to

A. Holes B. Atoms

C. Electrons D. Protons

Answer: A

Explanation: In a p-type semiconductor, the current conduction is due to holes.

5. What is the main function of a transistor?

A. Simplify B. Amplify

C. Rectify D. All of the above

Answer: B

Explanation: The main function of a transistor is to amplify.

6. In a semiconductor, what is responsible for conduction?

A. Electrons only B. Holes only

C. Both electrons and holes D. Neither electrons nor holes

Answer: C

Explanation: Both electrons and holes are responsible for conduction in a semiconductor.

7. What happens to the resistance of semiconductors on heating?

A. Increases B. Decreases

C. Remains the same D. First increases later decrease

Answer: B

Explanation: The resistance of semiconductors decreases on heating.

8. In intrinsic semiconductors at room temperature, the number of electrons and holes are

A. Unequal B. Equal

C. Infinite D. Zero

Answer : B

Explanation:

In intrinsic semiconductors, the number of electrons and holes are equal.

9. Which of the following is not a universal gate?

A. NOT B. AND

C. OR D. NAND

Answer : D

Explanation:

NAND is not a universal gate.

10. A p-type semiconductor is

A. Positively charged B. Negatively charged

C. Uncharged D. None of the above

Answer: C

Explanation:

A p-type semiconductor is electrically neutral that is uncharged.

11. In which range of temperature, freeze out point begins to occur?

A. Higher range B. Lower range

C. Middle range D. None

Answer: B

Explanation:

At lower range of temperature, the concentration and conductivity decrease with lowering of the temperature.

12. At what temperature the donor states are completely ionized?

A. 0 K B. ROOM

C. 300K D. 900K

Answer: B

Explanation:

At room temperature, the donors have donated their electrons to the conduction band.

13. What do you mean by the term ‘FREEZE-OUT’?

A. All the electrons are frozen at room temperature B. None of the electrons are thermally elevated to the

conduction band

C. All the electrons are in the conduction band D. All the holes are in the valence band

Answer: B

Explanation:

Freeze out means none of the electrons are transmitted to the conduction band.

14. Which of the following band is just above the intrinsic Fermi level for n-type semiconductor?

A. Donor band B. Valence band

C. Acceptor band D. Conduction band

Answer: A

Explanation:

For n-type semiconductors, the donor band is just above the intrinsic Fermi level.

15. For the below given figure, identify the correct option for satisfying the above semiconductor figure?

A. P type, A-Conduction band, B-donor energy band,

C- Valence band

B. P type, A-Conduction band, B-acceptor energy

band, C- Valence band

C. n type, A-Conduction band, B-donor energy band,

C- Valence band

D. n type, A-Conduction band, B-acceptor energy

band, C- Valence band

Answer: B

Explanation:

The given figure has B band below the intrinsic Fermi level, so that would be acceptor energy band and will be a p-type

semiconductor.

16. For which type of material, the number of free electron concentration is equal to the number of donor atoms?

A. P type semiconductor B. Metal

C. N-type semiconductor D. Insulator

Answer: C

Explanation:

The n-type semiconductor has equal concentration of free electron and donor atoms.

17. If Ef>Efi, then what is the type of the semiconductor?

A. n-type B. P-type

C. Elemental D. Compound

Answer: A

Explanation:

For n-type, the Fermi energy level is greater than the intrinsic Fermi energy level because in an energy band, Fermi level of donors

is always greater than that of the acceptors.

18. The f F

(E) decreases in which of the following band for p-type semiconductor?

A. Conduction band B. Donor band

C. Acceptor band D. Valence band

Answer: A

Explanation:

The probability of finding the electron in the conduction band decreases for a p-type semiconductor because in a p-type

semiconductor, the holes will be in conduction band rather than the electrons.

19. Which states get filled in the conduction band when the donor-type impurity is added to a crystal?

A. Na B. Nd

C. N D. P

Answer: B

Explanation:

When the donor-type impurity is added to a crystal, first Nd states get filled because it is of the highest energy.

20. When the temperature of either n-type or p-type increases, determine the movement of the position of the Fermi energy

level?

A. Towards up of energy gap B. Towards down of energy gap

C. Towards centre of energy gap D. Towards out of page

Answer: C

Explanation:

whenever the temperature increases, the Fermi energy level tends to move at the centre of the energy gap.

21. In a semiconductor which of the following carries can contribute to the current?

A. Electrons B. Holes

C. Both D. None

Answer: C

Explanation:

In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons

take part in the flow of the current.

22. The thermal equilibrium concentration of the electrons in the conduction band and the holes in the valence band depends

upon?

A. Effective density of states B. Fermi energy level

C. Both A and B D. Neither A nor B

Answer: C

Explanation:

The electrons and holes depend upon the effective density of the states and the Fermi energy level given by the formula,

n

i

2

= N

C

N

V

exp [−Eg/KT]

23. Which of the following expressions represents the correct distribution of the electrons in the conduction band? (gc(E) =

density of quantum states, fF(E) = Fermi Dirac probability

A. n(E) = gc(E) ∗ fF(E) B. n(E) = gc(−E)

∗fF(E)

C. n(E) = gc(E)

∗fF(−E)

D. n(E) = gc(−E)

∗fF(−E)

Answer: A

Explanation:

The distribution of the electrons in the conduction band is given by the product of the density into Fermi-dirac distribution.

24. What is the SI unit of conductivity?

A. Ωm B. (Ωm) − 1

C. Ω D. m

Answer: B

Explanation:

The formula of the conductivity is the σ = 1 /ρ.

So, the unit of resistivity is Ωm.

Now, the unit of conductivity becomes the inverse of resistivity.

25. What is the voltage difference if the current is 1 mA and length and area is 2 cm and 4 cm2 respectively? (ρ = 2Ωm)

A. 0. 025 V B. 25 V

C. 0. 25 V D. None

Answer: D

Explanation:

V = IR

R = ρl/(A) = 2 ∗ 2 / 4 = 100Ω

V = 1 mA × 100

= 0. 1 V.

26. In the below figure, a semiconductor having an area ‘A’ and length ‘L’ and carrying current ‘I’ applied a voltage of ‘V’ volts

across it. Calculate the relation between V and A?

A. V = (

(

ρ

∗

L

)

/A)

∗

1 B. V = (

(

ρ

∗

A

)

/L)

∗

C. V = ((ρ

∗

∣)/(A

∗

L)) D. V = ((ρ

∗

|

∗

A

∗

L)

Answer: A

Explanation:

Option A, satisfies the Ohm's law which is V = IR where R = (ρO)/A.

27. Which of the following expressions represent the correct formula for the density of electrons occupying the donor level?

A. N

d

• p = N

a

• n B. N

d

− p = N

a

• n

C. N

d

• p = N

a

− n D. N

d

− p = N

a

− n

Answer: A

Explanation:

The density of the electrons is equal to the electrons present in the substrate minus the number of donors present.

28. Calculate the number of electrons is the number of holes is 15 × 10

10

?

A. 15 × 10

10

B. 1. 5 × 10

8

C. 1. 5 × 10

9

D. 1. 5 × 10

10

Answer: C

Explanation:

n × p = ( 1. 5 × 10

10

)

2

n × 1 × 10

10

= 1. 5 × 1. 5 × 10

10 × 10

10

n = 1. 5 × 10

9

electrons.

29. Which of the following expression represent the correct formulae for calculating the exact position of the Fermi level for p-

type material?

A. E

F

= E

V

• kTln (N

D

/N

A

) B. E

F

= −E

V

• kTln (N

D

/N

A

)

C. E

F

= E

V

− kln (N

D

/N

A

) D. E

F

= −E

V

− kTln (N

D

/N

A

)

Answer: A

Explanation:

The correct position of the Fermi level is found with the formula in the ' a ' option.

30. Consider a bar of silicon having carrier concentration n0 = 10

15

cm

− 3

and ni = 10

10

cm

− 3

. Assume the excess carrier

concentrations to be n = 10

13

cm

− 3

, calculate the quasi-fermi energy level at T = 300 K?

A. 0 .2982eV B. 0 .2984eV

C. 0 .5971eV D. 1Ev

Answer: B

Explanation:

E

Fn

− E

Fi

= kTln (

n 0

• δn

n

i

)

= 1. 38

∗

− 23

× 300 × ln

(

13

15

/ 10

13

)

= 0 .2984eV.

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A COMPLETE PREPARATION FOR CBSE CLASS XII PHYSICS

CLASS XII ( PHYSICS )