Electronics: Physics unit 18 notes (FBISE class 10)

Introduction to electronics

Electronics studies and applies the principles of physics to design, create and operate active devices such as diodes, transistors and integrated circuits (ICs). These electronic devices control and amplify the flow of electric current and convert it from one form to another, for example from AC to DC or from analogue signals to digital signals.

Electronics: The branch of physics that helps to study the emission and behaviour of electrons with electronic devices.

Electronic devices play a major role in modern civilisation. The devices used every day include computers, mobile phones, the automated teller machine (ATM), the pen drive, the television and the digital camera. Electronic devices have hugely influenced many parts of modern society, such as telecommunications, education, health care, industry, security and entertainment.

The main component behind the advancement of electronics is the semiconductor industry. It has become one of the largest and most profitable sectors of the global economy. This unit builds your understanding of how physics is applied in modern technology.

Roadmap of the unit

Semiconductors (18.1 to 18.2): pure and doped silicon, n-type and p-type material.
Diodes (18.3 to 18.5): the PN junction, forward and reverse bias, the I-V curve and the light-emitting diode.
Transistors (18.6 to 18.8): structure, working, current gain, and the transistor as a switch; relays (18.7).
Digital electronics (18.9 to 18.13): binary states, logic gates, their uses, and analogue against digital.
Quantum computers (18.14): a new way of computing.

18.1Semiconductors

Materials whose electrical conductivity falls between that of conductors and insulators are called semiconductors. The atoms of group IV-A of the periodic table are semiconductors.

Semiconductor: A material whose electrical conductivity falls between that of conductors and insulators.
Intrinsic semiconductor: A pure semiconductor, such as silicon (Si) or germanium (Ge).
Extrinsic semiconductor: A semiconductor that is doped with some specific impurities.

Silicon: a pure semiconductor

Silicon is a semiconductor material. An atom of pure silicon has four electrons in its outer orbit (four valence electrons). In the crystal lattice of Si, the valence electrons of every Si atom are locked up in covalent bonds with the valence electrons of four neighbouring Si atoms, as shown in Figure 18.1.

Figure 18.1: 2-D crystal lattice of silicon. Bonds are formed by the sharing of electrons (the bonds are covalent).

In pure form a Si wafer does not contain any free charge carriers. Such pure semiconductors are called intrinsic semiconductors.

Making a useful semiconductor

To make useful semiconductor devices, materials such as phosphorus (P) and aluminium (Al) are added to Si. Semiconductors that are doped with some specific impurities are called extrinsic semiconductors. These impurities modify the electrical properties of the semiconductor and make it more suitable for electronic conduction. Extrinsic semiconductors are of two types: p-type and n-type (section 18.2).

Pure and doped semiconductors compared.
	Intrinsic semiconductor	Extrinsic semiconductor
What it is	Pure semiconductor	Semiconductor doped with a specific impurity
Examples	Pure Si or Ge	Si with P or Al added
Free charge carriers	None in pure form (all valence electrons are locked in bonds)	Supplied by the impurity atoms
Types	One	Two: n-type and p-type

18.2N-type and p-type semiconductorsOften examined

N-type semiconductor

When a pentavalent impurity (an element of group V-A) such as phosphorus is added to Si, the resulting mixture is called n-type. Four valence electrons of every phosphorus atom are locked up in covalent bonds with the valence electrons of four neighbouring Si atoms. The 5th valence electron of the phosphorus atom does not find a binding electron, so it remains free to float in the material (Figure 18.2).

Figure 18.2: Adding pentavalent impurities to Si yields free electrons.

P-type semiconductor

When a trivalent impurity (an element of group III-A) such as boron is added to Si, the resulting mixture is called p-type. Three valence electrons of every boron atom are locked up in covalent bonds with the valence electrons of three neighbouring Si atoms. The fourth electron of the remaining Si atom stays unpaired. A vacant spot, a hole, is created within the covalent bond between one boron atom and a neighbouring Si atom (Figure 18.3). These holes are considered to be positive charge carriers.

Figure 18.3: Adding trivalent impurities to Si yields holes.

N-type and p-type semiconductors compared.
	N-type	P-type
Impurity added to Si	Pentavalent (group V-A)	Trivalent (group III-A)
Valence electrons of the impurity atom	5	3
Examples	Phosphorus, arsenic, antimony, bismuth	Boron, aluminium, indium, gallium
What the impurity adds	A free electron (the 5th electron)	A hole (a vacancy in a bond)
Majority charge carriers	Electrons	Holes

Worked example: which type is it?

Silicon crystals are doped separately with arsenic, gallium, antimony and indium. For each impurity say how many valence electrons its atom has, whether the crystal becomes n-type or p-type, and what its majority charge carriers are.

Use the group of the impurity in the periodic table. Arsenic and antimony are group V-A elements, so they are pentavalent: five valence electrons. Four make bonds with Si neighbours and the fifth is free.
Gallium and indium are group III-A elements, so they are trivalent: three valence electrons. They can complete only three bonds, so each leaves a hole.
Pentavalent impurity gives an n-type crystal with electrons as the majority carriers. Trivalent impurity gives a p-type crystal with holes as the majority carriers.

Impurity	Valence electrons	Type	Majority carriers
Arsenic	5	n-type	Electrons
Antimony	5	n-type	Electrons
Gallium	3	p-type	Holes
Indium	3	p-type	Holes

As, Sb: n-type (electrons); Ga, In: p-type (holes)

18.3PN junction (or diode)Often examined

When a crystal of semiconductor is grown in such a way that one half is doped with a trivalent impurity (p-type) and the other half with a pentavalent impurity (n-type), a junction is formed between the two regions. Such a grown crystal is called a PN junction.

Formation of the depletion region

At the junction, randomly moving electrons from the n-type region diffuse into the p-type region and fill the vacancies (holes) there (Figure 18.4 a).
As a result of this diffusion, negative and positive ions are formed near the junction (Figure 18.4 b): negative ions on the p side and positive ions on the n side.
These ions create an electric field directed from the n-type to the p-type region. The region where this happens is called the depletion region.
Thus a potential barrier is formed at the junction. Its value is $0.3 V$ for a germanium crystal and $0.7 V$ for a silicon crystal.

Figure 18.4 (a) and (b): PN junction and formation of the depletion region. (a) randomly moving electrons from the n-type diffuse into the p-type region; (b) formation of the depletion region with its electric field.

Potential across the junction (silicon)

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

O

distance across the junction

V (V)

0.7 V

P-type

N-type

depletion layer

Figure 18.4 (c): the potential rises from the P side to the N side across the depletion region. The step is the potential barrier, about

0.7 V

for silicon and

0.3 V

for germanium.

PN junction: The junction formed between a p-type and an n-type region grown in one semiconductor crystal.
Depletion region: The region on both sides of the junction that is free of mobile charges, where the ions set up an electric field from n to p.
Potential barrier: The potential difference across the depletion region: $0.3 V$ for Ge and $0.7 V$ for Si.

The depletion region is free of mobile charges. The width of the depletion layer depends upon the carrier concentration. The depletion region stops further diffusion of electrons and holes and maintains the separation of the charge carriers.

The diode and its symbol

A PN junction diode allows current to pass in only one direction, so it is used as a rectifier in many electric circuits. In the symbol of a diode, the vertical line shows the n-region and the arrowhead represents the p-region (Figure 18.5).

Figure 18.5: PN junction diode. (a) Diode structure; (b) diode symbol. The conventional current flows in the direction of the arrowhead.

18.4Characteristics of a diode under biasingOften examined

When a diode is connected across an electric supply (a potential difference), it is said to be under biasing. A diode allows current to flow in only one direction, so a diode acts as a perfect insulator for currents trying to flow in one direction and as a perfect conductor for currents flowing through it in the other direction.

Biasing: Connecting a diode across an electric supply (a potential difference).
Forward bias: The biasing in which the diode allows current to flow.
Reverse bias: The biasing in which the current is resisted.

Forward and reverse bias compared (details in the next two sections).
	Forward bias	Reverse bias
Connection	Positive terminal of the supply to p-type; negative terminal to n-type	Positive terminal to n-type (cathode); negative terminal to p-type (anode)
Depletion region	Becomes narrower	Becomes wider
Potential barrier	Reduced	Increased
Current	Flows: a few milliamperes (mA)	Blocked: only a leakage current of a few microamperes ( $μ A$ )
Acts like	A conductor	An insulator

18.4.1Forward biased diodeOften examined

When the positive terminal of a DC source is connected to the p-type semiconductor and the negative terminal to the n-type semiconductor, as in Figure 18.6 (a), the junction is said to be in forward bias.

Figure 18.6 (a): Forward bias. The depletion region is narrow.

What happens in the junction

The holes on the p-side, being positively charged, are repelled from the positive terminal of the bias source and driven towards the junction.
The electrons on the n-side are repelled from the negative terminal and driven towards the junction.
The depletion region is reduced in width and the barrier potential is also reduced.
Once the barrier is reduced by the forward voltage, a conducting path is established. Electrons and holes combine and the current passes through the junction. This current is called the forward current and is of the order of a few milliamperes (mA).

The variation of forward current through the PN junction with the biasing voltage can be studied with the circuit of Figure 18.6 (b).

Figure 18.6 (b): Circuit to study forward bias. The milliammeter measures the forward current; the voltmeter measures the voltage across the diode.

18.4.2Reverse biased diodeOften examined

When the potential at the anode (p-type) is smaller than the potential at the cathode (n-type), the diode is said to be reverse biased, as shown in Figure 18.7 (a). In a reverse-biased diode the current is blocked.

Figure 18.7 (a): Reverse bias. The depletion region is wide.

In a reverse-biased connection the holes of the p-side are forced to the left and the electrons of the n-side are forced to the right, away from the junction. This leaves an empty zone around the PN junction that is free of charge carriers and creates a wider depletion region. This depletion region acts as an insulator, so no current flows through the junction due to the majority charge carriers.

Only a small current, of the order of a few microamperes ( $μ A$ ), flows due to the minority charge carriers. This small current is known as the reverse current or leakage current.

The variation of reverse current through the PN junction with the biasing voltage can be studied with the circuit of Figure 18.7 (b).

Figure 18.7 (b): Circuit to study reverse bias. The microammeter measures the tiny reverse (leakage) current.

18.4I-V characteristic of a diodeOften examined

The graph of current ( $I$ ) against voltage ( $V$ ) for a PN junction is shown in Figure 18.8.

I-V curve of a semiconductor diode

V_{d}

O

V

I

forward bias

reverse bias

breakdown

Figure 18.8: I-V curve for a semiconductor diode (a sketch: the current scale is in mA on the forward side and in

μ A

on the reverse side). The knee is at about

0.3 V

for germanium and

0.7 V

for silicon.

Reading the graph

Forward bias region: current and voltage are positive. While the forward voltage is less than $V_{d}$ , the forward current is nearly zero and the diode does not conduct. When the forward voltage equals $V_{d}$ the diode starts to conduct, and the current then rises steeply (the "knee"). $V_{d}$ is about $0.3 V$ for germanium and $0.7 V$ for silicon.
Reverse bias region: current and voltage are negative. The reverse current is tiny and almost constant. As the reverse bias voltage is raised, the width of the depletion region increases, and a point comes when the junction breaks down. This results in a large flow of current: the reverse breakdown voltage.

The four regions of the I-V curve.
Region	Voltage	Current
Forward, below $V_{d}$	Positive, less than $V_{d}$	Nearly zero: diode does not conduct
Forward, above $V_{d}$	Positive, more than $V_{d}$	Large and rising quickly (mA)
Reverse	Negative	Very small leakage ( $μ A$ )
Reverse breakdown	Negative, beyond the breakdown voltage	Large: the junction breaks down

Worked example: which lamp glows?

Two identical circuits each contain a battery, a lamp and a silicon diode. In circuit (a) the p side of the diode faces the positive terminal of the battery. In circuit (b) the diode is turned round. (i) Which lamp glows? Explain. (ii) In circuit (a) the battery voltage is raised slowly from zero. At about what voltage does the diode start to conduct? What would it be for a germanium diode?

Circuit (a) has the p side of the diode towards the positive terminal; circuit (b) has the diode reversed.

(i) Circuit (a). The p side is joined (through the lamp) to the positive terminal and the n side to the negative terminal, so the diode is forward biased. The depletion region narrows, the barrier falls, and electrons and holes cross the junction: a forward current of a few mA flows and the lamp glows.
Circuit (b). The p side is now towards the negative terminal, so the diode is reverse biased. The depletion region widens and acts as an insulator. Only a leakage current of a few microamperes flows, far too small to light the lamp, so the lamp stays off.
(ii) Below $V_{d}$ the forward current is nearly zero. The diode starts to conduct when the forward voltage reaches the potential barrier value $V_{d}$ .
$V_{d} (Si) \approx 0.7 V, V_{d} (Ge) \approx 0.3 V$
A germanium diode starts to conduct at the lower voltage, $0.3 V$ .

(a) glows, (b) off; V_{d} \approx 0.7 V (Si), 0.3 V (Ge)

18.5Light-emitting diode (LED)

A light-emitting diode (LED) emits energy in the form of light, and hence it glows when an electric current passes through it. These diodes are made from gallium arsenide phosphide (GaAsP) and gallium phosphide (GaP). The symbol of an LED and its forward-bias circuit are shown in Figure 18.9 (a) to (c).

Light-emitting diode (LED): A diode that emits energy in the form of light when electric current passes through it.

Figure 18.9 (a) and (b): the symbol of the LED and its forward-biased circuit with a series resistor.

How an LED gives out light

When the LED is forward biased, as in Figure 18.9, the potential barrier is lowered. The free electrons from the conduction band of the n-region cross the barrier and enter the p-region. As these electrons enter the p-region they fall into the holes lying in the valence band. They therefore fall from a higher energy level to a lower energy level, and in the process they radiate energy as light (Figure 18.9 d).

Figure 18.9 (d): action of a light-emitting diode. The energy lost by the falling electron leaves as light.

An LED is enclosed with a transparent cover so that the emitted light can be seen. LEDs can emit light of different colours such as red, green and blue. They are widely used in decorative lamps, optical communication and display devices.

18.6TransistorOften examined

A transistor is a device made of semiconductor material (for example germanium and silicon) that can both conduct and insulate. Transistors switch and modulate (control) electronic current. The transistor was the first device designed to act as both a transmitter and a resistor, converting sound waves into electronic waves and controlling electronic current.

The name transistor comes from the "trans" of transmitter and the "sistor" of resistor. These words describe the operation of a transistor, which is the transfer of an input signal from a low-resistance circuit to a high-resistance circuit.

Transistor: A semiconductor device that can both conduct and insulate; it switches and controls (modulates) electronic current, transferring an input signal from a low-resistance circuit to a high-resistance circuit.

18.6Regions and terminals of a transistorOften examined

In general, all transistors have three regions: the base (B), the collector (C) and the emitter (E).

The three regions of a transistor.
Region	Size	Doping
Emitter	Moderately sized	Heavily doped
Base	At the centre of the transistor; thin	Lightly doped
Collector	Larger than the emitter	Moderately doped

A transistor is a three-terminal semiconductor device that consists of two PN junctions. It is constructed in such a way that a lightly doped region called the base is sandwiched between two regions called the emitter and the collector. The function of a transistor is to amplify current. This makes transistors useful as switches or amplifiers. They are used extensively in electronic devices such as televisions, radio transmitters and mobile phones.

18.6.1Types of transistorOften examined

Transistors are made in two types: NPN and PNP.

NPN transistor: an NPN transistor is made by joining one P-type semiconductor in between two N-type semiconductors. A lightly doped P-type semiconductor is sandwiched between two heavily doped N-type regions.
PNP transistor: a PNP transistor is made by sandwiching an N-type semiconductor in between two P-type semiconductors. A lightly doped N-type semiconductor is sandwiched between two heavily doped P-type regions.

Figure 18.10: Notation and symbols used with the PNP transistor and the NPN transistor.

The two types of transistor.
	NPN transistor	PNP transistor
Emitter	N-type	P-type
Base (thin, lightly doped)	P-type	N-type
Collector	N-type	P-type
Emitter arrow in the symbol	Points away from the base	Points towards the base

18.6.2Working of a transistorOften examined

When no voltage is applied at the transistor base, the electrons in the emitter are prevented from passing to the collector side because of the PN junction. If a negative voltage is applied to the base of an NPN transistor, the PN junction between the base and the emitter becomes reverse biased. This forms a wider depletion region that prevents current flow.

Normal biasing of an NPN transistor

Usually two batteries are required for the normal operation of a transistor. Call them $V_{B E}$ and $V_{C B}$ (Figure 18.11).

Emitter-base junction forward biased. The negative terminal of the battery $V_{B E}$ is connected to the N-type emitter and its positive terminal to the P-type base.
Collector-base junction reverse biased. In the collector circuit, the N-type collector is connected to the positive terminal of the battery $V_{C B}$ and the P-type base to the negative terminal of the same battery.

Figure 18.11: An npn transistor under normal operation.

Movement of the charge carriers

Electrons enter the emitter from the negative terminal of the battery $V_{B E}$ and flow towards the junction.
The forward bias has reduced the potential barrier of the emitter-base junction. A few electrons combine with the hole carriers in the base to complete the emitter-base circuit.
The base is a very thin region (about $0.001$ inch). Most of the electrons flow on to the collector, because the collector is connected to the positive terminal of the battery $V_{C B}$ .
This reverse-biased potential is very large, so most of the majority charge carriers are attracted and cross the large base-to-collector depletion region.

Currents in a transistor

The direction of the conventional current in the external circuit is shown in Figure 18.11. The sum of the collector current ( $I_{C}$ ) and the base current ( $I_{B}$ ) is equal to the emitter current ( $I_{E}$ ):

I_{E} = I_{B} + I_{C}

(18.1)

$I_{E}$ : emitter current; $I_{B}$ : base current; $I_{C}$ : collector current

Approximately $95%$ to $98%$ of the current through the transistor goes from the emitter to the collector, and about $2%$ to $5%$ goes from the emitter to the base. The ratio of the collector current to the base current gives the current gain $β$ :

β = \frac{I _{C}}{I _{B}}

(18.2)

$β$ : current gain (a ratio with no unit)

Equation (18.2) gives the static or DC value of $β$ . The dynamic or AC value is calculated for the changes in the current values (the change in $I_{C}$ divided by the change in $I_{B}$ ). The ratio of the collector current to the emitter current gives the current gain $α$ :

α = \frac{I _{C}}{I _{E}}

(18.3)

$α$ : current gain (a ratio with no unit)

Example 18.1: emitter current

When the collector current in a junction transistor is

99.9 mA

and the base current is

0.1 mA

, how much is the emitter current?

Given

$I_{C} = 99.9 mA$
$I_{B} = 0.1 mA$

To find

emitter current $I_{E}$

The emitter current is the sum of the base current and the collector current (equation 18.1).
$I_{E} = I_{B} + I_{C}$
Put in the values.
$I_{E} = 99.9 mA + 0.1 mA = 100 mA$
Change to amperes.
$100 mA = 0.1 A$

I_{E} = 100 mA = 0.1 A

Example 18.2: base current needed

Find the value of the base current required to turn on a transistor for a collector current of

150 mA

, for a transistor having current gain

β = 190

.

Given

$I_{C} = 150 mA$
$β = 190$

To find

base current $I_{B}$

The current gain is $β = I_{C} / I_{B}$ (equation 18.2). Rearrange to find $I_{B}$ .
$I_{B} = \frac{I _{C}}{β}$
Put in the values.
$I_{B} = \frac{150 mA}{190} = 0.789 mA$

The base current needed to run the transistor is about $0.8 mA$ .

I_{B} = 0.789 mA \approx 0.8 mA

Worked example: currents and gains of an NPN transistor

In an NPN transistor the emitter current is

5.0 mA

and the base current is

0.15 mA

. Find (i) the collector current, (ii) the current gains

β

and

α

, and (iii) the percentage of the emitter current that reaches the collector. Is it within the usual

95%

to

98%

?

Conventional currents in the NPN transistor:

I_{B}

and

I_{C}

flow into the transistor and

I_{E}

flows out of it.

Given

$I_{E} = 5.0 mA$
$I_{B} = 0.15 mA$

To find

(i) $I_{C}$
(ii) $β$ and $α$
(iii) $I_{C}$ as a percentage of $I_{E}$

(i) The emitter current is the sum of the base and collector currents, so subtract to find $I_{C}$ .
$I_{C} = I_{E} - I_{B} = 5.0 - 0.15 = 4.85 mA$
(ii) Current gain $β$ is the collector current divided by the base current.
$β = \frac{I _{C}}{I _{B}} = \frac{4.85}{0.15} = 32.3$
Current gain $α$ is the collector current divided by the emitter current.
$α = \frac{I _{C}}{I _{E}} = \frac{4.85}{5.0} = 0.97$
(iii) $α$ as a percentage shows how much of the emitter current reaches the collector.
$0.97 \times 100 = 97%$
$97%$ lies between $95%$ and $98%$ , so the answer agrees with the usual range. The remaining $3%$ is the base current: $0.15/5.0 = 0.03$ .

I_{C} = 4.85 mA, β \approx 32, α = 0.97 (97%)

18.6.3Role of the transistor in the advancement of computer technology

The transistor is an important invention that changed the history of computers. Before transistors, digital circuits were composed of vacuum tubes. A vacuum tube was used to amplify voice, but the tubes consumed power, created heat, burned out rapidly and required high maintenance. Transistors are commonly used in digital devices because they are both economical and act as rapid-response switches.

Transistors transformed the world of electronics and had a huge impact on computer design. By replacing the bulky and unreliable vacuum tubes with transistors made of semiconductors, computers could perform the same functions using less power and space. With advances in engineering, the number of transistors that can be fitted per unit area onto a circuit board has continued to increase dramatically, and this has rapidly enhanced computing power.

The four generations of computers named in the book.
Generation of computers	Built with
First	Vacuum tubes
Second	Transistors
Third	Integrated circuits
Fourth	Microprocessors

Why transistors replaced vacuum tubes.
	Vacuum tube	Transistor
Size	Bulky	Small, so many fit on a circuit board
Power and heat	Consumes power and creates heat	Uses less power and space
Reliability	Burns out rapidly; high maintenance	Reliable
Cost and speed		Economical; rapid-response switch

18.7Relays and switching circuits

A relay is an electrically operated switch that opens and closes circuits by receiving electrical signals from outside sources. It consists of a set of input terminals for a signal and a set of operating contact terminals. For example, when you push the button on a TV remote to watch TV, it sends an electrical signal to the relay inside the TV, turning the main power ON (Figure 18.12).

Relay: An electrically operated switch that opens and closes circuits by receiving electrical signals from outside sources.

Figure 18.12: Block diagram of a relay used in a TV.

Relays are very important in electronics. They are used to control a high-power or high-voltage circuit (or device) with a low-power circuit. A relay embedded in an electrical product receives an electrical signal and sends the signal to other equipment by turning the switch ON and OFF. For example:

A timer circuit with a relay can switch power at a preset time.
Relays are used to protect electrical circuits from overload or faults.
Relays are used to turn ON or OFF high-power devices such as lamps or garage door motors with just a small DC voltage signal.

Types of relay

Various types of relay are used in many applications to control different amounts of current and different numbers of circuits. Relay technology can be divided into two main categories: movable contacts (mechanical relay) and no movable contacts.

Movable contacts (mechanical relay): this type has contacts that are mechanically moved by a magnetic force to switch signals, currents and voltages ON or OFF.
No movable contacts: the solid state relay (SSR) is an example. It has no moving part, which increases its long-term reliability. A solid state relay uses a semiconductor and electrical switching devices, activated by the control signal, to switch the controlled load. An optocoupler (a light-emitting diode coupled with a photo transistor or photo-TRIAC) can be used to isolate the control and the controlled circuits (Figure 18.13).

Figure 18.13 (b): Optocoupler, a light-emitting diode coupled with a photo-TRIAC (the solid state relay itself is shown in the book as a photograph in Figure 18.13 (a)).

The two types of relay.
	Mechanical relay	Solid state relay (SSR)
Contacts	Movable, moved by a magnetic force	None (no moving part)
Switching	Mechanical motion opens and closes the contacts	Semiconductor switching device, activated by a control signal
Why used	Switches signals, currents and voltages ON or OFF	Faster switching and durability (long-term reliability)

18.7.2Operating principles of solid state relays

Solid state relays operate according to the following principles.

The LED lights up when the current is connected at the input side (Figure 18.14 a).
The light sent by the LED is converted into a voltage when it is received by the photodiode array (Figure 18.14 b).
This voltage becomes the gate voltage that drives the transistor via the control circuit (Figure 18.14 c).

Figure 18.14: Operation of a solid state relay (a simplified block form of the three stages (a), (b) and (c) of the book figure).

By the operation of these electronic circuits, signals (currents and voltages) are switched ON or OFF electronically.

18.8Transistor as a switchOften examined

A transistor can be used as a switch if it is biased in the saturation and cut-off regions. The collector C and the emitter E of the transistor behave as the terminals of the switch. The base B and the emitter E act as the control terminal that decides the state of the switch, either ON or OFF.

The main reason behind using the transistor as a switch is that the current at the base (I_B) controls the current present at the collector (I_C), through the relation

I_C=\beta\,I_B

If the current at the base exceeds the minimum cut-off value, the transistor behaves like a closed switch; otherwise it behaves like an open switch.

Suppose a lamp is to be turned ON and OFF with a switch, as in Figure 18.15 (a). A transistor can be put in place of the switch, as in Figure 18.15 (b).

Figure 18.15: Transistor as a switch. (a) mechanical switch; (b) NPN transistor switch.

If the base of the transistor is not connected to a suitable voltage, the base current will be zero. Consequently the transistor cannot turn on. So a switch is connected in the base circuit (the book words this as "between the base and collector wires"), as in Figure 18.16.

Cutoff

If the switch is open, as in Figure 18.16 (a), there is no current through the base. According to the relation I_C=\beta I_B, no current flows from the collector to the emitter. In this situation the resistance between C and E becomes infinite and CE behaves as an open switch. The transistor is said to be in cutoff, and the lamp is OFF.

Saturation

If the switch is closed, as in Figure 18.16 (b), current is able to flow from the base to the emitter through the switch. This base current enables a much larger current to flow from the collector to the emitter because of the small resistance between C and E, lighting the lamp. In this state the circuit current is maximum and the transistor is said to be saturated.

Figure 18.16: NPN transistor switching. (a) Cutoff situation (lamp off); (b) saturated state (lamp on).

Cutoff and saturation compared.
State	Switch in base circuit	Base current I_B	Collector current I_C	Resistance between C and E	Lamp
Cutoff	Open	Zero	Zero (I_C=\beta I_B)	Infinite (open switch)	OFF
Saturated	Closed	Flows (above the minimum value)	Maximum	Small (closed switch)	ON

Worked example: will the lamp light?

A lamp needs a collector current of 120\,\mathrm{mA} to glow normally. It is switched by an NPN transistor of current gain \beta=150. (i) What is the smallest base current that will turn the lamp fully on? (ii) What are I_B, I_C and the state of the transistor when the base switch is open? (iii) The base switch is closed and a base current of 1.0\,\mathrm{mA} flows. Does the lamp light?

Given

I_C = 120\,\mathrm{mA}
\beta = 150

To find

(i) minimum I_B
(ii) I_B, I_C and state with the switch open
(iii) lamp state for I_B=1.0\,\mathrm{mA}

(i) Use I_C=\beta I_B and solve for the base current needed.
I_B=\dfrac{I_C}{\beta}=\dfrac{120\,\mathrm{mA}}{150}=0.80\,\mathrm{mA}
(ii) Switch open: no base current, so I_B=0.
I_C=\beta I_B=150\times0=0
With I_C=0 the resistance between C and E is infinite: the transistor is in cutoff, acts as an open switch and the lamp is OFF.
(iii) 1.0\,\mathrm{mA} is more than the minimum 0.80\,\mathrm{mA}, so the base current is large enough to make the transistor behave as a closed switch. It is saturated and the lamp is ON.

I_{B,\min}=0.80\,\mathrm{mA};\ \text{open: cutoff, lamp off};\ \text{closed: saturated, lamp on}

18.9Digital electronics

Digital electronic circuits deal with situations that have only two states: ON or OFF, TRUE or FALSE, HIGH or LOW, YES or NO and so on. For example, consider the circuit in Figure 18.17.

Figure 18.17: Switch and lamp circuit.

There are only two possible positions to flip a light switch: ON or OFF. (It is of no concern that there could be 110 volts, 120 volts or 220 volts, a fluctuating analogue "signal", running through the electrical wiring connected to the light switch.)

The state of the switch is called the input and the state of the current or lamp is called the output. When the switch is open, no current passes through the circuit and the lamp is OFF. In other words, when the input is 0 the output is also 0. When the switch is closed, current passes through the circuit and the lamp is ON. In other words, when the input is 1 the output is also 1. The possible combinations of the input and output states of this circuit are shown in Table 18.1.

Table 18.1: Switching.
Switch (input)	Lamp (output)	Input as a binary digit	Output as a binary digit
Open	OFF	0	0
Closed	ON	1	1

The same ON/OFF logic is used to build the digital electronic circuits inside your laptop computer, GPS device and smartphone. The two states are best operated on mathematically if they are numerically represented by the binary digits 1 and 0, also known as Boolean variables. English letters (A, B, C, etc.) are used to represent the Boolean variables. Boolean algebra is a formal way of describing logical operations. It is based on three basic logical operators: AND, OR and NOT.

Digital electronic circuit: A circuit that deals with situations having only two states (ON or OFF, TRUE or FALSE, HIGH or LOW, YES or NO).
Boolean variable: A binary digit, 1 or 0, representing one of the two states; written with a letter such as A, B or X.
Boolean algebra: A formal way of describing logical operations, based on the three basic operators AND, OR and NOT.

Different names for the two states.
State name	Written as 1	Written as 0
Switch	Closed	Open
Lamp or current	ON	OFF
Statement	TRUE	FALSE
Voltage level	HIGH	LOW
Answer	YES	NO

18.9.1Converting analogue data to digital data

The Analogue-to-Digital Converter (ADC) enables microprocessor-controlled circuits, such as Arduino and Raspberry Pi, to connect with the real world. In reality, analogue signals vary continuously and come from different sources and sensors that can detect sound, light, temperature or movement. Many digital systems engage with their surroundings by measuring these analogue signals from transducers.

So an electronic circuit is needed that can convert the continuously changing analogue signals into discrete digital signals. Such circuits are known as Analogue-to-Digital Converters. A Digital-to-Analogue Converter (DAC) does the reverse: it converts a discrete digital signal into an analogue signal.

Analogue-to-Digital Converter (ADC): An electronic circuit that converts continuously changing analogue signals to discrete digital signals.
Digital-to-Analogue Converter (DAC): An electronic circuit that converts a discrete digital signal to an analogue signal.

Analogue-to-digital and digital-to-analogue conversion: the ADC turns a continuous signal from a sensor into digital data, and the DAC turns digital data back into a continuous signal.

18.9.2Unit of data in computing

Analogue signals can be continuous and can provide an infinite number of different voltage values. Digital circuits, on the other hand, work with a binary signal that has only two discrete states: a logic "1" (HIGH) or a logic "0" (LOW). Each digit in binary (0 or 1) is called a bit, which is an abbreviation of binary digit.

Bit: Each digit in binary (0 or 1); an abbreviation of binary digit. It is the unit of data in computing.

Analogue and digital signals compared.
	Analogue signal	Digital signal
Values	Continuous: an infinite number of different voltage values	Only two discrete states
States	Not limited to two	Logic 1 (HIGH) or logic 0 (LOW)

18.9.3Logic gates: the building blocks of digital circuitsOften examined

Electronic devices are built from digital logic circuits. Logic gates are the main building blocks of all digital logic circuits. These circuits can act as switches and convert incoming voltage into binary electrical pulses of high (or 1) and low (or 0). Digital electronic circuits that are used to implement binary or logical operations are known as logic gates.

Logic gates can have one or more inputs but only one output. Electronic devices are usually made from large assemblies of logic gates, often packed in integrated circuits. The common logic gates are AND, OR, NAND, NOR and NOT. The set of inputs and outputs of a logic system in binary form is called a truth table.

Logic gate: A digital electronic circuit that implements a binary (logical) operation. It has one or more inputs and only one output.
Truth table: The set of inputs and outputs of a logic system written in binary form.

The five common logic gates (sections 18.10 and 18.11).
Gate	Number of inputs	Type
AND	Two or more	Fundamental
OR	Two or more	Fundamental
NOT	One	Fundamental
NAND	Two or more	Universal
NOR	Two or more	Universal

18.10Fundamental logic gatesOften examined

The three logic gates AND, OR and NOT are known as the fundamental logic gates. Their circuit symbols and Boolean expressions are shown below; each one is explained in the next three sections.

Symbols of the fundamental logic gates.

Summary of the fundamental gates.
Gate	Boolean expression	Output is 1 when	Output is 0 when
AND	X=A\cdot B	all inputs are 1	any input is 0
OR	X=A+B	at least one input is 1	all inputs are 0
NOT	X=\bar{A}	the input is 0	the input is 1

18.10.1AND gateOften examined

Consider Figure 18.18 (a), in which a lamp is connected in series with a battery using two switches (A and B) as two inputs. There are four possible states of the two switches:

When both switches A and B are open, the lamp is OFF.
When switch A is open but B is closed, the lamp is OFF.
When switch A is closed but B is open, the lamp is OFF.
When both switches A and B are closed, the lamp is ON.

It is clear that "the lamp will glow only if both the switches are closed". A logic gate that implements such an action is known as an AND gate. Its symbol is shown in Figure 18.18 (b).

Figure 18.18: AND gate. (a) switch depiction; (b) circuit symbol.

The Boolean expression for the AND operation is

X=A\cdot B

read as "X equals A AND B"; the dot (\cdot) represents the AND operation

Table 18.2: AND gate.
A (input)	B (input)	X=A\cdot B (output)
0	0	0
1	0	0
0	1	0
1	1	1

An AND gate has two or more inputs and only one output. The output of an AND gate is 1 only when all of its inputs are 1; in all other situations the output of an AND gate is 0.

18.10.2OR gateOften examined

Consider Figure 18.19 (a), in which a lamp is connected in parallel with a battery using two switches (A and B) as two inputs. There are four possible states of the two switches:

When both switches A and B are open, the lamp is OFF.
When switch A is open and B is closed, the lamp is ON.
When switch A is closed and B is open, the lamp is ON.
When both switches A and B are closed, the lamp is ON.

It is clear that "the lamp will glow if at least one of the switches is closed". A logic gate that implements such an action is known as an OR gate. Its symbol is shown in Figure 18.19 (b).

Figure 18.19: OR gate. (a) switch depiction; (b) circuit symbol.

The Boolean expression for the OR operation is

X=A+B

read as "X equals A OR B"; the plus sign (+) represents the OR operation

Table 18.3: OR gate.
A (input)	B (input)	X=A+B (output)
0	0	0
1	0	1
0	1	1
1	1	1

An OR gate has two or more inputs and only one output. The output of an OR gate is 0 only when all of its inputs are 0; in all other situations the output of an OR gate is 1.

18.10.3NOT gateOften examined

Consider Figure 18.20 (a), in which a lamp is connected in parallel with a switch A, with a battery. There are two possible states of the switch:

When the switch A is open, current passes through the lamp and it glows.
When the switch A is closed, no current passes through the lamp (the current takes the easy path through the closed switch) and it does not glow.

It is clear that "the lamp will glow if the switch is open". A logic gate that implements such an action is known as a NOT gate. Its symbol is shown in Figure 18.20 (b).

Figure 18.20: NOT gate. (a) switch depiction; (b) circuit symbol.

The Boolean expression for the NOT operation is

X=\bar{A}

read as "X equals A NOT"; a line (bar) over the symbol represents the NOT operation

Table 18.4: NOT gate.
A (input)	X=\bar{A} (output)
0	1
1	0

A NOT gate has one input and one output. The NOT operation inverts the state of a Boolean variable: the output of a NOT gate is 0 when its input is 1, and vice versa.

Worked example: truth table of a combination

A logic circuit passes input B through a NOT gate and then feeds A and the inverted B into an AND gate, giving X=A\cdot\bar{B}. (i) Complete its truth table. (ii) For which input combination is X=1? (iii) Why is this not a NAND gate?

The circuit: a NOT gate on B, then an AND gate.

Write all four combinations of A and B. Find \bar{B} first, because the NOT gate acts on B alone.
For each row apply the AND rule to A and \bar{B}: the output is 1 only when both of them are 1.

A	B	\bar{B}	X=A\cdot\bar{B}
0	0	1	0
1	0	1	1
0	1	0	0
1	1	0	0

(ii) X=1 only in the row A=1, B=0, where both inputs of the AND gate are 1.
(iii) In a NAND gate the NOT acts on the output of the AND gate, \overline{A\cdot B} (output 0 only when both inputs are 1). Here the NOT acts on the input B before the AND gate, so the truth table is different.

X=1\ \text{only for}\ A=1,\ B=0

18.11Universal logic gatesOften examined

The logic gates NAND and NOR are known as universal gates because they can be used to make all the other gates.

Universal logic gates: The logic gates NAND and NOR, called universal because they can be used to make all the other gates.

The two universal gates compared.
	NAND	NOR
Made from	AND gate followed by a NOT gate	OR gate followed by a NOT gate
Boolean expression	X=\overline{A\cdot B}	X=\overline{A+B}
Output is 0 when	all inputs are 1	any input is 1
Output is 1 when	any input is 0	all inputs are 0

18.11.1NAND gateOften examined

A NAND gate is obtained by coupling a NOT gate with the output terminal of the AND gate, as shown in Figure 18.21 (a). The NOT gate inverts the output of the AND gate. Thus the NAND gate is the reciprocal (the opposite) of the AND gate. The symbol of the NAND gate is shown in Figure 18.21 (b): the small circle shows the NOT operation.

Figure 18.21: NAND gate. (a) description; (b) circuit symbol.

The Boolean expression for the NAND operation is

X=\overline{A\cdot B}

read as "X equals A AND B, NOT": the bar covers the whole product

Table 18.5: NAND gate.
A (input)	B (input)	X=\overline{A\cdot B} (output)
0	0	1
1	0	1
0	1	1
1	1	0

18.11.2NOR gateOften examined

The NOR gate is obtained by coupling the output of the OR gate with a NOT gate, as shown in Figure 18.22 (a). The NOT gate inverts the output of the OR gate. The symbol of the NOR gate is shown in Figure 18.22 (b).

Figure 18.22: NOR gate. (a) description; (b) circuit symbol.

The Boolean expression for the NOR operation is

X=\overline{A+B}

read as "X equals A OR B, NOT": the bar covers the whole sum

Table 18.6: NOR gate.
A (input)	B (input)	X=\overline{A+B} (output)
0	0	1
1	0	0
0	1	0
1	1	0

So the output of a NOR gate is 1 only when all of its inputs are 0. If any input is 1, the output is 0.

18.12Uses of logic gates

Every day, many things around us work with the help of logic gates. Logic gates are used in electronic circuits to do useful tasks. They are also used for security purposes, for example the burglar alarm and the fire alarm (the book calls it the fire extinguisher).

Two security uses of logic gates.
Use	Gate used in the book	Idea
Burglar alarm	AND gate	Alarm rings only when the person sensor and the alarm switch are both ON
Fire alarm	NAND gates	A thermistor input switches a buzzer on, and the buzzer stays on until a reset switch is pressed

18.12.1Burglar alarm

A burglar alarm system detects movement or the opening of doors and windows using sensors and contacts, and a loud alarm is emitted to warn others. A simple burglar alarm circuit can be made using an AND gate. It has two inputs, the person sensor and the alarm switch, as shown in Figure 18.23.

If both the person sensor and the alarm switch are ON, the burglar alarm is activated.
If only one input is ON, the burglar alarm will not be activated.

Figure 18.23: Burglar alarm.

For example, if the person sensor is ON but the alarm switch is not, the burglar alarm will not be activated. When the person sensor detects a person, it gives the signal 1. Then the alarm switch will also give a signal 1. Thus "1" and "1" give an output "1", which makes the alarm ring.

The burglar alarm follows the AND truth table.
Person sensor A	Alarm switch B	Alarm X=A\cdot B
0 (no person)	0 (off)	0: no alarm
1 (person)	0 (off)	0: no alarm
0 (no person)	1 (on)	0: no alarm
1 (person)	1 (on)	1: alarm rings

18.12.2Fire extinguisher (fire alarm circuit)

A fire alarm circuit is a simple circuit that detects fire. Fire alarm circuits are very important devices to detect fire at the right time and prevent damage to people or property. The circuit is required to switch on when activated and then sound a buzzer continuously. A fire alarm circuit using NAND gates is shown in Figure 18.24.

Figure 18.24: Fire alarm circuit (the resistor above input 4 is shown as a pull-up to +6 V).

How the circuit works

No fire. The buzzer is initially off, so the output from A is 0. This requires the output from gate B to be 1. With no fire, the thermistor is cold and its resistance is high, so there is a voltage across it and the two inputs to gate A are both 1.
Fire. When the thermistor is warmed by a fire, its resistance falls, as does the voltage across it, and input 1 falls to 0. The output from A becomes 1, so there is a voltage across the buzzer, which sounds.
Memory. The input 3 to B becomes 1, and since input 4 is also 1, the output from B becomes 0. The inputs to A are 1 and 0, so its output stays at 1 even if the thermistor cools again.
Reset. To reset the alarm, the reset switch R must be pressed. This makes input 4 a 0, so the output from B is 1 and A has inputs of 1 and 1, so its output is 0 as originally.

States of the fire alarm. A NAND gate gives 0 only when both of its inputs are 1.
Situation	Input 1 (thermistor)	Input 4 (reset)	Output of A (buzzer)	Output of B (to input 2)
No fire	1	1	0 (off)	1
Fire	0	1	1 (sounds)	0
Fire gone, thermistor cool again	1	1	1 (still sounds)	0
Reset switch R pressed	1	0	0 (off)	1

18.13Analogue and digital electronics: a comparison

Based on the types of signals used, electronics can be broadly classified into two branches: analogue electronics and digital electronics.

Analogue electronics: The branch of electronics that deals with electronic systems and devices with continuous-time signals.
Digital electronics: The branch of electronics that deals with electronic systems and devices with digital signals, or discrete-time signals.
Analogue quantity (signal): One having continuous values over a limit, for example temperature, time, pressure, sound and distance.
Digital quantity (signal): One having a set of discrete values over a limit.

Analogue electronics

Analogue electronics can be used to amplify signals, filter noise and perform a wide variety of other functions. In analogue electronics, two types of components are used to design the systems: the active elements, such as diodes and transistors, and the passive elements, such as resistors, capacitors and inductors. Analogue electronics is widely used in radio and audio devices such as FM radios, TVs and telephones.

Digital electronics

Digital electronics uses binary logic functions to perform operations. The basic idea of a binary logic function is that it has only two states, "active high" and "active low". One of the most common applications of digital electronics is in computers. Other applications include data processing and storage, automation and digital watches. Digital electronic systems are usually made from a combination of logic gates, often packaged in an integrated circuit (IC). Digital electronics uses active components only.

A digital system has many advantages over an analogue one, some of which are given below:

A digital system has small size.
Digital systems are generally easier to design.
Accuracy and precision are greater.
Digital systems need less power to operate.
Operation can be programmed.
Digital circuits are not affected by noise (noise is fluctuation in voltage).

An analogue quantity (or signal) is one having continuous values over a limit, for example temperature, time, pressure, sound and distance. A digital quantity (or signal) is one having a set of discrete values over a limit. These quantities change in discrete steps. Graphs of both an analogue and a digital signal are shown in Figure 18.25.

Analogue signal

Figure 18.25 (left): an analogue signal takes continuous values as time passes (a smooth curve).

Digital signal

Figure 18.25 (right): a digital signal changes in discrete steps and has only a set of separate values.

Analogue and digital electronics compared.
	Analogue electronics	Digital electronics
Signals	Continuous-time (continuous values)	Discrete-time (a set of separate values)
Components	Active (diode, transistor) and passive (resistor, capacitor, inductor)	Active components only; combinations of logic gates in ICs
Examples	FM radio, TV, telephone, audio devices	Computers, data processing and storage, automation, digital watches
Effect of noise	Fluctuations in voltage affect the signal	Not affected by noise

Information: temperature as an analogue quantity

The temperature of air varies in a continuous fashion during the 24 hours of a day. A graph of temperature against time is a smooth curve, like the one below. The air temperature does not jump from, say, 75^\circ\mathrm{F} to 80^\circ\mathrm{F} instantly; it takes the infinite values in between. So temperature is an analogue quantity.

Temperature of air during a day

The air temperature changes continuously through the day, so it is an analogue quantity (a sketch of the graph in the book).

Worked example: analogue or digital?

(i) Say whether each of these is an analogue or a digital quantity and give a reason: the pressure in a bicycle tyre; the ON or OFF state of a lamp; the sound from a loudspeaker; the number shown on a digital watch. (ii) Sketch a graph of a signal that switches between HIGH and LOW, and a graph of a signal that varies continuously.

Digital and analogue signals

The digital signal (coloured) jumps between LOW and HIGH; the analogue signal (dashed) changes continuously.

Quantity	Type	Reason
Pressure in a tyre	Analogue	Pressure can take any value over a range (continuous values)
ON or OFF state of a lamp	Digital	Only two discrete states, 1 or 0
Sound from a loudspeaker	Analogue	Sound varies continuously
Number on a digital watch	Digital	It changes in separate steps (a set of discrete values)

(ii) The graph below shows a digital signal that is either HIGH (1) or LOW (0), with the continuous analogue signal for comparison. The digital signal has no values in between; the analogue signal passes through every value.

18.14Quantum computers

Quantum computing uses phenomena of quantum physics to create new ways of computing. Unlike traditional (classical) computers, which use bits to represent information as either a 0 or a 1, quantum computers use qubits (quantum bits), which can exist in multiple states simultaneously. This allows quantum computers to perform complex calculations at an exponentially faster rate than classical computers.

Quantum computing: Computing that uses phenomena of quantum physics to create new ways of computing.
Qubit (quantum bit): The quantum mechanical analogue of a classical bit; it can exist in multiple states simultaneously.

Classical and quantum computing compared.
	Classical computer	Quantum computer
Unit of information	Bit	Qubit (quantum bit)
States	Either 0 or 1	Multiple states at once (a coherent superposition)
Speed for complex calculations	Slower	Exponentially faster

Challenges of quantum computers

Quantum computers are still in the early stages of development and have to overcome manufacturing challenges: the core components function only at very cold temperatures, at almost absolute zero. Most qubits must be cooled to within a few thousandths of a degree of absolute zero to eliminate thermal noise and vibrations, which tend to destroy the information contained in the qubits.

Formula sheet

Emitter current of a transistor

I_E=I_B+I_C

I_E: emitter current; I_B: base current; I_C: collector current (equation 18.1)

\mathrm{A} or \mathrm{mA} (the same unit for all three)

Current gain beta

\beta=\dfrac{I_C}{I_B}

I_C: collector current; I_B: base current (equation 18.2); the DC (static) value. For changes use \Delta I_C/\Delta I_B

no unit (a ratio)

Current gain alpha

\alpha=\dfrac{I_C}{I_E}

I_C: collector current; I_E: emitter current (equation 18.3)

no unit (a ratio, less than 1)

Collector current in a transistor switch

I_C=\beta\,I_B

The base current I_B controls the collector current I_C; with I_B=0 the transistor is in cutoff

Unit: \mathrm{mA}

AND gate

X=A\cdot B

Output X is 1 only when all inputs are 1

Units: Boolean (0 or 1)

OR gate

X=A+B

Output X is 0 only when all inputs are 0

Units: Boolean (0 or 1)

NOT gate

X=\bar{A}

Output X is the inverse of the input A

Units: Boolean (0 or 1)

NAND gate

X=\overline{A\cdot B}

AND followed by NOT: output is 0 only when all inputs are 1

Units: Boolean (0 or 1)

NOR gate

X=\overline{A+B}

OR followed by NOT: output is 1 only when all inputs are 0

Units: Boolean (0 or 1)

Key terms

Electronics: The branch of physics that helps to study the emission and behaviour of electrons with electronic devices.
Semiconductor: A material whose electrical conductivity falls between that of conductors and insulators (atoms of group IV-A, such as silicon and germanium).
Intrinsic semiconductor: A pure semiconductor, such as Si or Ge. In pure form it has no free charge carriers.
Extrinsic semiconductor: A semiconductor doped with some specific impurities, which modify its electrical properties. It is n-type or p-type.
N-type semiconductor: Silicon doped with a pentavalent impurity such as phosphorus; the extra (5th) electron is free, so electrons are the majority carriers.
P-type semiconductor: Silicon doped with a trivalent impurity such as boron; each impurity atom leaves a hole, so holes are the majority carriers.
Pentavalent impurity: An impurity of group V-A (phosphorus, arsenic, antimony, bismuth) with five valence electrons.
Trivalent impurity: An impurity of group III-A (boron, aluminium, indium, gallium) with three valence electrons.
Hole: A vacant spot in a covalent bond, considered a positive charge carrier.
Majority charge carriers: Electrons in an n-type semiconductor and holes in a p-type semiconductor.
Minority charge carriers: The carriers of the other sign in a doped semiconductor; they cause the small reverse (leakage) current of a diode.
PN junction: The junction formed when one half of a crystal is doped p-type and the other half n-type; also called a diode.
Depletion region: The region near a junction that is free of mobile charges; its ions set up an electric field from n to p.
Potential barrier: The potential difference across the depletion region: 0.3\,\mathrm{V} for Ge and 0.7\,\mathrm{V} for Si.
Anode and cathode: The p-type side (anode) and the n-type side (cathode) of a diode.
Biasing: Connecting a diode across an electric supply (a potential difference).
Forward bias: The positive terminal of a DC source joined to the p-type and the negative terminal to the n-type side: the depletion region narrows and a forward current of a few mA flows.
Reverse bias: The p-type side at a lower potential than the n-type side: the depletion region widens and only a leakage current of a few \mu\mathrm{A} flows.
Reverse (leakage) current: The tiny current of the order of a few microamperes that flows through a reverse-biased diode because of minority carriers.
Light-emitting diode (LED): A diode that emits energy in the form of light when electric current passes through it.
Transistor: A three-terminal semiconductor device with two PN junctions that amplifies current and acts as a switch or amplifier.
Emitter, base and collector: The three regions of a transistor: emitter (heavily doped), base (thin, lightly doped, in the middle) and collector (larger, moderately doped).
NPN and PNP transistor: NPN: a p-type base between two n-type regions. PNP: an n-type base between two p-type regions.
Current gain: \beta=I_C/I_B (static DC value) and \alpha=I_C/I_E; the emitter current is I_E=I_B+I_C.
Cutoff: The state of a transistor with no base current: the resistance between collector and emitter is infinite and the transistor acts as an open switch.
Saturation: The state of a transistor with enough base current: collector current is maximum and the transistor acts as a closed switch.
Relay: An electrically operated switch that opens and closes circuits by receiving electrical signals from outside sources.
Solid state relay (SSR): A relay with no moving parts that uses semiconductor switching devices; it switches faster and lasts longer than a mechanical relay.
Optocoupler: A light-emitting diode coupled with a photo transistor or photo-TRIAC, used to isolate the control and the controlled circuits.
Digital electronics: The branch of electronics that deals with digital signals (discrete-time signals), using binary logic functions.
Analogue electronics: The branch of electronics that deals with continuous-time signals.
Bit: Each digit in binary (0 or 1); an abbreviation of binary digit.
Boolean algebra: A formal way of describing logical operations, based on the three basic operators AND, OR and NOT.
Analogue-to-Digital Converter (ADC): An electronic circuit that converts continuously changing analogue signals into discrete digital signals.
Digital-to-Analogue Converter (DAC): An electronic circuit that converts discrete digital signals into analogue signals.
Logic gate: A digital circuit that implements a binary operation; it has one or more inputs and only one output.
Truth table: The set of inputs and outputs of a logic system in binary form.
Fundamental logic gates: The three gates AND, OR and NOT.
Universal logic gates: The gates NAND and NOR, because they can be used to make all the other gates.
Memory (of a circuit): A circuit that keeps its state after receiving an input is said to exhibit memory, as in the fire alarm circuit.
Quantum computing: Computing that uses phenomena of quantum physics to create new ways of computing; it uses qubits instead of bits.
Qubit: A quantum bit: the quantum mechanical analogue of a classical bit, able to be in a superposition of multiple states at once.

Multiple choice questionsp. 255–256

Choose the best possible option in the following questions.

1

The approximate potential barrier for germanium and silicon transistors are ______ respectively?

A0.3\,\mathrm{V} and 0.7\,\mathrm{V}Answer
B0.7\,\mathrm{V} and 0.3\,\mathrm{V}
C0.7\,\mathrm{V} and 0.5\,\mathrm{V}
D0.5\,\mathrm{V} and 0.7\,\mathrm{V}

Why

The potential barrier is the voltage a pn junction must overcome before it conducts. It is about 0.3\,\mathrm{V} for germanium and 0.7\,\mathrm{V} for silicon (Section 18.3), so option A is correct.

Option B swaps the two values, and options C and D use 0.5\,\mathrm{V}, which is not the barrier for either material.

Note. The question says transistors, but the potential barrier belongs to the pn junction, which the book gives as 0.3\,\mathrm{V} (Ge) and 0.7\,\mathrm{V} (Si). The answer is the same for the junction in a transistor.

2

The current ratio of a beta is:

AI_C/I_E
BI_B/I_C
CI_E/I_B
DI_C/I_BAnswer

Why

The current gain \beta is the ratio of collector current to base current, \beta=\dfrac{I_C}{I_B} (equation 18.2), so option D is correct.

Option A (I_C/I_E) is the current gain \alpha (equation 18.3). Option B is the inverse of \beta, and option C (I_E/I_B) is not a gain of the transistor.

3

Which of the following relation is correct for a transistor?

AI_C=I_B+I_E
BI_B=I_C+I_E
CI_E=I_C+I_BAnswer
DI_E=I_C-I_B

Why

The emitter current is the sum of the base and collector currents, I_E=I_B+I_C (equation 18.1), so option C is correct.

Option A makes the collector current the largest, which is wrong because the collector current is the smaller part of I_E. Option B makes the base current the sum, and option D subtracts the base current, which is not the transistor relation.

4

The heavily doped region of the transistor is:

AEmitterAnswer
BCollector
CBase
DAll of these

Why

The emitter is the heavily doped region. It supplies a large number of charge carriers into the base, so option A is correct.

The collector is only moderately doped, and the base is thin and lightly doped. Option D is therefore wrong.

5

A transistor has ______ PN junctions.

A1
B2Answer
C3
D4

Why

A transistor has two pn junctions: the emitter-base junction (EBJ) and the collector-base junction (CBJ) (Figure 18.10), so option B is correct.

A single pn junction is a diode, so option A is wrong. A transistor has three regions but only two junctions, so options C and D are wrong.

6

If I_E=1\,\mathrm{mA} and I_C=0.25\,\mathrm{mA}, then the I_B value will be:

A0.75\,\mathrm{mA}Answer
B0.95\,\mathrm{mA}
C0.65\,\mathrm{mA}
D0.85\,\mathrm{mA}

Why

From I_E=I_B+I_C, the base current is I_B=I_E-I_C=1\,\mathrm{mA}-0.25\,\mathrm{mA}=0.75\,\mathrm{mA}, so option A is correct.

The other options do not come from this subtraction. Option B would need I_C=0.05\,\mathrm{mA}, and option C would need I_C=0.35\,\mathrm{mA}.

7

If the gain \beta of an NPN transistor is 200 and its collector current is 4 mA. The value of base current will be:

A20\,\mu\mathrm{A}Answer
B25\,\mu\mathrm{A}
C30\,\mu\mathrm{A}
D35\,\mu\mathrm{A}

Why

From \beta=I_C/I_B, the base current is I_B=\dfrac{I_C}{\beta}=\dfrac{4\,\mathrm{mA}}{200}=0.02\,\mathrm{mA}=20\,\mu\mathrm{A}, so option A is correct.

Option B would need a collector current of 5\,\mathrm{mA}, and options C and D would need 6\,\mathrm{mA} and 7\,\mathrm{mA}.

8

The majority charge carriers in the emitter of an NPN transistor are:

Apentavalent atoms
BelectronsAnswer
Ctrivalent atoms
Dholes

Why

The emitter of an NPN transistor is an N-type region. In an N-type semiconductor the majority charge carriers are electrons, so option B is correct.

Pentavalent atoms are the donor impurities that create the electrons, not the carriers themselves. Trivalent atoms and holes belong to P-type material, and holes are minority carriers in N-type material.

9FBISE SSC-II 2026 Annual I, Set M, MCQ 4

The output of a NAND gate is 0 when:

AA is '1' and B is '0'
BA is '0' and B is '1'
Cboth A and B are '0'
Dboth A and B are '1'Answer

Why

A NAND gate is an AND gate followed by a NOT gate, X=\overline{A\cdot B}. The AND output is 1 only when both inputs are 1, so the NAND output is 0 only when both inputs are 1 (Table 18.5), option D.

With A = 1 and B = 0, or A = 0 and B = 1, the AND output is 0, so the NAND output is 1. With both inputs 0 the NAND output is also 1.

10FBISE SSC-II 2026 Annual I, Set B, MCQ 4

If X=A+B, then X is 0 when:

Aboth A and B are '0'Answer
BA or B is '0'
CA is '0' and B is '1'
DA is '1' and B is '0'

Why

An OR gate gives X=A+B. Its output is 0 only when all inputs are 0 (Table 18.3), so option A is correct.

If one input is 0 and the other is 1, the OR output is 1, so options B, C and D are wrong.

11

The output of a NOR gate is '1' when:

Aboth of its inputs are '0'Answer
Bboth of its inputs are '1'
Cany of its inputs is '0'
Dany of its inputs is '1'

Why

A NOR gate is an OR gate followed by a NOT gate, X=\overline{A+B}. Its output is 1 only when the OR output is 0, which happens only when both inputs are 0 (Table 18.6), so option A is correct.

If any input is 1, the OR output is 1 and the NOR output is 0. This rules out options B, C and D.

12

The logical operation performed by this gate is:

Gate for question 12: an AND-shaped body with a small circle at the output.

AAND
BNOR
CNANDAnswer
DOR

Why

The gate has an AND-shaped body, so it is an AND gate, and the small circle at its output is a NOT. An AND gate followed by a NOT is a NAND gate, so option C is correct.

An AND gate alone (option A) has no circle at the output. A NOR gate (option B) and an OR gate (option D) have a curved, pointed body, not the AND shape.

Constructed response questionsp. 256–257

Follow the directions to respond to the following questions.

1

Consider the given logic circuit.

The given logic circuit: inputs A and B feed gate 1, whose output feeds gate 2, giving X.

Solution

The circuit is an AND gate followed by a NOT gate, so its output is X=\overline{A\cdot B}. Part (b) asks for a single gate with the same output.

(a) Identify the logic gates in the circuit. Name of the gate 1 is ______. Name of the gate 2 is ______.

Gate 1 is an AND gate. Its output is 1 only when both inputs A and B are 1.

Gate 2 is a NOT gate (inverter). The triangle with a small circle at its output gives the opposite of its input.

\text{Gate 1: AND gate}, \quad \text{Gate 2: NOT gate}

(b) Draw a single logic gate that can replace the given logic circuit.

The circuit gives X=\overline{A\cdot B}: an AND gate followed by a NOT gate. Joined together, these form one NAND gate, whose symbol is an AND shape with a circle at the output.

The single gate that replaces the circuit: a NAND gate, with X=\overline{A\cdot B}.

X=\overline{A\cdot B}\;\text{(NAND gate)}

2

Write the expression for the following combination of logic gates. Here A and B are inputs and X is output. Also complete its truth table.

The logic circuit: a NAND gate and a NOR gate take inputs A and B, and their outputs feed a NAND gate that gives X.

Truth table to complete
A	B	X =
0	0
1	0
0	1
1	1

Solution

Step 1: outputs of the first two gates

The top gate is a NAND gate with inputs A and B.
G_1=\overline{A\cdot B}
The bottom gate is a NOR gate with inputs A and B.
G_2=\overline{A+B}

Step 2: the output gate

The third gate is a NAND gate with inputs G_1 and G_2.
X=\overline{G_1\cdot G_2}=\overline{\overline{A\cdot B}\cdot\overline{A+B}}
Apply De Morgan's law, \overline{P\cdot Q}=\overline{P}+\overline{Q}, and remove the double bars.
X=AB+(A+B)
Simplify. AB is already contained in A+B, since any case with both inputs 1 also has A+B=1.
X=A+B

Step 3: complete the truth table

A	B	\overline{A\cdot B}	\overline{A+B}	X
0	0	1	1	0
1	0	1	0	1
0	1	1	0	1
1	1	0	0	1

Each row is checked by substitution. For A=B=0 both gate outputs are 1, so X=\overline{1\cdot1}=0. For A=1,B=0 the NOR output is 0, so X=\overline{1\cdot0}=1. For A=B=1 the NAND output is 0, so X=\overline{0\cdot0}=1.

X=\overline{\overline{A\cdot B}\cdot\overline{A+B}}=A+B

Note. The book prints no answer for this part; the truth table column is blank. The output is 1 whenever A or B is 1, which is the OR function.

Short response questionsp. 257

Give a short response to the following questions.

13 marks

How do N-type materials and forward biasing affect semiconductor devices?

Solution

N-type material. Adding a pentavalent impurity such as phosphorus to silicon gives each impurity atom one valence electron that does not take part in a covalent bond. This electron is free to move, so the N-type material has many free electrons. Electrons are its majority charge carriers, and the material conducts much better than pure silicon.

Forward biasing. When the positive terminal of a DC source is connected to the P-type side and the negative terminal to the N-type side, the holes and electrons are pushed towards the junction. The depletion region narrows and the potential barrier is reduced. Once the barrier is overcome, current flows through the junction. The current is of the order of milliamperes.

Together, the free electrons of the N-type layer supply the current, and forward biasing lowers the barrier so that the pn junction conducts. This is how a diode works as a rectifier.

23 marks

Why are pentavalent impurities called donors and trivalent impurities acceptors?

Solution

A pentavalent impurity (group V-A) has five valence electrons. Four of them form covalent bonds with neighbouring silicon atoms, and the fifth is left over. This extra electron is given (donated) to the crystal, where it becomes a free charge carrier. That is why the impurity is called a donor.

A trivalent impurity (group III-A) has three valence electrons, so one covalent bond is short of an electron. The impurity atom takes (accepts) an electron from a neighbouring bond to complete its bonds. This leaves a hole in the bond it took the electron from, so the impurity is called an acceptor.

33 marks

How do relays improve electrical systems? Give examples.

Solution

A relay is an electrically operated switch. A small electrical signal in its input circuit operates the contacts that switch a much larger current or voltage in the output circuit. A low-power circuit can therefore control a high-power or high-voltage device, and the two circuits are kept electrically separate.

Examples:

A timer circuit with a relay can switch the power on at a preset time.
Relays protect electrical circuits from overload or faults by opening the circuit.
A small DC signal can turn high-power devices such as lamps or garage door motors on and off.
In a TV, pressing a button on the remote control sends a signal to a relay, which turns the main power on or off (Figure 18.12).

43 marks

How do the three terminals of a transistor work together to amplify signals?

Solution

The emitter injects charge carriers into the base. The base is thin and lightly doped, so only a small part of the carriers (about 2 to 5 per cent of the emitter current) is lost to the base. Most carriers cross the base and are collected by the collector, which is reverse biased and draws them across.

A small change in the base current therefore produces a much larger change in the collector current. Since I_C=\beta I_B, the small signal in the base circuit appears multiplied in the collector circuit. This is how the transistor amplifies. The emitter current supplies both, with I_E=I_B+I_C.

53 marks

Why is one transistor region highly doped and another the smallest in size?

Solution

The emitter is heavily doped so that it supplies a large number of charge carriers into the base. Those carriers are the source of the collector current.

The base is the smallest region. It is thin (about 0.001 inch) and lightly doped. Because it is so thin, most carriers that enter from the emitter pass through it to the collector, and only a few combine with holes in the base. This keeps the base current small compared with the collector current.

63 marks

What does the arrow in a transistor symbol represent, and why is it important?

Solution

The arrow is on the emitter lead. It shows the direction of conventional current through the emitter, which is the direction in which the emitter-base junction conducts when it is forward biased. In an NPN transistor the arrow points outwards, away from the base. In a PNP transistor it points inwards, towards the base (Figure 18.10).

The arrow is important because it identifies the emitter, which tells us which lead is the emitter, and it shows whether the device is an NPN or a PNP transistor. A transistor connected with its emitter in the wrong place will not work as intended.

73 marks

Explain the normal biasing of an NPN transistor. What happens if it is reversed?

Solution

In normal operation the emitter-base junction is forward biased. The negative terminal of V_{BE} is connected to the N-type emitter and the positive terminal to the P-type base. The collector-base junction is reverse biased: the positive terminal of V_{CB} is connected to the N-type collector and the negative terminal to the P-type base.

If the biasing is reversed, the emitter-base junction becomes reverse biased. The depletion region widens and the potential barrier grows, so electrons cannot be injected into the base. Only a tiny leakage current flows, the transistor is switched off, and the collector current is almost zero. The transistor then cannot amplify.

83 marks

How does biasing (forward or reverse) change the depletion region in a diode?

Solution

Forward biasing pushes the majority carriers (electrons in the N-side, holes in the P-side) towards the junction. The depletion region becomes narrower, the potential barrier is reduced, and current can flow.

Reverse biasing pulls the holes and electrons away from the junction. The depletion region becomes wider, the barrier increases, and the majority carriers cannot cross. Only a very small reverse (leakage) current, of the order of microamperes, flows.

93 marks

Why does an n-type or p-type material have zero net charge?

Solution

A pure silicon crystal is electrically neutral. Doping adds a small number of impurity atoms, and each of them is itself neutral: a phosphorus or boron atom has as many protons as electrons.

In an N-type material each donor atom gives one extra electron. This electron is free, but it is balanced by the positive charge of the donor nucleus, so the total charge of the material is still zero. In a P-type material each acceptor atom takes an electron and creates a hole, and the positive and negative charges again balance. The material as a whole stays neutral; only the carriers move.

103 marks

Why are NAND and NOR gates called universal gates?

Solution

NAND and NOR gates are called universal gates because every other logic gate (NOT, AND and OR) can be built from either of them alone.

NOT from NAND: connect both inputs of a NAND gate together. \overline{A\cdot A}=\overline{A}.
AND from NAND: a NAND followed by a NOT (a second NAND with its inputs tied together) gives \overline{\overline{A\cdot B}}=A\cdot B.
OR from NAND: A+B=\overline{\overline{A}\cdot\overline{B}} by De Morgan's law, so two NOT gates (NAND with tied inputs) followed by a NAND give OR.

113 marks

For a NOR gate, how do you determine the output for different inputs?

Solution

A NOR gate is an OR gate followed by a NOT gate, so X=\overline{A+B}. To find the output, first find the OR output (1 if either input is 1), then invert it.

A	B	A+B	X=\overline{A+B}
0	0	0	1
1	0	1	0
0	1	1	0
1	1	1	0

The output is 1 only when both inputs are 0. If either input is 1, the output is 0 (Table 18.6).

123 marks

What makes digital systems better than analogue systems?

Solution

Digital systems have several advantages over analogue systems:

They are small in size.
They are generally easier to design.
Their accuracy and precision are greater.
They need less power to operate.
Their operation can be programmed.
They are not affected by noise (fluctuations in voltage).

The main reason is that a digital signal has only two states, 0 and 1. A small disturbance does not change the reading unless it is large enough to turn a 0 into a 1, whereas an analogue signal can take any value and so is altered by every small disturbance.

Long response questionsp. 257–258

Give a detailed response to the following questions.

1

What is an extrinsic semiconductor? How do impurities affect its conductivity?

Solution

Extrinsic semiconductor

An extrinsic semiconductor is a pure semiconductor (such as silicon or germanium) to which a small, controlled amount of a specific impurity has been added. This process is called doping. The impurity changes the electrical properties of the semiconductor so that it can be used in electronic devices.

Effect of impurities on conductivity

A pure silicon crystal has very few free charge carriers, so it is a poor conductor. Doping adds charge carriers, and this raises the conductivity greatly. The type of impurity decides which carriers are added:

Impurity	Group	Valence electrons	Effect	Type
Pentavalent (P, As, Sb, Bi)	V-A	5	gives extra free electrons	N-type
Trivalent (B, Al, In, Ga)	III-A	3	creates holes (a missing electron in a bond)	P-type

Majority carriers

In an N-type semiconductor the majority carriers are electrons, and in a P-type semiconductor they are holes. The other carriers in each material (holes in N-type, electrons in P-type) are present in much smaller numbers.

2

Describe how a PN junction forms and explain the effects of forward and reverse biasing on it.

Solution

Formation of a PN junction

When a crystal is grown so that one half is doped with a trivalent impurity (P-type) and the other half with a pentavalent impurity (N-type), a junction forms between the two regions. At the junction, randomly moving electrons from the N-side diffuse into the P-side and fill some holes. As a result, negative ions form near the junction on the P-side and positive ions on the N-side.

These ions set up an electric field directed from the N-side to the P-side. The region around the junction, which has no mobile charge carriers, is called the depletion region. The field forms a potential barrier of about 0.3\,\mathrm{V} for germanium and 0.7\,\mathrm{V} for silicon. The depletion region stops further diffusion of carriers.

Forward bias

When the positive terminal of a DC source is connected to the P-type side and the negative terminal to the N-type side, the holes and electrons are driven towards the junction. The depletion region narrows and the barrier is reduced. When the applied voltage exceeds the barrier (the knee voltage), a conducting path is set up and current flows, of the order of milliamperes.

Reverse bias

When the P-type side is connected to the negative terminal and the N-type side to the positive terminal, the holes are pulled away from the junction on the P-side and the electrons on the N-side. The depletion region widens and the barrier increases, so the majority carriers cannot cross. Only a very small reverse (leakage) current of the order of microamperes flows, due to the minority carriers.

Current-voltage graph

Current against voltage for a pn junction

Figure 18.8: I-V curve for a pn junction. The current is almost zero until the knee voltage, then rises sharply in forward bias. In reverse bias only a very small leakage current flows.

3

Using a diagram, explain how a solid-state relay works and its advantages over mechanical relays.

Solution

How a solid-state relay works

A solid-state relay (SSR) has no moving parts. A small control signal at its input switches a larger load circuit using semiconductor devices. An optocoupler (an LED coupled with a photodiode) keeps the control circuit and the load circuit electrically apart.

The control current lights the LED on the input side.
The light from the LED falls on the photodiode array, which converts it into a voltage.
This voltage acts as the gate voltage of the transistor through the control circuit, and the transistor switches the output circuit on or off.

Figure 18.13 and 18.14: block diagram of a solid-state relay. The optocoupler is the only link between the input and the output.

Advantages over mechanical relays

It has no moving parts, so it is more durable and reliable over a long period.
It switches faster, because no mechanical contacts have to open and close.
The optocoupler isolates the control circuit from the controlled circuit.

4

Describe how an NPN transistor operates, focusing on the movement of charge carriers.

Solution

Structure and biasing

An NPN transistor has a thin, lightly doped P-type base sandwiched between two heavily doped N-type regions, the emitter and the collector. The emitter-base junction is forward biased by V_{BE} and the collector-base junction is reverse biased by V_{CB}.

Figure 18.11: an NPN transistor under normal operation. Electrons leave the emitter, cross the thin base and are collected by the collector.

Movement of charge carriers

The forward bias lowers the barrier at the emitter-base junction, so electrons flow from the emitter into the base.
The base is very thin and lightly doped, so only a few electrons (about 2 to 5 per cent of the emitter current) combine with holes in the base. These form the base current I_B, which is supplied by the battery V_{BE}.
The collector is reverse biased, and the large reverse bias draws the electrons across the collector-base depletion region. About 95 to 98 per cent of the electrons pass through the base and are collected. This forms the collector current I_C.
The emitter current is the sum of the two, I_E=I_B+I_C.

Current gain

Because I_C is much larger than I_B, a small change in base current produces a large change in collector current. The current gain is \beta=\dfrac{I_C}{I_B} (static value) and \alpha=\dfrac{I_C}{I_E}.

5

How can a transistor function as a switch? Provide examples of its use in circuits.

Solution

Transistor as a switch

A transistor works as a switch when it is biased into its two extreme states: cut-off (switch open) and saturation (switch closed). The collector and emitter act as the two terminals of the switch, and the base current decides whether it is open or closed. Since I_C=\beta I_B, a small base current controls a much larger collector current. If the base current is below the minimum cut-off value, the transistor behaves as an open switch.

Cut-off: switch open

If the switch between the base and the battery is open, there is no base current. With I_B=0, no collector current flows, the resistance between C and E becomes very large, and the lamp is off.

Saturation: switch closed

If the base switch is closed, a base current flows. This allows a much larger collector current to flow from C to E, so the resistance between C and E is very small. The circuit current is then at its maximum, the lamp lights, and the transistor is said to be saturated.

State	Base switch	Base current	Collector current	Lamp
Cut-off	open	zero	almost zero	off
Saturation	closed	flows	maximum	on

Examples of use

Switching a lamp on and off with a small base current (Figures 18.15 and 18.16).
Digital devices use transistors as rapid-response switches, because they are both economical and fast (Section 18.6.3).

6

Explain the working of an OR gate with a circuit diagram. How does it process input signals?

Solution

Working of the OR gate

An OR gate gives an output of 1 when at least one of its inputs is 1. In the switch circuit, switches A and B are connected in parallel with each other, and this pair is connected in series with the battery and the lamp. The lamp lights if either switch is closed, or both are closed.

Figure 18.19: an OR gate. (a) Two switches in parallel control the lamp. (b) The OR symbol, with X = A + B.

How the inputs are processed

When both switches are open, both inputs are 0 and the lamp is off, so the output is 0. When switch A is closed and B is open, or B is closed and A is open, or both are closed, at least one input is 1 and the lamp lights, so the output is 1.

Table 18.3: truth table of the OR gate
A	B	X=A+B	Lamp
0	0	0	off
1	0	1	on
0	1	1	on
1	1	1	on

The Boolean expression is X=A+B, read as "X equals A OR B". The plus sign means OR, not addition: 1+1=1 in Boolean algebra.

7FBISE SSC-II 2026 Annual I, Set M, Q.6 OR

How are logic gates used in burglar alarms and fire extinguishers? Explain with examples.

Solution

Burglar alarm (AND gate)

A simple burglar alarm uses an AND gate with two inputs: a person sensor (input A) and an alarm switch (input B). The alarm rings only when both inputs are 1, so X=A\cdot B.

Figure 18.23: a simple burglar alarm with an AND gate. The alarm rings only when both the person sensor and the alarm switch are ON.

If the person sensor detects someone (1) but the alarm switch is off (0), the output is 0 and the alarm stays silent. When both inputs are 1, the output becomes 1 and the alarm rings.

Fire alarm circuit (NAND gates)

The book's fire alarm circuit (Figure 18.24) uses two NAND gates, A and B, as a latch. A thermistor and a variable resistor set the input voltages, and a buzzer is connected to the output of gate A. When there is no fire, the thermistor is cold and its resistance is high, so inputs 1 and 2 of gate A are both 1 and the buzzer is off.

When a fire warms the thermistor, its resistance falls, so input 1 of gate A falls to 0. The output of A becomes 1 and the buzzer sounds. Once the output of A is 1, the output of B falls to 0, which keeps the output of A at 1 even after the thermistor cools. The alarm is reset by pressing the reset switch R, which makes input 4 of gate B equal to 0. The output of B returns to 1, and the output of A returns to 0.

8

Discuss the challenges in development of quantum computers.

Solution

Why quantum computers are promising

Quantum computers use qubits, which can be in a coherent superposition of multiple states at the same time. This allows them to perform some complex calculations at an exponentially faster rate than classical computers.

Challenges

Very low temperatures: the core components work only at very cold temperatures. Most qubits must be cooled to within a few thousandths of a degree of absolute zero.
Noise: thermal noise and vibrations tend to destroy the information stored in the qubits, so the qubits must be isolated from them.
Manufacturing: the core components are difficult to make so that they function reliably, and engineers must overcome these manufacturing challenges.
Early stage: quantum computers are still in the early stages of development, so the technology is not yet ready for general use.

Numerical response questionsp. 258

Solve the questions given below.

1

In a junction transistor, emitter current is 5.82\,\mathrm{mA} and base current is 120\,\mu\mathrm{A}. Calculate the collector current.

Solution

Given

I_E = 5.82\,\mathrm{mA}
I_B = 120\,\mu\mathrm{A}=0.12\,\mathrm{mA}

To find

I_C (collector current)

Use the relation between the three currents and make I_C the subject.
I_E=I_B+I_C \quad\Rightarrow\quad I_C=I_E-I_B
Substitute the values, using I_B=0.12\,\mathrm{mA}.
I_C=5.82\,\mathrm{mA}-0.12\,\mathrm{mA}
Evaluate.
I_C=5.70\,\mathrm{mA}

I_C=5.7\,\mathrm{mA}

Book answer. 5.7 mA

Note. The working agrees with the book's answer.

2

For a 600\,\mu\mathrm{A} base current and 30\,\mathrm{mA} collector current, how much is the current gain \beta?

Solution

Given

I_B = 600\,\mu\mathrm{A}=0.6\,\mathrm{mA}
I_C = 30\,\mathrm{mA}

To find

\beta (current gain)

The current gain is the ratio of collector current to base current.
\beta=\dfrac{I_C}{I_B}
Use the same units for both currents: 600\,\mu\mathrm{A}=0.6\,\mathrm{mA}.
\beta=\dfrac{30\,\mathrm{mA}}{0.6\,\mathrm{mA}}
Evaluate.
\beta=50

\beta=50

Book answer. 50

3

Calculate the value of \beta gain if an emitter current is changed by 4\,\mathrm{mA} and the collector current changes by 3.5\,\mathrm{mA}.

Solution

Given

\Delta I_E = 4\,\mathrm{mA}
\Delta I_C = 3.5\,\mathrm{mA}

To find

\beta (ac current gain)

Since I_E=I_B+I_C, the changes obey the same relation: \Delta I_E=\Delta I_B+\Delta I_C. Make the change in base current the subject.
\Delta I_B=\Delta I_E-\Delta I_C
Substitute the changes.
\Delta I_B=4\,\mathrm{mA}-3.5\,\mathrm{mA}=0.5\,\mathrm{mA}
For changes in the currents, the gain is the ratio of the changes in collector and base current.
\beta=\dfrac{\Delta I_C}{\Delta I_B}
Evaluate.
\beta=\dfrac{3.5\,\mathrm{mA}}{0.5\,\mathrm{mA}}=7

\beta=7

Book answer. 7

Note. The question reads "the collector current is changes by 3.5 mA"; read it as "changes by". Because the currents are changes, this is the dynamic (ac) gain.

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 11 mark

The main function of a diode in a circuit is to:

AReduce voltage
BAmplify current
CStore charge
DRectify currentAnswer

Why

A diode conducts when it is forward biased and blocks current when it is reverse biased, so it lets current pass in one direction only. This is why it is used as a rectifier: it changes alternating current into one-way (direct) current. Option D is correct.

Reducing voltage is the job of a resistor, amplifying current is the job of a transistor, and storing charge is the job of a capacitor. None of these is the main function of a diode.

FBISE SSC-II 2026 Annual I, Set B, MCQ 41 mark

The output of an OR gate is LOW when:

AAny one input is LOW
BAny one input is HIGH
CAll inputs are LOWAnswer
DAll inputs are HIGH

Why

An OR gate has X=A+B. Its output is 1 (HIGH) if at least one input is 1, so it is 0 (LOW) only when all the inputs are 0 (Table 18.3). Option C is correct.

Option A is the tempting one: with one input LOW and the other HIGH the OR output is still HIGH. That rule (output LOW when any input is LOW) belongs to the AND gate, not the OR gate.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (ii) OR1.5+1.5 = 3 marks

State two functions of a diode in electronic circuits.

Solution

A diode is a PN junction that conducts when it is forward biased and blocks the current when it is reverse biased. Its two main functions follow from this.

Rectification (0.75 mark for the name, 0.75 for the reason). A diode is used as a rectifier. Because it allows current in one direction only, it changes alternating current (which reverses its direction) into current that flows in one direction only.
One-way switch (0.75 mark for the name, 0.75 for the reason). A diode acts like a closed switch (a conductor) when forward biased, with a forward current of a few milliamperes. It acts like an open switch (an insulator) when reverse biased, with only a leakage current of a few microamperes.

A light-emitting diode is a special diode with a further use: it gives out light when forward biased, so it is used in display devices and decorative lamps.

FBISE SSC-II 2026 Annual I, Set B, Q.6 OR5 marks

Describe the action of light emitting diode (LED) in passing current in one direction only and emitting light. Make diagram as well.

Solution

What an LED is

A light-emitting diode (LED) is a PN junction diode, made from gallium arsenide phosphide (GaAsP) or gallium phosphide (GaP), that emits energy in the form of light when an electric current passes through it. It is enclosed in a transparent cover so that the light can be seen.

Passing current in one direction only

Forward bias. The positive terminal of the supply is joined to the p-type side and the negative terminal to the n-type side, with a resistor in series to limit the current. The depletion region becomes narrow and the potential barrier is lowered, so electrons and holes cross the junction and a forward current of a few milliamperes flows.
Reverse bias. If the connections are reversed, the depletion region widens and the barrier rises, so the majority carriers cannot cross. Only a very small leakage current flows, and the LED does not glow.
So, like any diode, the LED passes current in one direction only, and it must be forward biased to work.

Emission of light

In forward bias the free electrons of the conduction band of the n-region cross the lowered barrier and enter the p-region. There they fall into holes lying in the valence band. Each electron therefore drops from a higher energy level to a lower energy level, and the energy it loses is radiated as light. LEDs can give red, green or blue light.

Diagram

Figure 18.9 (a) and (b): the symbol of the LED and its forward-biased circuit. The arrows pointing away from the symbol show the light that is emitted.

Figure 18.9 (d): inside a forward-biased LED. An electron falls from the conduction band into a hole in the valence band and the energy it loses leaves as light.

Note. The marks for this question are not clearly printed; 5 marks are assumed (Section C is 4 x 5).

FBISE SSC-II 2026 Annual I, Set D, MCQ 11 mark

A diode conducts electricity when it is:

AHeated
BReverse biased
CForward biasedAnswer
DUnbiased

Why

A diode conducts when it is forward biased: the positive terminal of the supply is connected to the p-type side and the negative terminal to the n-type side. The depletion region becomes narrow, the barrier is reduced and a forward current flows. Option C is correct.

Reverse bias widens the depletion region and blocks the current (only a microampere leakage current flows). With no bias the barrier of the junction stays in place, so there is no current, and heating is not the condition that makes a diode conduct.

FBISE SSC-II 2026 Annual I, Set D, MCQ 41 mark

For a NOR gate:

AX=\overline{A.B}
BX=A+B
CX=\overline{A+B}Answer
DX=A.B

Why

A NOR gate is an OR gate followed by a NOT gate, so X=\overline{A+B} (the bar covers the whole sum). Option C is correct.

Option A is the expression of a NAND gate, option B is an OR gate and option D is an AND gate. The bar in front of the whole sum is what makes it NOR.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (ii) OR2+1 = 3 marks

What is the role of a diode in rectification? Explain briefly diagram.

Solution

Role of the diode (2 marks)

Rectification means changing alternating current (ac) into current that flows in one direction only. A diode does this because it allows current in one direction only: it conducts when it is forward biased and blocks the current when it is reverse biased.

When an ac source is joined in series with a diode and a load resistor R, the diode is forward biased in the half-cycle in which its p-side (arrow side) is positive, and current flows through R. In the next half-cycle the polarity reverses, the diode is reverse biased and no current flows. The current through R therefore flows in pulses in one direction only (a pulsating direct current), which is why the diode is called a rectifier.

Diagram (1 mark)

Half-wave rectification: an ac source, a diode and a load resistor R in series. The diode conducts only in the half-cycles in which it is forward biased, so the current through R flows in one direction only.

FBISE SSC-II 2026 Annual I, Set D, Q.6 OR5 marks

Make truth table of following gates: AND, OR, NAND, NOR, NOT.

Solution

A truth table lists the output of a gate for every possible combination of its inputs. A gate with two inputs has 2^{2}=4 rows. The symbols and Boolean expressions are shown first, then the truth tables (1 mark for each gate).

Circuit symbols of the five gates. The small circle on the NAND and NOR symbols shows the NOT operation.

AND, OR, NAND and NOR (two inputs A and B)

Truth tables of the AND, OR, NAND and NOR gates (Tables 18.2, 18.3, 18.5 and 18.6).
A	B	AND X=A\cdot B	OR X=A+B	NAND X=\overline{A\cdot B}	NOR X=\overline{A+B}
0	0	0	0	1	1
1	0	0	1	1	0
0	1	0	1	1	0
1	1	1	1	0	0

NOT (one input A)

Truth table of the NOT gate (Table 18.4).
A	NOT X=\bar{A}
0	1
1	0

Rule for each gate

AND: the output is 1 only when all the inputs are 1.
OR: the output is 0 only when all the inputs are 0.
NAND: the AND output turned over, so the output is 0 only when all the inputs are 1.
NOR: the OR output turned over, so the output is 1 only when all the inputs are 0.
NOT: the output is the opposite of the input.

Note. The marks for this question are not clearly printed; 5 marks are assumed (Section C is 4 x 5).

FBISE SSC-II 2026 Annual I, Set M, MCQ 11 mark

A diode is commonly used as a:

AResistor
BTransformer
CRectifierAnswer
DCapacitor

Why

A diode passes current in one direction only, so it is commonly used as a rectifier, changing ac into one-way current. Option C is correct.

A resistor, a transformer and a capacitor each have a different job (limiting current, changing ac voltage and storing charge). None of them depends on the one-way action of a PN junction.

FBISE SSC-II 2026 Annual I, Set M, MCQ 41 mark

If both inputs of a NAND gate are HIGH, the output will be:

ASame as input
BHIGH
CUndefined
DLOWAnswer

Why

A NAND gate is an AND gate followed by a NOT gate, X=\overline{A\cdot B}. With both inputs 1 (HIGH) the AND output is 1, and the NOT gate turns it into 0 (LOW). This is the only row of Table 18.5 where the NAND output is 0, so option D is correct.

Option B (HIGH) is what an AND gate gives for these inputs. For a NAND gate the output is HIGH in the other three rows.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (ii) OR1.5+1.5 = 3 marks

Describe the working of a diode under forward bias. Also draw diagram.

Solution

Working under forward bias (1.5 marks)

A diode is forward biased when the positive terminal of a DC source is connected to its p-type side and the negative terminal to its n-type side.

The holes of the p-side are repelled by the positive terminal and driven towards the junction.
The electrons of the n-side are repelled by the negative terminal and driven towards the junction.
The depletion region becomes narrow and the potential barrier is reduced.
Once the applied voltage lowers the barrier enough (about 0.7\,\mathrm{V} for silicon, 0.3\,\mathrm{V} for germanium), electrons and holes combine at the junction and a current passes through the diode. This forward current is of the order of a few milliamperes, and the diode acts like a conductor.

Diagram (1.5 marks)

Figure 18.6 (a): a forward-biased diode. Positive terminal to the p-side, negative terminal to the n-side; the depletion region is narrow.

FBISE SSC-II 2026 Annual I, Set M, Q.6 OR3+2 = 5 marks

Explain the working of burglar alarm. Also draw its circuit diagram.

Solution

Principle (3 marks)

A burglar alarm detects an intruder and sounds a loud bell to warn others. The book's simple burglar alarm is built with an AND gate that has two inputs: a person sensor (input A) and an alarm switch (input B). The gate output is X=A\cdot B.

Working

When the person sensor detects a person it gives the signal 1. When there is nobody it gives 0. The alarm switch is 1 when the alarm system is switched ON and 0 when it is OFF.
An AND gate gives 1 only when both inputs are 1. So the alarm is activated only when the system is ON and the sensor detects a person.
X=A\cdot B
If only one input is 1 (for example the sensor detects someone but the alarm switch is off), the output is 0 and the alarm stays silent.
The output of the gate is only a small signal. When X=1 it supplies a base current I_B to an npn transistor used as a switch, so the transistor saturates (collector and emitter act as a closed switch) and a collector current I_C=\beta I_B flows through the coil of a relay.
The relay is an electrically operated switch: its coil closes the contacts of a separate circuit that holds a battery and the bell, so the bell rings. When X=0 the base current is zero, the transistor is in cutoff (open switch), the relay is off and the bell is silent.

The alarm follows the AND truth table.
Person sensor A	Alarm switch B	X=A\cdot B	Transistor	Relay and bell
0 (no person)	0 (off)	0	cutoff (open switch)	off, no alarm
1 (person)	0 (off)	0	cutoff	off, no alarm
0 (no person)	1 (on)	0	cutoff	off, no alarm
1 (person)	1 (on)	1	saturated (closed switch)	on, bell rings

Circuit diagram (2 marks)

Burglar alarm. The AND gate output X switches an npn transistor; the transistor feeds the relay coil; the relay contacts close the circuit of the bell. The relay is the part that lets the small signal switch the bell.

Block diagram of the burglar alarm: the bell rings only when the AND gate gives 1 and the transistor and relay pass the signal on.

Note. The marks for this question are not clearly printed; the split 3+2 (explanation and circuit diagram) is read from the paper. The book's burglar alarm is the AND-gate circuit; the transistor and relay output stage is added from sections 18.7.2 and 18.7.3.

Exam focus

Describe the action of a light-emitting diode: why it passes current one way only and how it emits light. Draw the symbol, the forward-biased circuit and the energy diagram. (long)
Set B Q.6 OR (5 marks) asked exactly this, and it is one of the unit's SLOs. The four-step answer (forward bias, electrons cross, fall into holes, energy leaves as light) is short and gets full marks.
Go to answer
Truth tables, symbols and Boolean expressions of the AND, OR, NOT, NAND and NOR gates. (long)
Set D Q.6 OR (5 marks) asked for all five truth tables. One logic-gate MCQ appeared in every set (Set B MCQ 4 on OR, Set D MCQ 4 on NOR, Set M MCQ 4 on NAND), and book exercise MCQs 9 to 12 and short question 11 test the same rules.
Go to answer
Explain the working of a burglar alarm and draw its circuit. (long)
Set M Q.6 OR (3+2 marks) and comprehensive question 7 of the book. Learn the AND gate circuit with the person sensor and alarm switch, its truth table, and the transistor and relay stage that rings the bell.
Go to answer
Describe the working of a diode under forward bias, with a diagram; compare it with reverse bias. (short)
Set M Q.2 (ii) OR (1.5+1.5 marks); also short question 8 and comprehensive question 2. Always state the connection, the effect on the depletion region and barrier, and the mA current.
Go to answer
The diode as a rectifier: functions of a diode and how one rectifies ac. (short)
Set B Q.2 (ii) OR asked for two functions of a diode and Set D Q.2 (ii) OR asked for its role in rectification with a diagram.
Go to answer
One-line diode MCQs: the diode as a rectifier, and when a diode conducts. (mcq)
MCQ 1 of all three 2026 sets was a diode question (main function, conducts when forward biased, commonly used as a rectifier).
Go to answer
The transistor as a switch (cutoff and saturation) and how relays improve circuits. (long)
Not asked on its own in 2026, but it forms the output stage of the burglar alarm answer in Set M Q.6 OR. Also comprehensive question 5 and short question 3 of the book.
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Transistor current numericals: I_E=I_B+I_C, \beta=I_C/I_B and the base current for a given gain. (numerical)
There was no numerical from this unit in the 2026 papers, but the book's three numericals and MCQs 2, 6 and 7 use these relations, so they remain a likely 3-mark short numerical.
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