Notes · ElectronicsUK · A-Levels

Semiconductor components

This chapter builds the discrete semiconductor devices that every analogue circuit relies on, starting from doped silicon and the p-n junction. It develops the diode, the bipolar transistor as a switch and as an amplifier, and the MOSFET, and shows how to select the resistors that set their operating conditions.

5 sections~20 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

T·0111 / 16

Exam profile

AO1 · Explain the structure and behaviour of doped semiconductors, the p-n junction, the diode, the bipolar transistor and the MOSFET.AO2 · Apply device models to calculate diode drops, transistor currents, current gain and the resistor values that set an operating point.AO3 · Select and justify a device (BJT or MOSFET) and its bias components for a switching or amplifying specification.

Operators:explaincalculatedeterminedesignsketchanalyseselect

basic level

At AS the emphasis is on the diode (including the 0.7 V forward drop, the LED and the zener) and the bipolar transistor used as a switch, with straightforward base- and collector-resistor calculations.

higher level

The full A-Level adds the transistor as a small-signal amplifier, the MOSFET as a voltage-operated switch, and reasoned device selection for a specification, integrated into complete sub-systems.

Depth

Text

Contents · 5 sections

Semiconductor components

Semiconductors and doping#

FoundationWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (semiconductor devices)

Intrinsic, n-type and p-type silicon

Fig. 1Doping fixes the majority-carrier population. The material stays neutral because each mobile carrier is balanced by a fixed impurity ion.

A semiconductor is a material such as silicon whose conductivity lies between that of a conductor and an insulator. In pure (intrinsic) silicon each atom shares four covalent bonds, so at absolute zero there are no free charge carriers and the material behaves as an insulator. As the temperature rises, thermal energy breaks a few bonds, releasing electrons into the conduction band and leaving behind positively charged holes. Crucially, the number of these thermally generated carriers rises with temperature, which is why a semiconductor's resistance falls as it warms — the opposite of a metal, and the physical origin of the thermistor recap you meet later in sensing.

Doping deliberately adds a tiny, controlled amount of impurity to fix the carrier population so it no longer depends only on temperature. Adding a group-V (pentavalent) element such as phosphorus, which has five outer electrons, leaves one spare electron per impurity atom that is free to move: this makes n-type material, in which the majority carriers are negative electrons. Adding a group-III (trivalent) element such as boron, with only three outer electrons, creates a vacancy — a hole — that behaves as a mobile positive carrier: this makes p-type material, in which the majority carriers are positive holes. Doped material is still electrically neutral overall, because each added electron or hole is balanced by the fixed charge of its parent impurity ion.

The distinction between majority and minority carriers matters for every device that follows. In n-type silicon electrons are the majority carriers and the few thermally generated holes are the minority carriers; in p-type silicon the roles are reversed. When you bring the two materials together to form a junction, it is the behaviour of these carriers at the boundary that produces the one-way conduction of a diode and the control action of a transistor. Understanding doping is therefore not an end in itself — it is the foundation on which the diode, the bipolar transistor and the MOSFET are all explained.

In the systems language of this course, the semiconductor devices are the building blocks that let a small input signal control a much larger flow of energy. A transistor lets a microampere base current or a fraction of a volt at a gate switch amperes in a load; a diode enforces a direction on current so that a rectifier can turn alternating current into direct current. Every later chapter — amplifiers, timing, power supplies, switching — is ultimately an arrangement of these doped-silicon building blocks around the same input-process-output pattern.

Worked example

Reasoning about carriers

Silicon is doped with phosphorus (group V). State the material type, identify the majority and minority carriers, and explain why the sample is electrically neutral.

Type
Phosphorus has five outer electrons; four form covalent bonds and one is left free. The spare electron is a negative carrier, so the material is n-type.
Carriers
The majority carriers are the donated electrons. A small number of holes are still generated thermally, so holes are the minority carriers.
Neutrality
Each free electron came from a phosphorus atom that is now a fixed positive ion. The mobile negative charge is exactly balanced by the fixed positive charge, so the sample carries no net charge.

Result: The material is n-type; majority carriers are electrons, minority carriers are holes, and the sample is neutral because every free electron is balanced by a fixed donor ion.

Typical mistakes

Saying that doped material carries a net charge — it does not; the mobile carrier is always balanced by the fixed impurity ion.
Confusing the effect of temperature on a semiconductor (resistance falls) with a metal (resistance rises).

Active revision

A sample of silicon is doped with boron. State the type of the material, name its majority carrier and explain, in terms of valency, why that carrier exists.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The p-n junction and the diode#

StandardWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (diode)

Silicon diode forward characteristic

Fig. 2The forward current stays negligible until about

0.7\,\text{V}

, then rises steeply. The marker shows the

0.7\,\text{V}

knee; beyond it a series resistor must set the current.

When p-type and n-type material are joined, electrons from the n-side diffuse across to fill holes on the p-side, and vice versa. This leaves a thin region either side of the junction that is stripped of mobile carriers — the depletion layer — and the exposed fixed ions set up an internal potential barrier that opposes further diffusion. For silicon this barrier corresponds to about

0.7\,\text{V}

. Nothing external has been connected yet; this built-in barrier is the reason a junction conducts in only one direction.

Under forward bias the p-side is made positive with respect to the n-side. Once the applied voltage exceeds the barrier, the depletion layer narrows and current flows freely: the diode is on and, for a silicon device, drops a nearly constant

V_F\approx 0.7\,\text{V}

almost regardless of current. Under reverse bias the polarity is reversed, the depletion layer widens, and only a negligible leakage current flows — until, at a high enough reverse voltage, the junction breaks down. The result is the classic diode I-V characteristic: a steep forward rise beyond about

0.7\,\text{V}

and a flat, near-zero reverse region.

The forward characteristic is exponential, so beyond the knee the current rises very steeply for a tiny further rise in voltage. This is why a diode is almost always used with a series resistor to limit the current: the resistor, not the diode, sets the operating current. In circuit analysis the standard model is to assume the diode drops a fixed

0.7\,\text{V}

when conducting and passes no current when reverse-biased. Forgetting to subtract this

0.7\,\text{V}

is the single most common slip in diode and transistor questions.

Two special diodes appear throughout the course. A light-emitting diode (LED) emits light when forward-biased and has a larger forward drop, typically

1.8

2.2\,\text{V}

depending on colour; it must always be run through a current-limiting resistor. A zener diode is designed to be operated in reverse breakdown at a precise, stable voltage

V_Z

, and is the basis of the simple voltage reference and shunt regulator you meet in the power-supply chapter. In every case the design task is the same: choose the series resistor so that the current sits at a safe, useful value.

To design the series resistor for an LED, apply Kirchhoff's voltage law around the loop. The supply voltage is shared between the resistor and the LED, so the resistor must drop the difference between the supply and the LED's forward voltage, and Ohm's law then fixes its value for the chosen current. This single calculation — supply minus device drop, divided by the wanted current — recurs for LEDs, transistor bias and zener references alike.

V_F \approx 0.7\,\text{V}\ \text{(Si)},\quad V_F \approx 2\,\text{V}\ \text{(LED)}

Forward drops

The near-constant conducting voltage assumed in analysis.

R = \dfrac{V_S - V_F}{I}

Series (limiting) resistor

Kirchhoff's voltage law: the resistor drops the supply minus the device forward voltage; Ohm's law then fixes R for the wanted current.

Worked example

Current-limiting resistor for an LED

A green LED with forward voltage $2.1\,\text{V}$ is to operate at $12\,\text{mA}$ from a $5.0\,\text{V}$ supply. Find the series resistor and the power it dissipates.

Resistor voltage
By Kirchhoff's voltage law the resistor drops the supply minus the LED: $V_R = 5.0 - 2.1 = 2.9\,\text{V}$ .
Resistance
Ohm's law: $R = V_R/I = 2.9/0.012 = 241.7\,\Omega$ . The nearest higher preferred value is $270\,\Omega$ , which gives a safe $I = 2.9/270 = 10.7\,\text{mA}$ .
$R = \dfrac{V_S - V_F}{I} = \dfrac{5.0 - 2.1}{0.012} = 242\,\Omega$
Power
The resistor dissipates $P = I^2 R = (0.0107)^2 \times 270 = 0.031\,\text{W}$ , so a standard $0.25\,\text{W}$ resistor is ample.

Result: A $270\,\Omega$ resistor gives about $10.7\,\text{mA}$ ; it dissipates roughly $31\,\text{mW}$ , well within a quarter-watt part.

Typical mistakes

Forgetting to subtract the $0.7\,\text{V}$ (or the LED's $\approx 2\,\text{V}$ ) forward drop when finding the resistor voltage.
Treating the forward characteristic as ohmic (a straight line) rather than an exponential curve with a knee.

Active revision

A red LED with a forward voltage of $2.0\,\text{V}$ is to be run at $15\,\text{mA}$ from a $9\,\text{V}$ supply. Calculate the required series resistor and choose a preferred value.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The bipolar transistor as a switch#

StandardWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (BJT switch)

NPN transistor switch driving a load

Fig. 3A base current through

R_B

saturates the transistor, so the load current flows from

+V_{CC}

through the load and the collector-emitter path to

0\,\text{V}

A bipolar junction transistor (BJT) has three terminals — base, collector and emitter — and, in the common NPN type, a small current into the base controls a much larger current from collector to emitter. The device has two junctions: the base-emitter junction behaves like a diode and must be forward-biased by about

0.7\,\text{V}

before any collector current flows at all. This single fact,

V_{BE}\approx 0.7\,\text{V}

, governs every bias calculation.

Used as a switch, the transistor is driven between two extreme states. With no base current the transistor is cut off: the collector-emitter path is effectively open and the load is off. With enough base current the transistor is driven into saturation: the collector-emitter path is nearly a short, dropping only

V_{CE(sat)}\approx 0.2\,\text{V}

, and the full load current flows. A transistor switch therefore spends its life at one of these two ends of its characteristic, never in between.

To turn the transistor fully on you must supply more base current than the bare minimum. The collector current wanted in the load, divided by the transistor's current gain

h_{FE}

, gives the minimum base current; in practice the designer provides several times this — an overdrive — so that even the worst-case (lowest)

h_{FE}

device saturates reliably. The base resistor is then found from Kirchhoff's voltage law: the input voltage minus the

0.7\,\text{V}

base-emitter drop, divided by the design base current. Choosing the base resistor is the core AS design skill.

The collector (load) resistor sets the on-state current. Applying Kirchhoff's voltage law around the collector loop, the supply is shared between the load, any series resistor and the small saturation voltage, so the collector resistor is the leftover voltage divided by the wanted collector current. When the load is a relay or a motor, a flyback (protection) diode must be connected across it, because switching off an inductive load produces a large back-emf that would otherwise destroy the transistor. This is a standard interfacing precaution rather than a calculation, but examiners expect it to be shown.

The transistor switch is the classic interface between a low-power control signal — a logic gate, a sensor sub-system, a microcontroller port — and a high-power output such as a lamp, relay or motor. It embodies the process stage of the input-process-output model: a tiny input decision commands a large output action. The same topology reappears, scaled up, in the high-power switching chapter where thyristors and triacs replace the transistor for mains loads.

I_B(\text{min}) = \dfrac{I_C}{h_{FE}}

Minimum base current

The least base current that supports the wanted collector current at the quoted gain.

R_B = \dfrac{V_{in} - V_{BE}}{I_B}

Base resistor

Kirchhoff's voltage law across the base loop; use a design base current several times the minimum.

Worked example

Choosing the base resistor to guarantee saturation

An NPN transistor drives a load needing $I_C = 15\,\text{mA}$ . The input is $5\,\text{V}$ , $V_{BE}=0.7\,\text{V}$ and the worst-case current gain is $h_{FE}=100$ . Choose a base resistor that firmly saturates the transistor.

Minimum base current
$I_B(\text{min}) = I_C/h_{FE} = 15/100 = 0.15\,\text{mA}$ .
Apply overdrive
To be sure of saturation for any device, design for about three times the minimum: $I_B \approx 0.45\,\text{mA}$ .
Base resistor
$R_B = (V_{in} - V_{BE})/I_B = (5 - 0.7)/0.00045 = 9556\,\Omega$ . Choose the nearest preferred value $9.1\,\text{k}\Omega$ .
$R_B = \dfrac{5 - 0.7}{0.45\times 10^{-3}} \approx 9.6\,\text{k}\Omega$
Check
With $9.1\,\text{k}\Omega$ , $I_B = 4.3/9100 = 0.47\,\text{mA}$ . The forced gain is $I_C/I_B = 15/0.47 = 32$ , well below $h_{FE}=100$ , so the transistor is firmly saturated.

Result: A $9.1\,\text{k}\Omega$ base resistor gives $I_B \approx 0.47\,\text{mA}$ and a forced gain of about 32, guaranteeing saturation ( $V_{CE}\approx 0.2\,\text{V}$ ).

Typical mistakes

Using $I_B = I_C/h_{FE}$ as the design base current with no overdrive, so a low-gain device fails to saturate.
Forgetting the $0.7\,\text{V}$ base-emitter drop when calculating the base resistor voltage.

Active revision

An NPN transistor with $h_{FE}=100$ is to switch a $12\,\text{V}$ , $120\,\text{mA}$ relay coil, driven from a $5\,\text{V}$ logic output. Choose a base resistor that guarantees saturation and state one other component the circuit needs.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The transistor as an amplifier#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (small-signal amplifier)

Common-emitter voltage amplifier

Fig. 4The input signal is coupled to the biased base; the collector load turns the amplified collector current into an inverted output voltage.

Between cut-off and saturation lies the active region, where the collector current is very nearly proportional to the base current:

I_C = h_{FE}\,I_B

. The constant of proportionality, the current gain

h_{FE}

(or

\beta

), is typically 100 to 400. In the active region a small change in base current produces a large, faithful change in collector current — this is amplification, and it is why the transistor was invented.

To amplify an alternating signal the transistor must first be biased to a steady operating point (the quiescent point) roughly halfway up its load line, so that the output can swing both up and down without clipping. A common-emitter stage does this with a base bias network and a collector load resistor

R_C

; the input signal is coupled to the base through a capacitor so it rides on top of the bias. The steady collector current sets the quiescent collector voltage through Kirchhoff's voltage law,

V_C = V_{CC} - I_C R_C

The voltage gain of a common-emitter stage with an emitter resistor

R_E

is, to a good approximation,

A_V \approx -R_C/R_E

. The minus sign records the phase inversion: as the input rises the collector voltage falls. The emitter resistor stabilises the operating point against temperature and gain spread, at the cost of some gain; bypassing it with a capacitor restores gain at signal frequencies. The essential design trade-off — gain against stability — is one you must be able to discuss, not just calculate.

An amplifier only stays linear while its output remains between the supply rails. If the input is too large the output tries to swing beyond the supply or below

0\,\text{V}

and is clipped flat at the rail, badly distorting the signal. The maximum undistorted output swing is set by the quiescent point and the supply voltage; choosing the bias to sit centrally maximises this headroom. Clipping is the analogue equivalent of saturation, and recognising it on a waveform is a standard exam skill that returns in the op-amp and audio chapters.

Compared with the switch, where the transistor is deliberately driven hard into its non-linear extremes, the amplifier keeps it carefully within its linear active region. The same device, the same

0.7\,\text{V}

base-emitter drop and the same

h_{FE}

relationship therefore serve two opposite purposes depending only on how it is biased — a neat illustration that in electronics behaviour follows from the surrounding component choices.

h_{FE} = \dfrac{I_C}{I_B}

Current gain

Collector current divided by base current in the active region.

A_V \approx -\dfrac{R_C}{R_E}

Common-emitter voltage gain

Approximate gain with an unbypassed emitter resistor; the minus sign is the phase inversion.

Worked example

Quiescent point and gain of a CE stage

A common-emitter amplifier runs from $V_{CC}=12\,\text{V}$ with $R_C=4.7\,\text{k}\Omega$ and $R_E=1.0\,\text{k}\Omega$ . The base current is $10\,\mu\text{A}$ and $h_{FE}=100$ . Find the collector current, the quiescent collector voltage and the voltage gain.

Collector current
$I_C = h_{FE} I_B = 100 \times 10\,\mu\text{A} = 1.0\,\text{mA}$ .
Quiescent collector voltage
$V_C = V_{CC} - I_C R_C = 12 - (1.0\,\text{mA})(4.7\,\text{k}\Omega) = 12 - 4.7 = 7.3\,\text{V}$ .
$V_C = V_{CC} - I_C R_C = 12 - 4.7 = 7.3\,\text{V}$
Voltage gain
$A_V \approx -R_C/R_E = -4.7\,\text{k}\Omega / 1.0\,\text{k}\Omega = -4.7$ . The output is 4.7 times larger than the input and inverted.
Swing check
The collector sits at $7.3\,\text{V}$ , so it can fall about $7.3\,\text{V}$ toward $0\,\text{V}$ and rise about $12-7.3=4.7\,\text{V}$ toward the rail; the smaller of these, $4.7\,\text{V}$ , limits the undistorted output amplitude.

Result: $I_C = 1.0\,\text{mA}$ , $V_C = 7.3\,\text{V}$ and $A_V \approx -4.7$ ; the output can swing about $\pm 4.7\,\text{V}$ before clipping on the upper rail.

Typical mistakes

Omitting the phase-inversion minus sign, or quoting a gain magnitude with the wrong sign.
Biasing the operating point too close to a rail, so the amplifier clips on one side of the waveform.

Active revision

A common-emitter amplifier has $V_{CC}=12\,\text{V}$ , $R_C=4.7\,\text{k}\Omega$ , $R_E=1.0\,\text{k}\Omega$ and a quiescent collector current of $1.0\,\text{mA}$ . Estimate the quiescent collector voltage and the voltage gain, and comment on the available output swing.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The MOSFET and device selection#

AdvancedWJEC/Eduqas A Level Electronics — Unit 1/Unit 3 (MOSFET, device choice)

n-channel enhancement MOSFET transfer characteristic

Fig. 5No drain current flows below the threshold at

2\,\text{V}

(dashed reference); above it the current rises as a square law. The marker is the operating point at

V_{GS}=4\,\text{V}

The metal-oxide-semiconductor field-effect transistor (MOSFET) is a voltage-operated device: the voltage on an insulated gate controls the current between drain and source, and because the gate is insulated by a thin oxide layer it draws essentially no steady current. In an n-channel enhancement MOSFET no drain current flows until the gate-source voltage exceeds a threshold

V_{GS(th)}

; above threshold a conducting channel is induced and the drain current rises approximately as the square of the gate overdrive.

This gives the MOSFET two great practical advantages over the bipolar transistor. First, its input is a near-perfect open circuit at DC, so it loads the driving stage hardly at all and needs no continuous base current — ideal for driving from a logic gate or microcontroller pin. Second, a power MOSFET fully turned on behaves like a low resistance (

R_{DS(on)}

), so it can switch large currents with low dissipation. The trade-off is that the gate has capacitance that must be charged and discharged, which matters at high switching speeds.

The transfer characteristic — drain current plotted against gate-source voltage — is the key graph. It sits flat on zero until

V_{GS(th)}

, then curves upward. For switching, you drive the gate well above threshold so the MOSFET is fully on; for a linear application the operating point is chosen on the rising part of the curve. Reading

V_{GS(th)}

off the graph, and recognising the square-law shape above it, are standard requirements.

Choosing between a BJT and a MOSFET is a genuine AO3 design decision. A BJT switch is cheap and simple but wastes power in its base drive and drops

V_{CE(sat)}

across the load; a MOSFET needs almost no drive current and drops very little when on, but needs its gate capacitance charged and can be sensitive to static damage. For a battery-powered switch driven from a microcontroller, the MOSFET's negligible drive current usually wins; for a small, slow, low-cost switch a BJT may be entirely adequate. The examiner rewards a justified choice, not merely a preference.

Whichever device is used, the systems role is identical: it is the controlled switch or amplifying element that lets a small input command a large output. Recognising that the diode enforces direction, the BJT is current-controlled and the MOSFET is voltage-controlled — and matching that behaviour to the demands of a specification — is the thread that ties this whole chapter to the analogue and power sub-systems that follow.

I_D \approx k\,(V_{GS} - V_{GS(th)})^2 \quad (V_{GS} > V_{GS(th)})

Enhancement MOSFET square law

Above threshold the drain current rises as the square of the gate overdrive; below threshold it is zero.

Worked example

Predicting MOSFET conduction and comparing devices

An n-channel enhancement MOSFET has $V_{GS(th)}=2.0\,\text{V}$ and, from its curve, $k=1.0\,\text{mA/V}^2$ . Find the drain current at $V_{GS}=1.5\,\text{V}$ and at $V_{GS}=4.0\,\text{V}$ , and justify choosing it over a BJT to switch a load from a $3.3\,\text{V}$ microcontroller pin.

Below threshold
At $V_{GS}=1.5\,\text{V}$ the gate is below the $2.0\,\text{V}$ threshold, so no channel forms and $I_D = 0$ .
Above threshold
At $V_{GS}=4.0\,\text{V}$ the overdrive is $4.0-2.0=2.0\,\text{V}$ , so $I_D = k(V_{GS}-V_{GS(th)})^2 = 1.0\times(2.0)^2 = 4.0\,\text{mA}$ .
$I_D = 1.0\times(4.0-2.0)^2 = 4.0\,\text{mA}$
Device choice
The microcontroller pin can source only a small current. The MOSFET's insulated gate draws essentially no steady current, so the pin easily drives it, whereas a BJT would demand a continuous base current the pin might not supply.

Result: $I_D = 0$ at $1.5\,\text{V}$ and $4.0\,\text{mA}$ at $4.0\,\text{V}$ ; the MOSFET is preferred because its gate needs no steady drive current from the microcontroller pin.

Typical mistakes

Thinking a MOSFET gate draws a steady current — at DC it draws essentially none; only the gate capacitance is charged.
Assuming drain current flows for any positive gate voltage, ignoring the threshold $V_{GS(th)}$ .

Active revision

An n-channel enhancement MOSFET has $V_{GS(th)}=2\,\text{V}$ . State whether it conducts when $V_{GS}=1.5\,\text{V}$ and when $V_{GS}=4\,\text{V}$ , and give one reason to prefer it over a BJT for switching a load from a microcontroller pin.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

Semiconductor components

5 sections~20 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

Semiconductor components

Key points

Reasoning about carriers

Key points

Current-limiting resistor for an LED

Key points

Choosing the base resistor to guarantee saturation

Key points

Quiescent point and gain of a CE stage

Key points

Predicting MOSFET conduction and comparing devices

Sources

Semiconductor components

Key points

Reasoning about carriers

Key points

Current-limiting resistor for an LED

Key points

Choosing the base resistor to guarantee saturation

Key points

Quiescent point and gain of a CE stage

Key points

Predicting MOSFET conduction and comparing devices

Sources