Semiconductor diodes and transistors

Diodes and bipolar junction transistors (BJTs) are the building blocks of almost every electronic circuit — power supplies, amplifiers, switches and logic. This topic starts with the PN junction, its I-V behaviour and diode models, moves to rectifiers and special-purpose diodes, and then covers the BJT: its structure, modes of operation, the three amplifier configurations, characteristic curves and biasing. NEC MCQs commonly test cut-in voltages, rectifier efficiency and ripple, PIV, current gain relations and which configuration has which impedance.

Key concepts

PN junction

• Joining p-type (holes majority) and n-type (electrons majority) silicon creates a depletion region with a built-in barrier potential: about 0.7 V for Si, 0.3 V for Ge.
• Forward bias (p to +): barrier reduces, current rises exponentially after the cut-in (knee) voltage.
• Reverse bias: barrier widens; only a tiny reverse saturation current $I_S$ (nA for Si, μA for Ge) flows until breakdown (avalanche or Zener).
• Reverse saturation current roughly doubles for every 10 °C rise.

Ideal vs practical diode

• Ideal: short circuit when forward biased, open when reverse.
• Practical (large-signal) models: constant-voltage drop (0.7 V), or piecewise linear (0.7 V + bulk resistance $r_f$).
• Small-signal model: dynamic resistance $r_d = \eta V_T/I_D$ (≈ 26 mV/$I_D$ at room temperature for η = 1).

Diode configurations: in series circuits, a diode conducts only if the applied voltage exceeds its drop in the forward direction; in parallel, the diode with the lower drop (e.g. Ge before Si) clamps the voltage; hybrid circuits are solved by assuming states and checking consistency.

Rectifiers

Parameter	Half-wave	Full-wave centre-tap	Bridge
$V_{dc}$	$V_m/\pi$	$2V_m/\pi$	$2V_m/\pi$
Ripple factor	1.21	0.482	0.482
Max efficiency	40.6%	81.2%	81.2%
PIV per diode	$V_m$	$2V_m$	$V_m$
Ripple frequency	f	2f	2f
Diodes	1	2	4

A capacitor filter reduces ripple: for full-wave, $V_{r(p-p)} \approx I_{dc}/(2fC)$.

Special diodes: Zener (voltage regulation in reverse breakdown), LED (light emission, forward biased; drop 1.5–3 V), photodiode (reverse biased, light-controlled current), varactor (reverse-biased junction capacitance for tuning), Schottky (metal-semiconductor, low drop ~0.3 V, fast switching), tunnel diode (negative resistance region, heavy doping).

BJT

• Three layers: NPN or PNP; terminals emitter (heavily doped), base (thin, lightly doped), collector (moderately doped, largest).
• In NPN, electrons are majority carriers; arrow on the emitter points out. In PNP the arrow points in.
• $I_E = I_B + I_C$; $\alpha = I_C/I_E$ (0.95–0.995); $\beta = I_C/I_B$ (20–500).

Modes of operation

Mode	BE junction	BC junction	Use
Active	forward	reverse	amplifier
Saturation	forward	forward	switch ON
Cut-off	reverse	reverse	switch OFF
Reverse active	reverse	forward	rarely used

Amplifier concept: a small input signal controls a larger output using power from the DC supply.

Configurations

Property	CB	CE	CC (emitter follower)
Current gain	α (<1)	β (high)	1 + β (high)
Voltage gain	high	high	≈ 1
Input impedance	lowest	medium	highest
Output impedance	highest	medium	lowest
Phase shift	0°	180°	0°

Characteristics: input characteristic ($I_B$ vs $V_{BE}$ for CE — like a diode); output characteristic ($I_C$ vs $V_{CE}$ at constant $I_B$ — shows active, saturation, cut-off regions and Early effect); current gain characteristic ($I_C$ vs $I_B$).

Biasing sets the Q-point. Methods: fixed bias (poor stability), collector-to-base feedback, emitter bias, voltage-divider bias (most stable, widely used). Stability factor $S = \partial I_C/\partial I_{CO}$; ideal S = 1, fixed bias S = 1 + β.

High-yield points

• Cut-in voltage: Si ≈ 0.7 V, Ge ≈ 0.3 V; $V_T \approx 26$ mV at 300 K.
• Half-wave: ripple factor 1.21, efficiency 40.6%; full-wave: 0.482 and 81.2%.
• PIV: half-wave $V_m$, centre-tap $2V_m$, bridge $V_m$.
• $\beta = \alpha/(1-\alpha)$ and $I_E = I_B + I_C$.
• Active mode: BE forward, BC reverse; saturation: both forward; cut-off: both reverse.

Formulas

$I_D = I_S\left(e^{V_D/\eta V_T} - 1\right),\quad V_T = kT/q \approx 26\text{ mV at 300 K}$

$\beta = \frac{\alpha}{1-\alpha},\quad \alpha = \frac{\beta}{1+\beta},\quad I_C = \beta I_B + (1+\beta)I_{CBO}$

Worked example

Voltage-divider bias: $V_{CC} = 12$ V, $R_1 = 40$ kΩ, $R_2 = 10$ kΩ, $R_E = 1$ kΩ, $R_C = 2$ kΩ, β large. Find $I_C$ and $V_{CE}$.

1. $V_B = 12 \times 10/(40+10) = 2.4$ V.
2. $V_E = 2.4 - 0.7 = 1.7$ V → $I_E = 1.7$ mA ≈ $I_C$.
3. $V_{CE} = 12 - 1.7(2 + 1) = 12 - 5.1 = 6.9$ V. Transistor is in the active region.

Exam tips

• Ripple factor: half-wave 1.21, full-wave 0.482; efficiency 40.6% vs 81.2%.
• PIV: centre-tap $2V_m$, bridge $V_m$.
• CC has highest input and lowest output impedance (buffer); CB has the opposite.
• Only CE gives 180° phase shift.
• If α = 0.98, β = 0.98/0.02 = 49.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. Two silicon diodes (0.7 V each, constant-voltage-drop model) are connected in series, both forward biased, with a 4.3 kΩ resistor across a 10 V dc source. What is the current?

• A. 2.0 mA
• B. 2.33 mA
• C. 2.16 mA
• D. 1.84 mA

Answer: A. $I = (10 - 0.7 - 0.7)/4.3\ \text{k}\Omega = 8.6/4.3 = 2.0$ mA.

2. For a BJT with $\beta = 99$, what is the current gain of the common-collector configuration ($I_E/I_B$)?

• A. 100
• B. 1
• C. 0.99
• D. 99

Answer: A. In the common-collector configuration the current gain is $I_E/I_B = 1 + \beta = 100$.

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