14.1 Electronics 1 & 2 — Semiconductor Devices & Analog Circuits

14.1 Electronics 1 & 2 — Semiconductor Devices & Analog Circuits

Topic D of the Electrical Engineering subject in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024) is Fundamentals of Electronic Communications, Electronics 1 and 2, weighted 2.25% of the whole examination and 5 of the 100 Electrical Engineering items. The predecessor 2011 syllabus called the same material Electronic Theory and Circuits, so this is long-standing REE content, not a new addition. Every variable-frequency drive, protective relay, UPS and smart meter in an electrical engineer's plant is built on the devices covered here.

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1. Semiconductor Fundamentals

Intrinsic silicon has four valence electrons and a band gap of about 1.1 eV (germanium, 0.67 eV). Doping creates the two extrinsic types:

At a PN junction, diffusion creates a depletion region and a built-in barrier potential of roughly 0.7 V for silicon and 0.3 V for germanium. Semiconductors have a negative temperature coefficient of resistance — resistance falls as temperature rises — the opposite of metallic conductors, and a favourite true/false item.

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2. Diodes and Diode Circuits

Shockley equation:

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

Models: the ideal diode is a closed switch when forward-biased; the practical model adds the 0.7 V drop; the complete model adds bulk resistance $r_d \approx 26\text{ mV}/I_D$.

Rectifier Circuits

$r = \frac{V_{ac,rms}}{V_{dc}} = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^{2} - 1}$

The bridge rectifier is preferred in practice because it achieves full-wave performance at only $V_m$ peak inverse voltage — half the PIV stress of the centre-tap arrangement.

Capacitor filtering reduces ripple; for light loading,

$V_{r(pp)} \approx \frac{I_{dc}}{f_{ripple}\,C}$

Other Diode Applications

• Clippers limit a waveform's amplitude; clampers shift its DC level without changing its shape.
• Zener diodes operate in reverse breakdown as voltage references; the series resistor must satisfy $I_{Z(\min)} \le I_Z \le I_{Z(\max)}$ across the full load and line range.
• Schottky diodes give a low 0.3 V drop and fast recovery; LEDs convert carrier recombination to photons; photodiodes and varactors exploit reverse-bias behaviour.

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3. Bipolar Junction Transistors

A BJT is a current-controlled device with three terminals and two junctions:

$I_E = I_B + I_C, \qquad \beta = \frac{I_C}{I_B}, \qquad \alpha = \frac{I_C}{I_E}, \qquad \beta = \frac{\alpha}{1 - \alpha}, \qquad \alpha = \frac{\beta}{1 + \beta}$

Operating Regions

Configurations

Only the common-emitter stage inverts. The emitter follower has near-unity voltage gain but high input and low output impedance, which is why it buffers a high-impedance sensor into a low-impedance load.

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4. Field-Effect Transistors

FETs are voltage-controlled and unipolar (one carrier type), with very high input impedance because the gate draws essentially no current.

JFET (depletion mode) — Shockley's equation:

$I_D = I_{DSS}\left(1 - \frac{V_{GS}}{V_P}\right)^{2}$

Enhancement MOSFET in saturation:

$I_D = k\left(V_{GS} - V_{TH}\right)^{2}$

The power MOSFET dominates switching below a few hundred volts, while the IGBT combines a MOSFET gate with a bipolar output stage for higher voltage and current — the standard device in motor drives and inverters, covered in the next section.

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5. Operational Amplifiers

The ideal op-amp: infinite open-loop gain, infinite input impedance, zero output impedance, infinite bandwidth, zero offset. With negative feedback, infinite gain forces the two inputs to the same potential — the virtual short — and infinite input impedance forces zero input current. Those two facts solve every configuration below.

The comparator exception is examined constantly: without negative feedback the virtual-short assumption is invalid and the output slams to a rail.

Common-mode rejection ratio:

$CMRR = 20\log_{10}\frac{A_d}{A_{cm}} \quad \text{dB}$

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Solved Board Exam Examples

Example 1: Bridge Rectifier Output

A bridge rectifier is fed from a transformer secondary of 24 V rms at 60 Hz. Using the practical diode model, find the DC output voltage and the ripple frequency.

Solution. Peak secondary voltage:

$V_m = 24\sqrt{2} = 33.94 \text{ V}$

In a bridge, two diodes conduct in series each half cycle:

$V_{m(\text{load})} = 33.94 - 2(0.7) = 32.54 \text{ V}$

$V_{dc} = \frac{2V_{m}}{\pi} = \frac{2(32.54)}{\pi} = \boxed{20.72 \text{ V}}$

The ripple frequency is $2f = \boxed{120 \text{ Hz}}$, because both half cycles are used. Ignoring the two diode drops would give 21.61 V — the intended distractor.

Example 2: BJT Current Relationships

A transistor has $\beta = 120$ and a base current of 40 µA. Find $I_C$, $I_E$ and $\alpha$.

Solution.

$I_C = \beta I_B = 120(40\ \mu\text{A}) = 4.80 \text{ mA}$ $I_E = I_B + I_C = 0.04 + 4.80 = \boxed{4.84 \text{ mA}}$ $\alpha = \frac{\beta}{1+\beta} = \frac{120}{121} = \boxed{0.9917}$

Note that $\alpha$ is always slightly less than unity, and the base current is never negligible in the emitter-current sum even though it is small.

Example 3: Op-Amp Gain Configuration

An op-amp has $R_f = 100$ kΩ and $R_{in} = 10$ kΩ. Compare the output for a +0.5 V input in the inverting and non-inverting configurations.

Solution. Inverting:

$V_o = -\frac{100}{10}(0.5) = \boxed{-5.0 \text{ V}}$

Non-inverting:

$V_o = \left(1 + \frac{100}{10}\right)(0.5) = 11(0.5) = \boxed{+5.5 \text{ V}}$

The non-inverting gain is always one greater in magnitude and preserves polarity. A candidate who writes $+5.0$ V for the non-inverting case has dropped the "1 +" term, which is the single most common op-amp error on the board exam.