JEE Physics Semiconductor Electronics Notes Set 03

Energy Bands in Solids

When atoms form a solid, their energy levels broaden into bands. The band formed by the outer electrons is the valence band and the next higher band is the conduction band; between them lies the forbidden gap \( E_g \).
Conductors: the bands overlap (or the conduction band is partly filled), so there is no gap. Insulators: a completely filled valence band, an empty conduction band and a large gap (more than about 5 eV). Semiconductors: the same arrangement with a small gap, about 0.7 eV for germanium and 1.1 eV for silicon.
At 0 K a pure semiconductor behaves as an insulator. As temperature rises, electrons jump to the conduction band leaving holes in the valence band, so conductivity rises and the resistance falls (negative temperature coefficient). In metals the resistance rises with temperature because the carrier number is fixed and collisions increase.
A photon can create an electron-hole pair if its energy is at least \( E_g \), that is \( \lambda\le\frac{12400}{E_g(\text{eV})}\ \text{Å} \).

Intrinsic and Extrinsic Semiconductors

In an intrinsic (pure) semiconductor \( n_e=n_h=n_i \) and the Fermi level lies in the middle of the gap. The conductivity is \( \sigma=e(n_e\mu_e+n_h\mu_h) \); electron mobility is greater than hole mobility.
Doping with a pentavalent atom (P, As, Sb) gives an n-type semiconductor: the extra electron occupies a donor level just below the conduction band, electrons are the majority carriers and the Fermi level moves towards the conduction band. Doping with a trivalent atom (B, Al, Ga, In) gives a p-type semiconductor: acceptor levels lie just above the valence band, holes are the majority carriers and the Fermi level moves towards the valence band. Both types remain electrically neutral.
Law of mass action: \( n_en_h=n_i^2 \). Drift current density \( J=\sigma E \). The drift velocity is \( v_d=\mu E \), and the total current is the sum of electron and hole currents, \( I=I_e+I_h=eA(n_ev_e+n_hv_h) \).

p-n Junction and Diode Characteristics

At a p-n junction, carriers diffuse across and leave behind fixed ions, forming a depletion region with a built-in field directed from n to p and a barrier potential (about 0.7 V for Si, 0.3 V for Ge). With no bias the diffusion and drift currents cancel.
Forward bias (p to positive): the barrier and the depletion layer shrink and a large current flows by diffusion of majority carriers once the voltage exceeds the knee voltage. Reverse bias: the barrier and the depletion layer widen, and only a tiny reverse (drift) current of minority carriers flows. The forward resistance is a few ohms to tens of ohms and the reverse resistance is very large, roughly \( 10^4 \) times larger.
Breakdown: in a lightly doped junction, high reverse voltage causes avalanche breakdown by collision ionisation; in a heavily doped junction, a thin depletion layer gives Zener breakdown by direct field ionisation. A Zener diode is operated in reverse breakdown as a voltage regulator and keeps a constant voltage across the load.

Rectifiers and Special Diodes

A half-wave rectifier uses one diode and conducts on alternate half cycles: output frequency equals the input frequency, \( V_{dc}=\frac{V_0}{\pi} \), \( V_{rms}=\frac{V_0}{2} \). A full-wave rectifier (two diodes with a centre-tapped transformer, or a four-diode bridge) delivers a pulse in every half cycle: output ripple frequency is twice the input frequency (100 Hz for 50 Hz mains), \( V_{dc}=\frac{2V_0}{\pi} \), \( V_{rms}=\frac{V_0}{\sqrt2} \). A capacitor across the load smooths the output towards the peak value.
Special diodes: a Zener diode (voltage regulation, reverse breakdown), a photodiode (reverse biased, current rises with light intensity), a light-emitting diode (forward biased, emits light with photon energy near the band gap) and a solar cell (generates voltage from light, no bias needed).

Transistors

A transistor has an emitter (heavily doped), a very thin lightly doped base and a collector (moderately doped, largest area). In active operation the emitter-base junction is forward biased and the collector-base junction is reverse biased. The currents obey \( I_E=I_B+I_C \).
Current gains: \( \alpha=\frac{I_C}{I_E} \) (common base, slightly less than 1), \( \beta=\frac{I_C}{I_B} \) (common emitter, typically 50 to 200), with \( \beta=\frac{\alpha}{1-\alpha} \) and \( \alpha=\frac{\beta}{1+\beta} \). Transconductance \( g_m=\frac{\Delta I_C}{\Delta V_{BE}} \).
In a common-emitter amplifier the output is 180 degrees out of phase with the input; the voltage gain is \( A_V=\beta\frac{R_L}{R_{in}} \) and the power gain is \( \beta A_V \). In a common-base amplifier input and output are in phase. The common-emitter configuration has the highest power gain; the common-collector (emitter follower) has voltage gain just below 1.
A transistor can also act as a switch (cut-off and saturation) and, with positive feedback, as an oscillator.

Logic Gates and Boolean Algebra

Basic gates and their Boolean expressions (inputs A, B, output Y):

NAND and NOR are universal gates: any other gate can be built from either of them. De Morgan's laws: \( \overline{A+B}=\bar A\cdot\bar B \) and \( \overline{A\cdot B}=\bar A+\bar B \). Other identities: \( A+A=A,\ A\cdot A=A,\ A+\bar A=1,\ A\cdot\bar A=0 \).

Worked Examples

Worked Example 1. The band gap of a semiconductor is 1.1 eV. Find the longest wavelength of light that can create an electron-hole pair.
Answer:
\( \lambda=\frac{12400}{1.1}\approx11270\ \text{Å} \), which is in the infrared.

Worked Example 2. In a transistor circuit \( I_E=5.00\ \text{mA} \) and \( I_C=4.90\ \text{mA} \). Find \( I_B \), \( \alpha \) and \( \beta \).
Answer:
\( I_B=5.00-4.90=0.10\ \text{mA} \); \( \alpha=\frac{4.90}{5.00}=0.98 \); \( \beta=\frac{4.90}{0.10}=49 \).

Worked Example 3. A silicon diode (drop 0.7 V) is in series with a 330 ohm resistor and a 5 V battery in forward bias. Find the current.
Answer:
\( I=\frac{5-0.7}{330}=13\ \text{mA} \).

Worked Example 4. A full-wave rectifier works from 60 Hz mains. What are the ripple frequency and the average output for a peak voltage of 10 V?
Answer:
Ripple frequency \( =2\times60=120\ \text{Hz} \). \( V_{dc}=\frac{2V_0}{\pi}=\frac{20}{\pi}=6.4\ \text{V} \).