JEE Physics Semiconductor Electronics Notes Set 05

Conductivity and the Mass-Action Law

Both carriers carry current, so \( \sigma=e(n_e\mu_e+n_h\mu_h) \). For a pure crystal \( n_e=n_h=n_i \) and \( \sigma=n_ie(\mu_e+\mu_h) \).
At a fixed temperature the product of the two carrier densities is constant: \( n_en_h=n_i^2 \). Doping a crystal heavily with donors raises \( n_e \) and lowers \( n_h \) so that the product stays the same.
In an N-type crystal with donor density \( N_D\gg n_i \), \( n_e\approx N_D \) and \( n_h\approx n_i^2/N_D \). In a P-type crystal \( n_h\approx N_A \) and \( n_e\approx n_i^2/N_A \).
Carrier density of an intrinsic semiconductor grows as \( T^{3/2}e^{-E_g/2kT} \), so its resistance falls rapidly on heating, unlike a metal.

Band Gap and the Threshold Wavelength

A photon can create an electron-hole pair only if \( h\nu\ge E_g \), that is \( \lambda\le\frac{hc}{E_g} \). With \( hc=1240\ \text{eV nm}=12400\ \text{eV\,\AA} \), the longest usable wavelength is \( \lambda_{max}=\frac{1240}{E_g(\text{eV})}\ \text{nm} \).
The same relation gives the colour of an LED: the emitted photon energy is about the band gap, so a visible LED needs a gap between roughly 1.8 eV and 3 eV. A donor or acceptor level lying a few tens of meV from a band needs only infrared radiation to ionise it.

Analysing Diode Circuits

Method for an ideal diode: assume a state (on or off), find the voltages, and check that the assumption is consistent. An ideal conducting diode is a wire; an ideal reverse-biased diode is an open circuit.
For a real silicon diode use a drop of 0.7 V while conducting (germanium 0.3 V), so \( I=\frac{V-0.7}{R} \).
Two diodes in parallel in the same direction: the one with the lower knee voltage conducts and clamps the voltage, so the other stays off. A diode in series with a source of opposite polarity larger than the supply stays off and the current is zero.
Dynamic resistance \( r=\frac{\Delta V}{\Delta I} \) is found from the slope of the forward characteristic; reverse resistance is of the order of \( 10^6\ \Omega \).

Zener Voltage Regulator

The series resistor \( R_s \) drops the excess voltage, and the zener holds the load voltage \( V_Z \): \( I=\frac{V_{in}-V_Z}{R_s}=I_Z+I_L \) with \( I_L=\frac{V_Z}{R_L} \).
If the load current falls, the zener current rises by the same amount so that \( I \) and the output voltage stay fixed. The zener current must never drop to zero, otherwise regulation is lost.

Transistor Amplifier Quantities

Relations: \( I_E=I_B+I_C \), \( \alpha=\frac{I_C}{I_E} \), \( \beta=\frac{I_C}{I_B} \), \( \beta=\frac{\alpha}{1-\alpha} \), \( \alpha=\frac{\beta}{1+\beta} \).
Common-emitter amplifier: voltage gain \( A_V=\beta\frac{R_L}{R_{in}} \), power gain \( A_P=\beta A_V=\beta^2\frac{R_L}{R_{in}} \), and the output is inverted (phase difference \( \pi \)). The common-base amplifier has no phase reversal and a current gain below 1.
Dynamic resistances: input resistance \( \frac{\Delta V_{BE}}{\Delta I_B} \) and output resistance \( \frac{\Delta V_{CE}}{\Delta I_C} \) at constant base current; the output resistance is much larger than the input resistance.

Logic Gate Identities

Basic identities: \( A+0=A,\ A+1=1,\ A\cdot0=0,\ A\cdot1=A,\ A+\bar A=1,\ A\cdot\bar A=0 \). De Morgan: \( \overline{A+B}=\bar A\cdot\bar B \) and \( \overline{AB}=\bar A+\bar B \).
NAND and NOR are universal. NOT: tie the two inputs of a NAND together. AND: NAND followed by a NOT. OR: invert both inputs and feed a NAND. XOR can be built from four NAND gates: with \( g=\overline{AB} \), \( Y=\overline{\overline{Ag}\cdot\overline{Bg}} \).
A diode OR gate and a diode AND gate need a pull-down or pull-up resistor at the output; a transistor switch behaves as a NOT gate.

Worked Examples

Example 1. A silicon crystal has intrinsic density \( 1.5\times10^{16}\ \text{m}^{-3} \). Doping raises the hole density to \( 4.5\times10^{22}\ \text{m}^{-3} \). Find the electron density and say whether the crystal is N-type or P-type.
Answer:
\( n_e=\frac{n_i^2}{n_h}=\frac{(1.5\times10^{16})^2}{4.5\times10^{22}}=5\times10^{9}\ \text{m}^{-3} \).
Holes are the majority carriers, so the crystal is P-type (acceptor doping).

Example 2. A 12 V supply feeds a 6 V zener diode and a 1.2 k\( \Omega \) load through a 600 \( \Omega \) series resistor. Find the current in the zener diode.
Answer:
Current through the series resistor: \( \frac{12-6}{600}=10\ \text{mA} \).
Load current: \( \frac{6}{1200}=5\ \text{mA} \).
Zener current: \( 10-5=5\ \text{mA} \).

Example 3. A common-emitter amplifier has \( \beta=100 \), input resistance 1 k\( \Omega \) and load resistance 5 k\( \Omega \). Find the voltage gain, the power gain and the output for a 2 mV input.
Answer:
\( A_V=\beta\frac{R_L}{R_{in}}=100\times5=500 \).
\( A_P=\beta A_V=100\times500=5\times10^{4} \).
\( V_{out}=500\times2\ \text{mV}=1\ \text{V} \), inverted relative to the input.

Example 4. A semiconductor diode is illuminated and its conductivity rises only for wavelengths below 827 nm. Find the band gap.
Answer:
\( E_g=\frac{1240}{827}=1.5\ \text{eV} \).