JEE Physics Semiconductor Electronics Notes Set 02

The Junction Transistor

A transistor has three doped regions and two junctions: emitter (heavily doped), base (thin and lightly doped) and collector (larger, moderately doped). An npn transistor has a thin p-type base between two n-type regions, and a pnp transistor has a thin n-type base between two p-type regions. The arrow on the emitter points along the conventional current: outwards in npn and inwards in pnp.

For normal (active) operation the emitter-base junction is forward biased and the collector-base junction is reverse biased. In an npn transistor, electrons from the emitter cross the thin base and are collected by the collector. A small fraction recombine in the base, giving the base current. The currents obey \( I_E = I_B + I_C \).

Regions of operation: cut-off (both junctions reverse biased, transistor off), active (amplifier) and saturation (both forward biased, transistor fully on). A switch uses only cut-off and saturation.

Current Gains

In the common base (CB) configuration the current gain is \( \alpha = I_C/I_E \), slightly less than 1. In the common emitter (CE) configuration the current gain is \( \beta = I_C/I_B \), usually 20 to 200. They are related by

\[ \beta = \frac{\alpha}{1-\alpha}, \qquad \alpha = \frac{\beta}{1+\beta}, \qquad I_E = (1+\beta) I_B \]

For small signals, the a.c. current gain is \( \beta_{ac} = (\Delta I_C/\Delta I_B)_{V_{CE}} \). The input resistance is \( r_i = (\Delta V_{BE}/\Delta I_B)_{V_{CE}} \) and the output resistance is \( r_o = (\Delta V_{CE}/\Delta I_C)_{I_B} \).

Transistor as an Amplifier

In a CE amplifier the input signal is applied between base and emitter, and the output is taken between collector and emitter across the load \( R_L \). A small change in base current produces a large change in collector current.

\[ A_I = \beta, \quad A_V = \beta\frac{R_L}{R_{in}}, \quad A_P = A_V A_I = \beta^2\frac{R_L}{R_{in}} \]

The transconductance is \( g_m = \Delta I_C/\Delta V_{BE} = \beta/R_{in} \), so \( A_V = g_m R_L \). The output voltage is 180 degrees out of phase with the input. A voltage gain in decibels is \( 20\log_{10}A_V \). The amplifier needs a dc bias so that the transistor stays in the active region for the whole signal cycle.

Transistor as a Switch and as an Oscillator

As a switch, the transistor is driven between cut-off and saturation. When the input voltage is below about 0.7 V (Si), the transistor is off and the output is high (near \( V_{CC} \)). When the input is high enough to saturate it, the output falls to nearly zero. So the CE switch also inverts the signal.

An oscillator is an amplifier with positive feedback: part of the output is fed back in phase with the input. An LC tank circuit fixes the frequency \( f = 1/(2\pi\sqrt{LC}) \), and the amplifier supplies the energy lost in the circuit.

Logic Gates and Truth Tables

A logic gate is a digital circuit whose output depends on its inputs according to a logical rule. Inputs and outputs take the values 0 (low) and 1 (high).

Boolean Laws and Universal Gates

Useful identities: \( A + 0 = A \), \( A \cdot 1 = A \), \( A + \bar{A} = 1 \), \( A \cdot \bar{A} = 0 \), \( A + AB = A \), and the distributive law \( (A+B)(A+C) = A + BC \). De Morgan's theorems are \( \overline{A+B} = \bar{A}\cdot\bar{B} \) and \( \overline{A \cdot B} = \bar{A} + \bar{B} \).

NAND and NOR are universal gates because every other gate can be built from either one. From NAND: NOT = NAND with both inputs joined (1 gate), AND = NAND followed by NOT (2 gates), OR = NAND of two inverted inputs (3 gates). From NOR: NOT = NOR with joined inputs, OR = NOR followed by NOT, AND = NOR of two inverted inputs (3 gates).

Integrated Circuits and Worked Examples

An integrated circuit (IC) is a complete circuit with many diodes, transistors, resistors and capacitors formed on one small chip of silicon. ICs may be small, medium, large or very large scale, depending on how many components they hold. They are cheap, reliable and consume little power.

Example 1. A CE transistor has \( \beta = 100 \), \( V_{CC} = 10 \) V, \( R_C = 4 \) k\( \Omega \) and the base current is 10 \( \mu \)A. Find \( V_{CE} \).
\( I_C = \beta I_B = 1 \) mA. \( V_{CE} = 10 - 1 \times 4 = 6 \) V.

Example 2. The alpha of a transistor is 0.96. Find beta.
\( \beta = 0.96/0.04 = 24 \).

Example 3. Show that two NOT gates joined to the inputs of a NAND gate behave like an OR gate.
\( Y = \overline{\bar{A}\cdot\bar{B}} = A + B \) by De Morgan's theorem.