Semiconductor Device Physics Chapter 10 BJT Fundamentals

Bipolar Junction Transistors (BJTs) Chapter 10 BJT Fundamentals Bipolar Junction Transistors (BJTs) Over the past decades, the higher layout density and low-power advantage of CMOS (Complementary Metal–Oxide–Semiconductor) has eroded away the BJT’s dominance in integrated-circuit products. Higher circuit density  better system performance BJTs are still preferred in some digital-circuit and analog-circuit applications because of their high speed and superior gain Faster circuit speed (+) Larger power dissipation (–) Transistor: current flowing between two terminals is controlled by a third terminal

Introduction There are two types of BJT: pnp and npn. Chapter 10 BJT Fundamentals Introduction There are two types of BJT: pnp and npn. The convention used in the textbook does not follow IEEE convention, where currents flowing into a terminal is defined as positive. We will follow the normal convention: . . . . . .

Circuit Configurations Chapter 10 BJT Fundamentals Circuit Configurations Common-Emitter I–V Characteristics Most popular configuration Active Mode Saturation Mode IC < bIB In active mode, bdc is the common emitter dc current gain

Modes of Operation Common-Emitter Output Characteristics Mode Chapter 10 BJT Fundamentals Modes of Operation Common-Emitter Output Characteristics Mode E-B Junction C-B Junction Saturation forward bias Active/Forward reverse bias Inverted Cutoff

Chapter 10 BJT Fundamentals BJT Electrostatics Under equilibrium and normal operating conditions, the BJT may be viewed electrostatically as two independent pn junctions. W : quasineutral base width

BJT Electrostatics Electrostatic potential, V(x) Electric field, E(x) Chapter 10 BJT Fundamentals BJT Electrostatics Electrostatic potential, V(x) Electric field, E(x) Charge density, ρ(x)

BJT Design Important features of a good transistor: Chapter 10 BJT Fundamentals BJT Design Important features of a good transistor: Injected minority carriers do not recombine in the neutral base region  short base, W << Lp for pnp transistor Emitter current is comprised almost entirely of carriers injected into the base rather than carriers injected into the emitter  the emitter must be doped heavier than the base pnp BJT, active mode

Base Current (Active Bias) Chapter 10 BJT Fundamentals Base Current (Active Bias) The base current consists of majority carriers (electrons) supplied for: Recombination of injected minority carriers in the base Injection of carriers into the emitter Reverse saturation current in collector junction Recombination in the base-emitter depletion region EMITTER BASE COLLECTOR 1 4 3 2 p-type n-type p-type

BJT Performance Parameters (pnp) Chapter 10 BJT Fundamentals BJT Performance Parameters (pnp) IEn ICn Negligible compared to holes injected from emitter IEp ICp Emitter Efficiency Base Transport Factor 5 1 2 Decrease relative to and to increase efficiency Decrease relative to to increase transport factor 1 2 Common base dc current gain:

Collector Current (Active Bias) Chapter 10 BJT Fundamentals Collector Current (Active Bias) The collector current is comprised of: Holes injected from emitter, which do not recombine in the base Reverse saturation current of collector junction 2 3 ICB0 :collector current when IE = 0 Common emitter dc current gain:

BJT Static Characteristics Semiconductor Device Physics Chapter 11 BJT Static Characteristics

Minority carrier constants Chapter 11 BJT Static Characteristics Notation (pnp BJT) Minority carrier constants

Emitter Region Diffusion equation: Boundary conditions: Chapter 11 BJT Static Characteristics Emitter Region Diffusion equation: Boundary conditions:

Base Region Diffusion equation: Boundary conditions: Chapter 11 BJT Static Characteristics Base Region Diffusion equation: Boundary conditions:

Collector Region Diffusion equation: Boundary conditions: Chapter 11 BJT Static Characteristics Collector Region Diffusion equation: Boundary conditions:

Ideal Transistor Analysis Chapter 11 BJT Static Characteristics Ideal Transistor Analysis Solve the minority-carrier diffusion equation in each quasi-neutral region to obtain excess minority-carrier profiles Each region has different set of boundary conditions Evaluate minority-carrier diffusion currents at edges of depletion regions Add hole and electron components together  terminal currents is obtained IE IC IB

Emitter Region Solution Chapter 11 BJT Static Characteristics Emitter Region Solution Diffusion equation: General solution: Boundary conditions: Solution

Collector Region Solution Chapter 11 BJT Static Characteristics Collector Region Solution Diffusion equation: General solution: Boundary conditions: Solution

Base Region Solution Diffusion equation: General solution: Chapter 11 BJT Static Characteristics Base Region Solution Diffusion equation: General solution: Boundary conditions: Solution

Base Region Solution Since We can write as Chapter 11 BJT Static Characteristics Base Region Solution Since We can write as

Chapter 11 BJT Static Characteristics Base Region Solution Since

Chapter 11 BJT Static Characteristics Terminal Currents Since Then

Simplified Relationships Chapter 11 BJT Static Characteristics Simplified Relationships To achieve high current gain, a typical BJT will be constructed so that W << LB. Using the limit value Due to VEB We will have Due to VCB

Performance Parameters Chapter 11 BJT Static Characteristics Performance Parameters For specific condition of “Active Mode”: emitter junction is forward biased and collector junction is reverse biased W << LB, nE0/pB0 = NB/NE

Chapter 11 BJT Static Characteristics Homework 7 1. (10.17) Consider a silicon pnp bipolar transistor at T = 300 K with uniform dopings of NE = 5×1018 cm–3, NB = 1017 cm–3, and NC = 5×1015 cm–3 . Let DB = 10 cm2/s, xB = 0.7 μm, and assume xB << LB. The transistor is operating in saturation with JP = 165 A/cm2 and VEB = 0.75 V. Determine: (a) VCB, (b) VEC(sat), (c) the number/cm2 of excess minority carrier holes in the base, and (d) the number/cm2 of excess minority carrier electrons in the long collector, take LC = 35 μm. 2. (10.14) Problem 10.4, Pierret’s “Semiconductor Device Fundamentals”. Due: Monday, 25.11.2013.