Lecture 17 Bipolar Junction Transistors (BJT): Part 1 Qualitative Understanding - How do they work? Reading: Pierret 10.1-10.6, 11.1
Looks sort of like two diodes back to back pnp mnemonic: Pouring N Pot npn mnemonic: Not Pouring N Voltage Nomenclature Standard V +-
Bipolar Junction Transistor Fundamentals Emitter emits holes Narrow Base controls number of holes emitted ollector collects holes emitted by the emitter Emitter emits electrons Narrow Base controls number of electrons emitted ollector collects electrons emitted by the emitter
V EB E + V B B + + V E 0
Both the input and output share the base in common Both the input and output share the emitter in common Both the input and output share the ollector in common
Active: s useful for amplifiers. Most common mode Saturation: Equivalent to an on state when transistor is used as a switch utoff : Equivalent to an off state when transistor is used as a switch nverted: Rarely if ever used.
Emitter Equilibrium Base ollector Emitter Saturation Base ollector Forward Forward Emitter utoff Base ollector Active (or Forward Active) Emitter Base Forward Accelerated by the Electric Field Reverse Reverse ollector Reversed
Operational modes can be defined based on base-emitter voltages and base-collector voltages When there is no base current, almost no collector current flows When base current flows, a collector current can flow The device is then a current controlled current device
: Electrostatics in Equilibrium Emitter is heavily doped Wwidth of the base quasi-neutral region W B Total Base width W EB Base-Emitter depletion width W B Base-ollector depletion width W EB < W B Base-Emitter built in voltage Emitter Doping >Base Doping>ollector Doping Base-ollector built in voltage Note: This slide refers to a pnp transistor
Equilibrium Active Mode Few electrons injected into the emitter Forward Many Holes injected into the base Narrow Base required to minimize recombination Reverse njected Holes diffuse through the base and are collected by the huge electric field at the collector Note: This slide refers to a pnp transistor
Note: Subscript indicates emitter/collector current (E/) and hole/electron contribution (p/n) E Ep + En Neglecting recombinationgeneration means p ~ Ep p + n Since emitter is more heavily doped than the base, En << Ep Since the base-collector junction is reverse biased, n << cp ~ E and ( B E - ) is small compared to and E Note: This slide refers to a pnp transistor
onsider a pnp Transistor: A small electron base current (flowing into the emitter from the base) controls a larger hole current flowing from emitter to collector. Effectively, we can have the collector-emitter current controlled by the base-emitter current. Note: This slide refers to a pnp transistor
: Performance Parameters () 1 γ Ep E Ep Ep + En Emitter Efficiency: haracterizes how effective the large hole current is controlled by the small electron current. Unity is best, zero is worst. ( 2) α T p Ep Base Transport Factor: haracterizes how much of the injected hole current is lost to recombination in the base. Unity is best, zero is worst. Note: This EE slide 3040 refers - Dr. to a Alan pnp transistor Doolittle
: Performance Parameters Active Mode, ommon Base haracteristics Forward Reverse Bo is defined as the collector current when the emitter is open circuited. t is the ollector-base junction saturation current. Bo Reverse fraction of emitter current making it across the base + leakage current 3 α + where α is the common base D current gain () E Bo ombining (1) and (2), p α T p + Ep Thus comparing this to (3), α γα T γα n T α T E Ep + Bo n and γα n T E + n Note: This slide refers to a pnp transistor
Reverse Forward Bipolar Junction Transistor Fundamentals: Performance Parameters Active Mode, ommon Emitter haracteristics Eo is defined as the collector current when the base is open circuited. multiple of the base current making it across the base + leakage current ( ) β B + Eo 4 where β is the common emitter D current gain But using E + B in (2), ( 5) α ( + ) Note: This slide refers to a pnp transistor ( 6) comparing (4) and solving for β α 1 α α 1 α B B and (6) and + Eo Bo Bo + 1 α Bo 1 α and β B