Chapter 4 Bipolar Junction Transistor REMEMBER THIS Current

Chapter 4 Bipolar Junction Transistor

REMEMBER THIS Current flow in the opposite direction of the electrons flow; same direction as holes e e e I h h h

Transistor Structures Ø The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions. Ø Bipolar transistor is a 3 -terminal device. Ø Emitter (E) Ø Base (B) Ø Collector (C) Ø The basic transistor principle is that the voltage between two terminals controls the current through the third terminal. Ø Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.

Transistor Structures Ø There are two types of bipolar junction transistor: npn and pnp. Ø The npn bipolar transistor contains a thin p-region between two n-regions. Ø The pnp bipolar transistor contains a thin n-region sandwiched between two pregions.

3 Regions of Operation Ø Active Operating range of the amplifier. Base-Emitter Junction forward biased. Collector-Base Junction reverse biased Ø Cutoff The amplifier is basically off. There is voltage but little current. Both junctions reverse biased Ø Saturation The amplifier is full on. There is little voltage but lots of current. Both junctions forward biased

OPERATIONS - npn ACTIVE MODE Ø The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased, . Ø Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base IE Ø Once in the base region, the electrons are quickly accelerated through the base due to the reversebiased C-B region IC + VBE - i. B ØSome electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current IB

TO ILLUSTRATE C B E VBE + • Imagine the marbles as electrons • A flat base region with gaps where the marbles may fall/trapped – recombine • A sloping collector region represents high electric field in the C-B region • Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector

MATHEMATICAL EXPRESSIONS IC IB + VBE IE - IE = IS [ e VBE / VT -1 ] = IS e VBE / VT Based on KCL: IE = IC + IB No. of electrons crossing the base region and then directly into the collector region is a constant factor of the no. of electrons exiting the base region IC = IB No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region. IC = IE Ideally = 1, but in reality it is between 0. 9 and 0. 998.

Based on KCL: IE = IC + IB IE = IB + IB = IB( + 1) IC = IB IC = IE IE = IB( + 1) Now With IC = IB = IC / Hence, IE = [ IC / ] ( + 1) IC = IE [ / + 1 ] Comparing with IC = IE = [ / + 1 ]

C OPERATIONS - pnp FORWARD ACTIVE MODE Ø The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased, . IE = IS [ e VEB / VT -1 ] = IS e VEB / VT **Notice that it is VEB Based on KCL: IE = IC + IB IB IC B VEB IE + E

pnp Transistor- Active mode

SUMMARY: Circuit Symbols and Conventions Based on KCL: IE = IC + IB npn bipolar transistor simple block diagram and circuit symbol. Arrow is on the emitter terminal that indicates the direction of emitter current (out of emitter terminal for the npn device) pnp bipolar transistor simple block diagram and circuit symbol. Arrow is on the emitter terminal that indicates the direction of emitter current (into of emitter terminal for the pnp device)

NPN PNP IE = IS [ e VBE / VT ] IE = IS [ e VEB / VT] IC = IB IC = IE IE = IB( + 1) = [ / + 1 ] = [ / 1 - ] Based on KCL: IE = IC + IB

EXAMPLE 4. 1 Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain and a base current of . Also assume that the transistor is biased forward in the forward active mode. Solution: The common-emitter current gain is The collector current is And the emitter current is

Examples • EXAMPLE 1 • Given IB = 6. 0 A and IC=510 A Determine , and IE l l EXAMPLE 3 PNP Transistor = 60, IC= 0. 85 m. A Determine , IE and IB • EXAMPLE 2 • NPN Transistor • Reverse saturation current Is = 10 -13 A with current gain, = 90. Based on VBE = 0. 685 V, determine IC , IB and IE

BJT: Current-Voltage Characteristic IC versus VCE

Common-Emitter Configuration - npn Ø The Emitter is common to both input (base-emitter) and output (collectoremitter). Ø Since Emitter is grounded, VC = VCE Ø With decreasing VC (VCE), the junction B-C will become forward biased too. § The current IC quickly drops to zero because electrons are no longer collected by the collector Node B 0 V

Characteristics of Common-Emitter - npn NOTE: VEC for PNP