BIPOLAR JUNCTION TRANSISTOR (BJT) CHAPTER 2 BIPOLAR JUNCTION TRANSISTOR (BJT)

Chapter Objectives Describe the basic structure of the BJT Explain how a transistor is biased and discuss the transistor currents and their application Discuss how a transistor is used as a voltage amplifier Discuss how a transistor is used as an electronic switch

2.0 Bipolar Junction Transistor (BJT) 2.1 Introduction 2.2 Structure of Bipolar Junction Transistor 2.3 Basic Transistor Operation 2.4 Transistor Characteristic and Parameter 2.5 Transistor Application 2.6 Transistor Packages and Terminal Identification 2.7 Typically Faults and Troubleshooting 2.8 DC Operating Point 2.9 Voltage-Divider Bias 2.10 Other Bias Method

2.1 Introduction Transistor can be used as amplifier or switch. It’s act as a current controlling device. Transistor is to amplify signals, this could be an audio signal or perhaps some high frequency radio signal. It has to be able to do this without distorting the original input.

2.2 Structure of Bipolar Junction Transistor BJT is constructed with three doped semiconductor regions called emitter, base and collector. These three regions are separated by two PN junctions. There are two types of BJT, NPN where two N regions separated by a thin P region and PNP, consist of two P regions separated by a thin N region. Each region is called emitter, base and collector. Emitter function as current supplier or carrier. Collector role is to collect charge for circuit operation Base act as junction that control the current flow.

N P More Prevalent

There is one PN junction in diodes. While in bipolar junction transistors (BJT), there are three layers and two PN junctions, base-emitter junction and base-collector junction. Base region is lightly doped, emitter heavily and collector moderate. Fig 4-1 b & c Basic BJT Construction

2.2 Structure of Bipolar Junction Transistor PN junction must be supplied with external dc bias voltage and for both NPN and PNP BE junction is forward-bias and BC junction is reverse-bias, but the bias voltage polarities and current direction are reversed between two types. Bipolar = ??? Refers to the use of both holes and electrons as carriers in the transistor structure

For NPN transistor, electron at emitter is being pushed by negative terminal power supply toward junction into P type base. Base is lightly doped and very narrow, so only small percentage of electron flow from BE junction can combines with hole, flow out of base lead as valence electron forming small base current, IB. Most electron will diffuse into BC depletion region. They are pulled through reverse-bias BC junction by attraction between positive and negative ions i.e. by positive collector voltage and forming collector current, IC which is larger than base current.

Illustration of BJT action

Transistor Current Emitter current is sum of collector current and base current. IE = IC + IB IB is small compare to IE and IC, that we assume IE ≈ IC.

2.2 Transistor Characteristic and Parameters When transistor connected to DC bias voltages (as shown in figure below), VBB forward-bias base-emitter junction and VCC reverse-bias the base-collector junction. Normally VCC direct from supply and VBB (which is smaller) from voltage divider. When transistor operate within it’s linear limit, IC = ßIB, where dc beta, ßDC is current gain of transistor and usually designated as hFE. The value typically from 20 to 200. Fig 4-6a ßDC = hfe with range 20 to 200

Ratio of IC and IE is alpha dc, αDC = IC / IE Ratio of IC and IE is alpha dc, αDC = IC / IE . Typically, αDC range from 0.95 to 0.99 and always less than 1. For proper operation the base-emitter junction is forward biased by VBB and conducts just like a diode. Collector-base junction is reverse biased by VCC and blocks current flow through it’s junction just like a diode. Current flow through the base-emitter junction will help establish the path for current flow from the collector to emitter. Exercise : a. Calculate α and ß and IE when IB = 8µA and IC = 3mA. b. Calculate α, IC and IE when IB = 100 µA and ß= 220.

IC = ßIB αDC = IC / IE IE = IC + IB Exercise : Calculate α and ß and IE when IB = 8µA and IC = 3mA. Answer: α= 0.997 ß= 375 IE= 3.008mA b. Calculate α, IC and IE when IB = 100 µA and ß= 220. Answer: α= 0.995 IC= 22mA IE= 22.1mA IC = ßIB αDC = IC / IE IE = IC + IB

Current and Voltage Analysis There are three key dc voltages and three key DC currents to be considered. IB: dc base current IE: dc emitter current IC: dc collector current VBE: dc voltage across base-emitter junction VCB: dc voltage across collector-base junction VCE: dc voltage from collector to emitter Fig 4-7

Analysis of this transistor circuit to predict the DC voltages and currents requires use of Ohm’s, Kirchhoff’s and transistor β. VBE = 0.7V VBB – VBE = VRB VRB/RB = IB VCE = VCC – VRC = VCC – ICRC IC = β IB VCB = VCE - VBE Fig 4-7