Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.chapter 17electronics fundamentalscircuits, devices, and applicationsTHOMAS L. FLOYDDAVID M. BUCHLAElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Bipolar junction transistors (BJTs)The BJT is a transistor with three regions and two pn junctions. The regions are named the emitter, the base, and the collector and each is connected to a lead.There are two types of BJTs – npn and pnp.nnpppnE (Emitter)B (Base)C (Collector)EBCSeparating the regions are two junctions. Base-Collector junctionBase-Emitter junctionElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.BJT biasingFor normal operation, the base-emitter junction is forward-biased and the base collector junction is reverse-biased. npnBE forward- biasedBC reverse- biasedFor the npn transistor, this condition requires that the base is more positive than the emitter and the collector is more positive than the base.++For the pnp transistor, this condition requires that the base is more negative than the emitter and the collector is more negative than the base.pnp++Electronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.BJT currentsA small base current (IB) is able to control a larger collector current (IC). Some important current relationships for a BJT are:E C BI I I= +C DC EαI I=C DC BβI I=IIIIBIEICElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Voltage-divider biasBecause the base current is small, the approximation2B CC1 2RV VR R� �=� �+� �is useful for calculating the base voltage.R1R2RCREVBVEAfter calculating VB, you can find VE by subtracting 0.7 V for VBE. Next, calculate IE by applying Ohm’s law to RE: C EI I@Then apply the approximation Finally, you can find the collector voltage fromC CC C CV V I R= -VCEEEVIR=Electronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Voltage-divider biasCalculate VB, VE, and VC for the circuit.2B CC1 26.8 k15 V = 27 k + 6.8 kRV VR R� �W� �= =� �� �+ W W� �� �R1R2RCREVE = VB  0.7 V =C E2.32 mAI I@ =( ) ( )C CC C C15 V 2.32 mA 2.2 kV V I R= - = - W =EEE2.32 V 2.32 mA1.0 kVIR= = =W27 k6.8 k 1.0 k2.2 k+15 V2N39043.02 V2.32 V9.90 VElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Collector characteristic curvesThe collector characteristic curves are a family of curves that show how collector current varies with the collector-emittervoltage for a given IB.ICVCE0IB6IB5IB4IB3IB2IB1IB = 0The saturation region occurs when the base-emitter and the base-collector junctions are both forward biased.The curves are divided into three regions:The active regionactive region is after the saturation region. This is the region for operation of class-A operation.The breakdownbreakdown regionregion is after the active region and is is characterized by rapid increase in collector current. Operation in this region may destroy the transistor.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Load linesA load line is an IV curve that represents the response of a circuit that is external to a specified load. For example, the load line for the Thevenin circuit can be found by calculating the two end points: the current with a shorted load, and the output voltage with no load.+12 V2.0 k004 8 12246I (mA)V (V)ISL = 6.0 mAVSL = 0 VINL = 0 mAVNL = 12 VLoad lineElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Load linesThe IV response for any load will intersect the load line and enables you to read the load current and load voltage directly from the graph.+12 V2.0 k004 8 12246I (mA)V (V) Read the load current and load voltage from the graph if a 3.0 k resistor is the load.3.0 kRL =IV curve for 3.0 k resistorVL = 7.2 V IL = 2.4 mAQ-pointElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Load linesThe load line concept can be extended to a transistor circuit. For example, if the transistor is connected as a load, the transistor characteristic+12 V2.0 k004 8 12246I (mA)V (V)curve and the base current establish the Q-point.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Load linesLoad lines can illustrate the operating conditions for a transistor circuit.004 8 12246I (mA)V (V)If you add a transistor load to the last circuit, the base current will establish the Q-point. Assume the base current is represented by the blue line.+12 V2.0 kFor this base current, the Q-point is:Assume the IV curves are as shown:The load voltage (VCE) and current (IC) can be read from the graph.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Load lines For the transistor, assume the base current is established at 10 A by the bias circuit. Show the Q-point and read the value of VCE and IC.004 8 12246IC (mA)VCE (V)+12 V2.0 kBias circuitIB = 5.0 AIB = 10 AIB = 15 AIB = 20 AIB = 25 AThe Q-point is the intersection of the load line with the 10 A base current.VCE = 7.0 V; IC = 2.4 mAElectronics Fundamentals 8th edition Floyd/BuchlaChapter 17Chapter 17© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Signal operationWhen a signal is applied to a