Presentation on theme: "Recommended Books Robert Boylestad and Louis Nashelsky, “Electronic Devices and Circuit Theory”, Prentice Hall, 7th Edition or Latest. Thomas L. Floyd,"— Presentation transcript:
1Recommended BooksRobert Boylestad and Louis Nashelsky, “Electronic Devices and Circuit Theory”, Prentice Hall, 7th Edition or Latest.Thomas L. Floyd, “Electronic Devices”, Prentice Hall, 7th Edition or Latest, ISBN:
2This LectureCurrent and Voltage Analysis of BJT – A Review
3Types of Bipolar Junction Transistors npnpnpnpnpnpCECCCross SectionBBBBSchematic SymbolSchematic SymbolEECollector doping is usually ~ 106Base doping is slightly higher ~ 107 – 108Emitter doping is much higher ~ 1015
5DC Beta and DC AlphaDC Beta (dc) : The ratio of the dc collector current (Ic) to the dc base current (IB) is the dc beta. It is also called the dc current gain of a transistor.Typical values of dc range from less than 20 to 200 or higher.If temperature goes up, dc goes up and vice versa.DC Alpha (dc): It is the ratio of dc collector current (Ic) to the dc emitter current (IE).Typically values of dc range from 0.95 to 0.99, but it is always less than unity.
6Relationship between dc and dc For an NPN transistor Dividing each term by IC we get orSimilarly, we can prove that
7Problems on dc and dcDetermine dc and IE for IB = 50A and IC = 3.65 mA.Solution:
8Problems on dc and dc2. What is the dc when IC = 8.23mA and IE = 8.69 mA.Solution:A certain transistor exhibits an dc of Determine IC when IE = 9.35 mA.
9Current and Voltage Analysis IB: dc base currentIE: dc emitter currentIC: dc collector currentVBE: dc voltage across base-emitter junctionVCB: dc voltage across collector-base junctionVCE: dc voltage from collector to emitterTransistor bias circuit.
10Current and Voltage Analysis When the BE junction is forward-biased, it is like a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3)From KVL, the voltage across RB isBy Ohm’s law;Solving for IB
11Current and Voltage Analysis The voltage at the collector is;The voltage drop across RC isVCE can be rewritten asThe voltage across the reverse-biased CB junction is
12ProblemsDetermine IB, IC, IE, VBE, VCB and VCE in the circuit. The transistor has a dc = 150. Solution:
13ProblemsA base current of 50A is applied to the transistor in the adjacent Fig, and a voltage of 5V is dropped across RC. Determine the dc and dc of the transistor. Solution:
14ProblemsFind VCE, VBE and VCB in the given circuit.Solution:
15Problems: Homework Find IB, IE and IC in Fig.1. dc = 0.98. Fig. 1 Ans: IE = 1.3 mA, IB = 30,IC = 1.27 mA.2. Determine the terminal voltages of each transistor with respect to ground for circuit in Fig. 2. Also determine VCE, VBE and VBC.Ans. VB = 10 V, VC = 20 V, VE = 9.3 V, VCE = 10.7, VBE = 0.7 V, VBC = -10 V.Fig. 1Fig. 2
16Modes of Operation BJTs have three regions of operation: Active: BJT acts like an amplifier (most common use)Saturation - BJT acts like a short circuitCutoff - BJT acts like an open circuitBJT is used as a switchBy switchingbetween thesetwo regions.
17More about Transistor Regions Cutoff: In this region,IB = 0 and VCE = VCC.That is, both the base-Emitter and the base-collector junctions arereversed biased.Under this condition, there is a very small amount of collector leakage current ICE0 due mainly to thermally produced carriers. It is usually neglected in circuit analysis.
18More about Transistor Regions Saturation: When theBase-emitter junction isforward biased and thebase current is increased,The collector current alsoIncreases (IC = dcIB) and VCEDecreases (VCE = VCC – ICRC). When VCE reaches its saturation, there is no further change in IC.
19DC Load LineThe bottom of the load Line is at ideal cutoff where IC = 0 and VCE = VCC. The top of the load line is at saturation where IC = IC(sat) and VCE = VCE (sat).
20Quiescent-Point (Q-Point) Operating point of an amplifier to state the values of collector current (ICQ) and collector-emitter voltage (VCEQ).Determined by using transistor output characteristic and DC load line.Quiescent means quiet, still or inactive.
21ExampleThe transistor shown in Figure (a) is biased with variable voltages VCC and VBB to obtain certain values of IB, IC, IE and VCE. The collector characteristic curves are shown in Figure (b). Find Q-point when: (a) IB = 200A (b) 300A (c) 400A.
22Solution:IC = dcIB = 100200 10-6= 20 mAVCE = VCC – ICRC =10 – 2010-3220 = 5.6 VThis Q-Point is shown as Q1.(b) IC = dcIB = 100300 10-3= 30 mA10 – 3010-3220 = 3.4 VThis Q-Point is shown as Q2.(c) IC = dcIB = 100400 10-6VCE = VCC – ICRC= 10 – 4010-3220 = 1.2 VThis Q-Point is shown as Q3.
23ProblemDetermine the interceptpoints of the dc load line onThe vertical and horizontalAxes of the collectorcharacteristic curves in theFig.(b) Assume that you wish to bias the transistor with IB = 20A. To what voltage must you change the VBB supply. What are IC and VCE at the Q-point , given that dc = 50. VBE =0.7