1、Bipolar Junction Transistors,ET 212 Electronics,Electrical and Telecommunication Engineering TechnologyProfessor Jang,Acknowledgement,I want to express my gratitude to Prentice Hall giving me the permission to use instructors material for developing this module. I would like to thank the Department
2、of Electrical and Telecommunications Engineering Technology of NYCCT for giving me support to commence and complete this module. I hope this module is helpful to enhance our students academic performance.,Objectives,Introduction to Bipolar Junction Transistor (BJT),Basic Transistor Bias and Operatio
3、n,Amplifier or Switch,Parameters, Characteristics and Transistor Circuits,ET212 Electronics BJTs Floyd 2,Key Words: BJT, Bias, Transistor, Amplifier, Switch,Introduction,A transistor is a device which can be used as either an amplifier or a switch. Lets first consider its operation in a more simple
4、view as a current controlling device.,ET212 Electronics BJTs Floyd 3,Basic Transistor Operation,Look at this one circuit as two separate circuits, the base-emitter(left side) circuit and the collector-emitter(right side) circuit. Note that the emitter leg serves as a conductor for both circuits.The
5、amount of current flow in the base-emitter circuit controls the amount of current that flows in the collector circuit. Small changes in base-emitter current yields a large change in collector-current.,ET212 Electronics BJTs Floyd 4,Transistor Structure,The BJT (bipolar junction transistor) is constr
6、ucted with three doped semiconductor regions separated by two pn junctions, as shown in Figure (a). The three regions are called emitter, base, and collector. Physical representations of the two types of BJTs are shown in Figure (b) and (c). One type consists of two n regions separated by a p region
7、s (npn), and other type consists of two p regions separated by an n region (pnp).,ET212 Electronics BJTs Floyd 5,ET212 Electronics BJT Prof. Jang 6,Transistor Currents,The directions of the currents in both npn and pnp transistors and their schematic symbol are shown in Figure (a) and (b). Notice th
8、at the arrow on the emitter of the transistor symbols points in the direction of conventional current. These diagrams show that the emitter current (IE) is the sum of the collector current (IC) and the base current (IB), expressed as follows:IE = IC + IB,Transistor Characteristics and Parameters,Fig
9、ure shows the proper bias arrangement for npn transistor for active operation as an amplifier.Notice that the base-emitter (BE) junction is forward-biased and the base-collector (BC) junction is reverse-biased. As previously discussed, base-emitter current changes yields large changes in collector-e
10、mitter current. The factor of this change is called beta(). = IC/IB,The ratio of the dc collector current (IC) to the dc emitter current (IE) is the alpha. = IC/IE,ET212 Electronics BJTs 7,Ex 3-1 Determine DC and IE for a transistor where IB = 50 A and IC = 3.65 mA.,IE = IC + IB = 3.65 mA + 50 A = 3
11、.70 mA,ET212 Electronics BJTs Floyd 8,Transistor Characteristics and Parameters,The collector current is determined by multiplying the base current by beta.,Analysis of this transistor circuit to predict the dc voltages and currents requires use of Ohms law, Kirchhoffs voltage law and the beta for t
12、he transistor. Application of these laws begins with the base circuit to determine the amount of base current. Using Kichhoffs voltage law, subtract the .7 VBE and the remaining voltage is dropped across RB. Determining the current for the base with this information is a matter of applying of Ohms l
13、aw. VRB/RB = IB,ET212 Electronics BJTs 9,.7 VBE will be used in most analysis examples.,Transistor Characteristics and Parameters,What we ultimately determine by use of Kirchhoffs voltage law for series circuits is that in the base circuit VBB is distributed across the base-emitter junction and RB i
14、n the base circuit. In the collector circuit we determine that VCC is distributed proportionally across RC and the transistor(VCE).,ET212 Electronics BJTs Floyd 10,Current and Voltage Analysis,There are three key dc voltages and three key dc currents to be considered. Note that these measurements ar
15、e important for troubleshooting.,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,ET212 Electronics BJTs Floyd 11,Current and Voltage Analysis-cont
16、inued,When the base-emitter junction is forward-biased,VBE 0.7 VVRB = IBRB : by Ohms lawIBRB = VBB VBE : substituting for VRBIB = (VBB VBE) / RB : solving for IBVCE = VCC VRc : voltage at the collector withVRc = ICRC respect to emitterVCE = VCC ICRC The voltage across the reverse-biased collector-ba
17、se junction VCB = VCE VBE where IC = DCIB,ET212 Electronics BJTs Floyd 12,Ex 3-2 Determine IB, IC, VBE, VCE, and VCB in the circuit of Figure. The transistor has a DC = 150.,When the base-emitter junction is forward-biased,VBE 0.7 VIB = (VBB VBE) / RB = (5 V 0.7 V) / 10 k = 430 A,ET212 Electronics B
18、JTs Floyd 13,IC = DCIB = (150)(430 A) = 64.5 mAIE = IC + IB = 64.5 mA + 430 A = 64.9 mA,VCE = VCC ICRC = 10 V (64.5 mA)(100 ) = 3.55 VVCB = VCE VBE = 3.55 V 0.7 V = 2.85 V,ET212 Electronics BJT Prof. Jang 14,Collector Characteristic Curve,Collector characteristic curves gives a graphical illustratio
19、n of the relationship of collector current and VCE with specified amounts of base current. With greater increases of VCC , VCE continues to increase until it reaches breakdown, but the current remains about the same in the linear region from .7V to the breakdown voltage.,Ex 3-3 Sketch an ideal famil
20、y of collector curves for for the circuit in Figure for IB = 5 A increment. Assume DC = 100 and that VCE does not exceed breakdown.,IC = DC IB,ET212 Electronics BJTs 15,IB IC 5 A 0.5 mA 10 A 1.0 mA 15 A 1.5 mA 20 A 2.0 mA 25 A 2.5 mA,Transistor Characteristics and Parameters-Cutoff,With no IB the tr
21、ansistor is in the cutoff region and just as the name implies there is practically no current flow in the collector part of the circuit. With the transistor in a cutoff state the the full VCC can be measured across the collector and emitter(VCE),ET212 Electronics BJT Prof. Jang 16,Cutoff: Collector
22、leakage current (ICEO) is extremely small and is usually neglected. Base-emitter and base-collector junctions are reverse-biased.,Transistor Characteristics and Parameters - Saturation,Once this maximum is reached, the transistor is said to be in saturation. Note that saturation can be determined by
23、 application of Ohms law. IC(sat)=VCC/RC The measured voltage across this now seemingly “shorted” collector and emitter is 0V.,Saturation: As IB increases due to increasing VBB, IC also increases and VCE decreases due to the increased voltage drop across RC. When the transistor reaches saturation, I
24、C can increase no further regardless of further increase in IB. Base-emitter and base-collector junctions are forward-biased.,17,Transistor Characteristics and Parameters - DC Load Line,The dc load line graphically illustrates IC(sat) and Cutoff for a transistor.,Floyd 18,DC load line on a family of
25、 collector characteristic curves illustrating the cutoff and saturation conditions.,Ex 3-4 Determine whether or not the transistors in Figure is in saturation. Assume VCE(sat) = 0.2 V.,First, determine IC(sat),Now, see if IB is large enough to produce IC(sat).,ET212 Electronics BJTs Floyd 19,Transis
26、tor Characteristics and Parameters Maximum Transistor Ratings,A transistor has limitations on its operation. The product of VCE and IC cannot be maximum at the same time. If VCE is maximum, IC can be calculated as,Ex 4-5 A certain transistor is to be operated with VCE = 6 V. If its maximum power rat
27、ing is 250 mW, what is the most collector current that it can handle?,ET212 Electronics BJTs Floyd 20,First, find IB so that you can determine IC.,The voltage drop across RC is.,PD = VCE(max)IC = (15V)(19.5mA) = 293 mW,VCE(max) will be exceeded first because the entire supply voltage, VCC will be dr
28、opped across the transistor.,VRc = ICRC = (19.5 mA)(1.0 k) = 19.5 V VRc = VCC VCE when VCE = VCE(max) = 15 V VCC(max) = VCE(max) + VRc = 15 V + 19.5 V = 34.5 V,Ex 3-5 The transistor in Figure has the following maximum ratings: PD(max) = 800 mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the ma
29、ximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?,Floyd 21,The Transistor as an Amplifier,Amplification of a relatively small ac voltage can be had by placing the ac signal source in the base circuit. Recall that small changes in the base curr
30、ent circuit causes large changes in collector current circuit.,The ac emitter current : Ie Ic = Vb/re The ac collector voltage : Vc = IcRc Since Ic Ie, the ac collector voltage : Vc IeRc The ratio of Vc to Vb is the ac voltage gain : Av = Vc/Vb Substituting IeRc for Vc and Iere for Vb : Av = Vc/Vb I
31、cRc/Iere The Ie terms cancel : Av Rc/re,ET212 Electronics BJTs Floyd 22,Ex 3-6 Determine the voltage gain and the ac output voltage in Figure if re = 50 .,The voltage gain : Av Rc/re = 1.0 k/50 = 20 The ac output voltage : AvVb = (20)(100 mV) = 2 V,ET212 Electronics BJTs Floyd 23,The Transistor as a
32、 Switch,ET212 Electronics BJTs Floyd 24,A transistor when used as a switch is simply being biased so that it is in cutoff (switched off) or saturation (switched on). Remember that the VCE in cutoff is VCC and 0V in saturation.,Conditions in Cutoff & Saturation,A transistor is in the cutoff region wh
33、en the base-emitter junction is not forward-biased. All of the current are zero, and VCE is equal to VCCVCE(cutoff) = VCCWhen the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated. The formula for collector sa
34、turation current is,The minimum value of base current needed to produce saturation is,ET212 Electronics BJTs Floyd 25,Ex 3-7 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V? (b) What minimum value of IB is required to saturate this transistor if DC is 200? Neglect VCE(sat). (c)
35、Calculate the maximum value of RB when VIN = 5 V.,(a) When VIN = 0 VVCE = VCC = 10 V(b) Since VCE(sat) is neglected,(c) When the transistor is on, VBE 0.7 V.VRB = VIN VBE 5 V 0.7 V = 4.3 VCalculate the maximum value of RB,Floyd 26,Troubleshooting,Internal opens within the transistor itself could als
36、o cause transistor operation to cease. Erroneous voltage measurements that are typically low are a result of point that is not “solidly connected”. This called a floating point. This is typically indicative of an open. More in-depth discussion of typical failures are discussed within the textbook.,O
37、pens in the external resistors or connections of the base or the circuit collector circuit would cause current to cease in the collector and the voltage measurements would indicate this.,ET212 Electronics BJTs Floyd 27,Troubleshooting,Testing a transistor can be viewed more simply if you view it as testing two diode junctions. Forward bias having low resistance and reverse bias having infinite resistance.,ET212 Electronics BJTs Floyd 28,
