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A simple NPN bipolar transistor circuit that consists of an NPN bipolartransistor, RBand RCis given in Fig.2.25. If VCC= 15 V, VIN= 5 V andb= 100,calculate IB, IC. What mode will this transistor operate in?Thefirst task is to write Kirchoff’s Voltage Law (KVL) around the BE and CE junctionsas shown in Eqs.2.6and2.7. Therefore,VIN¼RBIBþVBEð2:6ÞVCC¼RCICþVCEð2:7ÞTest for cut-off mode:Set IB= IC= 0 A. This produces VBE= 5 V and VCE= 15 V, and it proves that thetransistor does not operate in cut-off region.Test for active mode, set VBE= 0.7 V and solve IBfrom Eq.2.6. Thus,IB¼ðVIN±VBEÞRB¼ð5±0:7Þ10 KX¼0:43 mAThen,IC=bIB= 100 (0.43 mA) = 43 mA. Substitute IC= 43 mA into Eq.2.7and solve forVCE. Thus,VCE= VCC-RCIC= 15-0.1 KΩ(43 mA) = 15-4.3 = 10.7 VThis value is greater than 0.2 V and proves that the transistor operates in the active regionas shown in Fig.2.26. In thisfigure, Eq.2.7is superimposed on top of the transistor IC-VCEcharacteristics in order to show where exactly this transistor operates at (quiescent point, Q).VCC= 15VRC= 100ΩVCEVIN= 5VVBEIBICRB= 10KΩFig. 2.25A simple NPN bipolar circuit462Diode and Bipolar Transistor Circuits
However, the location of the Q point is not optimal in thisfigure, and as it needs to becloser to the midpoint between 0 and VCCsuch as VCEQ= VCC/2 = 7.5 V to be moresuitable for analog applications.From Fig.2.26, when IB= 0.43 mA and VCE= 7.5 V, ICbecomes 43 mA. Thus,RC¼ðVCC±VCEQÞIC¼ð15±7:5Þ0:043¼174Xinstead of 100X:The new load line is plotted in Fig.2.26with dashed lines.Example 2.2Another bipolar transistor circuit is given in Fig.2.27. In this schematic, ICisselected to be 20 mA. The circuit still uses VIN= 5 V and VCC= 15 V for its operation.What are the values of RBand RCsuch that this transistor operates in the middle of the activeregion at VCEQ= 7.5 V?ICVCE0IB= 0.43mA15VQ10.7VIC= 43mAICMAX= 15V/100Ω = 150mAVCE= VCC–RCIC(load line)7.5VRC = 100ΩRC = 174ΩFig. 2.26The quiescent point of the transistorVCC= 15VRCVCEVIN= 5VVBEIBIC= 20mARBFig. 2.27A bipolar circuit whose IC= 20 mA, but its RCand RBare unknown2.7Bipolar Transistor Circuits47
For this design question, one must refer to the manufacturer’s datasheet and fetch thetransistor IC-VCEcharacteristics as shown in Fig.2.28.The load line from Eq.2.7is superimposed on top of the IC-VCEcharacteristics, and IBbecomes 2 mA at VCEQ= 7.5 V when IC= 20 mA. Since the transistor operates in activeregion,b= IC/IBbecomes 10. Rewriting Eq.2.7yields:RC¼ðVCC±VCEQÞIC¼ð15±7:5Þ0:020¼375XRewriting Eq.2.6yieldsRB¼ðVIN±VBEÞIB¼ð5±0:7Þ0:002¼2:15 KXHowever, there may be an instance where the load line may not necessarily intersect anyof the IC-VCEcurves at VCEQ= 7.5 V provided by the manufacturer, and the situationbecomes similar to the case in Fig.2.29. In this instance, RCis still calculated using Eq.2.7since VCEQand ICQhave not changed.