BJT REgions

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Post on 11-Apr-2015

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Name of the file gives good info. This file intended for any ECE students wants to brush-up/refresh/read MOST basics of a BJT.

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<p> LP4 1TransistorsThey are unidirectional current carrying devices like diodes with capability to control the current flowing through them The switch current can be controlled by either current or voltage Bipolar Junction Transistors (BJT) control current by current Field Effect Transistors (FET) control current by voltageThey can be used either as switches or as amplifiersDiodes and Transistors are the basic building blocks of the multibillion dollar semiconductor industries LP4 2NPN Bipolar Junction TransistorOne N-P (Base Collector) diode one P-N (Base Emitter) diode LP4 3PNP Bipolar Junction TransistorOne P-N (Base Collector) diode one N-P (Base Emitter) diode LP4 4Analogy with Transistor :Fluid-jet operated Valve LP4 5NPN BJT Current flow LP4 6BJT and From the previous figure iE = iB + iCDefine = iC / iE Define = iC / iBThen = iC / (iE iC) = /(1- )Then iC = iE ; iB = (1-) iE Typically 100 for small signal BJTs (BJTs thathandle low power) operating in active region (region where BJTs work as amplifiers) LP4 7BJT in Active Region Common Emitter(CE) Connection Called CE because emitter is common to both VBB and VCC LP4 8Analogy with Transistor in Active Region: Fluid-jet operated ValveIn active region this stopper does not really havenoticeable effect on the flow rate LP4 9BJT in Active Region (2)Base Emitter junction is forward biasedBase Collector junction is reverse biasedFor a particular iB, iC is independent of RCC transistor is acting as current controlled current source (iC is controlled by iB, and iC = iB) Since the base emitter junction is forward biased, from Shockleyequation]]]</p> <p>,`</p> <p>.| 1VVexp I iTBEES E LP4 10BJT in Active Region (3)]]]</p> <p>,`</p> <p>.| 1VVexp I ) - (1 iTBEES BSince iB = (1-) iE , the previous equation can be rewritten asNormally the above equation is never used to calculate iB Since for all small signal transistors vBE 0.7. It is only usefulfor deriving the small signal characteristics of the BJT.For example, for the CE connection, iB can be simply calculated as,BBBE BBBRV Vi or by drawing load line on the base emitter side LP4 11Deriving BJT Operating points in Active Region An ExampleIn the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5 k, RCC = 1 k, VCC = 10V. Find IB,IC,VCE, and the transistor power dissipation using the characteristics as shown belowBBBE BBBRV VI By Applying KVL to the base emitter circuitBy using this equation along with the iB / vBE characteristics of the base emitter junction, IB = 40 A LP4 12Deriving BJT Operating points in Active Region An Example (2)By using this equation along with the iC / vCE characteristics of the base collector junction, iC = 4 mA, VCE = 6VBy Applying KVL to the collector emitter circuitCCCE CCCRV VI 100A 40mA 4IIBC Transistor power dissipation = VCEIC = 24 mWWe can also solve the problem without using the characteristicsif and VBE values are known LP4 13iB100 A05V vBEiC10 mA020V vCE100 A80 A60 A40 A20 ADeriving BJT Operating points in Active Region An Example (3)Output CharacteristicsInput Characteristics LP4 14BJT in Cutoff Region Under this condition iB= 0As a result iC becomes negligibly smallBoth base-emitter as well base-collector junctions may be reverse biasedUnder this condition the BJT can be treated as an off switch LP4 15Analogy with Transistor CutoffFluid-jet operated Valve LP4 16BJT in Saturation Region Under this condition iC / iB &lt; in active regionBoth base emitter as well as base collector junctions are forward biasedVCE 0.2 VUnder this condition the BJT can be treated as an on switch LP4 17A BJT can enter saturation in the following ways (refer to the CE circuit)For a particular value of iB, if we keep on increasing RCCFor a particular value of RCC, if we keep on increasing iBFor a particular value of iB, if we replace the transistor with one with higher BJT in Saturation Region (2) LP4 18Analogy with Transistor in Saturation Region: Fluid-jet operated Valve(1) This stopper is almost closed; thus valve position does not have much influence on the flow rate LP4 19Analogy with Transistor SaturationFluid-jet operated Valve (2) The valve is wide open; changing valve position a little bit does not have much influence on the flow rate. LP4 20In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5 k, RCC = 100 k, VCC = 10V. Find IB,IC,VCE, and the transistor power dissipation using the characteristics as shown belowBJT in Saturation Region Example 1Here even though IB is still 40 A; from the output characteristicsIC can be found to be only about 1mA and VCE 0.2V( VBC 0.5V or base collector junction is forward biased (how?)) = IC / IB = 1mA/40 A = 25&lt; 100 LP4 21iB100 A05V vBEInput CharacteristicsBJT in Saturation Region Example 1 (2) iC10 mA020V vCE100 A80 A60 A40 A20 A LP4 22BJT in Saturation Region Example 2In the CE Transistor circuit shown earlier VBB= 5V, RBB= 50 k, RCC = 1 k, VCC = 10V. Find IB,IC,VCE, and the transistor power dissipation using the characteristics as shown belowHere IB is 80 A from the input characteristics; IC can be found to be only about 7.9 mA from the output characteristics and VCE 0.5V( VBC 0.2V or base collector junction is forward biased (how?)) = IC / IB = 7.9 mA/80 A = 98.75 &lt; 100Note: In this case the BJT is not in very hard saturationTransistor power dissipation = VCEIC 4 mW LP4 2310 mAOutput CharacteristicsiC020V vCE100 A80 A60 A40 A20 AiB100 A05V vBEInput CharacteristicsBJT in Saturation Region Example 2 (2) LP4 24In the CE Transistor circuit shown earlier VBB= 5V, VBE = 0.7V RBB= 107.5 k, RCC = 1 k, VCC = 10V, = 400. Find IB,IC,VCE, and the transistor power dissipation using the characteristics as shown belowBJT in Saturation Region Example 3A 40RV VIBBBE BBB By Applying KVL to the base emitter circuitThen IC = IB= 400*40 A = 16000 A and VCE = VCC-RCC* IC =10- 0.016*1000 = -6V(?)But VCE cannot become negative (since current can flow only from collector to emitter).Hence the transistor is in saturation LP4 25BJT in Saturation Region Example 3(2)Hence VCE 0.2VIC = (10 0.2) /1 = 9.8 mAHence the operating = 9.8 mA / 40 A = 245 LP4 26BJT Operating Regions at a Glance (1) LP4 27BJT Operating Regions at a Glance (2) LP4 28BJT Large-signal (DC) model LP4 29BJT Q Point (Bias Point)Q point means Quiescent or Operating point Very important for amplifiers because wrong Q point selection increases amplifier distortion Need to have a stable Q point, meaning the the operating point should not be sensitive to variation to temperature or BJT , which can vary widely LP4 30 Four Resistor bias Circuit for Stable Q PointBy far best circuit for providing stable bias point</p>