BJT STRUCTUREBasic structure of the bipolar junction transistor (BJT) determines itsoperating characteristics.Constructed with 3 doped semiconductor regions called emitter, base,and collector, which separated by two pn junctions.2 types of BJT;(2)npn: Two n regions separated by a p region(3)pnp: Two p regions separated by an n region.BIPOLAR:refers to the useof both holes &electrons ascurrent carriersin the transistorstructure.
Base-emitter junction: the pn junction joining the base region & theemitter region.Base-collector junction: the pn junction joining the base region & thecollector region.A wire lead connects to each of the 3 regions. These leads labeled as; E: emitter B: base C: collector
BASE REGION: lightly doped, & very thinEMITTER REGION: heavily dopedCOLLECTOR REGION: moderately doped
BASIC BJT OPERATIONFor a BJT to operate properly as an amplifier, the two pn junctionsmust be correctly biased with external dc voltages.Figure: shows a bias arrangement for npn BJTs for operation as anamplifier.In both cases, BE junction is forward-biased & the BC junction isreverse-biased. called forward-reverse bias. 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 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.
The heavily doped n-type emitter region has a very high density ofconduction-band (free) electrons.These free electrons easily diffuse through the forward-based BEjunction into the lightly doped & very thin p-type base region(indicated by wide arrow).The base has a low density of holes, which are the majority carriers(represented by the white circles).A small percentage of the total number of free electrons injected intothe base region recombine with holes & move as valence electronsthrough the base region & into the emitter region as hole current(indicated by red arrows).
When the electrons that have recombined with holes as valenceelectrons leave the crystalline structure of the base, they become freeelectrons in the metallic base lead & produce the external basecurrent.Most of the free electrons that have entered the base do not recombinewith holes because the base is very thin.As the free electrons move toward the reverse-biased BC junction,they are swept across into the collector region by the attraction of thepositive collector supply voltage.The free electrons move through the collector region, into the externalcircuit, & then return into the emitter region along with the basecurrent.The emitter current is slightly greater than the collector currentbecause of the small base current that splits off from the total currentinjected into the base region from the emitter.
Transistor CurrentsThe directions of the currents in both npn and pnp transistors and theirschematic symbol are shown in Figure (a) and (b). Arrow on the emitterof the transistor symbols points in the direction of conventionalcurrent. These diagrams show that the emitter current (IE) is the sum ofthe collector current (IC) and the base current (IB), expressed as follows: I E = IC + I B
BJT CHARACTERISTICS & PARAMETERSFigure shows the proper biasarrangement for npntransistor for activeoperation as an amplifier.Notice that the base-emitter(BE) junction is forward-biased by VBB and the base-collector (BC) junction isreverse-biased by VCC. The dccurrent gain of a transistor isthe ratio of the dc collectorcurrent (IC) to the dc base The ratio of the dc collector current (IC)current (IB), and called dc beta to the dc emitter current (IE) is the dc(β DC). alpha. α DC = IC/IE
Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.
Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA. I C 3.65mAβ DC = = = 73 IB 50 µA IE = IC + IB = 3.65 mA + 50 μA = 3.70 mA I C 3.65mAα DC = = = 0.986 I E 3.70mA
Analysis of this transistor circuit to predict the dc voltages and currentsrequires use of Ohm’s law, Kirchhoff’s voltage law and the beta for thetransistor;Application of these laws begins with the base circuit to determine theamount of base current. Using Kichhoff’s voltage law, subtract the VBE=0.7 V, and the remaining voltage is dropped across RB .Thus, VRB = VBB - VBE.Determining the current for the base with this information is a matter ofapplying of Ohm’s law. VRB/RB = IBThe collector current isdetermined bymultiplying the basecurrent by beta.Thus, IC= βDC * IB
What we ultimatelydetermine by use ofKirchhoff’s voltage lawfor series circuits is that,in the base circuit, VBB isdistributed across thebase-emitter junctionand RB in the basecircuit. In the collectorcircuit we determine thatVCC is distributedproportionally acrossRC and thetransistor(VCE).
BJT Circuit AnalysisThere are three key dc voltages and three key dc currents to beconsidered. Note that these measurements are important fortroubleshooting.IB: dc base currentIE: dc emitter currentIC: dc collector currentVBE: dc voltage acrossbase-emitter junctionVCB: dc voltage acrosscollector-base junctionVCE: dc voltage fromcollector to emitter
When the base-emitter junction is forward-biased, VBE ≅ 0.7 V VRB = IBRB : by Ohm’s law IBRB = VBB – VBE : substituting for VRB IB = (VBB – VBE) / RB : solving for IB VCE = VCC – VRc : voltage at the collector with respect to the grounded emitter VRc = ICRC VCE = VCC – ICRC : voltage at the collector with respect to the emitter The voltage across the reverse-biased collector-base junction
Ex 4-2 Determine I , I , I , VB C E , VCE, and VCB in the circuit of Figure. The BEtransistor has a βDC = 150.
Ex 4-2 Determine I , I , I , VB C E , VCE, and VCB in the circuit of Figure. The BE transistor has a βDC = 150.When the base-emitter junction is forward-biased, IC = βDCIB = (150)(430 μA) VBE ≅ 0.7 V = 64.5 mA IB = (VBB – VBE) / RB IE = IC + IB = 64.5 mA + 430 μA = (5 V – 0.7 V) / 10 kΩ = 430 μA = 64.9 mAVCE = VCC – ICRC = 10 V – (64.5 mA)(100 Ω) = 3.55 V VCB = VCE – VBE = 3.55 V – 0.7 V = 2.85 VSince the collector is at a highervoltage than the base, the collector-base junction is reverse-biased.
Collector Characteristic CurvesGives a graphicalillustration of therelationship of collectorcurrent and VCE withspecified amounts ofbase current. Withgreater increases of VCC ,VCE continues to increaseuntil it reachesbreakdown, but thecurrent remains about thesame in the linear regionfrom 0.7V to thebreakdown voltage.
Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.
Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5μA increment. Assume βDC = 100 and that VCE does not exceed breakdown. IC = βDC IB IB IC 5 μA 0.5 mA 10 μA 1.0 mA 15 μA 1.5 mA 20 μA 2.0 mA 25 μA 2.5 mA
CutoffWith no IB , the transistor is in the cutoff region and just as thename implies there is practically no current flow in thecollector part of the circuit. With the transistor in a cutoff state,the full VCC can be measured across the collector andemitter(VCE). Cutoff: Collector leakage current (ICEO) is extremely small and is usually neglected. Base-emitter and base-collector junctions are reverse-biased.
Saturation Once VCE reaches its maximum value, the transistor is said to be in saturation.Saturation: As IB increases due to increasing VBB, IC also increases and VCEdecreases due to the increased voltage drop across RC. When the transistor reachessaturation, IC can increase no further regardless of further increase in IB. Base-emitter and base-collector junctions are forward-biased.
DC Load LineThe dc load line graphically illustrates IC(sat) and cutoff for a transistor. Active region of the transistor’s operation. DC load line on a family of collector characteristic curves illustrating the cutoff and saturation conditions.
Ex 4-4 Determine whether or not the transistors in Figure is insaturation. Assume VCE(sat) = 0.2 V.
Ex 4-4 Determine whether or not the transistors in Figure is in saturation. Assume VCE(sat) = 0.2 V.First, determine IC(sat) VCC − VCE ( sat )I C ( sat ) = RC 10 V − 0.2 V = = 9.8 mA 1.0 kΩNow, see if IB is large enough to produce IC(sat). VBB − VBE 3V − 0.7 V 2.3VIB = = = = 0.23 mA Thus, IC greater than RB 10kΩ 10 kΩ IC(sat). Therefore, theI C = β DC I B = (50)(0.23 mA) = 11.5 mA transistor is saturated.
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 PD (max) IC = VCEEx 4-5 A certain transistor is to be operated with VCE = 6 V. Ifits maximum power rating is 250 mW, what is the most collectorcurrent that it can handle? PD (max) 250 mW IC = = = 41.7 mA VCE 6V
Ex 4-6 The transistor in Figure has the following maximum ratings: P D(max)= 800mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCCcan be adjusted without exceeding a rating. Which rating would be exceeded first?
Ex 4-6 The transistor in Figure has the following maximum ratings: P D(max)= 800 mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?First, find IB so that you can determine IC. VBB − VBE 5V − 0.7V IB = = = 195µA RB 22kΩ I C = β DC I B = (100)(195µA) = 19.5mAThe voltage drop across RC is. VRc = ICRC = (19.5 mA)(1.0 kΩ) = 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 PD = VCE(max)IC = (15V)(19.5mA) = 293 mWVCE(max) will be exceeded first because the entire supply voltage, VCC will bedropped across the transistor.
Derating PD(max) P D(max) is usually specified at 25°C. At higher temperatures, P D(max) is less. Datasheets often give derating factors for determining P D(max) at any temperature above 25°C.Ex 4-7A certain transistor has a P D(max) of 1 mW at 25°C. The deratingfactor is 5 mW/ °C. What is the P D(max) at a temperature of 70°C?
Transistor DatasheetRefer Figure 4-20 (a partial datasheet for the 2N3904 npntransistor).The maximum collector-emitter voltage (VCEO) is 40V.The CEO subscript indicates that the voltage is measured fromcollector to emitter with the base open. VCEO= VCE(MAX)The maximum collector current is 200 mA.* Other characteristics can be referred from the datasheet.
A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCCcan be adjusted without exceeding a rating. Which rating would be exceeded first?
A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?PD(max) = 800 mWVCE(max) = 15 VIC(max) = 100 mA. I B = 195µ A I C = β DC I B = 19.5mAVCC(max) = VCE(max) + VRc = 40 V + 19.5 V = 59.5 VHowever at the max value of VCE, the power dissipation is PD = VCE(max)IC = (40V)(19.5mA) = 780 mW Power Dissipation exceeds the maximum of 645 mW specified on the datasheet.
THE BJT AS AN AMPLIFIERAmplification of a relativelysmall ac voltage can be had byplacing the ac signal source inthe base circuit.Recall that small changes in thebase current circuit causes largechanges in collector currentcircuit. The ac emitter current : Ie ≈ Ic = Vb/r’e 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 Ier’e for Vb : Av = Vc/Vb ≈ IcRc/Ier’e The Ie terms cancel : Av ≈ Rc/r’e
Ex 4-9 Determine the voltage gain and the ac outputvoltage in Figure if r’e = 50 Ω. The voltage gain : Av ≈ Rc/r’e = 1.0 kΩ/50 Ω = 20 The ac output voltage : AvVb = (20)(100 mV) = 2 V
THE BJT AS A SWITCHA transistor when used as a switch is simply being biased so that itis in cutoff (switched off) or saturation (switched on). Rememberthat the VCE in cutoff is VCC and 0V in saturation.
Conditions in Cutoff & SaturationA transistor is in the cutoff region when the base-emitter junction is notforward-biased. All of the current are zero, and VCE is equal to VCC VCE(cutoff) = VCCWhen the base-emitter junction is forward-biased and there is enough basecurrent to produce a maximum collector current, the transistor is saturated.The formula for collector saturation current is VCC − VCE ( sat ) I C ( sat ) = RCThe minimum value of base current I C ( sat ) I B (min) =needed to produce saturation is β DC
Ex 4-10 (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 βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.
Ex 4-10 (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 βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.(a) When VIN = 0 V VCE = VCC = 10 V(b) Since VCE(sat) is neglected, VCC 10 V I C ( sat ) = = = 10 mA RC 1.0 kΩ I C ( sat ) 10 mA I B (min) = = = 50 µA β DC 200(c) When the transistor is on, VBE ≈ 0.7 V. VR = VIN – VBE ≈ 5 V – 0.7 V = 4.3 V B Calculate the maximum value of RB VRB 4.3V RB (max) = = = 86 kΩ I B (min) 50 µA
Transistor ConstructionThere are two types of transistors: • pnp • npn pnpThe terminals are labeled: • E - Emitter • B - Base • C - Collector npn 42
Transistor OperationWith the external sources, VEE and VCC, connected as shown: • The emitter-base junction is forward biased • The base-collector junction is reverse biased 43
Currents in a TransistorEmitter current is the sum of the collector andbase currents: IE = IC + IBThe collector current is comprised of twocurrents: IC = IC + I CO majority minority 44
Common-Base ConfigurationThe base is common to both input (emitter–base) andoutput (collector–base) of the transistor. 45
Common-Base Amplifier Input CharacteristicsThis curve shows the relationshipbetween of input current (IE) to inputvoltage (VBE) for three output voltage(VCB) levels. 46
Common-Base AmplifierOutput CharacteristicsThis graph demonstratesthe output current (IC) toan output voltage (VCB) forvarious levels of inputcurrent (IE). 47
Operating Regions• Active – Operating range of the amplifier.• Cutoff – The amplifier is basically off. There is voltage, but little current.• Saturation – The amplifier is full on. There is current, but little voltage. 48
ApproximationsEmitter and collector currents: I ≅I C EBase-emitter voltage: VBE = 0.7 V (for Silicon) 49
Alpha (α )Alpha (α ) is the ratio of IC to IE : IC αdc = IEIdeally: α = 1In reality: α is between 0.9 and 0.998Alpha (α ) in the AC mode: mode ΔI C αac = ΔI E 50
Transistor AmplificationCurrents and Voltages: Voltage Gain: V 200mV VL 50VI E = Ii = i = = 10mA Av = = = 250 Ri 20Ω Vi 200mVI ≅I C EI ≅ I = 10 mA L iV = I R = (10 ma )(5 kΩ) = 50 V L L 51
Common–Emitter ConfigurationThe emitter is common to both input(base-emitter) and output (collector-emitter).The input is on the base and theoutput is on the collector. 52
Common-Emitter Characteristics Collector Characteristics Base Characteristics 53
Common-Emitter Amplifier CurrentsIdeal Currents IE = IC + IB IC = α IEActual Currents IC = α IE + ICBO where ICBO = minority collector current ICBO is usually so small that it can be ignored, except in high power transistors and in high temperature environments.When IB = 0 µA the transistor is in cutoff, but there is some minoritycurrent flowing called ICEO. I CBO I CEO = I B = 0 μA 1− α 54
Beta (β )β represents the amplification factor of a transistor. (β issometimes referred to as hfe, a term used in transistor modelingcalculations)In DC mode: IC βdc = IBIn AC mode: ∆IC βac = VCE =constant ∆IB 55
Beta (β )Determining β from a Graph (3.2 mA − 2.2 mA) β AC = (30 μA − 20 μA) 1 mA = V = 7.5 10 μA CE = 100 2.7 mA β DC = VCE = 7.5 25 µA = 108 56
Beta (β )Relationship between amplification factors β and α β α α= β= β+1 α −1Relationship Between Currents I C = βI B I E = (β + 1)I B 57
Common–Collector ConfigurationThe input is on thebase and the output ison the emitter. 58
Common–Collector ConfigurationThe characteristics aresimilar to those of thecommon-emitterconfiguration, except thevertical axis is IE. 59
Operating Limits for Each ConfigurationVCE is at maximum and IC is atminimum (ICmax= ICEO) in the cutoffregion.IC is at maximum and VCE is atminimum (VCE max = VCEsat = VCEO) inthe saturation region.The transistor operates in the activeregion between saturation and cutoff. 60
Power DissipationCommon-base: PCmax = VCB I CCommon-emitter: PCmax = VCE I CCommon-collector: PCmax = VCE I E 61