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Week 4: Bipolar Junction Transistor

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Week 4: Bipolar Junction Transistor ELE 2110A Electronic Circuits Week 4: Bipolar Junction Transistor ELE2110A © 2008 Lecture 04 - 1 Topics to cover … z Physical Operation and I-V Characteristics z BJT Circuit Models z DC Analysis z BJT Biasing z Reading Assignment: Chap 5.1-5.3, 5.5-5.9, 5.11, 10.1-10.3 of Jaeger & Blalock, or Chap 5.1-5.5 of Sedra & Smith ELE2110A © 2008 Lecture 04 - 2 Bipolar Junction Transistor (BJT) z npn transistor: z 3 terminals: – emitter, base, and collector z 2 pn junctions: – emitter-base junction (EBJ) – collector-base junction (CBJ) z Emitter: heavily doped z Base: very thin Vertical npn ELE2110A © 2008 Lecture 04 - 3 Regions of Operation z Depends on the biasing across each of the junctions, different regions of operation are obtained: Base-Emitter Base-Collector Junction Junction Reverse Bias Forward Bias Forward Bias Forward-active region Saturation region (Active region) (Closed switch) (Good amplifier) Reverse Bias Cutoff region Reverse-active region (open switch) (Poor amplifier, rarely used) ELE2110A © 2008 Lecture 04 - 4 BJT in Active Region E-field 1 Biasing: 2 E-B: Forward 3 C-B: Reverse 4 Operation: 1. Forward bias of EBJ causes electrons to diffuse from emitter into base. 2. As base region is very thin, the majority of these electrons diffuse to the edge of the depletion region of CBJ, and then are swept to the collector by the electric field of the reverse-biased CBJ. 3. A small fraction of these electrons recombine with the holes in base region. 4. Holes are injected from base to emitter region. (4) << (1). ELE2110A © 2008 Lecture 04 - 5 Minority Concentration Profiles E-field z Current dominated by electrons from emitter to base (by design) b/c of the forward bias and minority carrier concentration gradient through the base – some recombination causes bowing of electron concentration (in the base) ELE2110A © 2008 Lecture 04 - 6 Diffusion Current Through the Base iC z Diffusion of electrons through the base is set by concentration profile at the EBJ vBE /VT n p (0) = n p 0e z Diffusion current of electrons through the base is (assuming an ideal straight line case): dn (x) n (0) I = A qD p = A qD (− p ) n E n dx E n W z The collector current:i C = −I n (- sign: current into collector) ELE2110A © 2008 Lecture 04 - 7 Collector Current z The collector current: 2 v /V BE T qAE Dnnp0 qAE Dnni iC = ISe where I S = = W N AW z Note that iC is independent of collector voltage. z The current at the collector is controlled by the voltage across the other two terminals – a voltage controlled current source. This control is the basic transistor action. z Saturation current IS is – inversely proportional to W and directly proportional to AE z Want short base and large emitter area for high currents 2 – dependent on temperature due to ni term ELE2110A © 2008 Lecture 04 - 8 Base Current 1 2 3 4 Base current consists of two components: iB1 and iB2: z iB1 , due to forward bias of EBJ, is an exponential function of vBE. z iB2 , due to recombination, is directly proportional to the numbers of electrons injected from the emitter, which in turn is an exponential function of vBE. vBE /VT iB ∝ e ELE2110A © 2008 Lecture 04 - 9 The Beta (β) z Can relate iB and iC by the following equation i I i = C = S e v BE /VT B β β z β=iC/iB – β is constant for a particular transistor – On the order of 100-200 in modern devices (but can be higher) – called common-emitter current gain. – also denoted as βF. ELE2110A © 2008 Lecture 04 - 10 Emitter Current 1 2 3 4 Treat transistor as a super-node, by KCL we obtain 1 1+ β β i = i + i = i + i = i or i = i E B C β C C β C C 1+ β E β α ∴ i =α i where α = or β = C E 1+ β 1−α α is called common-base current gain. α < 1 but very close to 1. ELE2110A © 2008 Lecture 04 - 11 Circuit Symbols and Conventions pnp transistor structure: ELE2110A © 2008 Lecture 04 - 12 I-V Characteristics z Collector current vs. vCE shows the BJT looks like a current source (ideally) – Plot only shows values where BCJ is reverse biased and so BJT in active region z However, real BJTs have non-ideal effects ELE2110A © 2008 Lecture 04 - 13 Early Effect iC vBE /VT vCE iC = ISe (1+ ) VA z Early Effect: – Current in active region depends (slightly) on vCE – Account for Early effect with additional term in collector current equation –VA is a parameter for the BJT (50 to 100) and called the Early voltage – Nonzero slope means the output resistance is NOT infinite. z At low value of vCE, the CBJ becomes forward-biased and the transistor enters the saturation region. ELE2110A © 2008 Lecture 04 - 14 What causes the Early effect? z When VCB increases: – depletion region of CBJ widens – so the effective base width decreases (base-width modulation) z Shorter effective base width → higher dn/dx ELE2110A © 2008 Lecture 04 - 15 BJT Breakdown z If reverse voltage across either of the two pn junctions in the transistor is too large, corresponding diode will break down. z Emitter is the most heavily doped region and collector is the most lightly doped region. z Due to doping differences, base-emitter diode has relatively low breakdown voltage (3 to 10 V). Collector- base diode can be designed to break down at much larger voltages. z Transistors must be selected in accordance with possible reverse voltages in circuit. ELE2110A © 2008 Lecture 04 - 16 Topics to cover … z Physical Operation and I-V Characteristics z BJT Circuit Models z DC Analysis z BJT Biasing ELE2110A © 2008 Lecture 04 - 17 Equivalent Circuits for Active Mode vBE /VT i = β i iC = ISe C B z Reverse saturation of DB = Is/β, because: i I i = C = S e v BE /VT B β β ELE2110A © 2008 Lecture 04 - 18 Equivalent Circuits for Active Mode vBE /VT iC =α iE iC = ISe z Reverse saturation of DE = Is/α, because: i I i = C = S e vBE / VT E α α ELE2110A © 2008 Lecture 04 - 19 Simplified Model Simplified model: z Base-emitter diode is replaced by a constant voltage drop (VBE = 0.7 V) since it is biased in forward-active region. ELE2110A © 2008 Lecture 04 - 20 Simplified Circuit Model for Saturation Mode EBJ Forward-biased; CBJ Forward–biased. 0.5V VCEsat = 0.2V 0.7V z Forward voltage drop VBC is small b/c collector doping level is low (by design) z No expressions for terminal currents other than iC + iB = iE. z The ratio of IC to IB is called the forced current gain, ˆ β = I Csat / I Bsat which is much smaller than β Æ next slide. ELE2110A © 2008 Lecture 04 - 21 Forced Beta Assume BJT works in active mode: z Increase IB (or VI) causes – IC to increase proportionally (IC = βIB) – VC to drop z VC will drop to a point where BCJ becomes forward-biased, then: – VCE clamps to VBE - 0.5V, or 0.2V – Further increases of IB will not increase IC – IC is said “saturated” – IC/IB becomes smaller than β and is called “forced β”. ELE2110A © 2008 Lecture 04 - 22 Simplified Circuit Model for Cutoff Mode Cutoff Region: EBJ Reverse-biased CBJ Reverse-biased iB = 0, iE = 0, iC = 0 : No expression for terminal voltages. ELE2110A © 2008 Lecture 04 - 23 Models for pnp Transistors PNP model in active mode VEB ≅ 0.7 V PNP model in saturation mode VEB ≅ 0.7 V VEC ≅ 0.2 V ELE2110A © 2008 Lecture 04 - 24 Topics to cover … z Physical Operation and I-V Characteristics z BJT Circuit Models z DC Analysis z BJT Biasing ELE2110A © 2008 Lecture 04 - 25 Example 1 • Problem: Determine the Q-point. • Given data: β = 100 • Assumptions: VBE = 0.7 V • Analysis: VE = 4 −VBE ≅ 4 − 0.7 = 3.3 V Assume active VE 3.3 IE = = = 1 mA mode of operation, RE 3.3 For active mode operation, IC = αIE β 100 Q α = = ≅ 0.99 β +1 101 ∴ IC = 0.99 ×1 = 0.99 mA VC = 10 − IC RC = 10 − 0.99 × 4.7 ≅ +5.3 V VBC= –1.3 V. The CBJ is reverse- biased and the transistor is indeed in the active mode as assumed. I 1 I = E = ≅ 0.01 mA B β +1 101 ELE2110A © 2008 Lecture 04 - 26 Example 2 • Problem: VB changes to 6V. Find Q-point. • Analysis: Assume active mode operation, we have VE = 6 −VBE ≅ 6 − 0.7 = 5.3 V 6 v VE 5.3 IE = = = 1.6 mA RE 3.3 For active mode operation, IC = αIE = 0.99 ×1.6 ≅ 1.6 mA VC = 10 − IC RC = 10 −1.6 × 4.7 ≅ +2.48 V Since VBC= 6 – 2.48 = 3.52 V, the CBJ is forward-biased and the original assumption of active mode operation is INCORRECT. A Second guess must be tried. ELE2110A © 2008 Lecture 04 - 27 Example 2 (Cont.) Assume the transistor is in saturation region. Use the saturation model, we have VE = 6 −VBE ≅ 6 − 0.7 = 5.3 V VE 5.3 IE = = = 1.6 mA RE 3.3 VC = VE +VCEsat ≅ +5.3 + 0.2 = 5.5 V 10 − 5.5 I = = 0.96 mA C 4.7 IB = IE − IC = 1.6 − 0.96 = 0.64 mA Thus the transistor is operating at a forced β of IC 0.96 β forced = = = 1.5 IB 0.64 Since βforced << normal β of 100, the transistor is indeed saturated.
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