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BJT’s Emitter-Base Junction C iC n p E iE n iB B Collector-Base Junction Emitter Base Collector Active Base Region Figure 5.1(a) - Simplified cross-section of an npn transistor with currents that occur during "normal" operation p + n to r ec o ll + C p + n + p r it te n Em ne pi ta x y iC p n + p + i se Ba E xy pi ta ne + n b u ri p + p i ed B l ay er p p Active Transistor R egion + p p Figure 5. 1(b) - Cross-section of an integrated npn bipolar junction transistor iC Collector (C) n vBC iC Collector iB iB p Base (B) Base v iE n BE Emitter Emitter (E) (b) (a) iE Figu r e 5 . 2 - (a ) Id e a lize d n p n tr a n s is tor s tr u c tu r e for a ge n e r a l b ia s c on d ition (b ) Cir c u it s ym b ol for th e n p n tr a n s is tor Forward Characteristics C iC n iF Collector iF iB B v BE iF F p Base Emitter iF F n iE E Figure 5. 3 - npn transistor with vBE applied and vBC = 0. Reverse Characteristics C iC n v Collector BC iB B p iR R iR Base Emitter n iE E F igu r e 5 .4 - Tr a n s is tor with vBC a p p lied a n d vBE = 0 . Table 5.1 Common-Emitter and Common-Base Current Gain Comparison F or R 0.1 0.5 0.9 0.95 0.99 0.998 F F R or R 1 F 1 R 0.11 1 9 19 99 499 ENGR 311 – Electronic Devices and Circuits October 26, 2000 Transistor Model: Current Amplifier A Summary For Clarification (assume npn for the following general rules/properties – for pnp reverse polarities) Rules / Properties 1 – The collector must be positive than the emitter. 2 – The base-emitter and base-collector circuits behave like diodes. Normally the base-emitter diode is conducting and the base-collector diode is reverse-biased 3 – When 1 and 2 are obeyed Ic is proportional to Ib (Ic = beta . Ib) Both Ib and Ic follow to the emitter. Note: the collector current is not due to forward conduction of the base-collector diode; that diode is reversebiased. Just think of it as “transistor action.” Property 3 gives the transistor its usefulness: a small current flowing into the base controls a much larger current flowing into the collector. Note the effect of property 2. This means you can’t go sticking a voltage across the base-emitter terminals, because an enormous current will flow if the base is more positive than the emitter by more than about 0.6 to 0.8 volt. This rule also implies that an operating transistor has Vb = ~ Ve + 0.6 (Vb = Ve + Vbe) (for an npn). Let me emphasize again that you should not try to think of the collector current as diode conduction. It isn’t, because the collector-base diode normally has voltages applied across it in a reverse direction. Furthermore, collector current varies very little with collector voltage (it behaves like a not-too-great current source), unlike forward diode conduction, where the current rises very rapidly with applied voltage. Current flow The forward bias on the base-emitter junction will cause current flow across this junction. Current will consist of two components: electrons injected from the emitter into the base, and holes from the base into the emitter. The electrons injected from the emitter into the base are minority carriers in the p-type base region. Because the base is usually very thin the excess minority carriers (electron) concentration in the base will have an almost straight-line profile. The electrons will reach the boundary of the collector-base depletion region. Because the collector is more positive than the base these electrons will be swept across the CB junction region into the collector. They are then “collected” to constitute the collector current. By convention the direction of ic will be opposite to that of the electron flow; thus ic will flow into the collector terminal. Ic – Vce Characteristic for an npn Transistor Ic- Vbe Characteristics Biasing For common emitter amplifier ENGR 311 - BJTs – Exercises - October 29, 2001 Examples Example 1 - Beta = 100, vBE = 0.7V at iC = 1mA. Design circuit so that a current of 2mA flows through the collector and a voltage of +5V appears at the collector. Example 2 - In the circuit below vC = 0.7V. If Beta = 50, find IE, IB, IC and VC. Solution Example 3 – In the circuit below, Vb = 1V, VE = 1.7V. What are alfa and beta for this transistor? What voltage VC do you expect at the collector. Example 4 - Beta = 100 – Determine all node voltages and branch currents. Example 5 Determine the voltages at all nodes and current through all branches. Assume beta 1 and beta2 = 200. Assume Q1 is in the active mode. ENGR 311 - Graphical Representation of Transistor Characteristics - October 31, 2001 Conceptual circuit for measuring the iC-vCE characteristics of the BJT. (b) The iC-vCE characteristics of a practical BJT. The iC-vCB characteristics for an npn transistor in the active mode Determine the voltages at all nodes and the currents at all branches in the circuit below. Solution The Transistor As An Amplifier – DC Conditions (a) Conceptual circuit to illustrate the operation of the transistor of an amplifier. (b) The circuit of (a) with the signal source vbe eliminated for dc (bias) analysis. The Collector Current and The Transconductance The Base Current and the Input Resistance at the Base Transistor as An Amplifier - Small Signal Approximation Transconductance (gm), Input Resistance at the Base (r), Input Resistance at the Emitter (re), Voltage Gain Exercise 4.22 and 4.23 Small-Signal Equivalent Circuits Models Amplifier Circuit Without DC Sources Hybrid- The T Model Application of the Small-Signal Equivalent Circuits 1 2 3 4 5 Example 4.9 DC Analysis Small-Signal Analysis Example 4.11 Determine voltage gain in the circuit below DC Analysis Small-Signal Model Small-Signal Analysis Directly on Circuit Graphical Analysis Graphical determination of the signal components vbe, ib, ic, and vce when a signal component vi is superimposed on the dc voltage VBB. Biasing The BJT For Discrete-Circuit Design Basic Single-Stage BJT Amplifier Configurations Minority-Carrier Transport In the Base Region + + iB v v BE IF F N i BC I REC IR R N P (p , n bo ) i bo E C iT (a) Emitter Base Collector Space Charge Regions n( x) n(0) dn i qAD n T dx n(W) (b) 0 x W F igu re 5.15 - (a ) Cu rren t s in t h e ba s e region of a n pn t ra n s is t or (b) Min orit y ca rrier con cen t ra t ion in t h e ba s e of t h e n pn t ra n s is t or iT = qADn. dn/dx = -qADnn. (nbo/Wb). [exp(vbe/Vt) – exp(vbc/Vt)] Is = qADn.(nbo/Wb) = (qADn.ni^2 )/Nab.Wb nbo = equilibrium electron density A = cross-sectional area of the base region Wb = base width Dn = diffusity (cm^2/s) Nab = doping concentration in base of transistor’ ni = intrinsic-carrier concentration (10^10.cm^3) nbo = ni^2 / Nab Base Transit Time To turn the BJT minority-carrier charge must be introduced into the base to establish the gradient. The forward transit time tau-f represents the time constant associated with storing the required charge Q in the base region and is defined by Q/IT Diffusion Capacitance For the base-emitter voltage and hence the collector current in the BJT to change, the charge stored in the base region also must change. n( x) n(0) Q n(W) = n bo n bo x 0 W F igu r e 5 . 1 6 (a ) - E xc e s s m in or ity c h a r ge Q s tor e d in th e b ip ola r b a s e r e gion n( x) n(0, V BE2 ) Q n(0, V BE1 ) Q n bo n(W) = n bo x 0 W F igu r e 5 .1 6 (b ) - S t o r ed ch a r ge Q ch a n ges a s vBE ch a n ges This change in charge with vbe can be modeled by a capacitance CD CD = (Ic/VT). f Frequency Dependence of the Common-Emitter Current Gain iB C C B iC B i = i + C F B v iB BE v i = I S exp ( C - i BE ) VT i = ( + 1) i B E F E E E Figure 5.20 - Simplified model for the npn transistor for the forward-active region iB iC C B + - 0.7 V i = i v C BE F B iE E Figure 5. 21 - Further simplification of the npn model for the forward-active region Beta-cutoff Frequency 10 3 Common-Emitter Current Gain 10 2 10 1 10 0 f T 10 -1 10 4 10 5 10 6 10 7 10 8 10 9 Frequency (Hz) Figure 5.22 - Common-e mitte r curre nt ga in vs . fre que ncy Transconductance Relates changes in ic to changes in vbe gm = dic/dvbe (@Q-point) gm = Ic /VT CD = gm.f The Early Effect (James Early form Bell Labs) Experimentally demonstrated that when the output curves are extrapolated back to a point of zero collector current, the curves all intersect at a constant voltage point vce = -VA I = 100 uA B 4.0mA C o l l e c t o r 2.0mA I = 80 uA B I = 60 uA B C u r r e n t I = 40 uA B I = 20 uA B 0A -V A -15V -10V -5V 0V 5V 10V 15V Collector-Emitter Voltage F igu r e - 5 . 3 0 Tr a n s is t or ou t p u t c h a r a c t e r is t ic s id e n t ifyin g t h e e a r ly volt a ge V A Modeling the Early Effect ic Betaf ib Tolerances in Bias Circuits Worst Case Analysis V = +12 V IC CC 22k R 2 R R C EQ R C 22k IB 36 k V 12 k VEQ Q1 R 4V 18 k R E +12 V 16 k R 1 16 k CC IE E Study the operation of the transistor considering tolerances (worst case anaysis) in the circuit. Assume that the 12V power supply has a 5% tolerance and the resistors have 10% tolerance. Assume also that the voltage drop in REQ can be neglected, and beta is large. VEQ (max, min) IC (max, min) VCE (max, min) Monte Carlo Analysis Perform Monte Carlo Analysis on previous circuit assuming the random values to Vcc, R1, R2, Rc, Re, and beta. (Use Excel and/or Pspice). Calculate VEQ REQ IB IC IE VC = VCC – IC.RC – IE.RE Electronic Devices and Circuits – 11/5/00 Monte Carlo Analysis – Using Pspice Probe Output Ic(Q), Ib(Q), Vce