ChargeStorage Elements: BaseCharging Capacitance C b


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1 ChargeStorage Elements: BaseCharging Capacitance C b * Minority electrons are stored in the base  this charge q NB is a function of the baseemitter voltage * base is still neutral... majority carriers neutralize the injected electrons q PB = q NB C b = q PB v BE Q EE 105 Spring 2000 Page 1 Week 10, Lecture 20
2 Base Transit Time * The electron charge in the base is found by integrating the electron concentration in the base  the area is A E (under the emitter): q PB W B = q NB = qa E n ()dx x = pb qa 2 E W B n pbo e v BE V th * The stored charge is proportional to the collector current: 1 q PB W 2 B ( W B D nb ) qa E D nb npbo e v 2 BE V th W B = = ic W B 2D nb * The proportionality constant looks like a diffusion time (it is) and is defined as the base transit time: 2 W B τ F = D nb A typical transit time is τ F = 10 ps for an oxideisolated npn BJT. * The basecharging capacitance is: C b q PB = = v BE Q g m τ F EE 105 Spring 2000 Page 2 Week 10, Lecture 20
3 Complete SmallSignal Model * Add the depletion capacitance from the baseemitter junction to find the total baseemitter capacitance: C π = C je C b C je 2C jeo C jeo is proportional to the emitterbase junction area (A E ) * Depletion capacitance from the basecollector junction: C µ = C µo C µ = V CB φ Bc C µo is proportional to the basecollector junction area (A C ) * Depletion capacitance from collector (n buried layer) to bulk: C cs C cs = C cso 1 V CS φ Bs C cso is proportional to the collectorsubstrate junction area (A S ) EE 105 Spring 2000 Page 3 Week 10, Lecture 20
4 npn BJT SPICE model Close correspondence to EbersMoll and smallsignal models Name Parameter Description Units IS transport saturation current [I S ] Amps BF ideal maximum forward beta [β F ] None VAF forward Early voltage [V An ] Volts BR ideal maximum reverse beta [β R ] None RB zero bias base resistance [r b ] Ohms RE emitter resistance [r ex ] Ohms RC collector resistance [r c ] Ohms CJE BE zerobias depletion capacitance [C jeo ] Farads VJE BE builtin potential [φ Be ] Volts MJE BE junction exponential factor None CJC BC zerobias depletion capacitance [C µo ] Farads VJC BC builtin potential [φ Bc ] Volts MJC BC junction exponential factor None CJS substrate zerobias depletion capacitance [C cso ] Farads VJS substrate builtin potential [φ Bs ] Volts MJS substrate junction exponential factor None TF ideal forward transit time [τ F ] Seconds.MODEL MODQN NPN IS=1E17 BF=100 VAF=25 TF=50P CJE=8E15 VJE=0.95 MJE=0.5 CJC=22E15 VJC=0.79 MJC=0.5 CJS=41E15 VJS=0.71 MJS=0.5 RB=250 RC=200 RE=5 EE 105 Spring 2000 Page 4 Week 10, Lecture 20
5 A Simple MOSFET Amplifier Amplify = make something larger... something = current, voltage, or power V SUP (positive DC supply) i RD R D R S _ v OUT v s _ v IN V BIAS _ V SUP (negative DC supply) V BIAS is selected so that V OUT is centered between V SUP and V SUP : V OUT = 0 V... NOT v OUT = 0 V! Find the DC drain current: I RD = (V SUP  V OUT ) / R D = V SUP / R D I RD = I D = I DSAT... verify that MOSFET is saturated after finding V BIAS (to find V BIAS, solve saturation current equation for V GS... the result is that for normal device dimensions and DC drain currents, V GS = V Tn (0.25 to 0.5) V) EE 105 Spring 2000 Page 5 Week 10, Lecture 20
6 MOSFET Amplifier Now consider the effect of the small signal voltage: v IN = V BIAS v s so v GS = V BIAS ( V SUP ) v s = V GS v s Let v s () t = vˆs cos( ωt) Approach 1. Just use v IN in the equation for the total drain current and find v OUT W v OUT = V SUP R D i D V SUP R D ( µ n C ox ) L ( VGS v V ) 2 s Tn W v OUT V SUP R D ( µ n C ox ) ( VGS V 2L Tn ) 2 v s 1 ( = V GS V Tn ) I D v s v OUT V SUP R D I D 1 ( = V GS V Tn ) V SUP Expand (1 x) 2 = 1 2x x 2 v s V GS V Tn 1 2v s v s 2 = V GS V Tn V GS V Tn EE 105 Spring 2000 Page 6 Week 10, Lecture 20
7 Special Case: v s is Small What s small? 2v s v s 2» V GS V Tn V GS V Tn... true if 2v s «1 V GS V Tn For this case, the total output voltage is 2v s v OUT V SUP R D I D 1 ( R D I D v s = V V GS V Tn ) SUP V SUP ( V GS V Tn ) The average output voltage V OUT = 0 V so the total output voltage is the smallsignal voltage in this special case: 2R D I D v OUT v out ( V GS V Tn ) v 2V SUP = = = s ( V GS V Tn ) v = A s v v s v s (t) v out (t) v s (t) t EE 105 Spring 2000 Page 7 Week 10, Lecture 20
8 Is There a Better Way? Approach 2. Do problem in two steps. 1. DC voltages and currents (ignore small signals sources): set bias point of of the MOSFET... we had to do this to pick V BIAS already 2. Substitute the smallsignal model of the MOSFET and the smallsignal models of the other circuit elements R S C gd R D v s _ C gs g m v s r o v out Where are the DC supplies?? EE 105 Spring 2000 Page 8 Week 10, Lecture 20
9 SmallSignal Models of TwoTerminal Circuit Elements i R SmallSignal Model v R  i R R v R i SUP i SUP V SUP _ v SUP v SUP  i SUP I SUP i SUP v SUP  v SUP EE 105 Spring 2000 Page 9 Week 10, Lecture 20
10 SmallSignal Output Voltage Don t bother with capacitors... wait until Chapter 10 to put them in v out = g m v s ( R D r o ) Smallsignal voltage gain: A v = g m ( R D r o ) Transconductance of MOSFET in saturation g m W = µ n C ox  ( VBIAS V L Tn ) = 2I DSAT V BIAS V Tn Smallsignal voltage gain: A v = 2I DSAT ( RD r V BIAS V o ) Tn... almost identical to bruteforce result, but the smallsignal model includes the effect of channellength modulation (through r o = (1 / λ n I DSAT )) EE 105 Spring 2000 Page 10 Week 10, Lecture 20
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