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chap12 - .1(a Under the quasistatic assumption the carriers...

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Unformatted text preview: CHAPTER 12 12.1 (a) Under the quasistatic assumption the carriers and hence the device under analysis are assumed to respond to a time-vaying signal as if it were a dc. bias. In the derivation of the generalized two—port model, one specifically equates the total time-varying terminal currents (i3, ic) to the dc. currents that would exist under equivalent biasing conditions. (b) Two separate definitions are necessary because, contrary to the polarities assumed in the development of the generalized small—signal model, the [B and [C currents were previously taken to be positive flowing out of the base and the collector terminals in a pup BIT. (As noted in Section 10.1, the direction of positive current was so chosen to avoid unnecessary complications, serious sign—related difficulties, in the physical description of current flow inside the BJT when operated in the standard amplifying mode.) (c) The Hybrid—Pi model gets its name from the n—like arrangement of circuit elements with “hybrid” (a combination of conductance and resistance) units. ‘ ((1) Names (see the first paragraph in Subsection 12.1.2): gm...transconductance r0...output resistance ruminput resistance r”...feedthrough resistance (e) The capacitors model the collector—base and emitter-base pn junction capacitances which cannot be neglected at higher frequencies. (f) The minority carrier concentration in the base continues to increase as pictured in plot (iii) of Fig. 12.4(d) until a maximum build-up consistent with the applied biases is attained. The base current varies as QB/ 13 and therefore also continues to increase toward a saturating maximum value. (In the quantitative analysis, 1‘}; increases from ICCTt at the start of saturation to a saturating value of [3313.) Once saturation biased, ic remains essentially constant at ic 5 [CC = VCCJRL. (g) In words, the base transit time is the average time taken by minority carriers to diffuse acrOss the quasineutral base. Mathematically (see Eq. 12.22), ft = W2/2DB. (Mfidc = [din = Tia/T: (i) An £13 < 0 aids the widthdrawal of stored charge from the quasineutral base, which in turn reduces both the storage delay time and the fall time. (i) A Schottky diode clamp is a circuit arrangement where a Schottky diode is connected between the collector and base of a BJT as pictured in Fig. 12.7(a). The Schottky diode conducts at a lower forward bias than a pn junction and therefore minimizes the forward (saturation-mode) bias that is applied to the BIT under turn-on conditions. This reduces the stored charge and speeds up the tum-off transient. (Also see Subsection 12.2.4.) 12—1 12.2 The BJT viewed as a two-port network and connected in the common-base configuration is pictured below. iEZIE’He i =1 +i C C CC + VCB+vcb :9 out in 2') VEB+ veg, Invoking the quasistatic assumption we can write iE(VEB+Deb,VCB+vcb) E IE(VEB+veb,VCB+vcb) = IE(VEB,VCB) +ie iC(VEB+veb,VCB+ch) E 1C(VEB+veb,VCB+vcb) = iC(VEB,VCB) +1}: 01‘ 1e = IE(VEB+veb,VCB+vcb) ~IE(VEB,VCB) ic = 1C(VEB+veb,VCB+vcb) _1C(VEB,VCB) Next performing a‘Taylor series expansion of the first term on the right—hand side of the above eqiiations, and keeping only first order terms, we obtain 81 BI IE(VEB+veb,VCB+vcb) = 1E(VEB,VCB) + —£’ Deb +J~l vcb aVEB VCB aVCB V53 31 81' [C(VEB+Ueb,VCB+Ucb) = [C(VEB,VCB) + —9‘i Deb +_—C—{ Deb aVEB VCB (Bi/CB VEB which when substituted into the preceding equations gives I :6 z 8% wii Deb aVEB VCB aVCB VEB 1c = a_IC_i Deb +fi—i Deb aVEB VCB aVCB VEB 12-2 If the direction of positive current flow is as defined in Fig. 10.2 (+15 out and +10 in for an npn BIT, +1}; in and Hg out for a pup BIT), then introducing yield the emitter and collector ac, current node equations is = giiveb+gizvcb ic = gziveb+g220cb The low—frequency small—signal equivalent circuit characterizing the ac. response of the BJT connected in the common base configuration is therefore concluded to be 12-3 12.3 .. From an inspection of Fig. 11.5(d), one concludes 1C E 1.1 mA at the specified operating point. Given the BIT is to be modeled using the simplified equivalent circuit of Fig. 12.2(a), and assuming T = 300 K, one computes (referring to Eqs. 12.9), ._ __ __ _____._._ = -2 gm [CT 00259 4.25 x 10 S = kT/q = 0.0259. = 5.18 x 103 (2 18 5x106 12.4 . The node equations appropriate for the B and C terminals in the Hybrid~Pi model (Fig. 12.2b) assume the form ib z ”be/rwt+vbc/ru ic = gm%e+UCb/rtt+vce/ro But vbc 2 ~ch = vbe ~ 006. Thus = unim— we) i0 = Mgm~%)+v %+%) A comparison of the preceding equations With text Eqs. (12.6) leads to the conclusion .. l 1 = "1 £11 ._ ,7: Jr,” . 812 r” _ 1 1 1 _. “M x ———+—-— 821 gm r“ 822 r” r0 Clearly r“ = —l/g 12. Moreover, substituting 1/ru = —g12 into the other three expressions allows us‘"t0 solve for the remaining Hybrid-Pi parameters in terms of the generalized model parameters“ Specifically, r1; : l/(g11+812) rs = “1/812 gm = 821— 812 r0 = 1/(£22 + 812) Although in a somewhat different order, the preceding are Eqs. (12.10). 124 12¢ Computations were first performed to determine the VEB values required to obtain an IC = 1 mA with and without accounting for base width modulation. These VEB voltages were then incorporated directly into the final program (P__12_05.m on the Instructor’s disk). In the MATLAB program, the user is asked whether he/she wishes to input VEB and . VEC or to use the preset values. The small incremental voltage deviations from the dc. voltage values used in approximating the partial derivatives appearing in Eqs. (12.5) were varied until a factor of two change in the incremental values led to no change to five significant places in the computed ggj parameters. The 5' parameters were in turn used to compute the Hybrid-Pi parameters employing Eqs. (12. 0). Sample results with and without accounting for base width modulation are tabulated below. In both cases there is at most a third«place difference between the gm and r,t computed from first principles and the gm and r,c computed using Eqs. (12.9). As expected, g12 and g2; are approximately zero when base width modulation is assumed to be negligible, and therefore r0 and r” become infinite. Finite values are obtained for r0 and r” when base width modulation is included. Note that base width modulation has little effect on gm but leads to a significant increase in r“. An increase in lec E gmrn is of course expected when base width modulation is included. No base—width modulation gm = 3.8685X10'2 S VEB = 0.67416 V r0 = co VEC = 10 v r1: = 4.5960X103 o [C = 1.0000 mA r“ : 00 gm = 3.8612X10'2 S ...using Eq. (12.9) ’"n = 4.6047X103 £2 With base-width modulation included gm = 3.8510><10~2 S VEB = 0.66961 V tr,t = 5.9530X103 £2 - 1C : 1.0000 mA r“ = 7.5141X107 o gm = ,I3.86ll><l()'2 S ...using Eq. (12.9) r,t = 5.9761X103 {2 12—5 MATLAB program scn'pt... %Computation of the Hybrid Pi Parameters (Problem 12.5) %Initialization clear; close format compact; format short e bw=input('Include base—width modulation? l—Yes, 2—No...'); s=input(‘Manually input VEB and VEC? l~Yes, 2-No...‘); %Input Eber—Moll Parameters BJTO %Voltages used in Calculation VbiE=kT*log(NE*NB/ni“2); VbiC=kT*log(NC*NB/ni“2); if 3:: , VEBO=input('Input VEB in volts, VEB '); VECO=input('Input VEC in volts, VEC - v | else VECO=10 if bw== , 30:0.669606 else VEBO=O.674162 end; end %iB and.iC Calculations VEB=VEBO: VEC=VECO: iB=;- iC=: for i=lz5, if bw==l, VCB=VEB-VEC: BJTmod else end IBO=(1-aF).*IFO+(1~aR).*IRO; IBl=(l—aF).*IFO+(l—aR).*IRO.*exp(—VEC/kT); IB=(IB1.*exp(VEB/kT)—IBO)i IC=( (aF. *IFO—IRO . *exp (—VEC/kT)) .* (IB+IBO) ../IBl+IRO-—aF. *IRO) ; %Reset voltages V if 1:: , VEB=VEBO—0.000l; else; end if i== , VEB=VEBO+0.000l; else; end if i== , VEB=VEBO; VEC= VECO—0.0l; else; end if '==4, VECzVECO+0.0l: else; end iB=[iB,IB]; iC=[iC,IC]; end 12-6 %Compute Generalized Two—Port Model Parameters g11=(iB(3)—iB(2))/0.0002; g12=(iB(5)—iB(4))/0.02; g21=(iC(3)—iC(2))/0.0002; gZ2=(iC(5)—iC(4))/0.02; fprintf('\nHybrid—Pi Model Parameters\n‘) gm=g21~g12 if g22+g12==0 ro=inf else ro=l/(g22+g12) end rpi=l/(gll+g12) if g12==0, rmu=inf else, rmu=—1/g12 end fprintf('\ngm and rpi computed using Eqs.(12.9)\n') gmziC(1)/0.0259 rpi=0.0259/1B(1) 12.6 (a) The high—frequency equivalent circuit of Fig. 12.2(0) with vce = O can be manipulated into the form ib » ic where :5 H k‘ + S... 5% 8- 12—7 Combining node and loop analysis we note ib = Ylvbe‘ — szcb' (l) ic = gmvbg' + Y Zucb' + veg/r0 (2) 1en: + vce' + (ib+ic)re = 0 (3) ”be! ‘ ”CC. + ”ch! = 0 (4) ,Eq. (4) is used to eliminate Deb'in Eqs. (1) and (2). Eqs. (1) and (2) are then combined to eliminate vbea Next Eqs. (3) is used to eliminate 006'. Finally, the idib ration is formed g1v1ng (Y2wgm ic _ Y1+Y2 ib Yz-gm 1 Y — Y +—-—~ —1 (music (2 roe Using the MATLAB program to compute lip/£1)! versus frequency, one determines an fry 2 235 MHz . Data sheets list the ff of the 2N3906 pnp BJT to be approximately 200 MHz. (It should be noted that the Electronics Workbench software program was used to determine the dc operating point that produced an IC = 1 mA. The series resistances listed in the problem statement were those quoted by the EW program. Zero—bias capacitance values employed in computing the Hybrid—Pi parameters were also extracted from the Electronics Workbench program.) A plot of lidibl versus frequency, and the MATLAB m—file constructed to generate the plot and determine fr, are reproduced on the next page 12-8 f(Hz) MATLAB program scfipt... %Problem.12.6...fT determination %Initialization clear; close %Parameters %|ic/ibl vs. frequency gm=3.869~2; f=logspace(4,9,200); rpi=4.65e3; w=2.*pi.*f; ro=2.00e4; Yl=l/rpi+j.*w.*Ceb; rmu=3.59e6; Y2=l/rmu+j.*w.*ch; Ceb=23.6e~12; R:(Y2~gm)./(Y1+Y2); CCb=2.32e—12; Den=R~*rC.*Y2 w (Y2+1/ro).*rc ~ 1: rb=10; beta=abs(R./Den); %beta=iiC/ibl rc=2.8; %Plot re=0; loglog(f,beta); grid xlabel('f (Hz)‘); ylabe1(‘l ic / ib 1‘) 12-9 12.7 The Eqs. (6.68)/(6.69) solution for the IDIFF flowing in a narrow base diode is 2 « (IA [Aria—00811070 /LP) (6 qVA/kT__ 1) [DIFF = LP ND sinh(xC'/Lp) For application to a BJT we make the symbol replacements..l)p ~4) DB, Lp ——> L3, ND n) NB, xc‘ —) W, and VA —) VEB- Then 2 q A 93:1. cosh(W/LB) (e qVEB/kT._ 1) I = ”FF LB NB sinh(W/LB) Since W/LB << 1 in a standard transistor, the cosh/sinh factor can be expanded as noted in the problem statement to obtain WC?Sh(W/LB) :- L—B—[l + lPKfl W/LB << 1 smh(W/LB) W 3 LB and n2 .2 1 == A—1+l(—W—H quEB/kT~—1 ”FF: (‘10 W BN3“ 3 LB ( ) Introducing the substitutions cited in Subsection 7.3.2, that is, 2 andLl—Wf: 2) W (1 +jcorB) 08:3 DBT TB nd(e qVEB/kT .i 1) z; (qveb/kne qVEB/kT yields the corresponding a.c. relationship DB "‘1?” 12m: q WEI; Finally, by definition, 1+_l_ W2 +jw wW__2_)(qveb quEB/kT 3 DBTB 3133 kT YD = GD+ijD = idiff/Ueb and therefore 12-10 1L8 The pictured "on" point in Fig. 12.3(b) lies right on the I B = Vs/RS line. Therefore [BB E VS/RS = 30flA. Inspecting the plot we find ICC E VCdRL = 5.0 mA. We know fidc = [d] B = M1}. Although base width modulation clearly causes fidc to vary somewhat depending on the dc. operating point, it is reasonable to employ a median value in obtaining the desired estimate. Specifically, using the point where the load line crosses the I B = 15 ,uA characteristic, we obtain 13 = [C :- (0.624)(Vcc/RL) : (0.624)(5X10'3) 1, = 208 t 13 IB 15><10~6 'Thus -3 ICC:I : (5x10 ) = 0.80 13313 — (30x10-6)(208) 12—11 l___2.9 (a)/(b) The required plots and the generating MA'ILAB m—file are reproduced below The computational relationships used m producing the plots were a; z "W 111%) if a: = 0 581: TB 2 . _ 1H) ...1f!,=-1 where x = [cots/1313113 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 {CC taut! lBB tauB 12- 12 0 , . , _ 0 0.1 0.2 (13 0.4 0.5 0.6 07 0.8 09 1 ICC taut/188 tauB MATLAB program script... %Rise and Storage—Delay Time plots (Prob. 12.9) %Initialization clear; close %Rise time computation x=linspace(0.01,0.99); rise=log(l./(1mx)); %rise=tr/tauB plot(x,rise); grid xlabel('ICC.taut / IBB tauB‘); ylabel(‘tr / tauB‘) pause %Storage-Delay Time computation delayOzlog(l./x); %delay0=tsd/tauB, xi=0 delay1=loq(2./(1+x)); %delayl=tsd/tauB, xi=l plot(x,delay0,x,delayl); grid xlabel(‘ICC taut / IBB tauB‘); ylabel(‘tsd / tauB') text(0q08,2.8,‘xi=0‘); text(0.08,0.8,‘xi=li) 12—13 12.10 (a) Let t1 be the time when ic = 0.9ICC and t2 the time when ic r: 0.11CC. Making use of Eq. (12.31b), we can then write ic(t1) = 091cc = 133%[(1+§)e41/TB— 5] t ic(t2) = 011cc 133%l<1+€>e"flTB—€] Solving for the t’s yields m( l + 5 ) TB m 0‘9ICCTt/IBBTB + <5 “( 1 + g ) 1B1 ____________ 0.11CCz'f/IBB’EB + 6 and per the measurements—based definition 11 I! I2 09k0fl$gm+f):qfl{afi+§) If = I241 : r3 ln( 0.11CC1JIBBIB +6 0.1x + 6 where x = ICCTt/[BBTB (b) With 6 = 0 and 5 = l, the part (a) relationship simplifies to E zfm9 .“n§=o m \ 09x+1 - - The requested lf/TB versus x plot is displayed on the next page along with the script of the MATLAB m—file used to generate the plot. Consistent with the analysis in Subsection 12.2.3, the plotted fall times decrease when 6 > 0. This occurs because an i}; < 0 aids the withdrawal of charge from the quasineutral base. If the x—ratio increases either due to an increase in ICC or a decrease in 18B, the charge storage is enhanced relative to the charge removal capability of the base 12-14 current. Thus, the tf/‘l‘B ratio for the {3: 1 curve increases with increasing x. When (5 = 0, the charge removal from the base occurs only by recombination and the fall—time collector current assumes the simple form, ic = Aexp(—t/m). Since tf is always evaluated employing the same relative iC values, ic(t1)/ic(t2) = constant = CXp(If/’Z‘B), and If/TB is seen to be a constant independent of x. 2.5 1‘5 ........ ......... ttttttttt ......... .......... .......... ........ :3 :9 ’2: ‘ ............................................................................................... 0.5 ........ ......... ttttttttt .......... .......... tttttttttt tttttttttt .......... ..... 3 ........ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0,9 1 ICC taut] [as tauB MATLAB program script... %Fa11 Time (Problem 12.10) %Initialization clear; Close %Fall Time computations x0=[0,1]; y0={log(9),log(9)]; %tf/tauB when xi=O xl=1inspace(0,l); yl=lOg((0.9.*x1+l)./(O.l.*xl+1)); %tf/tauB when xi=l plot(x0,y0,x1,yl); grid xlabel('ICC taut / IBB tauB‘); ylabel(‘tf / tauB‘) text(0.47,2.l,‘xi=0‘); text(0.47,0.4,‘xi=l‘) 12-15 ...
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