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Chapter 12 - CHAPTER 12 m(3 Under the quasistatic...

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Unformatted text preview: CHAPTER 12 m. (3.) 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, is) 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 IB and 1C currents were previously taken to be positive flowing out of the base and the collector terminals in a pnp BJT. (As noted in Section 10.1, the direction of positive current was sochosen 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 white arrangement of circuit elements with “hybrid” (a combination of conductance and resistance) units. (d) Names (see the first paragraph in Subsection 12.1.2): gm...transconductance rc_...output resistance ruminput resistance r”...feedthrough resistance (e) The capacitors model the collectonbase 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 Q3113 and therefore also continues to increase toward a saturating maximum value. (In the quantitative analysis, i3 increases from ICCTt at the start of saturation to a saturating value of [3313.) Once saturation biased, 1]: remains essentially constant at 5C 5 [CC = VCdRL- (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), 1.: WZIZDB. (h) the =IdIB = TB/Tt (i) An £3 < 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 BJT 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 BIT viewed as a two-port network and connected in the common-base configuration is pictured below. iE=IE+ie i :1 H C E C c in 2: VEB+Ueb _ VCB+ch => out Invoking the quasistatic assumption we can write iE(VEB+veb,VCB+vcb) E IE(VEB+veb,VCB+vcb) = IE(VEB:VCB) + ie ic(VEB+veb.Vcn+vcb) E [C(VEBWeuVCBch) = IC(VEB.VCB) + ic 01' is = IE(VEB+veb.VCB+vcb)JEWEBNCB) ic = IC(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 equations, and keeping only first order teams, we obtain a! a; 1E(VEB+veb.VCB+vcb) = IEWEBNCB) + £4 Deb +—E-‘ Deb . . BVEB VCB BVCB VEB a: a! [C(VEB+Ueb.VCB+Ucb) = 1C(VEB.VCB) + 4(3' Deb +~—C—l Deb aVEB veg aVcn V53 which when substituted into the preceding equations gives . a! E 313 1e 3 veb + vcb aVEB VCB aVCB VEB it: = __131C Deb +‘_‘31C vcb aVEB VCB aVCB VEB 12-2 If the direction of positive current flovi' is as defined in Fig. 10.2 (+15 out and +Ic in for an npn BIT, +1]; in and He out for a pup BIT), then introducing 311;, 315 all; 3]}; 811 E = ; 812 E = aVBEiVBC aVEBiVCB aVBCiVBE aVCBIVEB H II 11 fl npn pnp npn pnp +15 out +15 in +1}; out +1}; in _ 81C 31C _ 31C 31c 821 = = . ; 322 = = aVEB VCB aVBE VBC aVCB VEB aVBC VBE fl 1'? 1'! fl npn pnp npn pnp HQ in +Ic out «Hg in +Ic out yield the emitter and collector ac. current node equations is = Enveb+glzvcb is = gziveb +822vcb The low—frequency small-signal equivalent circuit characterizing the ac. response of the BIT connected in the common base configuration is therefore concluded to be 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), n=k_T/i= 0- 259 = 5.18x 10352 13 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 = vbi‘jrtt+ ”be/”u it: = gmvbe + “Deb/”1.1 + Ucdro But vbc =—va = vbe - use. Thus = area aa miracles A comparison of the preceding equations with text Eqs. (12.6) leads to the conclusion -1 L = _L 811 ‘rnJrru 812 m _ _L mil. .1. 821 -gm ’11 8'2 -’u+’o Clearly r” = -l/g12. Moreover, substituting l/rp = -—glz into the other three expressions allows us to solve for the remaining Hybrid-Pi parameters in terms of the generalized model parameters. Specifically, II I c: 00 s N G: = M8]! +812) ’11 8m = 821-812 r0 = I/(s’22+312) Although in a somewhat different order, the preceding are Eqs. (12.10). 12-4 1.2.5 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 gij parameters. The gi' parameters were in turn used to compute the Hybrid-Pi parameters employing Eqs. (12. 1‘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 r5: computed from first principles and the gm and r,t computed using Eqs. (12.9). As expected, g1; and g2; are approximately zero when base width modulation is assumed to be negligible, and therefore re 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 rm. An increase in fidc s gmrfi is of comse expected when base width modulation is included. No base-width modulation gm = 3.8685x10-2 s VEB = 0.67416 v r7; = 4.596OX103 .0. ‘ 1C = 1.0000 IDA r!l z: 00 gm = 3.8612X10’2 S ...using Eq. (12.9) r,[ = 4.6047X103 Q . With base-width modulation included gm = 3-8510><10-2 8 V3; = 0.66961 v r0 = 1.4932x105 Q VEC =10 v rIt = 5.9530X103 Q [C = 1.0000 mA r” = 7.5141x107 52 gm == 3.8611X10'2 S ...using Eq. (12.9) rn = 5.9761x103 Q ‘ 12-5 MATLAB program script... %Computation of the Hybrid Pi Parameters (Problem 12.5) %Initialization clear; close format compact: format short e bw=input('Inc1ude base—width modulation? l-Yes, Z-No...'): s=input('Manually input VEB and VEC? 1-Yes, 2-No...'): %Input Eber—Moll Parameters BJTO %Voltages used in Calculation VbiEnkT*log(NE*NB/ni‘2); VbiC=kT*log(NC*NB/ni‘2); if s== VEBO=inputt'Input VEB in volts, VEB VECO=input('Input VEC in volts, VEC else - VECO=10 if bw==1, VEBO=O.669606 else VEBO=0.674162 end: end '): '); %iB and iC Calculations VEB=VEBO: ‘ VEC=VECO: iB=; ic=; for i=l:5, if bw==1, VCB=VEB*VEC: BJTmod else end IBO= (l— aF). *IFO+(1- aR). *IRO: IB1= (l- aF). *IFO+(l-aR). *IRO. *exp(—VEC/kT): IB=(IBl. *exp(VEB/kT) —IBO); IC:((aF. *IFO—IRO. *exp(—VEC/kT)).*(IB+IBO)./IB1+IRO“aF.*IRO); %Reset Voltages if i==l, VEB=VEBD- 0. 0001; elSe; end if i==2, VEB=VEBO+0.0001: else: end if i==3, VEB=VEBO: VEC= VECO-0.01; else: end if i==4, VEC=VECC+0.01: else: end iB=[iB,IB]; iC=[iC,IC]; end 12- 6 %Compute Generalized TworPort Model Parameters 911=(iB(3)-iB(2))/0.0002; 912=(iB(5)-iB(4))/0.02: 921=(iC(3)—iC(2))/0.0002; 922=(iC(5)—iC(4))/0.02; fprintf('\nHybridrPi Model Parameters\n') gm:g21—glz if g22+912==0 ro=inf else ro=1/(922+g12) end rpi=1/(g11+g12) if g12==0, rmu=inf else,' rmu=-1/g12 end fprintf('\ngm and rpi cemputed using Eqs.(12.9)\n') gm=iC (1) /o . 0259 rpi=0.0259/iB(1) 12.6 (a) The high-frequency equivalent circuit of Fig. 12.2(c) with vce = 0 can be manipulated into the form where Y] = #Hmcd, Y = —L+ 'coC 2 m J (in 12-7 Combining node and loop analysis we note = Ylvbe' — szcb' (I) IDc: = gmvbeI + YZch' '1” ”Ce/1'0 (2) icrc + 1)ce' + (fb+ic)re = 0 (3) ”be. — 19cc:r + 1)cb' = 0 (4) Eq. (4) is used to eliminate Ucb' in Eqs. (1) and (2). Eqs. (1) and (2) are then combined to eliminate ”be" Next Eqs. (3) is used to eliminate vce'. Finally, the idib ration is formed guns +gm—Y2 Yr—l ++Y1+Y2(26 ) 2+;n—(n+%)m+m—1 {YZ—gm) £9: Y1+Y2 ib Yz—gm 1 Y — Y +— *1 Y1+Y2},62 (2 r0 0 Using the MA'I‘LAB program to compute lidibl versus frequency, one determines an fT = 235 MHz . Data sheets list the fr of the 2N3906 pnp BIT 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 MATLAB program script... %Problem 12.6...fT determination %Initialization clear; close %Parameters gm=3.86e—2; rpi=4.65e3: ro=2.00e4: rmu=3.59e6: Ceb=23.6e-12: ch=2.32e—12: rble: rc=2.8: re:0: %|ic/ibI vé. frequency f=logspace(4,9,200): w=2;*pi.*f: Y1=1/rpi+j.*w.*Ceb; Y2=1/rmu+j.*w.*ch; R= (YZ-gm) ./ (Y1+Y2) ; Den=R.*rc.*¥2 — (Y2+l/ro).*rc — 1: beta=abs(R./Den); %Plot loglog(f,beta); x1abel('f (Hz)'): %beta=[ic/ibl grid ylabel('| ic / ib l') 12,] The Eqs. (6.68)](6.69) solution for the 11311:}: flowing in a narrow base diode is ggigogtIMpqvA/ml) IDIFF = (IA LP ND sinh(xc'/Lp) For application to a BJT we make the symbol replacements...Dp ——> DB, LP —> L3, ND -—> N13, xc‘ ——) W, and VA m) VEB. Then 2 I [DH-7F = qA D._B 1.2L C'OSth/Li) (e qVEB/kT _ 1) LB NB $1nh(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 mm 3 1% , £2] . Sinh(W/LB) "— W I + 3 (LB) lV/LB << 1 and 2 .. DB ”i N 1(W)2 V B/kT I = A————— 1+~——— qu -1 Dm- (q ME 3L3 ]( ) Introducing the substitutions cited in Subsection 7.3.2, that is, B 133113 DBTB (I + jcorB) and {ngEBIkT_ 1) =5 (qveb/kjjquEB/kT yields the con'esponding a.c. relationship 2 -. .. DB "i) 1 = A—-—— mn (q WNB Finally, by definition, 1 + 1- ——-W2 + jco *WZ ) (quit) quEB/kT 3 D313 31)}; H" YD = GD +J'wCD = idifr/veb and therefore 12-10 12.3 The pictured "0n" point in Fig. 12.3(b) lies right on the [3 = Vs/Rs line. Therefore [BB E Vs/Rs = 30pA. Inspecting the plot we find ICC :—: VcdRL = 5.0 mA. We know Bdc = IdIB = 13/11. Although base width modulation clearly causes fidc to vary someth 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]; = 15 pA characteristic, we obtain ' a_m=ttwn=w = 208 13 13 15 >(10'6 my:I : (5x103) _ = 0.80 [BBTB (30X10‘6)(208) 12— 11 ....—_....__———_..T..___..-..-.~_..v.__. . . . 12,2 . (a)/(b) The required plots and the generating MA’ILAB m-file are reproduced below. The computational relationships used in producing the plots were A; =1 _.1__ 1:3 1-—x _1. V ._ Ea: 111x) mung—o “‘8 2 . _ l l-I-x) "'lféfl where x = ICCTI/IBBTB 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ICC mull IBB tauB 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ICC taut] IBB 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./(1—x)); %rise=tr/tauB plot(x,rise): grid x1abe1{'ICC taut / IBB tauB’); ylabel('tr / tauB') pause %St0rage-Delay Time computation delay0=log(1./x); %delay0=tsd/tauB, xi=0 delayl=log(2./(1+x}): %delay1=tsd/tauB, xi=1 plot(x,delay0,x,delayl); grid xlabel('ICC taut / IBB tauB'); ylabel(‘tsd / tauB') text(0.08,2.8,'xi=0'); text(0.08,0.8,'xi=l') 12-13 12.19 (a) Let :1 be the time when ic 2' 0.9Icc and :2 the time when ic = 0.1Icc. Making use of Eq. (12.31b), we can then write I.CUI) = 0.9ICC = 133%[(1+§)e41/TB_§] iCUz) = 0.11cc =IBB%[(1+§)6"7JTB- 5] Solving for the t’s yields 1% 1 + 5 ) m — 0-91CCTt/1BB’FB + .5 1 +5 ‘2 3 113 w 0.lI(jcq/IBBTB +§ and per the measarements— based definition tl If = tg—tl = mln( 0.91ccw133m + 5) _ a3 I“(0.9x + 5 ) OJICCTt/IBB’IB + if 0.11: + f where x = [cc‘fi/IBBTB (b) With g: 0 and 1;: 1, the part (a) relationship simplifies to f_r : Iln9 ...if§=0‘ TB ‘ 0.9x+1 - _ menu) "‘lfg‘rl The requested tf/t‘g 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 t: > 0. This occurs because an if; < 0 aids the withdrawal of charge from the quasineutral base. If the x—ratio increases either due to an increase in 1 cc or a decrease in 1313, the charge storage is enhanced relative to the charge removal capability of the base 12-14 current. Thus, the tf/‘IB ratio for the 5: 1 curve increases with increasing x. When E = 0, the charge removal from the base occurs only by recombination and the fall-time collector current assumes the simple form, ic = Aexp(—t/tj3). Since If is always evaluated employing the same relative ic values, ic(t1)/ic(t2) = constant 2 CXp(If/TB), and If/TB is seen to be a constant independent of x. 2.5 15 a: E E 1 Q5 0 0 DJ 02 03 04 as as Q? as as 1 ICC taut] IBB tauB MA'I‘LAB 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=0 x1=linspace(0,l): _ yl=log((0.9.*xl+1)./(0.1.*xl+1)}; %tf/tauB when xi=1 plot(x0,y0,x1,yl); grid ‘ xlabel('ICC taut / IBB tauB'): ylabel('tf / tauB') text(0.47,2.1, 'Xi=0'): text(0.47,0.4, 'xi=1': 12—15 ...
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