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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 timevaying signal as if it were a dc. bias. In the derivation of the
generalized two—port model, one speciﬁcally equates the total timevarying 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 emitterbase 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 buildup 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
(saturationmode) bias that is applied to the BIT under turnon conditions. This reduces
the stored charge and speeds up the tumoff transient. (Also see Subsection 12.2.4.) 12—1 12.2 The BJT viewed as a twoport network and connected in the commonbase 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 122 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 123 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 HybridPi 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 HybridPi 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 basewidth 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, 2No...‘);
%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=VEBVEC:
BJTmod
else
end
IBO=(1aF).*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 126 %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 Yzgm 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 128 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‘) 129 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 1210 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 — (30x106)(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 %StorageDelay 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 1214 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‘) 1215 ...
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 Spring '07
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