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Unformatted text preview: If?) ﬁlmsy, 7017+ ; : (347:0 @ Low mam;ch MODEL. In
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40min 4+: ' 9.38 For this circuit, ‘
13 = (VCC  Vf)/RB = 11.3 pA =: I." = TNT/[gs 2.2 kg
Find the midband gain: 7
‘ For this circuit,
Vn/Vin = (rullRB)/(Rs + TullRB) " ’n/(Rs + tax
where RB >> rn.‘ With CC set to a short, RL is connected
directly to the collector of the BIT. so that Vent/Vin = ‘grnrnmCRQ/(RS + tn) = “BJRCHRQ/(Rs + tn)
For B0 = 100, van/Vin = —100(5.07 ten/(1m + 2.2 kn) as —160 E 44 dB
Evaluate the low frequency pole due to Cs:
‘ (OL = + Hr“) = rad/S
where RB Hr“ v: 1'“. Similarly, evaluate the high frequency
pole due to CL: ‘
00L = 1/CL(RCR,) ~ 14 Mrad/s .
t where RLIIRC = RC. Here is a Bode plot of the response: . .. h._._..‘_......,.. .. 13.5.5.5 .L‘ NE NM (233‘ " ;, , 05310945.
.922 05:06. :oomumr Pow! Minus C, =, lo F “ﬂh—ﬁ c~ = 4/»?
338 . COHPUW L0 FREQ I .
:2 (“3",”) Cs I CE Low We; Poua SoLquuf SM? ‘
:72: ‘ Cs. ' . . Ci . v 1:, ~ r = :9 mi (Mama
(59 4 ‘ ” J:c ' W
x?) a .
= ,Gr‘..2'W.!’~. 4 ... _‘ @ I .Cc: F’ande Low F2611 IPOLe" m. t? “N, V . m: .twéen W117"
Vc luk—L '1 9.42 Examine the circuit of Fig. 9.76 (pg 653) with a BJT
(BF = [30 = 50) in place. a To seth = IE = 2 mA with VB = 0. choose 'R4 = (VB — VEEyIE = [—(—15 V)]/(2 mA) = 7.5 kg.
Approximately divide VCC between R3 and Q1
= VC = 7.5 V, hence choose
R3 = (15 V  7.5 V)/(2 mA) = 3.75 k9.
Round this value dewn to 3.7 k9 = VC = 7.6 V.
Note that VB = VE + Vf = 0 + 0.7 = 0.7 V. From
approximate voltage division,
. R2Vcc + RIVEE
V,3 s —————————
R1 4» R2 This expression can be rearranged, yielding R2 VB — VEE 0.7 V — (—15 V) + = —— = ———————— = '
15 V  0.7 V The resistor values R2 = 22 k9; R1 = 20 it!) will provide
this ratio. b. With CE effectively bypassing the emitter of Q1 to
ground in the midband, and neglecting the loading of R1
and R2 on the vsRS source, the midband gain becomes A. = 45.0%, "Ra/(r, + R5) = 50(2.7 ICED/(625 Q + 2 1:52) = —51
where r,I = [So/gm = 50(0.025 V)/(2 mA) = 625 Q, and
RgllRL = 2.7 k9. Similarly, rm = R1 [lellgt + RS = 2.6
kg, and to!“ = = 2.7 kg. c. Assume that CE bypasses the emitter to ground well
above the breakpoint of CS = the Rm seen by Cs is (RS '4' I." + (Bo + 1)R4), and
f5 = 1/2n(1 pF)[2 k9 + 590 Q + (51)(7.5 1(9)] ) = 0.41 Hz
d. Setting the pole of CE to 20 Hz conﬁrms the
assumption of part (c) that CE is the dominant low
'frequency capacitor. Hence CS will behave as a open
circuit at the pole of CE = The Rm seen by CE becomes
;R4ll [(r,t + R, "Kg/(BO + 1)] = .212 Q. Choose
j CE = 1/21t(20 Hz)(212 Q) = 38 pF.
(Choose closest standard value of 39 pF.) 9.56 Here is the desired ordering of poles:
Cs: 0.5 Hz; CC: 2 Hz; CE: 20 Hz. This order implies that CE will behave as an open circuit
at the pole of CS, and CS will be have as a short at the
pole of CE.
For CS: The smallsignal Thévenin resistance seen by CS.
computed with CE = open, becomes rth = R5 + R1 lle ll [rn + ([30 + 1)R4]
Assume that [r1I + ([30 + 1)R4] >> R1 le : rms ’5 + = 4.3 kg.
Choose C5 = 1/21I:(0.5 Hz)(4.3 k9) = 74 pF
For CC: The Thévenin resistance seen by Cc becomes rmc = ‘l' RL = k9. Choose Cc = l/22t(2 Hz)(15.6 k9) = 5.1 uF For CE: The Thévenin resistance seen by Cc, taken with
CS = short, becomes
rm = R4“ (r,‘t + RS)/(l3°+1)
where R1 lle H R3 = R5. For r1: >> RS
=5 I’m]; ‘3 R4(rn/Bo) E R4(1/gm)
For an estimated IC = 1 mA (VCC and VE are not given),
gm=40mA/V=:1/gm=259.
=> CE '= 1/21t(20 Hz)(24 Q) = 326 pF,
where R4l(1/gm) ~ 24 9. Choose closest standard value
of CE = 330 pF. 9.69 Analysis of the bias conﬁguration using the constant
voltage approximation yields VB = 2.1 V, VE = 1.4 V, IC
—= 1.4 mA, gm = 56 mA/V, and r“: [iolgm = 0.9 kg to 3.6 If CE is made to be the dominant pole, then
f3 = l/21tiREll [lat/(i5o + 1)]}C5
= lam/Bows = zm/zncE
(C3 = short at f“). Similarly, if CS is made to be the
dominant pole, then
f}, = 1/2n{[r,. + (13., + l)RElllR1lR2}Cs
The second expression depends directly on Bo, whereas the
first does not. Bo is not know precisely, hence set the
dominant pole via CE:
CE: gm/‘anH = (56 mA/V)/(21t)(20 Hz) = 446 uF
(Choose next highest standard value of 500 pF.)
The pole of Cs must be set to a frequency well below 20
Hz. Arbitrarily choose f3 = 0.2 Hz. Assume a lowend [3°
(yielding the lowest expected value of the Thévenin
resistance seen by C5) = I ,,
C5 = 1/21t{[r,t + ([30 + 1)RE]'[Rl llelfS = l/(21t){[0.9 k9
+ (51)(1 km] "(100 kQ)(2'7 1(9)}(02 Hz) = 53 FR
(Choose closest standard value of 50 pF.) a 9.66 Here is a picture of one possible circuit topology that
1% will work. A typical R5 = 50 9 source resrstance (e.g., as
l
3, ' ' ' ' ' enerator) is
one might see looking into a funcuon g
assumed. The circuit’s high—frequency smallSignal model,
in which CS is set to a short circuit, is also shown.  Vac: Iov
, "‘th 1.0.2 MA
t IS'kIl as c. Q Cour RC of 9.112 a. Biasing VOUT at 5 V requires a drop across
5 V => ID = 1 mA. This current requires that
1/.
VGS = + = [(1 mA)(1 mA/V2)]* .+ 2 z t 3 V
, Setting VGS to this value requires a —
R A/RB = VG/(VDD _ VG) = (3 V)/(lO v — 3 V) _ 0.43 d
where RA is the resistor connecting the gate to group , Choose, for example, the values RA = 430 k9, B .
l b. Note that CS effects the droop, but not the use d falcll
3 times of VOUT. The latter are determinedby CL, gs, at; I
‘ C d. Examine the circuit before the droopsn v0m caused“);
l C‘3 develops. The model for the circuit over this _ 0 reigime becomes that of Figs. 9.67 and 9.68 with R5 
i and r = on. For {A
‘ g; = map)” = 2[(1 mA/V2)(1 mA)] = 2 mA/V
The Millermultipligdcslalue of C8d becomes
C = C do. + gm
A= (15mm + (2 mA/V)(5 kn)1_= 11 pF ' e
1‘ For an estimated r,L of, say, 50 0 (high end; longest um Without the external capacitor CX, the highfrequency
behavior of this circuit is governed by C,t and C". The
5 'I'hévenin resistances seen by these capacitances, as
determined from Eqs. (9.140) and (9.150) with RC = 0.
become
rm = (R5’ + RE)rn/[Rs’ + ([30 + DRE + rﬁ], and
v rm” = [Cu + (Bo +, where R3’ = R5 + rx. The resistance rm“ will have the
same order of magnitude as RS, and rum will be lower in
magnitude. With Cu and CF in the picofarad range, and
assuming RS to lie in the kﬂ range, the values of the
intrinsic poles (those due to C,r and CP only) will lie in at
least the tens to hundreds of MHz range. An fH of 1 MHz
will be most easily realized by adding the external
capacitor Cx to the circuit. One approach involves biasing
the BIT appropriately, computi g r and r , then
selecting the value of Cx needeth‘o seihﬁl to 1 Set the bias current: (Use the constantvoltage
approximation.) We arbitrarily choose IC = IE = 0.2 mA
and the value RE = 10 kg for good bias stability =5 The
drop across RE must be 2 V = VB = 2.7 V. Using
approximate voltage division,
V13 = VccR2/(R1 + R2)
= (2.7 V)/(10 V — 2.7 V) = 0.37 .
= Choose, for example, R1 = 15 k0, R2 = 5.6 1:9.
For the chosen bias current, '
gm = IC/(nVT) = (0.2 mA)/(I)(0.025 V) = 8 mA/V, and constant), the time constant ass0ciated with the charging of
this capacitance becomes i = (go the pF + 11 pF) = 0.65 _ns
t . . . t assoc!
2 Similarly, the time constan
% signal charging and discharging of CL through RG becomes
' rL=RCCL= (5 kn)(8p1=)=4ons .
. where RC << ro has been assumed. Th1s_ second time
constant IL is much longer than 1A and dominates the rise
and fall time of vom, hence the exact value of rx is not crucial to a determination of the time. 51%ng = 63 £0133 1‘9 , f =1/2m c =1/2n(122 o 1.2 =11 GHZ
Now compute the Thévenm poles 0f the “V0 Internal I (so high it cant’linevgn be measured)t(1singptlt:c)>nnal labOratory
capacrtances. instruments). Similarly, 1
c1: = gm/21th  cl1 rm”: Rs'll tr. + (Bo + 1)RE] i
= (8 mA/V)/(21t)(400 MHz) — 2 pF = 12 DF = (100 Q)ll[12.5 k9 + (101)(10 km] .s 100 Q
rm = (R5? + RE)rn/[Rs’ + ([30 + DRE + tn] fp = l/Zrtr‘hpC = l/21t(100 Q)(2 pF) = 796 MHz '
= (100 o + 10 k9)(12.5 my These two intean poles are much higher than the desired
I utnm (lOleOlQ) + 125 ml fH. Setting fH to 1 MHz can be accomplished by
= 122 Q connecting an external capacitor Cx in parallel with C , as
where R5’ = R5 + rx = 50 $2 + 50 Q = 100 Q. and an _ shown in the circuit diagram. It’s value must be: p average [30 has been assumed. Thus. . g ~ Cx = 1/21:erth = 1/0700 MHlem Q) = 16 “F 6m Fluo mLLen CAQACITWCE 7 ll: Cﬁ=osf= C 3F

. g”: IJ'DIHA/V NH“. ’54“? f ﬁﬂ'M'E Vocmaetlum VI 7 O 50 SHIITS 11142 loo SHEETS
22144 200 SHEETS 2214] .® 9.74 First find the bias point of Q1:
V3 = Voch/(Rt + R2)
.= (12 V)(40 kin/(80 k9 + 40 k9) = 4 V
=5 [C = 1,; = (VB — Vf)/R4 = (3.3 V)/(3.3 k9) = 1 mA
=> gm = IcmVT = (1 mA)/(l)(0.025 V) = 40 mA/V
2:) In = [Sng =100/(40 mA/V) = 2.5 kg 
a. ; In + R1 ﬂ°+1 b. The magnitude of 1/ijE will equal 1'“IE at i .
' fE = l/21cCErmE = 274 Hz '7
The emitter of Q1 will be bypassed well above this 3
frequency. ‘
c. Find the poles of the other capacitors.
Pole of Cs: Assume CE to behave as an open circuit at this
pole frequency =>
=> rms = R5 + R1R2[r1t + ([30 +1)R4] = 29 k9
= fS = lflnCSr'ms =‘5.5 Hz '
(conﬁrmed to be lower in frequency thanf5) =58!) Yuan = R4 «5611 339%? 50 SHEETS 22142 100 SHEETS
' 22144 200 SHEETS 22141 @ 9.92 At t = 0+ (just after the step function), the capacitor will behave as a short circuit, so that VOUT = 0. After _. ' many time constants, the capacitor will behave as an open . ' circuit, allowing the circuit to yield a voltage of '
v VOUT = VoRz/(Ri '* R2)
= (5 V)(10 kQ)/(5 k9 + 10 k9) = 3.33 V 
where R1 = 5 kg and R2 = 10 k9. The transition will be
exponential, with time constant determined by the Thévenin
resistance seen by the capacitor, in this case R—Hl = R1 [le
= 3.33 kn. Thus
yoUT = voa e“), where T = RThC = (3.3 k9)(10 11F) = 33 ms. vom will
reach (1  l/e) = 0.63 of itsxtinal value after one time
constant I. Here is a plot of VOUT versus time: 000., (v) V 55.4.0» 7C . C, ' (in, _= Ar lh' F .C,‘ L/Mfrs 50 SHEETS 22142 100 SHEETS
22144 200 SHEETS 22141 > RMARntswﬂngi 1/44 = w“ CL]? 5m: C: =» [0"];
I: l (A V
45914;". ﬁf'DG "SELL0.6 _ 4.53444,
»  _ A2141}; Luv . lab/4g :saQW. um, owoverm .2: _=_ 423'” i: 5.5m mew 8.“, 3.7%,.
Emma 2: «can mEmzm on 57% A ...
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 Spring '06
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