fall 2002 ch9b

fall 2002 ch9b - If films-y 7017(347:0 Low mam;ch MODEL In...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: If?) films-y, 7017+ ; : (347:0 @ Low mam;ch MODEL. In I: in 2 In 0 m am hi :1 mm 3 2§ 33 m. at? N“ 22-lé'l @ Q1? CL . a 3 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(RC||R,) ~ 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. :oomum-r 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" V-c l-uk—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 vs-RS 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 small-signal 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 R4|l(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)RElllR1|lR2}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 low-end [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 small-Signal 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 Miller-multipligdcslalue 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 high-frequency 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 constant-voltage 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 = 6-3 £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 = 1-2 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) = 1-6 “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 11-142 loo SHEETS 22-144 200 SHEETS 22-14] .® 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 + R1||R2||[r1t + ([30 +1)R4] = 29 k9 = fS = lflnCSr'ms =‘5.5 Hz ' (confirmed to be lower in frequency than-f5) =58!) Yuan = R4 «5-611 339%? 50 SHEETS 22-142 100 SHEETS ' 22-144 200 SHEETS 22-141 @ 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 22-142 100 SHEETS 22-144 200 SHEETS 22-141 > RMARntswflngi 1/44 = w“ CL]? 5m: C: =» [0"]; I: l (A V 45914;". fif'D-G "SELL-0.6 _ 4.53444, » - _ A2141}; Luv . lab/4g :saQW. um, owover-m .2: _=_ 423'” i: 5.5m mew 8.“, 3.7%,. Emma 2: «can mEmzm on 57% A ...
View Full Document

This note was uploaded on 01/25/2012 for the course EE 3220 taught by Professor Audiffred during the Spring '06 term at LSU.

Page1 / 9

fall 2002 ch9b - If films-y 7017(347:0 Low mam;ch MODEL In...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online