MICROELECTRONIC CIRCUIT DESIGN

by

RICHARD C. JAEGER


First Printing Errata - Updated 3/10/01

Note: The second printing took place in mid-July 1997.


Page number: Correction

29: The values of the constant B for Si and Ge are reversed in the caption of Fig. 2.4.

83: Eq. 3.64 - 16.7

143: Middle of the page - VDS = 15 - 5 = 10V

154: Eq. 4.100 VSG ? -VP

177: The arrow points in the wrong direction in the iT source in Fig. 5.9(a)

188: Eq. 5.34 - In the equation for iE, the plus sign should be a minus sign.

191: Four lines above the figure - IB = 1.01 mA / (50+1) = ...

199: Below Eq. 5.51 - BF+1 = 75

201: Ex. 5.8 - BF+1 = 75

257: In Table 7.2 and 4 lines above the table - 2.07 V should be 1.99 V

258: Fig. 7.16 (W/L)L = 1/9.03 & also in text nearby

261: In the equations at the top of the page, the placement of the two "and ..." clauses is reversed.

262: Eq. 7.34 VIH = VTNS + 2vO = ...

264: Fig. 7.20(c) (W/L)L = 1/9.03

266: Fig. 7.22 MA is an enhancement-mode transistor

291: Eq. 7.60 should be tPLH. Equation for Rons should use KS not KL

299: Prob. 7.41 - Y + ... should be Y = ...

308: The subscripts on MN and MP are reversed in both regions 2 and 4 in Fig. 8.4.

326: The values of both Kn and Kp should be 2 times larger in Fig. 8.29

366: The lowest black dot is in the wrong location in Fig. 9.40. It should be on the line to the data buffer.

408: 5th line of text - "is forced to become 0.1 V"

409: Last line should be = -1.03 V

410: Eq. 10.43: iE2 = (vB2-vBE2-(-VEE))/REE

411: Fig. 10.25: The 3.3 kohm resistor should be a 0.3 mA current source; Fig. 10.26: The 1 kohm resistor should be a 5.6 kohm resistor,

436: Near bottom: iB2 = iB1. not (bR+1)iB since Schottky diode bypasses base current around the collector-base junction.

437: Fig. 10.72 - B1 should be iB1, since iB2 = iB1.

468: Exercise 2 - vS = 0.001 sin 2000*pi*t

505: Above Eq. 12.56 should refer to Fig. 12.19

528: Ex. 12.10 - RID = 1 Gohm, RO = 50 ohms

556: Fig. 12.65 - vS =-5 sin 120*pi*t

741: Eq. 15.53: r'o = (2/3) ro2 - Eq. 15.54: u'f = uf2 / 3.  Text just above Fig. 15.13 - " ... reduced by a factor of 3."

769: Exercise - I2 = 500 uA, Overall voltage gain = 2410

781: Above Eq. 15.174 refers to 15.67

908: Fig. 17.4 - RG = 243 kohms

909: Eq.17.31 - R4 should be R3

910: wp1: 430 kohms --> 243 kohms ; wp1 = 41.0 ; fL = 31.5 Hz

912: Eq. (17.35) R2 should be RS

915: Fig. 17.11 - RG = 243 kohms

918: Fig. 17.20 - RG = 243 kohms

931: Fig. 17.36 - RG = 243 kohms ; Rth = 996 ohms

1033: Eq. 18.143 and 18.144 - The sgmC3 terms should be sGC3

1034: Eq. 18.146 - The gmC3 term should be GC3 ; gmR = C3/C1 ; gmR * C3/C1
 

Problem Statement Updates

3.55 Use f = 60 Hz 3.61 Use vs = 10 sin 120*pi*t 3.79 delta = 3/4 3.100 Use C = 100 uF
4.11 VS = 10 V 4.46 (a) W/L = 20/1 (b) VDD = 15V 4.65 R = 100 kohms
5.50 VCC = -15V 5.59 VCC = +10 V 5.64 Use bF = 100 5.68 Fig. 5.35
5.74 bR = 0.5
7.12 Use Gamma = 0 7.71 VTO = -0.7V, K'p=10uA/V2
8.52 Ignore body effect
9.24 Use W/L = 50/1 9.30 VDD = 5V 9.34 (b) Connect substrates to +5V
9.48 Do not use (b)
10.12 bF = 35 10.17 Fig. 10.10 10.58 IBR = -0.5mA 10.78 Redesign RB
11.52 Refer to 11.42 11.53 Refer to 11.40 11.54 Refer to 11.41
12.67 RO = 100 ohms 12.99 Use C1 = C2 12.109 Use 20 dB gain 12.129 Fig. 12.126
13.9 Bypass R1 13.68 2vgs/(VGS-VTN) 13.118 1 kHz 13.120 10 kHz
13.131 Refer to 13.9 13.132 Refer to 13.13 13.133 Refer to 13.17 13.134 Refer to 13.21
13.135 Refer to 13.25
14.7 (a) -10 (b) -20 14.32 P14.1(g) 14.35 VTP = -1 V 14.60 KP = 200uA/V2 VTP = -1V
14.68 Lambda = 0.02 14.69 Use L = 1H 14.70 Lambda = 0.02 14.76 C1 = 47uF C2 = 4.7uF f = 10kHz
15.27 IC2 = bFIC1 15.66 VDD = VSS = 12V VP = +2V 15.71 RD = 75 kohms 15.72 Sources of M1 and M2 should be connected
15.92 RE = 1 kohm 15.114 P15.107 15.137 VTN = +1V
16.16 8V-->10V 5V-->10V 16.22 VA = 70V bF = 100 16.24 -VEE =- 5V 16.64 (W/L)7 = 15/1
16.93 5% tolerances 16.98 R= 1.2188 kohms 16.106 R = 500 ohms 5 mm --> 5 um 16.114 single-slope converter
17.32 CGS = CGD = 2.5 pF 17.41/17.42 VCC = +12 V 17.43 rx = 300 ohms 17.66 RL = 100 kohms
17.67 RL = 100 kohms 17.70 VDD = VSS = 10V 17.73 I3 = 500 uA
18.27 AV = RL[A/(1+AB)] can be represented 18.33 Fig. 18.13 with C1 = C2 = 1 uF 18.46 C1 = C2 = 5 nF, R1 = R2 = 2 kohms
18.67 gm = 10 mS, bo = 100 18.68 Neglect rpi 18.73 VDD = 10V


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