MICROELECTRONIC CIRCUIT DESIGN
by
RICHARD C. JAEGER
Second Printing Errata - Updated 04/05/02
Page number: Correction
29: The values of the constant B for Si and Ge are reversed in the caption of fig. 2.4.
97: Fig. 3.78 - The polarity of VC2 should be reversed in order to agree with the output voltage waveform.
162: Prob. 4.59 - The gate of the lower FET should be connected to its drain.
173: Eq. (5.13) - The equation for iB should have "exp" in the last term: [exp(vBC/VT) -1]
brackets299: Prob. 7.41 - Y + ... should be Y = ...
262: Eq. 7.34 VIH = VTNS + 2vO = ...
299: Prob. 7.41 - Y + ... should be Y = ...
310: Line between Eq. (8.4) and (8.5) - ... and vDS = vO.
360: Row 0 - A2, A1 and A0 should each be individually complemented; Row 1 - A2 and A1 be individually complemented
402: Eq. 10.34 and 10.35 - vOH an vOL should be VOH and VOL
409: Last line should be = -1.03 V
436: Near bottom: iB2 = iB1, not (bR+1)iB, since the Schottky diode bypasses the base current around the collector-base junction.
437: Fig. 10.72 - B1 should be iB1, since iB2 = iB1.
459: Footnote 2: Common IF frequencies include 262 kHz, 455 kHz, 10.7 MHz, 21.4 MHz, 30 MHz, 45 MHz, 60 MHz, 160 MHz.
578: Prob. 12.99 - Calculate the sensitivity of wo with respect to C2 rather than Q.
586: Next to last line: (VGS = 3.5 V)
589: Third answer to second exercise should be ic(t) = -0.26 sin 2000pt mA
609: Eq. 13.66 " = iD" should be removed in the equation for y22
615: Eq. (13.92) - IG = - ISG
681: The value of Rth sybstituted in the C-G voltage gain expression should be 1.71 kW, not 18 kW.
698: Eq. (14.119) - VTH should be VTN
701: Three lines below Eq. (14.126) - Fig. 14.39(a) should be Fig. 14.42(a)
741: Text just above Fig. 15.13 - " ... reduced by a factor of 3."
745: The second equals sign in Eq. (15.65) should be removed.
749: Eq. (15.84) - eliminate "Acd =" It should only
read Acc = ...
Eq. (15.85) - Acd should
be Acc
The text below Eq.
(15.85) should read - "... the common-mode gain is determined.... supply
voltages, Acc = 0.5."
750: Top of page - "stage characterized by Eq. (12.79), CMRR is defined
in Eq. (12.83) as"
Eq. (15.88) - CMRR =|(Add/2)/Acc|
= ...
757: Eq. (15.110) - Acc =(voc/vic) = -gmRD/..
Eq. (15.112) - CMRR =|(Add/2)/Acc|
= ...
788: Eq. (15.194) - The factor of two should be in the numerator, not the
denominator. This error propagates into Eq. (15.195)
in which the answer should be 2.49
mA/V2. In Eq. (15.196), the result should be 5.00 V, and in
Eq. (15.197), R3 = 250 kW
which
gives R4 = 500 kW.
799: Prob. 15.63 - Kn should be Kp
820: Eq. (16.23) - h11 = 1/(gm1+gpi1+go1+gpi2) which is approximately equal to 1/gm1
821: Fig. 16.12 should include resistor ro1 across the input port.
840: The polarity of the source attached to v2 in Fig. 16.38 should be reversed.
847: Eq. (16.102) The output resistance of the Darlington should be 2ro7/3 which changes the value of AV2 to uf7/5.
847: Eq. (16.102) The denominator of Eq. (16.103) then should be 21.7 and the gain 4.15 x 105.
909: Eq. (17.31) - R4 should be R3
912: Eq. (17.35) - R2 should be RS
914: Example 17.3 - R3S = 22.7 W, C3 = 3.50 mF
943: Fig. 17.56 - RG = 243 kW
951: Next to last paragraph - Reference to Fig. 17.66(b) should be Fig. 17.65(b) on page 952.
952, 953, 955: Figures 17.65-17.68 - M1 should be an NMOS depletion-mode MOSFET.
978: Problems 17.41 & 17.42 - Use VCC = +12 V.
980: Problem 17.66 - Use RL = 100 kohms.
997: Middle of page in B: y22 is found with v1 = 0.
998: A = -119 kW. b = -10-5 S. The numerator of the calculation of ATR below Fig. 18.17 is missing a minus sign as is the answer. The answer should be -54.4 kW.
1011: Table 18.2 - The values of ro should be 155 kW and 64.8 kW
1102: Numerator of Eq. (18.123) should be 2 x 1024.
1033: Eq. 18.143 - The second sGC3 term should be sgmCGD.
1045: Problem 18.33 - Use C1 = C2 = 1 mF.
1098: First answer to Prob. 2.7 should be negative.
1102: Prob. 10.103 - Y = A+B+C; 0 V, -0.90 V; 0 V, -0.45 V
1105: Prob. 14.47 - ... C-B or C-G...; Prob. 14.53 - -ufvs, R5 + ro(1+gmR5) which is approximately equal to ro(1+gmR5)
