MICROELECTRONIC CIRCUIT DESIGN

RICHARD C. JAEGER and ARTHUR T. BRADLEY

Errata - Updated 5/11/02

Problem 2.13, Page 11: The velocity should be 107 cm/s yielding 16,000 A/cm2.

Prob. 2.25, Page 13: upp = 8.32 x 1018/V-cm-s and the problem only has one solution - the first bold entry in the table with upp = 8.4 x 1018.

Prob. 3.20, Page 22: VT = 0.0264 V

Prob. 3.34, Page 26:  The load line should intersect the vD axis at +5 V.

Prob. 3.41, Page 29: Q-Point for CVD  model should be (40.0 mA, 0.6 V).

Prob. 3.54(a), Page 33: The denominator of the IZ expression should be 50.  The answer is correct.

Prob. 7.16, Page 110: IDS = 33 uA

Prob. 7.36, Page 114: (W/L)L = 1/2.06

Prob. 13.33(b), Page 286: C3 is a bypass capacitor, and R1 should be replaced by a short circuit in both schematics.

Prob. 14.12, Page 319: The expression for AI should be multiplied by an additional current division factor at the collector of 8.2kW/(8.2kW+100kW) so that AI = -4.62.

Prob. 14.21, Page 322: A typo occurred in the last line. It should be vbe = 0.985 vs ...

Prob. 15.31: Note in the (b) part that v1 = v2 = 5 V exceeeds the common-mode input range for the amplifier.  The input transistors are saturated.
One possible modification would be to change VCC to +18 V.

Prob. 15.54: RSS appearing in the denominator of the last equation should be IDS.

Prob. 15.69: The dc half-circuit should use an ideal current source symbol.

Prob. 15.79, Page 380: The expression for Av1 is incorrect.

Prob. 15.91, Page 386: RD = 3.83 kohms which yields Adm = 528

Prob. 15.93, Page 387: VCE = VCC - VEB3 - (-VBE2) = 5.00 V

Prob. 15.95, Page 388: RL should be 5 kohms which yields emitter follower gain = 0.980 and Adm = 249

Prob. 16.19, Page 413: The last line ROUT2 should be ROUT3.

Prob. 16.48, Page 422: The last equality should be IC6 = 5.42 uA.

Prob. 16.61, Page 428: IC2 = 200 uA-... ; At bottom , VBE3 = VEB4 = ...

Prob. 16.64: Table values for VDS7 and VDS8 should be 7.25 V and 2.75 V respectively.

Prob. 16.75: First line - IC44 should be IC14.  Note that two transistors are accidentally labeled as Q6.

Prob. 17.7, Page 453: A division symbol should follow the (1/2p) term in the fH expression

Prob. 17.23, Page. 461: The value of Cp in the first row should be 0.773 pF

Prob. 17.69, Page 482: The wrong value of capacitance was used in the calculations.  It should be 7.5 pF yielding answers of 10.6 MHz and 33.3 V/ms.

Prob. 17.75, Page 484: The value of ro is calculated incorrectly and affects the rest of the answers. ro = 6.99 kohm, AV = -41.1, BW = 7.74 MHz, Q = 2.91

Prob 18.26, Page 507: The positive directions for v1 and io are both backwards.  This error cancels in the evaluation of A.  However, vo = -5000 io, so Av = -1.

THE FOLLOWING PROBLEMS WERE ADDED TO THE SECOND PRINTING AND SOLUTIONS ARE INCLUDED IN THE SOLUTIONS MANUAL.

CHAPTER 12

12.134 Two diodes are added to the circuit in Fig. P12.131 to convert it to a monostable multivibrator similar to the circuit in Fig. 12.76, and the power supplies are changed to ±10V. What are the pulse width and recovery time of the monostable circuit?

12.135 Design a monostable multivibrator to have a pulse width of 10 us and a recovery time of 5 us. Use the circuit in Fig. 12.76 with ±5 V.

Problem Correction

12.114 (a) What is the output voltage at the end of one clock cycle of the SC integrator in Fig. 12.53 if C1 = 1 pF, C2 = 0.2 pF, vs = 1 V, and there is a stray capacitance CS = 0.1 pF between each end of capacitor C1 and ground. What are the gain and gain error of this circuit? (b) Repeat for the integrator in Fig. 12.57.

CHAPTER 17

Light bulb: 17.85 Change the two capacitor values in the circuit in Fig. P17.84(a) to give the same center frequency as in Fig. P17.84(b). What are the Q and mid-band gain for the new circuit?

SPICE: 17.86 (a) Simulate the circuit in Prob. 17.84(a) and compare the results to the hand calculations in Prob. 17.84. (b) Simulate the circuit in Prob. 17.84(b) and compare the results to the hand calculations in Prob. 17.84. (c) Simulate the circuit in Prob. 17.85 and compare the results to the hand calculations in Prob. 17.85.

*17.87 (a) Derive an expression for the input admittance of the common-emitter circuit in Fig. 17.21 and show that the input capacitance and input resistance can be represented by the expressions below for wCuRL << 1.

Cin = Cpi + Cu (1+gmRL)

Rin = rpi ||{RL/[(1+gmRL)(wCuRL)2]}

(b) A MOSFET has CGS = 6 pF, CGD = 2 pF, gm = 5 mS and RL = 10 kohms. What are the values of Cin and Rin at a frequency of 5 MHz?