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

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

Prob. 3.20, Page 22: V_T = 0.0264 V

Prob. 3.34, Page 26:  The load line should intersect the v_D 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 I_Z expression should be 50.  The answer is correct.

Prob. 7.16, Page 110: I_DS = 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 A_I should be multiplied by an additional current division factor at the collector of 8.2kW/(8.2kW+100kW) so that A_I = -4.62.

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

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 V_CC to +18 V.

Prob. 15.54: R_SS appearing in the denominator of the last equation should be I_DS.

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

Prob. 15.79, Page 380: The expression for A_v1 is incorrect.

Prob. 15.91, Page 386: R_D = 3.83 kohms which yields A_dm = 528

Prob. 15.93, Page 387: V_CE = V_CC - V_EB3 - (-V_BE2) = 5.00 V

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

Prob. 16.19, Page 413: The last line R_OUT2 should be R_OUT3.

Prob. 16.48, Page 422: The last equality should be I_C6 = 5.42 uA.

Prob. 16.61, Page 428: I_C2 = 200 uA-... ; At bottom , V_BE3 = V_EB4 = ...

Prob. 16.64: Table values for V_DS7 and V_DS8 should be 7.25 V and 2.75 V respectively.

Prob. 16.75: First line - I_C44 should be I_C14.  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 f_H expression

Prob. 17.23, Page. 461: The value of C_p 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 r_o is calculated incorrectly and affects the rest of the answers. r_o = 6.99 kohm, A_V = -41.1, BW = 7.74 MHz, Q = 2.91

Prob 18.26, Page 507: The positive directions for v₁ and i_o are both backwards.  This error cancels in the evaluation of A.  However, v_o = -5000 i_o, so A_v = -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, v_s = 1 V, and there is a stray capacitance C_S = 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 wC_uR_L << 1.

C_in = C_pi + C_u (1+g_mR_L)

R_in = r_pi ||{R_L/[(1+g_mR_L)(wC_uR_L)²]}

(b) A MOSFET has C_GS = 6 pF, C_GD = 2 pF, g_m = 5 mS and R_L = 10 kohms. What are the values of C_in and R_in at a frequency of 5 MHz?

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