Analog Integrated Circuit Design Problem Answers

Analog Integrated Circuit Design

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Analog Book---Answers Analog Integrated Circuit Design --- Problem Answers Textbook by David Johns and Ken Martin, Wiley, 1997. Solutions by Khoman Phang and Ali Sheikholeslami ==================== ANSWERS TO PROBLEMS ==================== Chapter 1: ========== 1) n:10e25 carriers/m^3, p:3.6e8 carriers/m^3 2) 0.87V, decreases 3) 140fC 4) t_fall=0.37ns, t_rise=0.48ns 5) t_fall=0.37ns, t_rise=0.44ns 7) 143uA 8) r_ds=167kohms, lambda=0.3 9) r_ds=182kohms, gm=230uA/V, gs=44uA/V 10) Cgs=86fF, Cgd=10fF, Cdb=60fF, Csb=74fF 11) 0.976V 12) 1.25ns, 3.33ns 13) 1.35ns, 6.1ns 14) gm=3.8mA/V, r_pi=26kohms, r_e=260ohms, r_o=800kohms, gmr_o=3000 15) t_fall=690ps, t_rise=7.8ns 16) t_fall=850ps, t_rise=10.3ns Chapter 2: ========== 8) 0 lambda 12) Cdb=0.019pF, Csb=0.034pF 13) Cdb=0.016pF, Csb=0.44pF 14) y2=2.415x1, x2=0.631x1 15) -0.6%, +0.15% 16) 350 Ohm 17) 379.26 Ohm, height=24um, width=152um Chapter 3: ========== 1) I_out=20uA, R_out=640kohms, Max. Vout=170mV 2) gain = -4800*sqrt(2*un*Cox*W*L/Ibias) 3) w_3dB = Ibias/(4800*L*C_l) 5) 2*PI*1.05MHz 6) 1/R_out = gm1 + gs + g_ds1 + g_ds2 7) 1/R_out = gm1 + gs http://www.eecg.toronto.edu/~johns/nobots/Book/book_ans.html (1 of 6) [1/28/2008 3:58:00 PM]
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Analog Book---Answers 8) a) w_o=2*PI*35MHz, Q=0.332, w_z=2*PI*844MHz b) w_o=2*PI*33MHz, Q=0.570, w_z=2*PI*844MHz 9) a) C_1=0.125pF, R_1=4780ohms, p_1=2*PI*3.61MHz, p_2=2*PI*281MHz b) C_1=0.144pF, R_1=4780ohms, p_1=2*PI*3.61MHz, p_2=2*PI*244MHz 10) 2*PI*1.3MHz 11) R_out = 2*r_ds1 12) V_out >= 1.33V 13) R_out = r_ds4*[1+ r_ds2(gm4+gs4+g_ds4)] 16) 2*PI*57kHz 17) 2*PI*80MHz 22) R=22kohms, R_out=400kohms 23) gain=-V_a/[2*V_t] 25) gain=0.9993, R_out=52ohms 26) gain=beta*R_e*(R_l//R_c)/[(beta+1)*(re*Rs+Re*Rs+re*Re)], Rin=re//Re 31) gain = 116, w_3dB=2*PI*10kHz Chapter 4: ========== 1) dBm(50 Ohm)=dBm(75 Ohm)+1.76 2) a)-18.24dBm, b)-17.16dBm, c)-15.35dBm, d)-30.67dBm 3) -44.77dBm, 0.058mV/root_Hz 4) -31.6dBm 6) they are equal.
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