A common-emitter amplifier is shown below. The DC bias gives Ic = 0.2 mA (you don't...

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A common-emitter amplifier is shown below. The DC bias gives Ic = 0.2 mA (you don't need to calculate the DC bias.) The BJT has = 100, C = 20 pF and C = 2 pF, and the load has CL= 10 pF. You can ignore ro and rx in this problem. Assume the coupling and bypass capacitors are large. 92 (10) a) Calculate f. You can find the OCTC's, or use the dominant pole expression from the book, but show your work clearly. G = 20pF fH = 789 MHZ th= == {G+Zz) an. z05-20005) = 78901Hz = Ti = Go [4+Cp (inger)] _gine = 40-Ie [20pF+2pf (1+85,102)] = 845 = 1.62X10-9 = =1.4215 2=RL Cefr = 1024 (205) - 2005 cepz = CM (1. =) = 2pF (1) FOUTINE. - ZpF vi 2k VCC 100k 100k 25k 10k Cu ZpF 8-100 100k CL 10pF A common-emitter amplifier is shown below. The DC bias gives Ic = 0.2 mA (you don't need to calculate the DC bias.) The BJT has = 100, C = 20 pF and C = 2 pF, and the load has CL= 10 pF. You can ignore ro and rx in this problem. Assume the coupling and bypass capacitors are large. 92 (10) a) Calculate f. You can find the OCTC's, or use the dominant pole expression from the book, but show your work clearly. G = 20pF fH = 789 MHZ th= == {G+Zz) an. z05-20005) = 78901Hz = Ti = Go [4+Cp (inger)] _gine = 40-Ie [20pF+2pf (1+85,102)] = 845 = 1.62X10-9 = =1.4215 2=RL Cefr = 1024 (205) - 2005 cepz = CM (1. =) = 2pF (1) FOUTINE. - ZpF vi 2k VCC 100k 100k 25k 10k Cu ZpF 8-100 100k CL 10pF

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