Question

Question 14-16 Two amplifiers A and B are cascaded through capacitor coupling. The amplifier A has gain of 16dB, input impedance of 100Kohm, and output impedance of 5Kohm. The amplifier B has gain of 20dB, input impedance of 20Kohm and output impedance of 1 kohm. The lower corner frequency f -2.OKHz. 14. If the capacitor is chosen for the dominant pole, the value of the capacitor should be close to: a)3pF; b)300pF; c)3nF; d) 30nF; e) cannot be determined. 15. If the capacitor is not for dominant pole, the value of the capacitor can be: a)3pF; b)300pF; c)3nF; d) 30nF; e) cannot be determined. 16. The impedance is seen by the capacitor is: a)20Koh; b) 25Kohm; c)100Kohm; d) 101 kohm; e) 105Kohm

Answer 1

5K 20k /onks Ž IK for finding value of Cepecitoo is dominont Pole or non dominent pole we weat to find first Empe dence is Reen by the @paitor 9, impe dence Seen by Cepeiitor (R) = 5k+0K R = 25K * b= 25 ku is the Coorect option. (4). for domi nent pole Cepecitoo . ft- then 2TIRC =3-18 X10T ITRF C = 3 8nf C= 30F is the Corect option. C greeter (15). 6o non dominent- pole fi shnild be then 2rec C>>3·1&nF d is Coorect. Ontim


5K 20K IK for finding value of Cepecitoo êg dominent Pole or non dominent pole we Went to find first Empe dence s Seen by the apacitoo (16). impe dence een by Cepeitor (R) = 5kt20K R = 25k is the (orrect option. % b = 25 kn for dominent pole Cepecitoo - (4). f- then QTTRC =318 Xio 9 ITRF - 3.18nF C= 30F is the Corect option. C (15). for non dominent pole fe shnild be greeler 2Tec C>>3·1&nF i ortion d is Coorect.