Question

How many times the output voltage size of the -6dB filter compare with the input voltage? -14 1 3 1 2 What is the intensity of the current at the diode D1?find a closest answer. 15 V www 2k D 5 Vok svo D. 1.2 mA 2.4 mA 3 3.6 mA 4 4.8 mA 6.0 mA What the intensity of the current flowing through diode D1 in the following circuit? Diode voltage drop is 0.6V. find a closest answer. D D2 D 12 V Ik 2 0 5.1 mA > 10.8 mA 3 11.4 mA 4 15.9 mA 27.3 m

Answer 1

(1) Vout = -6 dB Vin 20 logo = -6 → xx -6120 10 0.5012 ~ 0.50 Vout Vin = 0.50 Vout - ya Vpn output voltage size is į times the input 15V (2) VI Assume Diodes are K ideal 5Vat the Is Volcon) VDI CON) = OV 5vott * Iq Let assume both Diodes are both Diodes are in on state. voltage drop across two diodes is same after ON and current I will equally shared between Di and D2.j.es, I= Iz 15 V It I = 2K 15-5 2 = 5mA 5 Va & Il I = I1+Ig = 211 5V * Iz :: I1= 512 = 2.5 mA

current at diode D2 is 2 2.4MA Cclosest answer) D2 * Vac (3) D2 D3 Vy 와 3 Vz & 12V 1K ww :2K www 1k Let assume Di is ON → 12-V2 = 0.6 z ve = 11.4 V If DI Dl On then D2 will be ON Vx- Vy = 0.6 → Vy = 21.4 -0.6 = 10.8V If D2 On the D3 will also ON Vy-Vz = 0.6 → V2 = Vy-0.6 = 10.8 -0.6=10.2 y 0.6V 0.6V vy Vz 0.6V H > ID3 >IDI I'D2 v Ix lk ly & Iz 2k 12 V 16 10.2 ID3 = I2 = VZ ak = 5.2 mA 2 Iy = Vy 1K 10.8 mA It = Val 2 11.4 mA By KVL at Vy node, IDE = Iyt Iz 15.9 MA By kul at Va mode, ID1 = Ix+ ID2 = 11.4+ 15.9 = 27.3 MA current flowing through Di is Ipl = 27.3 mA