Question

3. Analog CKT: Use the following circuit to answer questions from a) to c). (15 points) Vin Vout a) Deal with the circuit below (a) When Vin is positive and the diodes are assumed to be ideal (no drop), Vout is equal to (a)-Vin (b) 0 (c) Vin (d) Vin/ 2 (e)-Vin/ 2 b) If Vin is 5 V and the diodes are assumed to be ideal (no drop), the voltage across the top diode is (a) 0 (b) 2.5 V

Answer 1

Given circuit Diodes are Ideal, Basics of diode (Ideal) A rightarrow Anode (+ ve) k rightarrow Cathode (-ve) A: - Given Input supply is +ve voltage. Top diode will conduct. Bottom diode will not conduct. \r\n Equivalent circuit: - Apply voltage division Rule V_out = R/R + R times V_in V_out = R/2R V_in V_out = V_in/2 Ans - d b: - Given supply voltage V_in = 5V Top diode will conduct. Bottom diode will not conduct. Voltage drop across top diode will be 0 volt because diode will be short circuited. Voltage across top diode = 0 Volt \r\n C: - Given input voltage V_in = A sin wave Let V_in = V_m sin wt (1) During +ve half of supply voltage: - Top diode will conduct Bottom diode will not conduct Equivalent circuit: - V_out = (R/R + R) (V_m sin wt) = R/2R (V_m sin wt) V_out = V_m/2 sin wt During +ve half of supply (ii) - during negative half