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Electronics Report

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Electrical Engineering
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College of Banking and Financial Studies
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Q1 a). Figure 1 Voltage drop across Germanium diode = 0.3V Voltage drop across the Silicon Diode = 0.7V 9.7-V0 = I2*(4.3*103) (1) 9.3-V0 = I1*(4.7*103) (2) V0 = 3.3*103I+0.7 (3) Substituting (3) to (1); 9.7-[(3.3*103I ] +0.7} = I2*(4.3*103) 9.3-[(3.3*103I ] +0.7} = I1C Since, I= I1+I2 I = {9.7-[(3.3*103I ] +0.7}/(4.3*103) +{9.3-[(3.3*103I ] +0.7} }/(4.7*103) (4) By Solving for I; I = 1.6mA Substiting I to (3); V0= 5.98V b). Figure 2 Let’s assume I1, ID, I2, I exists. Then, V1=VO2= 0.7V (Voltage gap across a current flowing silicon diode is 0.7V) So I2=0 I= I1 + ID (1) V01-0.7 = I *1000 (2) 20-V01 = ID*2000 + 0.3 (3) 20-V01 = I1*2000 + 0.7 (4) From (1), (2), (3), (4); {(V01-0.7)/1000} = {(19.7-V01)/2000} + {(19.3-V01)/2000} V01= 10.1 V ID = 4.8 mA I = 9.4 mA b). Figure 3 Figure 4 T= (0 - T/2) D1 is reverse biased. So; Figure 5 Current cannot flow through until Vi = 0.3. VoMAX = (14.7/2) V = 7.35 V Vo = (Vi-0.3)/2 The output voltage in the 0-T/2 Time interval; Figure 6 T= (T/2-T) D2 is reverse biased. So the circuit will be simplified to; Figure 7 VoMAX = -(14.7/2) V = -7.35 Vo = (Vi-0.3)/2 The output voltage in the (T/2-T) Time interval; Figure 8 PIV of D1= (15-7.35) V = 7.65 V PIV of D2 = [-15-(-7.35)] = -7.65 V PIV of D2= 7.65V d). Current flow across Zenor diodes when they are revers biased. When the Zenor diode’s reverse bias voltage is equal to the Zenor voltage drop current pass through keeping the Voltage between the Zenor diode keep ...
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