Question

The circuit below was assembled and the obtained waveforms can be seen in the graph (Vsecpico: 17V / 60Hz, C = 653 uF, R = 60Ω, RL = 60Ω, Vz = 5V and
rz = 5Ω). Use the constant voltage drop model for diode VDon = 0.7V

D1, 110 Vac Vsec 这个字: Det

a) Analyzing the circuit and waveforms, which components are open (burned)? Justify.
When replacing the burnt components with new ones

b) What is the peak voltage (Vp) in the capacitor and in the zener? Justify.
c) What is the amplitude of the ripple voltage (Vrp) in the capacitor and in the zener. Justify.

Answer 1

R Re Vsee -Tc Hovac from the given graphs, The peale value of secondary voltage = 14V dlodes ideal, the constant voltage As the are not drop acron cach diode is notv in full Bridge rectifier, in each half cycle two diodes will be conduction so the peale voltage available for the capacitor will be : 1 V-8 (0.7) 1506 V This is confirmed through the graph of capacitor voltage. so, capacitor is working fine. Now, lets asume that zener o diode is open. So Volaage are R & Re across By Voltage division across Voltage R = 60+60) 7.8V Vollege across 15.6.7.80 T.8v. ener the voltage across Breakdown value, it is greater than its should act y

Source SV constant vo Hage sv (": N2=50 But, from the graph 3 zenes voltage, it is clear that is varying Voltage porn 7.88 to 6.30 Thus, open (burned) Zener diode is is 15.65 bcoz (b) Peak voltage in capacitor ideal there be drop of o. ale not as diodes in each diode. (Vsec) peak - (drop) * no a diodes in conduction 17-2 (0.7) = 15.64 As Zener diode QC3083 R & RL Peale vorage across zener is 7.8V. open by appleging voltage division get 7.8V 3r lo (C) Ripple voltage Amplitude in :- is capacitor = 15.6 V - 12.6 V = (ii) in tener = 7.fr Vo man - (Ve) min) 1.5V ((V3) man 6.3V (vDonin) from ous but it analysis, the Answers (a) Jener diode is open. Because voltage across should be is not. zener (b) Vp in capacitor Vo is Jeney 15.6V 7.8V 3V (C) Vop in capacitor Vop io zenes = 1.54