Question

(3) a) For the circuit shown in Figure Q3, calculate the voltage across the load RL. Assume the load is a welding system; calculate the voltage across the load considering the diodes are silicon (8 marks) b) Explain the role of the capacitor connected across the load. (2 marks) c) If the diodes D and D. are reversed, explain the consequences. (2 marks) Illustrate the waveform across the capacitor for Figure Q4 with respect to the secondary winding voltage (5 marks) e) If the diodes are replaced with germanium diodes, can the voltage across the load will remain same? Please justify your answer. (3 marks) D: 20:4 D3 230V,60H RL 10 De 1000 F

Answer 1

* It is a center tap full-wave rectifier.

a) From the diagram primary to secondary turns N₁ / N₂ = 20 / 4 = 5

RMS primary voltage =250V

RMS secondary voltage = 250 ( N₂ / N₁ )=250 ( 1 / 5) = 50V

Max. voltage across secondary = V_m = 50 ( )=70.7V

V2

Idc 2Vm ARL 2 X 70.7 3.14X10X103 4.5m A

$\therefore$ DC output voltage

Vdc = Idc.RL = 4.5X10-3X10 X 10% = 45V

Hence voltage  across load =45V

b) * Capacitor is used to get consistent d.c. signal. Capacitor is used to remove variations in DC signal. Thus giving us an almost pure DC signal from an AC signal.

* Capacitor minimises the ripple content in output waveform.

c) When D₁ and D₄ reversed

During +ve half cycle D₂ and D₄ are forward biased

D₃ and D₁ are reverse biased.

During -ve half cycle D₃ and D₄ are forward biased,

D₂ and D₄ are reverse biased.

d)

Waveform across capacitor

  

e) Doesn't remain same.

  The Basic difference between silicon and germanium diode is the voltage needed to become “forward-biased”. Silicon diode require 0.7 volts to become forward-biased, whereas germanium diodes require only 0.3 volts to become forward-biased.

As the Ge diode has low potential barrier voltage ( 0.3V). It conducts more current in forward bias than Si.Formula for potential drop is V=IR.As current increases more potential drop occurs.