0

enter image description here

Boylested says that during the positive voltage half cycle diode D2 is cut off.

But why can't there exist a current flowing through the outer loop via C1, the secondary of transformer coil, C2, and D2 in the positive cycle when the input voltage begins to decrease and the capacitor voltage is greater than the transformer voltage?

ocrdu
  • 8,705
  • 21
  • 30
  • 42
Kashmiri
  • 167
  • 9

1 Answers1

1

Why Can't Current Flow In This Circuit?

Well, it can - except if the voltage waveform is a square wave. I suspect that, if you look at the text accompanying the figure you've shown, that that is exactly what is happening.

When the voltage is positive, it stays at one voltage. As a result, the capacitor never starts to discharge, as it would during the second half of a sine wave.

When the voltage goes negative, it does so "infinitely fast" to a negative voltage and stays there.

So, check the waveform which is supposed to be driving your circuit.

WhatRoughBeast
  • 59,978
  • 2
  • 37
  • 97
  • Due to presence of transformer I suspect it isnt a square wave – Kashmiri Jan 25 '21 at 17:51
  • Why in the world would the presence of a transformer suggest it's not a square wave? Stop suspecting and start reading. – WhatRoughBeast Jan 25 '21 at 17:58
  • since V=L d i/ d t so current cannot abruptly jump in an inductor as in a square wave. – Kashmiri Jan 26 '21 at 14:30
  • @YasirSadiq - That's not an inductor as you think of it - it's a transformer winding. Depending on construction and core material, you can get extremely fast transitions. As a matter of fact, such high-speed transformers are a critical component of switching power supplies - such as the one you show. Furthermore, it is almost certain that the textbook you got the schematic from was discussing the behavior of an ideal switcher, which has ideal components like transformers with no associated inductance. – WhatRoughBeast Jan 26 '21 at 20:40