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Why does the output voltage drop by 2VD? Whats troubling me is the ground placed after the load resistor. If we were to remove that ground and measure Vout as shown in fig 4.25a, i understand that there would be a 2VD drop. Like this though, it seems like there should only be a drop of 1VD, no?

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    \$\begingroup\$ It's got zilch to do with the gnd symbol - it's just a symbol naming the node as ground - it doesn't bring anything to the party. \$\endgroup\$ – Andy aka Apr 14 '16 at 18:08
  • \$\begingroup\$ @Andyaka so we can place ground anywhere in the circuit and the circuit would still operate the same? \$\endgroup\$ – Georan Apr 14 '16 at 18:21
  • \$\begingroup\$ In that schematic shown, yes, put that ground where ever you want at the secondary side of the circuit. \$\endgroup\$ – soosai steven Apr 14 '16 at 18:37
  • \$\begingroup\$ @soosaisteven Right. Usually we assume that the lowest node of a circuit is ground. However, this is not the case in this circuit, correct? \$\endgroup\$ – Georan Apr 14 '16 at 18:45

you forgot about the the 1VD drop from the - node of the secondary winding to the GND node so you get a net of 2DV drop. Basically, the - node of vs is at 1VD below GND.

If you trace out the circuit loop, it goes through 2 diodes.

  • \$\begingroup\$ Just drew it out, makes sense thx \$\endgroup\$ – Georan Apr 14 '16 at 18:15

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