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Why does it hold the voltage constant? Isnt it supposed to act as a ideal diode during forward bias ie behave as a closed switch?

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closed as unclear what you're asking by brhans, Marcus Müller, Dmitry Grigoryev, Nick Alexeev Sep 11 '18 at 17:38

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    \$\begingroup\$ Welcome to EE.SE. Please edit your title to give a summary of the question. "This is very confusing" is not a good title for a question about diode forward voltage drop. \$\endgroup\$ – Transistor Sep 10 '18 at 18:29
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    \$\begingroup\$ Hello and welcome! To get a good answer, I recommend you edit your question: 1) Crop the image to focus on the relevant info 2) Add more detail to your description of what the circuit is supposed to do, and what you don't understand about how it works. 3) Edit your title to be more informative \$\endgroup\$ – Selvek Sep 10 '18 at 18:29
  • \$\begingroup\$ Please post text as text. Please don't post text as screenshots. Text is searchable, but screenshots aren't. \$\endgroup\$ – Nick Alexeev Sep 11 '18 at 17:37
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Well, no, this is not an example of an ideal diode.

Real diodes have a forward voltage drop of approximately 0.7 V (for silicon). When the diode is forward biased, its low impedance (relative to the series resistor) keeps the output voltage at 0.7 V relative to the 0 V at the cathode.

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  • \$\begingroup\$ Real diodes have a Vf that varies with the forward current through them, so it WILL NOT clamp at 0.7V since the current follows the sine wave input. \$\endgroup\$ – Jack Creasey Sep 10 '18 at 18:58
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    \$\begingroup\$ @JackCreasey: You should write your own answer that explains that. I felt that it was more detail than the OP was looking for. \$\endgroup\$ – Dave Tweed Sep 10 '18 at 19:02
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Isnt it supposed to act as a ideal diode during forward bias ie behave as a closed switch?

No. It's supposed to behave as a real diode that drops 0.7 V when forward biased.

The graph shows this clearly. On positive half cycles the diode is forward biased and the output is clamped at 0.7 V.

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