While solving these problems I'm confused about sign conventions of output voltages of IDEAL Diodes. Actually I can't get between which nodes the V is to be taken? Is it between +5 and -V? I solve them like this

a) V = +5 (Between +5 and 0) since diode is conducting.

but answer is -5V? Can any one explain a bit?

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  • \$\begingroup\$ Hint: Either the diode dominates, or the resistor dominates. They are wired to be mutually exclusive. \$\endgroup\$ – Sparky256 Oct 28 '18 at 3:53
  • \$\begingroup\$ For (a), you should merely notice that the diode is arranged so that it conducts (is on.) Assuming an ideal diode here, conduction means "zero voltage drop across the diode." If there isn't any voltage drop across the diode, then the voltage on both sides of the diode must be the same. That's how they came up with -5 V as the answer for (a). \$\endgroup\$ – jonk Oct 28 '18 at 3:59
  • \$\begingroup\$ @jonk I'm asking about where this -5 appears? Is it between Positive end of diode and -5? \$\endgroup\$ – Saqeeb Oct 28 '18 at 4:03
  • \$\begingroup\$ @FanBoy It appears where the letter \$V\$ appears and it is relative to the same point that the two power supplies are also relative to. \$\endgroup\$ – jonk Oct 28 '18 at 4:14
  • \$\begingroup\$ All voltages are measured relative to "Zero volts", which isn't shown in those drawings, but is implied by the "+5V" and "-5V" labels. \$\endgroup\$ – Peter Bennett Oct 28 '18 at 4:41

If the answer to a) is -5V, then your textbook is assuming an ideal diode with no forward drop. An ideal diode is a perfect conductor when forward-biased, meaning there is no voltage drop across it. Therefore V must equal -5V.

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  • \$\begingroup\$ Across which terminals the V is to be collected? \$\endgroup\$ – Saqeeb Oct 28 '18 at 3:54
  • \$\begingroup\$ What makes you think these are ideal diodes? \$\endgroup\$ – Sparky256 Oct 28 '18 at 3:55
  • \$\begingroup\$ Is this V between +ve terminal of diode and -5V? \$\endgroup\$ – Saqeeb Oct 28 '18 at 3:56
  • \$\begingroup\$ @Sparky256 Book has mentioned it \$\endgroup\$ – Saqeeb Oct 28 '18 at 3:56
  • \$\begingroup\$ @Sparky256 Because the correct answer, supposedly, is -5 V. Only way that happens is when the diode is "ideal." \$\endgroup\$ – jonk Oct 28 '18 at 3:56

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