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The question asks to solve for I and V.

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Why the voltage is assumed to be zero?

Here is the solution

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How did he arrive that V = 0 volts. If he says it’s 2 volts would it be wrong? Why? I am confused.

This example is from microelectronic circuits by sedra smith sixth edition page#182. Just confused how is V = 0 volts

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    \$\begingroup\$ If both diodes are ideal, and both of them are on, what's the voltage across each diode? \$\endgroup\$
    – Hearth
    Commented Jan 10, 2021 at 16:56
  • \$\begingroup\$ Precisely @Hearth \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2021 at 16:57
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    \$\begingroup\$ When analysing diode circuits you start with an assumption of which diodes are on and which are off. After you used this assumption in your calculation you can validate the assumption from the results you got. Then you either found a valid solution or you have to Run calculations again using another assumption. The assumption in this example is that both diodes are on. The voltage at node B ist therefore 0V thus making V=0V. \$\endgroup\$ Commented Jan 10, 2021 at 17:02
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    \$\begingroup\$ @OMAR Not if you assume that both diodes are conducting it's not. \$\endgroup\$
    – Hearth
    Commented Jan 10, 2021 at 17:04
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    \$\begingroup\$ A real diode with a real forward volt drop will produce a "several hundred" millivolt drop that is negative at VB. \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2021 at 17:17

1 Answer 1

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If an ideal diode is conducting, the voltage across it is 0V (by the definition of ideal that this book seems to use). That means that the diode can be viewed as a short circuit, and the ground node on the left diode's anode is "shorted" through the two ideal diodes to the node at the right diode's anode. Hence \$V = 0\ \mathrm V\$.

This actually also works if you assume the other common definition of an ideal diode, that a conducting ideal diode has exactly 0.7 (or 0.65, or 0.6, or whatever) volts across it, because you'll then have two identical voltage sources in series, but in opposite directions, so the overall voltage comes out to be zero.

With real diodes, the forward voltage depends (slightly) on the forward current, and the two diodes in this circuit have different currents through them, so the voltage at \$V\$ will come out to be a few millivolts, though small enough to ignore for many purposes.

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