Why the voltage across an ideal pn junction diode remains 0.7 volts, even when the applied voltage exceeds 0.7 volts? Shouldn't it be the applied voltage minus the 0.7 V?
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By definition, the voltage across an ideal junction diode cannot be greater than 0.7V. The operating point of an ideal diode must always lie somewhere on the red line in the graph of voltage vs. current below:
Either the voltage is exactly 0.7V, in which case, the current can be any non-negative value, or else the voltage is less than 0.7V, in which case the current must be zero.
If you place an ideal, 0.7V diode in parallel with an ideal constant voltage source whose value is greater than 0.7V, then the behavior of the circuit is undefined.
The voltage vs. current graph for a real diode is different:
If you apply a voltage less than the diode's $V_f$ (0.7V in the picture) then a tiny amount of current will flow, but as you raise the voltage above 0.7V (which you can do) then the current starts to rise very quickly.
Very quickly. If you raise the voltage by more than a fraction of a volt above the diode's $V_f$ then there probably will be enough current to destroy the diode.
In the negative quadrant (which is shown at different voltage and current scales from the positive quadrant), there will be a tiny "leakage" current when the diode is reverse biased. I don't know numbers, but it will be microamperes or nanoamperes.
If you increase the reverse voltage to the diode's "breakdown voltage," (shown as -100V in this case), then it will be just like the forward case, the current will start to rise very quickly. (But note! Operating in the reverse breakdown region may damage the diode even if the current is limited.)
