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I was studying Diodes and I came across the information that when the diode is reversed biased the depletion region is increased. Now the depletion region is composed of neutral atoms that are combined and some layer is formed that further halts the movements of electrons and holes. When the diode is forward biased the depletion region is shrinked and when the diode is reversed biased the depletion region is widened as more neutral atoms are created near the layer. This is perfectly understandable why this phenomena happened. The next thing is the diode required some voltage in order for the electrons to cross the depletion region after biasing. This voltage is 0.7 volts or nearly 1 volts in case of silicon. When the depletion region is widened obviously more voltage is required by electrons to cross the depletion region.

So my question is that when the biasing is off (Electric source) what happened to the depletion region after reverse biasing which cause the depletion region to be widened, will it come back to the normal where it was initially and why and by which mechanism it will come back, or if it was not suppose to come back to the normal then it means that voltage required to cross the depletion region will be changed depending upon the thickness ?

Most electronics book (Which I studied) mentioned the phenomena of forward and reversed biased and mention their change on the deplition region but didnt mention that when the biasing is off after applying the reverse or forward biasing then what would happened.

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  • \$\begingroup\$ No biasing (biasing off) is the same as biasing being present but with a voltage of 0 volts. \$\endgroup\$ – Andy aka Jan 12 '18 at 14:03
  • \$\begingroup\$ But what would happened to depletion region when biasing is Off ? \$\endgroup\$ – Muhammad Usman Jan 12 '18 at 14:51
  • \$\begingroup\$ This voltage is 3 volts or nearly 3 volts in case of silicon. Oh, and there is me thinking all silicon diodes have a forward voltage of around 0.6 to 0.8 V. <Measuring a silicon diode in forward mode> yep, still 0.7 V in this part of the universe. Where does your 3 Volts come from? \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 14:54
  • \$\begingroup\$ Most books will start the discussion of a PN juntion in the unbiased situation, so perhaps you should go back a bit in your book to the section about PN junctions but just before the effect of external biasing is discussed. \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 14:57
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    \$\begingroup\$ It was mistake for 3 volts I corrected it. But the question is the depletion region after the external biasing not before. \$\endgroup\$ – Muhammad Usman Jan 12 '18 at 15:00
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Depletion region exists without external connections or as well with external 0V bias. It's a balance state caused by different electron energy state structures in different materials and thermal motion (=diffusion).

Reverse bias widens the depletion region. We can easily think that nothing happens if we simply remove the connection to external voltage supply, the depletion region stays widened because no charge is inserted nor taken off. But nothing is fully ideal. There are minority carriers and some leakage current. The situation returns to 0V balance - not in a second in every case, but finally it happens.

I have met junction FETs where the gate charge stays at least several minutes. That means the depletion region is several minutes substantially wider than in 0V state. In diodes the depletion region isn't practically observable, but in junction FETs it is. Simply connect a resistance meter between drain and source. Then apply a voltage between the gate and source, minus to the gate if the FET is N-channel type. Disconnect the gate, the resistance between drain and source can be high for several minutes until leakage currents discharge the gate.

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  • \$\begingroup\$ I rephrased it, Please check again \$\endgroup\$ – Muhammad Usman Jan 12 '18 at 16:32
  • \$\begingroup\$ I understand bro I receive the similar reactions that's why I realize that Question requires more clarity therefore I edit. If you help me out with your knowledge with this one, I will be very thankful. \$\endgroup\$ – Muhammad Usman Jan 12 '18 at 16:40
  • \$\begingroup\$ I am very thankful for help but the part of question is still needs some answer that OK the situation returns to 0 volt but by which phenomena because if the depletion region shrinked by itself (to 0 Volts after some time) then there is a question that why 0 volt thickness is not shrinking more to become negligible. \$\endgroup\$ – Muhammad Usman Jan 12 '18 at 16:52
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    \$\begingroup\$ @MuhammadUsman You should see that there's two fighting things. Thermal statistics (=everything sooner or later gets kicked up from the rest state if material has non-zero K temperature) and the attraction of the electrons to fall to as low energy state as possible in the atomic structure. The depletion region is the consequence of the balance between these fighting forces. Derivation of the formulas need statistical physics, electrodynamics and diffusion differential equations. \$\endgroup\$ – user287001 Jan 12 '18 at 17:34
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By "remove bias," do you mean, turn down the power supply voltage connected to the reversed diode?

Or do you mean, suddenly disconnect the diode entirely?

A reverse-biased diode behaves as a charged capacitor (and is used as 'varicap' diodes in rf tuners.) If we suddenly disconnect the diode from the circuit, then the depletion region initially stays wide, the voltage across the diode stays initially high, and the "capacitor" remains charged.

But the depletion region is an imperfect dielectric, with leakage caused by thermally-generated electron/hole pairs. So, the "capacitor" will self-discharge, and the depletion region will rapidly shrink with time. Someone correct me, but I recall that it's changing with a constant di/dt slope rather than exponential, since the shrinking depletion zone produces a "constant current" effect, rather than constant resistance. At least initially, when Vd is far from zero.

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  • \$\begingroup\$ Hi wbeaty, the depletion region is formed of neutral atoms, not the ionized one, and also please can you give any reference. Thanks \$\endgroup\$ – Muhammad Usman Jan 13 '18 at 9:53
  • \$\begingroup\$ @MuhammadUsman check any semicon physics text: reverse-biased pn junctions, the origin of leakage current. (Also see: temperature dependence of intrinsic currents.) Depletion regions are not neutral atoms. If they were, they'd have infinite resistance, zero leakage current, and any charges placed on floating diode terminals would persist forever. (Note that technician textbooks may claim that reverse-biased diodes have zero leakage current. They're wrong, see engineering texts instead.) \$\endgroup\$ – wbeaty Jan 15 '18 at 5:59
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If no voltage is applied across the diode, then the built-in potential (typ. around 0.7V) is still there across the depletion region.

The reason you can't see this built-in potential is because the metal-contacts to the semiconductor also have a bias voltage that cancel out the built-in potential.

This is why you will not be able to measure a current if you short-circuit a diode.

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  • \$\begingroup\$ Can you explain more by what you mean when you say the metal contacts cancel out the potential? \$\endgroup\$ – evildemonic Jan 12 '18 at 15:49
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    \$\begingroup\$ @evildemonic When you form a metal-semiconductor junction you form another type of diode. This diode has a built in potential as well. If you have a p-n junction and two metal-semiconductor junctions you have a total of three diodes. The sum of all their built in potentials is zero. \$\endgroup\$ – Matt Jan 12 '18 at 16:22
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    \$\begingroup\$ If you were to sandwich a PN junction between equally doped semiconductors, you'd see that the built in potential gets canceled right? Well, when connecting metal contacts something similar happens. \$\endgroup\$ – Sven B Jan 12 '18 at 16:49
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Diode forward voltage Diode forward current

0.7v 10mA

0.682v 5mA

0.664v 2.5mA

0.646v 1.25mA

0.642v 1mA

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