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This question already has an answer here:

What will happen to the current if a diode is placed the wrong way round?

Should the current be 0?

I need to measure the current when the diode is the wrong way round.

Thanks

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marked as duplicate by winny, Olin Lathrop, JRE, pipe, Wesley Lee Aug 9 '17 at 14:52

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    \$\begingroup\$ Take a couple of minutes to look up the function and operation of a diode, there are a few graphs that will be informative to you! \$\endgroup\$ – MrPhooky Aug 9 '17 at 7:20
  • \$\begingroup\$ Take care that the battery voltage does not exceed diode reverse breakdown voltage. Easy enough with a 1N4148, not so easy with a LED. Some of the latter will take only a few volts. \$\endgroup\$ – Neil_UK Aug 9 '17 at 7:31
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    \$\begingroup\$ @Ben: What is the wrong way? Sometime your wrong way is the right way (photodiodes, zener diodes). I think what you mean is commonly called reverse. \$\endgroup\$ – Curd Aug 9 '17 at 7:38
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    \$\begingroup\$ I'm sorry, but this really "Doesn't show any research effort" in my view. \$\endgroup\$ – pipe Aug 9 '17 at 12:08
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    \$\begingroup\$ Exactly what kind of diode are you using? \$\endgroup\$ – The Photon Aug 9 '17 at 14:21
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Diodes have a small leakage current when reverse biased and that is the current you will measure - leakage current: -

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  • \$\begingroup\$ I have looked it up. And when a diode is reversed the current is 0. Because it breaks the circuit. \$\endgroup\$ – Kyle Anderson Aug 9 '17 at 11:41
  • \$\begingroup\$ Does 53.291uA sound like a leakage current? \$\endgroup\$ – Kyle Anderson Aug 9 '17 at 11:42
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    \$\begingroup\$ @Ben That sounds very much like a leakage current. \$\endgroup\$ – Harry Svensson Aug 9 '17 at 12:07
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    \$\begingroup\$ 50 uA seems like a very large leakage current to me, at least for a standard signal or low-power rectifier. For a Schottky it would not be surprising. \$\endgroup\$ – The Photon Aug 9 '17 at 15:01
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Normally there will be reverse leakage when a diode is reverse biased. Due to the increased potential, holes on the n side of semiconductor are dragged to negative terminal of the battery and electrons on the p side is dragged to the positive terminal of the battery. This constitute a reverse a current which is a also called as reverse saturation current. This is because even though increase in reverse potential, current will not be increased that much for a certain limit. This leakage current varies with diodes according to the spec sheet.

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