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What I am talking about can be seen in this graph of current against voltage of a silicon diode. Why does the silicon diode have varying properties depending on the voltage across it? (regions of $0.60-0.65~\mathrm{V}, 0.65-0.75~\mathrm{V}, <0.75~\mathrm{V}$)

enter image description here

Thanks in advance

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  • $\begingroup$ I edited your title to better represent the question. $\endgroup$ – garyp Sep 3 '16 at 12:27
  • $\begingroup$ Both @MikeDunlavey and I are curious about this graph. Where did it come from? I though it was meant to be a rough sketch of diode behavior, as it is not quite right in the details. But as MikeDunlavey points out, it might be some other device. $\endgroup$ – garyp Sep 3 '16 at 15:27
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What I understand about diodes is that the current is exponential in the voltage**, and the changing slope is the changing conductance. It's no more complicated than that.

If I'm right about that, your plot is not familiar to me, both because it goes to zero on the left, and is linear on the right. So it appears to be for some device other than a simple diode.

** voltage as measured by a voltmeter across the diode.

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  • $\begingroup$ Hi, thanks for your input first of all. Do you mind explaining to me what imposed voltage means? Is that the region where the gradient is not constant or 0? I have never come across that term before.. I have posted an answer that I think covers my question. Thanks for your input though :). $\endgroup$ – nyxaria Sep 3 '16 at 14:30
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I have found the answer to my question on this website, and it goes as follows:

                                              enter image description here

In the forward direction (forward biased) it can be seen that very little current flows until a certain voltage has been reached. This represents the work that is required to enable the charge carriers to cross the depletion layer. This voltage varies from one type of semiconductor to another. For germanium it is around 0.2 or 0.3 volts and for silicon it is about 0.6 volts.

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    $\begingroup$ That explanation is ok as far as it goes. The graph is not quite right, or does not show enough detail to be right. The correct curve does not go through the origin with zero slope, and there is current through the diode when the voltage is negative. $\endgroup$ – garyp Sep 3 '16 at 15:29
  • $\begingroup$ @garyp if you read the website, you would see that the graph does in fact show both of these things. $\endgroup$ – nyxaria Sep 11 '16 at 20:16
  • $\begingroup$ The graphic there is the same as the graphic here. Neither point is shown clearly enough to make it evident. $\endgroup$ – garyp Sep 12 '16 at 0:09

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