I don’t know whether there is already a question and answer to this but I couldn’t find one yet. Regarding the following information from a text:

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The equation for emitter reverse saturation current Ies shows that the Ies is directly caused by Vbe in non-linear way. But where does -1 in the parenthesis come from?

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    \$\begingroup\$ Without the "-1", what would be the current with 0 applied voltage? Do you think that would be physically possible? \$\endgroup\$ – The Photon May 15 at 18:34
  • \$\begingroup\$ No not possible. So is that the reason to zero the left side when Vbe is zero? \$\endgroup\$ – panic attack May 15 at 18:35
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    \$\begingroup\$ More importantly, what would be the current if the \$V_{be}\$ were just slightly negative? What would that mean about the power consumption of the device? \$\endgroup\$ – The Photon May 15 at 18:37
  • \$\begingroup\$ @ThePhoton Sadly I don’t have an answer to this one:( \$\endgroup\$ – panic attack May 15 at 18:39
  • \$\begingroup\$ Try drawing the I-V curve of the device with the "-1" removed. What is the difference between quadrants I and III of an I-V graph vs quadrants II and IV? \$\endgroup\$ – The Photon May 15 at 18:46

That is the Schockley Diode Equation, if you want to look it up. If you're an EE student sit up and pay attention in your solid-state quantum mechanics class -- you'll end up deriving this one there.

The 1 comes from the rate of electron-hole pair generation in the junction; it is constant over voltage (although not over temperature -- I was in the industry for about 5 years before I realized that \$I_{ss}\$ is temperature dependent. I felt cheated that I was not told sooner).

The \$e^\frac{V_{BE}}{V_T}\$ part is from the rate of recombination of hole-electron pairs; it is exponential with voltage, and exactly matches electron-hole pair generation at \$V_{BE} = 0\$ (as one would hope it would).


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