0
\$\begingroup\$

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:

enter image description here

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?

\$\endgroup\$
  • 1
    \$\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
  • 1
    \$\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
2
\$\begingroup\$

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).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.