Can I use the analogy of combination of two batteries to get the potential difference between two points(by applying Kirchhoff's law) in case ,if a pn junction diode is used in the place of other battery?

Consider a simple case shown in figure,there in situation 1 potential due to two batteries are added up and in situation two the the second battery is replaced by a diode in forward bias, if I apply Kirchhoff's law then moving from point AI am gaining a potential of 2V and a gain in potential V(s) in diode . In doing so what is the basic mistake I am doing?
As per Wikipedia in forward bias net voltage = 2V-Vs, but if I follow the above approach iam getting 2V+Vs. I think about this a lot but could not get any idea.

schematic representation of my question

  • $\begingroup$ Hi, strictly speaking that's not a homework question. It is big doubt of mine $\endgroup$
    – JM97
    Jan 2, 2016 at 12:36
  • $\begingroup$ Hello, and welcome to Stack Exchange. It isn't clear what you're asking here. Perhaps if you transcribed and elaborated the text in your image, you'd have a better chance of getting a good answer. $\endgroup$ Jan 7, 2016 at 3:39
  • $\begingroup$ @DanielGriscom I have transcribed and elaborated the text in the image , hope this helps. $\endgroup$
    – JM97
    Mar 13, 2016 at 1:02
  • $\begingroup$ You need to read the answers to the question "pn junction voltage drop?" physics.stackexchange.com/q/86843 $\endgroup$
    – Farcher
    Mar 22, 2016 at 7:18
  • $\begingroup$ @Farcher Iam not getting satisfactory answer. $\endgroup$
    – JM97
    Mar 23, 2016 at 9:16

1 Answer 1


See this image from http://www.allaboutcircuits.com/textbook/semiconductors/chpt-2/the-p-n-junction/:

P-N junction

By looking at graph shown, we can say that in forward bias, PN junction is working as passive device (V/I > 0). So, it causes a voltage drop in this case. Thus, net voltage will be 2V-Vs.

When we talk about reversed bias, then net voltage will be 2V+Vs.

  • $\begingroup$ @JM97 my actual field is biology. But as you requested me to answer this, I tried my best. Hope this helps :) $\endgroup$ Mar 24, 2016 at 11:59

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