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In case of an n-p-n transistor, the Vbe is more like voltage across a p-n diode. Now, since the direction of electric field in the depletion-region is from n to p, how come the voltage across the p-n junction is +0.7 V and not -0.7V?

In other words, why n-region is not at a higher potential (as it should be from the diagram below)?

A p-n junction

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  • \$\begingroup\$ The voltage drop is 0.7V. This matches your observation, rather than refuting it. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 21 '14 at 7:36
  • \$\begingroup\$ Long story short, 0.7 V is the external voltage required to counter this built in potential ( AKA junction potential ). Naturally it will be in opposite direction to that of junction potential. \$\endgroup\$ – Plutonium smuggler Sep 23 '14 at 0:45
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Short summary,

  • A diode can be forward or reversed biased. This concept only makes sense in the presence of external voltage bias.
  • The intrinsic potential difference of the p-n junction is -0.7V. The negative sign is by convention as it goes against turning on/forward biasing the diode.
  • By applying an external bias of +0.7V we counteract the intrinsic potential difference. The positive sign of the bias is by convention as it helps turning on the diode.

In detail,

  • p-n junction diodes are one of the most important Silicon building blocks, so they come with their own set of conventions:

When a diode is externally biased by applying a more POSITIVE potential to the p end of the junction than to the n end of the junction, we say that it is FORWARD biased and we define that bias voltage with a POSITIVE sign.

When a diode is externally biased by applying a more NEGATIVE potential to the p end of the junction than to the n end of the junction, we say that it is REVERSED biased and we define that bias voltage with a NEGATIVE sign.

Following the previous, we say than the forward current is POSITIVE and the reverse current is NEGATIVE. A positive current is defined as the current in which the holes (h+) move in the same sense and/or when the electrons (e-) move in the opposite sence.

  • You can think/visualize the p-n junction diode as having an internal or intrinsic potential difference of -0.7V (referenced from p to n). This will help you when analyzing any circuit!
  • Without external biasing, you can think/visualize the p-n junction as being "intrinsecally reversed".
  • Now, if you overcome the intrinsic potential difference with an external positive voltage bias higher that 0.7V, you will have reduced to zero or reversed the intrinsic potential difference, effectively turning on the diode and making it forward biased.

Note that, in electronics, the way voltage and current signs are defined is simply a convention,

  • It's key to internalize these conventions as soon as possible during your learning process, as most people in the field, textbooks and articles adhere to these conventions.
  • It is a sign of deeper interest/curiosity/knowledge trying to understand the rationale behing some of these conventions. It'll surely will help you to step on more advanced concepts in the future. However, note that some of these conventions have no deep rationale behind, instead are a mixture of historical reasons or, in some cases, were defined arbitrarily.
  • Communication is a process inherent to human beings so, unless you want to avoid communication with others I suggest you to follow these sort of pre-established "rules".
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A diode(eg. 1N4007) can drop 0.7V across it when it is forward biased. And when it is reverse biased it drop 0V. So, -0.7V is not possible.

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