Why potential of n side is higher in unbiased p-n junction

[Rishav sapahi]

why the potential of n side is higher than that of the p side when no battery is connected to the p-n junction?

 

[MPKU]

Well if there is no battery, the circuit is pretty much useless... The p-n junction is typically associated with a voltage drop across the p-n junction (somewhere around .5-3V for LEDs). However with no battery in the circuit, there will be no current flowing and thus no voltage across the junction, so they should be at the same potential.

 

[Rishav sapahi]

but in my book they are stating that n-side is at higher potential without any explanation

 

[MPKU]

Is the battery attached to the n-side? If so, then yes, the n-side would be at a higher potential and the p-side would be a 0V, since there in no (practically none) current flowing through the diode. This is known as the reverse-bias direction. The voltage at the n-side would be at the same voltage as the battery (assuming it is just connected with a wire).

 

[gneill]

Look up "PN junction depletion region". The natural potential that develops spontaneously across a PN junction is due to the diffusion of charge carriers across the junction. P type material has "holes" in its bond structure that can be "filled" by accepting electrons, while the N side has more than enough conduction electrons to satisfy its bond requirements.

However, every electron that crosses over leaves a small net positive charge in the N material, while the P side gains a small negative charge. These charges build up until they suppress further movement --- they raise a potential barrier against further diffusion of charges. This is what leads to the "turn on" bias requirement to get current moving across the junction (about 0.7V for silicon PN junctions).

 

[Rishav sapahi]

so Is high potential just means that given side is more positively charged than other

 

[gneill]

so Is high potential just means that given side is more positively charged than other

Yes. A potential develops across the junction.