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[SOLVED] Confused with PN Junction, Help please!

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Onigece

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In a PN Junction with no bias, why the free Electrons from the N-type material do not come close or be attracted to the positively charged side of the depletion region which is also on the N-type? Isn't it that positive and negative charges attracts each other?
 

When no Bias is applied to the PN junction, the temperature of the depletion region will be low. That's the reason a free electron will not readily fill the hole (hole is actually absence of electron). And when forward biased, the energey level of electrons will increase and the conduction will take place.. Please refer **broken link removed** for a very nice description..

Hope this helps
 
i think when there is no bias/potential difference across the PN junction,the electrons does't overcome the barrier ,but when the pd is applied the movement of electrons to holes accelerate.
 
in electronics many times what they teach u is partially correct/ or wrong and when u study further they say we actually taught u wrong this is the correct reason . learning electroncis frm last 4 years and 2 weeks ago i learnt why only " 0.7 v " and not any other value to forward bias a Si pn junction . its actually the work function difference that plays the game .
 
cks3976 said:
Well... Diode equation relates the Diode current with Device temperature. Please have a look at Diode Equation | pveducation.org.. I think my explanation is still valid....
Click to expand...

No, it is not - sorry.
Your diode reference is valid for an external voltage. However, the original question requires no bias.
Read the reference mentioned by me carefully and try to understand.
You will see that diffusion takes place as soon as the p-doped and the n-doped parts come close together - without any external voltage.
Thus, across the pn-junction a so called "diffusion voltage" builds up (approx. 0.7 volts).
This voltage acts as a diffusion barriere and stopes continuation of the diffusion process.
 
when there is no bias the the electrons and holes will move through the junction(because of diffusion),but as they move to the other side an electric field will start to be produced and this field will create a force against the the electrons movement .this field get stronger by moving more electrons , until the time that the electric field will become as strong as that it will not permit more electrons to move to other side.unless we apply an external electric field against the intrinsically produced electric field to weaken it and let electrons pass the junction.
hope to be useful
 
Thank you is_razi - your explanation is better than mine in posting #7.
Finally - at the end of the diffusion process - there is a equilibrium between two forces: The force caused by the electric field (in one direction) and the force caused by the "wish" of the free electrons and holes (something like a pressure) to move to the other shore.
 
Oh yes.... that's crystal clear now :) Thanks is_razi & LvW
 
The diffusion of negative charge from N to P make the depletion region, this also make hole region in n type and electron region in p type. Now, why electron from P does no come to positive region of N type? If it is attached back, it’s same with the 2 original P and N ( whole electrons back to N and whole holes back to P). The natural diffusion will be stopped when hole region in N is large enough to make the barrier that does no permit diffusing.
File:pn-junction-equilibrium.png - Wikipedia, the free encyclopedia
And electrons from the right hand side is not attracted to positive region because there does not have a boundary, you can get it through the concentration of h and e change uniformly. Therefore at that “ boundary” does not have a real attraction.
regard.
 
Onigece said:
In a PN Junction with no bias, why the free Electrons from the N-type material do not come close or be attracted to the positively charged side of the depletion region which is also on the N-type? Isn't it that positive and negative charges attracts each other?
Click to expand...

The impurity atoms are present uniformly throughout the device (assuming uniform doping). Since the concentration of holes (electrons) in p+(n) region near the junction is less due to diffusion, they become uncompensated and therefore give rise to electric field. Holes (electrons) in deeper p+(n) regions does not feel attaction forces as all the flux lines (equal to charge) emanting from the one type of uncompensated ions to the other type.
 

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