Pn junction under forward bias

  • #1
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If two batteries are connected such that positive terminal of one is in contact with negative terminal of other then we add the potentials so the potential will then be their sum.

Similarly if a battery is connected to a pn junction diode such that positive terminal is connected to negative side of junction potential (ie.connected to p type) then why don't we add their potentials? Isn't junction potential similar to the second battery as in the first case?

Please explain it at level of a high school student.
 

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  • #2
BvU
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why don't we add their potentials
Who says we don't ? Certainly in the case the other side of the diode is hanging in the air, adding the potentials is just fine. But things only become interesting once that other side is connected to something and current starts flowing. If the depletion layer thickens, the diode blocks and if it gets thinner, it conducts. Read on, McDuff !
 
  • #3
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Who says we don't ? Certainly in the case the other side of the diode is hanging in the air, adding the potentials is just fine. But things only become interesting once that other side is connected to something and current starts flowing. If the depletion layer thickens, the diode blocks and if it gets thinner, it conducts. Read on, McDuff !
But since it is the opposing potential generally we subtract them.
 

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  • #4
BvU
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you sure about that ? :wink:
 
  • #5
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you sure about that ? :wink:
if we forward bias it, +ve terminal of battery gets - potential of diode and -ve of battery gets +ve potential of diode.(-,+) of battery(-,+), this doesn't seem to be a barrier to the battery.
 
  • #6
BvU
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Nothing happens until the other side of the diode is also connected.
Check the current starts flowing video.
 
  • #7
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if we forward bias it, +ve terminal of battery gets - potential of diode and -ve of battery gets +ve potential of diode.(-,+) of battery(-,+), this doesn't seem to be a barrier to the battery.
 

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  • #8
BvU
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if we forward bias it, ... -ve of battery gets +ve potential of diode.(-,+) of battery(-,+)
and a lot of things happen ! The loose pn junction was steady state thanks to this small potential difference. That is now reduced and inverted by the battery so charge starts hopping over the pn boundary with gusto. The boundary goes thinner and etc.

See the video (are you such a quick viewer ?)
 
  • #10
BvU
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Any further questions ? Methinks you understand how it works, right ?
 
  • #11
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Any further questions ? Methinks you understand how it works, right ?
please check new reply. :)
 
  • #12
BvU
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I only see a yes.png. Does that mean: yes I understand ?
 
  • #13
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I only see a yes.png. Does that mean: yes I understand ?
That means, in the 1st video she said we have E(field) from n side to p side. We apply larger E( from battery) against it (during forward biasing).
In 2nd video, she uses +ve potential at p and -ve potential at n side. No reason provided. Back to original question.
As I said earlier barrier potential is +ve on p side.
(Since last 10 hours I am stuck on this.)
 
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  • #14
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That means, in the 1st video she said we have E(field) from n side to p side. We apply larger E( from battery) against it (during forward biasing).
In 2nd video, she uses +ve potential at p and -ve potential at n side. No reason provided. Back to original question.
As I said earlier barrier potential is +ve on p side. (reverse of potential used in your mage earlier)
(Since last 10 hours I am stuck on this.)

1 https://electronics.stackexchange.c...tion-region-which-side-is-at-higher-potential
2 https://electronics.stackexchange.c...e-a-battery?noredirect=1#comment909858_375761
3 https://electronics.stackexchange.com/questions/256295/pn-junction-under-forward-bias/256303

related and helpful links
 
  • #15
See Wikipedia:
Valance and condition bands.
 

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