Why Don't We Add Potentials in a Forward-Biased PN Junction?

AI Thread Summary
In a forward-biased PN junction, the positive terminal of a battery is connected to the p-type material, while the negative terminal connects to the n-type material. This configuration does not simply add potentials like in a series battery connection because the junction potential acts as a barrier that opposes current flow. When the battery is applied, it reduces this barrier, allowing current to flow once the depletion layer becomes thinner. The key point is that the effective potential difference is influenced by the junction's characteristics and the external battery, leading to a complex interaction rather than a straightforward addition of voltages. Understanding this interaction is crucial for grasping how diodes function in circuits.
Topsykret
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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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Topsykret said:
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, https://www.electronics-tutorials.ws/diode/diode_2.html 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 !
 
BvU said:
Who says we don't ? Certainly in the case the other side of the diode is hanging in the air, https://www.electronics-tutorials.ws/diode/diode_2.html 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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https://www.electronics-tutorials.ws/diode/diode4.gifyou sure about that ? :wink:
 
BvU said:
https://www.electronics-tutorials.ws/diode/diode4.gifyou 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.
 
Nothing happens until the other side of the diode is also connected.
Check the current starts flowing video.
 
Topsykret said:
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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Topsykret said:
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 ?)
 
BvU said:
Nothing happens until the other side of the diode is also connected.
Check the current starts flowing video.
 

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  • #10
Any further questions ? Methinks you understand how it works, right ?
 
  • #11
BvU said:
Any further questions ? Methinks you understand how it works, right ?
please check new reply. :)
 
  • #12
I only see a yes.png. Does that mean: yes I understand ?
 
  • #13
BvU said:
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
Topsykret said:
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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