L.E.Ds producing voltage without power supply except for white light incident on it

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  • #26
chroot
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ryan750 said:
oh yes and jus to clear up - when the valence electron moves thru the conduction band to fill the positive holes in the valence band of the p - type semiconductor - the excited electron relaxes and emits light yes.
The electrons emit photons as the slide down the potential hill in the junction. The energy they gain in "rolling down the hill" is converted to light.

There is no relaxation involved, since the electrons in question are not bound to any atom. They are, approximately, a free electron gas.

- Warren
 
  • #27
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also last thing - the potential difference occurs in the l.e.d because of the energy changes to the electrons. The electrons move from higher to lower and from lower to higher levels - so these energy changes are the potential difference in the l.e.d.

i know thats basic stuff but i just want to be sure of my definitions for friday
 
  • #28
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chroot said:
The electrons emit photons as the slide down the potential hill in the junction. The energy they gain in "rolling down the hill" is converted to light.

There is no relaxation involved, since the electrons in question are not bound to any atom. They are, approximately, a free electron gas.

- Warren
okay so no relaxation either - man - all of my course terminology has gone so out of the window - but im glad that im gonna get it right. and thankyou for welcoming me - i can't wait to get stuck into all of these rooms and all of the interesting topics out there.
 
  • #29
ZapperZ
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chroot said:
The electrons emit photons as the slide down the potential hill in the junction. The energy they gain in "rolling down the hill" is converted to light.

There is no relaxation involved, since the electrons in question are not bound to any atom. They are, approximately, a free electron gas.

- Warren
Er.. hang on. The created photons in LED's are due to the recombination process, i.e. electron recombining with holes. Electrons rolling downhill does not create any photons. This is a continuous energy that is absorbed by the lattice.

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/led.html

Zz.
 
  • #30
chroot
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Sorry, ZapperZ, I stand corrected. That makes complete sense, now that I think about it. If the electrons emitted as they rolled downhill, the resulting light would not be monochromatic.

- Warren
 
  • #31
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ok - i just need yto clear this subject up once and for all -

the l.e.d has no power supply - just a voltmeter to take readings with.

so if there was no power supply why would the electrons move from the p - type material to the n - type material when this type is more negative right. As in the response that Zz gave he said they they would 'see' or be attracted by the E field - which i assume isnt there without a power supply.

i thought that it would be that valence electrons in the n type are excited and move into the conduction band which enables them to move freely to the holes in the p type semiconductor due to their positive attraction.
 
  • #32
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ryan750 said:
ok - i just need yto clear this subject up once and for all -

the l.e.d has no power supply - just a voltmeter to take readings with.

so if there was no power supply why would the electrons move from the p - type material to the n - type material when this type is more negative right. As in the response that Zz gave he said they they would 'see' or be attracted by the E field - which i assume isnt there without a power supply.
No no! The electric field in the DEPLETION zone is there whether you supply power to it or not! It exists due to the migration of holes and charges when the PN semiconductors come in contact with each other. The accumulation occurs until the coulombic forces prevents any further migration and an equilibrium condition is set up. There is ZERO applied external power or field here.

This is why I said that I assumed you understand the physics of PN junction in the first place.

Zz.
 
  • #33
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ZapperZ said:
No no! The electric field in the DEPLETION zone is there whether you supply power to it or not! It exists due to the migration of holes and charges when the PN semiconductors come in contact with each other. The accumulation occurs until the coulombic forces prevents any further migration and an equilibrium condition is set up. There is ZERO applied external power or field here.

This is why I said that I assumed you understand the physics of PN junction in the first place.

Zz.
sorry - so electrons from type move top n tyoe and holes move from n tyoe to p type.

can i just ask what holes are? i know they are positive but thats it.
 
  • #34
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ryan750 said:
sorry - so electrons from type move top n tyoe and holes move from n tyoe to p type.

can i just ask what holes are? i know they are positive but thats it.
Holes are "bubbles in the electron sea". Rather than trying to write the dynamics of ALL the electrons in the say, you can get all the same relevant info by renormalizing the sea to zero potential and treat the bubbles as positive charges.

I think you need to review the physics of PN junctions.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html

Zz.
 
  • #35
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ZapperZ said:
Holes are "bubbles in the electron sea". Rather than trying to write the dynamics of ALL the electrons in the say, you can get all the same relevant info by renormalizing the sea to zero potential and treat the bubbles as positive charges.

I think you need to review the physics of PN junctions.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html

Zz.
so the light provides the energy to valence electrons in the n type to be excited and they are attracted to the p type and combine with holes. This process carried out produces energy changes to the electrons and holes which creates the potential difference seen.

is this right?
 
  • #36
ZapperZ
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ryan750 said:
so the light provides the energy to valence electrons in the n type to be excited and they are attracted to the p type and combine with holes. This process carried out produces energy changes to the electrons and holes which creates the potential difference seen.

is this right?
Er.. no!

First of all, you need to understand PN junction! This is INDEPENDENT of LED and the photodiode condition that you asked at the beginning of this thread. You are now mixing the physics of PN junction with the physics of that question you asked AND the emission of light in LEDs. It has gotten utterly confusing.

The depletion zone (and the build up of internal E-field) occurs simply because the p-type and n-type semiconductors are in contact with each other. Whether you are going to use this for LED's, or photodiode, is irrelevant.

In the photodiode, the electron that migrates does NOT recombine with the hole! If it does, this is creates light and thus, the LED situation. This is not what you're asking for and it is why I said you're mixing different stuff into one scenario.

I don't want to cut-and-paste what I have said earlier, but I'm going to:

1. In the P-type semiconductor, an electron is excited from the valence band into the conduction band, leaving behind a positive hole in the valence band. If this is close to the depletion zone, it sees an electric field that will force it into the N-type semiconductor (similar to a forward bias). The hole, on the other hand, stays in the P-type. As more of these occur, there will be an additional accumulation of holes in the P-type and more electrons in the N-type.

2. In the N-type semiconductor, the same excitation occurs, but this time, it is the holes that see an E-field that will cause it to migrate over to the P-type. The electrons stay in the N-type.

The combination of 1 and 2 will force the accumulation of a potential bias (similar to a forward bias) between the P and N-type semiconductors. It is only sustainable with continued light source. If you cut the light source, the equilibrium is destroyed.
Notice that I said nothing about any recombination of electrons and holes. Such process is the source of light in LED's and that's not what I was describing! Instead, it is the accumulation of charges on each side of the PN junction that is the source of the potential difference that you detected.

Zz.
 

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