Understanding the Mechanics of Light-Induced Voltage in L.E.D.s

In summary: LED as the electrons make their transitions.Does ur explanation allow for this - do u mean that the electrons released would move to the holes in the p type material.I don't see anything wrong with my explanation.
  • #1
ryan750
23
0
hello - i feel confident that i understand the principles of how L.E.D's work. I can grasp the concept of the depletion zone and how this leads to conduction in one direction.

But how does light from an ordinary filament lamp cause a voltage in an L.E.D to be produced when the L.E.D has no obvious power supply?

I know of the principles of the photoelectric effect and realize it has something to do with this but need a concise explanation of how the voltage is produced.

i have been discussing this with sum1 else and have come up with this explanation:

light from the lamp when incident on the semiconductor in the L.E.D causes valence electrons in the n type semiconductor to be released into the conduction/ depletion layer. these are attracted to the p type semiconductor and they move into the 'holes'. this causes a drop in energy level for the electron which is called relaxation and causes light to be emitted.
this process means that a current is flowing from the n to the p type semiconductor which means that a voltage is present in this mini - circuit.

now the fact that white light has a low photon energy means that if it were the photoelectric efect that causes this process - the material in the L.E.D would need to have a low work function or sum other proerty that allows this to happen.

anyone who can correct any details - add more information - or completely change the explanation to the question - please feel free

thanks
 
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  • #2
If passing a current through a diode creates light, then shining light on a diode should create current. Most physical processes, like this one, are symmetric.

When you shine light on a diode, the photons kick electrons out of atoms, creating an electron-hole pair (this is the photoelectric effect). When this occurs in the depletion region (the area arround the p-n junction which is normally devoid of charge carriers), the hole is rapidly swept away to the p-type semiconductor, and the electron to the n-type semiconductor. This is a current.

- Warren
 
  • #3
brilliant - will light be emitted during this process though?
 
  • #4
I would guess that the light-induced current would, indeed, cause the diode to produce some additional light, but it would be a very tiny fraction of the incident light and probably would be negligible.

- Warren
 
  • #5
chroot said:
I would guess that the light-induced current would, indeed, cause the diode to produce some additional light, but it would be a very tiny fraction of the incident light and probably would be negligible.

- Warren
well the thing is - i have seen this experiment done and the light level produced by the diode seemed to be at a relatively high intensity that u would expect from an L.E.D in a circuit. so this would mean that the voltage is being created by a process that has a large amount of relaxation occurring. Does ur explanation allow for this - do u mean that the electrons released would move to the holes in the p type material.
 
  • #6
So ideally you would want monocromatic light of the same frequency as that that is normally emitted from the LED. That way, when you shine the light onto the LED it hits the gas contained inside and the atoms absorb the photons and the electrons transition up to higher orbitals. They then cascade down, and release a photon of equivalent wavelength which is what induces a current. Just trying to see if that's how they work.. I never really thought about it, and I've done very little work with electricty, only work funtions and general magnetism and ac circuits.

Edit: the way my explanation goes, beacues the photons are quantized in energy, you would see light emitted from the LED as the electrons make their transitions.
 
  • #7
ryan750 said:
Does ur explanation allow for this - do u mean that the electrons released would move to the holes in the p type material.
I don't see anything wrong with my explanation. No, the released electrons move toward the n-type, and the holes move toward the p-type. Remember that in a forward-biased pn junction, the p-type has the more positive potential, and the electrons are moving due to the externally-applied electric field.

- Warren
 
  • #8
Hybird said:
So ideally you would want monocromatic light of the same frequency as that that is normally emitted from the LED. That way, when you shine the light onto the LED it hits the gas contained inside and the atoms absorb the photons and the electrons transition up to higher orbitals. They then cascade down, and release a photon of equivalent wavelength which is what induces a current. Just trying to see if that's how they work.. I never really thought about it, and I've done very little work with electricty, only work funtions and general magnetism and ac circuits.

Edit: the way my explanation goes, beacues the photons are quantized in energy, you would see light emitted from the LED as the electrons make their transitions.

gases arent involved. it is a semiconductor consisting of two doped materials that form a p - n junction.
 
  • #9
Hybird said:
So ideally you would want monocromatic light of the same frequency as that that is normally emitted from the LED. That way, when you shine the light onto the LED it hits the gas contained inside and the atoms absorb the photons and the electrons transition up to higher orbitals. They then cascade down, and release a photon of equivalent wavelength which is what induces a current. Just trying to see if that's how they work.. I never really thought about it, and I've done very little work with electricty, only work funtions and general magnetism and ac circuits.

Edit: the way my explanation goes, beacues the photons are quantized in energy, you would see light emitted from the LED as the electrons make their transitions.
There's no gas in a diode, and the rest of your explanation is pretty much entirely wrong. Please don't post on topics with which you are unfamiliar.

- Warren
 
  • #10
chroot said:
I don't see anything wrong with my explanation. No, the released electrons move toward the n-type, and the holes move toward the p-type. Remember that in a forward-biased pn junction, the p-type has the more positive potential, and the electrons are moving due to the externally-applied electric field.

- Warren

no chroot. there is no electric field supplied. It is simply an L.E.D connected to a voltmeter. then a normal filament lamp is shone onto the L.E.D and a voltage is read on the meter - this increases with decreasing distance of the filament lamp to the L.E.D ie. increasing light intensity. i know that this will follow an inversely proportional relationship by the Intensity equation of a point source.

so no power supply. just the movement of the released electrons causes the current flow and hence voltage created. how do u explain that?
 
  • #11
ryan750 said:
so no power supply. just the movement of the released electrons causes the current flow and hence voltage created. how do u explain that?
Sorry, I misread the question. If the pn junction is in equilibrium, then yes, the electrons will "roll down-hill" into the n-type semiconductor, while the holes will "roll up-hill" into the p-type semiconductor.

- Warren
 
  • #12
chroot said:
There's no gas in a diode, and the rest of your explanation is pretty much entirely wrong. Please don't post on topics with which you are unfamiliar.

- Warren

Take it easy there man, I was asking a question not trying to prove someone wrong. Ya know ask, so then maybe I will understand..
 
  • #13
Hybird said:
Take it easy there man, I was asking a question not trying to prove someone wrong. Ya know ask, so then maybe I will understand..
There wasn't a single question mark in your entire post. It sounded as though you were making a statement.

- Warren
 
  • #14
chroot said:
Sorry, I misread the question. If the pn junction is in equilibrium, then yes, the electrons will "roll down-hill" into the n-type semiconductor, while the holes will "roll up-hill" into the p-type semiconductor.

- Warren

wot explains the low work funciont if the material in the diode
 
  • #15
I'm not sure I understand why you think anything is special about the work function.

- Warren
 
  • #16
chroot said:
I'm not sure I understand why you think anything is special about the work function.

- Warren


the work function would be low because this preocess occurs in a l.e.d with visible light - how dus the material hav sucj a low work function
 
Last edited:
  • #17
ryan750 said:
well that's what the photoelectric effect depends on - why would ithave no relevance here?

the work function would be low because this preocess occurs in a l.e.d with visible light

Because what you described is not the photoelectric effect, but rather a "photodiode" characteristics. What's the difference? A photoelectric effect is defined as the process whereby electron emission INTO THE VACUUM STATE occurs due to absorption of light. The problem you described is the formation of a CURRENT or a potential different IN THE MATERIAL due to absorption of light. The electrons are not released into the vacuum state, but rather went from the valence to the conduction band. This process simply overcomes the band gap, and has nothing to do with the work function, the latter is defined as the energy difference between the top of the valence band and the vacuum state.

Zz.
 
  • #18
ZapperZ said:
Because what you described is not the photoelectric effect, but rather a "photodiode" characteristics. What's the difference? A photoelectric effect is defined as the process whereby electron emission INTO THE VACUUM STATE occurs due to absorption of light. The problem you described is the formation of a CURRENT or a potential different IN THE MATERIAL due to absorption of light. The electrons are not released into the vacuum state, but rather went from the valence to the conduction band. This process simply overcomes the band gap, and has nothing to do with the work function, the latter is defined as the energy difference between the top of the valence band and the vacuum state.

Zz.

ok - u seemed pretty clued up about this subject: please could u give a concise account to this question - it would be greatly appreciated - even if it means repeating some of the details that have been said on this thread already - pls:

Light from a filament lamp is incident on a l.e.d that is connected to a voltmeter only.

1. explain how the voltage is generated in the l.e.d

could u give as detailed response as possible pls - sorry to be so demanding but its just that i am to be examined on this topic on friday. thankyou so much
 
  • #19
ZapperZ said:
Because what you described is not the photoelectric effect, but rather a "photodiode" characteristics. What's the difference? A photoelectric effect is defined as the process whereby electron emission INTO THE VACUUM STATE occurs due to absorption of light. The problem you described is the formation of a CURRENT or a potential different IN THE MATERIAL due to absorption of light. The electrons are not released into the vacuum state, but rather went from the valence to the conduction band. This process simply overcomes the band gap, and has nothing to do with the work function, the latter is defined as the energy difference between the top of the valence band and the vacuum state.
Aha! Now I understand what ryan750 was asking. I was wondering how in the world the work function was relevant in the first place.

- Warren
 
  • #20
chroot said:
Aha! Now I understand what ryan750 was asking. I was wondering how in the world the work function was relevant in the first place.

- Warren

its ok i have just emailed a source that is expert on the subject - it is a diode company and they should be able to explain all - thanks for all ur help
 
  • #21
ryan750 said:
ok - u seemed pretty clued up about this subject: please could u give a concise account to this question - it would be greatly appreciated - even if it means repeating some of the details that have been said on this thread already - pls:

Light from a filament lamp is incident on a l.e.d that is connected to a voltmeter only.

1. explain how the voltage is generated in the l.e.d

could u give as detailed response as possible pls - sorry to be so demanding but its just that i am to be examined on this topic on friday. thankyou so much

I will assume that you know what a PN junction is and all its characteristics at thermal equilibrium.

When you have light of the "right" energy (and a white should contain within it the frequency that has the same value as the color that the LED emits) then two things can happen:

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.

Zz.
 
  • #22
Well, I work for one of the largest semiconductor companies on the planet, ryans, and ZapperZ's a professional physicist.

I believe we've already explained it satisfactorily -- the light stimulates electron-hole pairs in the depletion region, which then cross the junction and build up on the opposite sides of the semiconductor, creating a potential difference.

- Warren
 
  • #23
right i really think I've got it now - thanks for persevering with me - i probably would have lostthe marks in the exam if i had said it was due to the photoelectric effect - I am so glad i have researched it now. oh and sorry for doubting u guys - i had no idea that u were expert in the subject - lucky me tho to have found this site tonight and chatted to the experts just before my exam. thanks
 
  • #24
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.
 
  • #25
ryan750 said:
right i really think I've got it now - thanks for persevering with me - i probably would have lostthe marks in the exam if i had said it was due to the photoelectric effect - I am so glad i have researched it now. oh and sorry for doubting u guys - i had no idea that u were expert in the subject - lucky me tho to have found this site tonight and chatted to the experts just before my exam. thanks
And I apologize for misusing the term "photoelectric effect," which should really only be used to describe photons kicking electrons completely out of the metal. I didn't do a very good job of isolating your misconception!

Welcome to PF, by the way.

- Warren
 
  • #26
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
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 that's basic stuff but i just want to be sure of my definitions for friday
 
  • #28
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 I am glad that I am going to 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
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
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
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 isn't 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
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 isn't 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
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 that's it.
 
  • #34
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 that's 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
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?
 
<h2>What is the mechanism behind light-induced voltage in L.E.D.s?</h2><p>The mechanism behind light-induced voltage in L.E.D.s is known as the photovoltaic effect. This effect occurs when photons of light interact with the semiconductor material in the L.E.D., causing the release of electrons and the generation of a voltage difference.</p><h2>How does the structure of an L.E.D. contribute to its light-induced voltage?</h2><p>The structure of an L.E.D. plays a crucial role in its light-induced voltage. The semiconductor layers within the L.E.D. are designed to have a specific bandgap, which allows for the efficient conversion of light energy into electrical energy. Additionally, the p-n junction within the L.E.D. helps to create a built-in electric field that aids in the separation of electrons and holes, further contributing to the voltage generation.</p><h2>What factors affect the magnitude of light-induced voltage in L.E.D.s?</h2><p>The magnitude of light-induced voltage in L.E.D.s can be affected by several factors, such as the intensity and wavelength of the incident light, the material and structure of the L.E.D., and the temperature of the device. Higher light intensity, shorter wavelengths, and lower temperatures typically result in a higher voltage output.</p><h2>Can light-induced voltage in L.E.D.s be used for practical applications?</h2><p>Yes, light-induced voltage in L.E.D.s has several practical applications, such as in solar cells, photodetectors, and optoelectronic devices. These devices utilize the photovoltaic effect of L.E.D.s to convert light energy into electrical energy, making them highly efficient and sustainable sources of power.</p><h2>Are there any challenges in understanding the mechanics of light-induced voltage in L.E.D.s?</h2><p>While the basic principles behind light-induced voltage in L.E.D.s are well understood, there are still ongoing research and developments to improve the efficiency and performance of these devices. Additionally, the complex interactions between light, materials, and electric fields make it a challenging area of study, requiring advanced techniques and equipment for accurate measurements and analysis.</p>

Related to Understanding the Mechanics of Light-Induced Voltage in L.E.D.s

What is the mechanism behind light-induced voltage in L.E.D.s?

The mechanism behind light-induced voltage in L.E.D.s is known as the photovoltaic effect. This effect occurs when photons of light interact with the semiconductor material in the L.E.D., causing the release of electrons and the generation of a voltage difference.

How does the structure of an L.E.D. contribute to its light-induced voltage?

The structure of an L.E.D. plays a crucial role in its light-induced voltage. The semiconductor layers within the L.E.D. are designed to have a specific bandgap, which allows for the efficient conversion of light energy into electrical energy. Additionally, the p-n junction within the L.E.D. helps to create a built-in electric field that aids in the separation of electrons and holes, further contributing to the voltage generation.

What factors affect the magnitude of light-induced voltage in L.E.D.s?

The magnitude of light-induced voltage in L.E.D.s can be affected by several factors, such as the intensity and wavelength of the incident light, the material and structure of the L.E.D., and the temperature of the device. Higher light intensity, shorter wavelengths, and lower temperatures typically result in a higher voltage output.

Can light-induced voltage in L.E.D.s be used for practical applications?

Yes, light-induced voltage in L.E.D.s has several practical applications, such as in solar cells, photodetectors, and optoelectronic devices. These devices utilize the photovoltaic effect of L.E.D.s to convert light energy into electrical energy, making them highly efficient and sustainable sources of power.

Are there any challenges in understanding the mechanics of light-induced voltage in L.E.D.s?

While the basic principles behind light-induced voltage in L.E.D.s are well understood, there are still ongoing research and developments to improve the efficiency and performance of these devices. Additionally, the complex interactions between light, materials, and electric fields make it a challenging area of study, requiring advanced techniques and equipment for accurate measurements and analysis.

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