1 electron = 1 photon?

Started Apr 12, 2013 | Discussions
Jack Hogan
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1 electron = 1 photon?
Apr 12, 2013

Simplifying, assuming a modern B&W silicon sensor without a CFA or other filtering and 100% fill factor, if we see one electron produced at the output of a photosite, can we assume that it was the result of one photon making it through to silicon - and that the probability of it dislodging an electron is the Charge Collection Efficiency (QE) of the semiconductor at the wavelength of the incoming photon, so that for a given Exposure

Electrons produced = photons(wavelength) hitting silicon * QE(wavelength)?

If the visible spectrum is between 380 and 760 nm, with a near-infrared photon having half the frequency/energy of a near-ultraviolet photon therefore burying deeper into silicon, does the following responsivity curve merely represent QE at various silicon depths?

Relative Number of electrons Generated as a Function of Impinging Photon Wavelength

Or could the fact that the responsivity at 760nm is more than 3 times that at 380nm mean that, for instance, sometimes 1 photon produces two electrons?

Jack

John Siward
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There is only one electron...
In reply to Jack Hogan, Apr 12, 2013

From Richard Feynman's Nobel Lecture:

I received a telephone call one day at the graduate college at Princeton from Professor Wheeler, in which he said, "Feynman, I know why all electrons have the same charge and the same mass" "Why?" "Because, they are all the same electron!"

J.

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alanr0
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Re: 1 electron = 1 photon?
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

Simplifying, assuming a modern B&W silicon sensor without a CFA or other filtering and 100% fill factor, if we see one electron produced at the output of a photosite, can we assume that it was the result of one photon making it through to silicon - and that the probability of it dislodging an electron is the Charge Collection Efficiency (QE) of the semiconductor at the wavelength of the incoming photon, so that for a given Exposure

Electrons produced = photons(wavelength) hitting silicon * QE(wavelength)?

If the visible spectrum is between 380 and 760 nm, with a near-infrared photon having half the frequency/energy of a near-ultraviolet photon therefore burying deeper into silicon, does the following responsivity curve merely represent QE at various silicon depths? Or could the fact that the responsivity at 760nm is more than 3 times that at 380nm mean that, for instance, sometimes 1 photon produces two electrons?

https://upload.wikimedia.org/wikipedia/commons/thumb/4/41/Response_silicon_photodiode.svg/500px-Response_silicon_photodiode.svg.png

The response is shown in A/W.  For each electron released, the charge is constant, but the photon energy is proportional to the photon frequency, and inversely proportional to the wavelength.

Efficiency in A/W = (electron charge x wavelength)/(planck's const x velocity of light)

At 100% quantum efficiency (1 electron per photon), you get 0.645 A/W at 800 nm, but only 0.323 A/W at 400 nm.  Your graph shows the quantum efficiency falling from around 84% at 600-800 nm, to around 53% at 400 nm.

In other words, less than 100% QE, with peak performance at 600-800 nm.

HTH

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Alan Robinson

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looper1234
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Re: 1 electron = 1 photon?
In reply to alanr0, Apr 12, 2013

photo-sites don't really distinguish color, things like filters are used, these cuts out photons that are not "lucky"

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John Siward
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Re: 1 electron = 1 photon?
In reply to Jack Hogan, Apr 12, 2013

Optical power in Watts = photon energy * no. of photons / sec.

Photocurrent in Amps = electron charge * no. of electrons generated / sec. = electron charge * (no. of photons / sec) * efficiency.

Response (A / W) as plotted = photocurrent (A) / optical power (W)

= electron charge * efficiency / photon energy

Photon energy = h * frequency = h * c / wavelength (where h is Planck's constant and c is speed of light)

So you can use that to work out the efficiency from the curve you showed.

Silicon is what they call an "indirect bandgap" material, which means that an electron jumping from the valence band to the conduction band also exchanges momentum with the lattice in the form of a phonon.  This represents a loss of energy, so you can't get to 100% efficiency.  This is in contrast to "direct bandgap" materials such as GaAs, InP, etc.

Your basic point is right: a red photon is much more likely to penetrate deeper into the Si than a blue photon.  This is the principle of operation of Sigma's "Foveon" sensor which stacks three photodiodes vertically to allow the sensor to measure colour without using a colour filter array.

J.

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FractalFlame
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Re: There is only one electron...
In reply to John Siward, Apr 12, 2013

John Siward wrote:

From Richard Feynman's Nobel Lecture:

I received a telephone call one day at the graduate college at Princeton from Professor Wheeler, in which he said, "Feynman, I know why all electrons have the same charge and the same mass" "Why?" "Because, they are all the same electron!"

J.

Someone theorised that the entire universe is the result of one atom, moving at infinity time the speed of light, moving through every point in the entire univers so fast it appears real and solid..

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Allan Olesen
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Repeating the question: Can 1 photon sometimes produce two electrons?
In reply to Jack Hogan, Apr 12, 2013

It seems to me that nobody is answering the very interesting question at the end of Jack Hogan's original post. So I will repeat it because I am curious about that too:

Can 1 photon sometimes produce two electrons?

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Jack Hogan
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Re: 1 electron = 1 photon?
In reply to alanr0, Apr 12, 2013

alanr0 wrote:

The response is shown in A/W.  For each electron released, the charge is constant, but the photon energy is proportional to the photon frequency, and inversely proportional to the wavelength.

Efficiency in A/W = (electron charge x wavelength)/(planck's const x velocity of light)

At 100% quantum efficiency (1 electron per photon), you get 0.645 A/W at 800 nm, but only 0.323 A/W at 400 nm.  Your graph shows the quantum efficiency falling from around 84% at 600-800 nm, to around 53% at 400 nm.

In other words, less than 100% QE, with peak performance at 600-800 nm.

HTH

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Alan Robinson

Very helpful as always, Alan, I assume this holds on average.   Can there ever be a case where a single particularly energetic photon results in two electrons being produced by the photodiode?

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looper1234
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Re: Repeating the question: Can 1 photon sometimes produce two electrons?
In reply to Allan Olesen, Apr 12, 2013

Allan Olesen wrote:

It seems to me that nobody is answering the very interesting question at the end of Jack Hogan's original post. So I will repeat it because I am curious about that too:

Can 1 photon sometimes produce two electrons?

from the graph the response is never exceeding 1.

Also photos are created by electrons dropping from higher energy shells to lower ones.

their energy levels are fix, the same exact amount of energy would be needed to reverse this process at the other end.

but this never happens at 100%, it would seem that more energetic photos have better success rate, but never will the exceed the 100% rate.

also, electron on electron shells are expressed in percentage of existence, we can never be sure as to say an event took place, we can only express it as a percentage, such is the natural of things at the quantum level.

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Jack Hogan
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Re: 1 electron = 1 photon?
In reply to John Siward, Apr 12, 2013

John Siward wrote:

Optical power in Watts = photon energy * no. of photons / sec.

Photocurrent in Amps = electron charge * no. of electrons generated / sec. = electron charge * (no. of photons / sec) * efficiency.

Response (A / W) as plotted = photocurrent (A) / optical power (W)

= electron charge * efficiency / photon energy

Photon energy = h * frequency = h * c / wavelength (where h is Planck's constant and c is speed of light)

So you can use that to work out the efficiency from the curve you showed.

Silicon is what they call an "indirect bandgap" material, which means that an electron jumping from the valence band to the conduction band also exchanges momentum with the lattice in the form of a phonon.  This represents a loss of energy, so you can't get to 100% efficiency.  This is in contrast to "direct bandgap" materials such as GaAs, InP, etc.

Your basic point is right: a red photon is much more likely to penetrate deeper into the Si than a blue photon.  This is the principle of operation of Sigma's "Foveon" sensor which stacks three photodiodes vertically to allow the sensor to measure colour without using a colour filter array.

J.

Thank you for walking me through it, John, quite clear.  So could we rearrange the Responsivity diagram's x-axis to represent logarithmic penetration depth, the curve then representing QE at the relative depth?  If so, I would have expected more discontinuities in it, per this interesting introduction.

Jack
PS Re: Feynman

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looper1234
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Re: 1 electron = 1 photon?
In reply to Jack Hogan, Apr 12, 2013
Very helpful as always, Alan, I assume this holds on average.   Can there ever be a case where a single particularly energetic photon results in two electrons being produced by the photodiode?

the sensor material must have being chosen to respond well to visible lights, a much higher energy photon will react differently to this material. that's, no transference of energy would take place, photon is not destroyed its energy not released to push electrons to a higher shell.

and hence the graph shows drop off in response in those undesirable wave lengths.

also it is impossible for one photo to push out two electrons.

quantum physics states photons have a fixed energy state, they are discrete and not analogs. An interaction will release all of its energy not part of it.

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Jack Hogan
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Integer vs Floating Point
In reply to looper1234, Apr 12, 2013

looper1234 wrote:

also, electron on electron shells are expressed in percentage of existence, we can never be sure as to say an event took place, we can only express it as a percentage, such is the natural of things at the quantum level.

Good point, looper1234, it's a stochastic process, right?  So does it really make sense to talk about photons as integers?  Shouldn't we really be perfectly happy with floating point, as in 8.62 photons producing 5.85 electrons, then converted to 1.26 ADUs?

And what does this say for Unity Gain proponents ?

Jack

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olliess
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Re: Integer vs Floating Point
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

looper1234 wrote:

also, electron on electron shells are expressed in percentage of existence, we can never be sure as to say an event took place, we can only express it as a percentage, such is the natural of things at the quantum level.

Good point, looper1234, it's a stochastic process, right?  So does it really make sense to talk about photons as integers?  Shouldn't we really be perfectly happy with floating point, as in 8.62 photons producing 5.85 electrons, then converted to 1.26 ADUs?

There might a certain chance of an additional photon reaching the sensor or not... but when you go to measure it, either it was there or it wasn't.

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Allan Olesen
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Re: Repeating the question: Can 1 photon sometimes produce two electrons?
In reply to looper1234, Apr 12, 2013

looper1234 wrote:

Allan Olesen wrote:

It seems to me that nobody is answering the very interesting question at the end of Jack Hogan's original post. So I will repeat it because I am curious about that too:

Can 1 photon sometimes produce two electrons?

from the graph the response is never exceeding 1.

But that graph shows an average for many photons. It doesn't show the possible outcomes for just one photon. If it did, there would not be any values between 0 and 1 since you can't get a half electron from exactly one photon.

Also photos are created by electrons dropping from higher energy shells to lower ones.

their energy levels are fix, the same exact amount of energy would be needed to reverse this process at the other end.

And that answers the question much better.

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PhotonTrapper
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Re: 1 electron = 1 photon?
In reply to looper1234, Apr 12, 2013

looper1234 wrote:

also it is impossible for one photo to push out two electrons.

This is not correct if you state it as a general rule. There are cases of double-electron emission by single photon.

Example with one circularly polarized photon:

http://iopscience.iop.org/0953-4075/32/2/034

More here:

http://onlinelibrary.wiley.com/doi/10.1002/pssb.200945203/abstract

http://www2.mpi-halle.mpg.de/exp_department_1/research_projects/electron_pair_correlation/double_photoemission/

For the time being, I don't think DPE has been achieved on CCD or CMOS though.

Regards,

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PhotonTrapper
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Re: 1 electron = 1 photon?
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

Simplifying, assuming a modern B&W silicon sensor without a CFA or other filtering and 100% fill factor, if we see one electron produced at the output of a photosite, can we assume that it was the result of one photon making it through to silicon - and that the probability of it dislodging an electron is the Charge Collection Efficiency (QE) of the semiconductor at the wavelength of the incoming photon, so that for a given Exposure

Electrons produced = photons(wavelength) hitting silicon * QE(wavelength)?

If the visible spectrum is between 380 and 760 nm, with a near-infrared photon having half the frequency/energy of a near-ultraviolet photon therefore burying deeper into silicon, does the following responsivity curve merely represent QE at various silicon depths?

Relative Number of electrons Generated as a Function of Impinging Photon Wavelength

Or could the fact that the responsivity at 760nm is more than 3 times that at 380nm mean that, for instance, sometimes 1 photon produces two electrons?

Jack

I think you would have to include interaction cross section coefficients into your model. They are not part of the charge collection efficiency but they are a function of the material (Z atomic number) and photon energy (i.e. wavelength).

Regarding double electron emission from on single photon, yes it is possible, but I don't think it has been done on CCD or CMOS material.

Regards,

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alanr0
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APD, EMCCD and Compton scattering
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

alanr0 wrote:

The response is shown in A/W.  For each electron released, the charge is constant, but the photon energy is proportional to the photon frequency, and inversely proportional to the wavelength.

Efficiency in A/W = (electron charge x wavelength)/(planck's const x velocity of light)

At 100% quantum efficiency (1 electron per photon), you get 0.645 A/W at 800 nm, but only 0.323 A/W at 400 nm.  Your graph shows the quantum efficiency falling from around 84% at 600-800 nm, to around 53% at 400 nm.

In other words, less than 100% QE, with peak performance at 600-800 nm.

HTH

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Alan Robinson

Very helpful as always, Alan, I assume this holds on average.   Can there ever be a case where a single particularly energetic photon results in two electrons being produced by the photodiode?

If the photon has sufficient energy, more than one electron can be released.  PhotonTrapper has identified possible mechanisms, such as double electron photoemission.  I don't believe this or other mechanisms make a significant contribution when conventional silicon photodiodes exposed to visible light.

At much higher (X- and gamma-ray) photon energies, Compton scattering can release multiple electrons as a photon propagates through the material.  At each scattering event, an electron-hole pair is created, and the photon loses some energy.  Consult Storm & Israel's tables for the dependence on atomic number and photon energy if you don't have access to code like this.

A different mechanism operates in an avalanche photodiode (APD).  Arguably there are two processes operating sequentially here.  An electron-hole pair is first generated by photo-absorption.  APDs operate at much higher bias voltages than conventional photodiodes, and the carriers are accelerated by the bias field, releasing further electron-hole pairs by avalanche multiplication.

Impact ionisation is also exploited by electron-multiplying EMCCD detectors, to produce multiple electrons from a single absorbed photon.

Cheers

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Alan Robinson

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mike703
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Re: 1 electron = 1 photon?
In reply to Jack Hogan, Apr 12, 2013

There is a nice explanation of the shape of the current vs. wavelngth curve in this article: see figures 5 - 7.  It doesn't require production of 2 electrons by one photon.

www.aphesa.com/downloads/download2.php?id=1

Best wishes

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alanr0
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Internal vs external QE
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

John Siward wrote:

Your basic point is right: a red photon is much more likely to penetrate deeper into the Si than a blue photon.  This is the principle of operation of Sigma's "Foveon" sensor which stacks three photodiodes vertically to allow the sensor to measure colour without using a colour filter array.

J.

Thank you for walking me through it, John, quite clear.  So could we rearrange the Responsivity diagram's x-axis to represent logarithmic penetration depth, the curve then representing QE at the relative depth?  If so, I would have expected more discontinuities in it, per this interesting introduction.

Jack:

I suspect it will be difficult to get a reliable indication of penetration depth.  In particular, there may be more going on than simple electron-hole creation at wavelength shorter than 400 nm.

To make a start, I would first calculate QE from the responsivity to remove the inverse wavelength dependence of photon energy.  This will have a much 'flatter' shape than the curve you presented.  You also need to account for incident light which is reflected from the surface of the chip or the package window.  Silicon has a rather high refractive index of around 4 , so there is strong Fresnel reflection unless a suitable anti-reflection coating is applied.  Even with an AR coating, the external quantum efficiency can be significantly lower than the internal QE.

HTH

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Alan Robinson

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alanr0
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Multiple Exciton Generation
In reply to Jack Hogan, Apr 12, 2013

Jack Hogan wrote:

Very helpful as always, Alan, I assume this holds on average.   Can there ever be a case where a single particularly energetic photon results in two electrons being produced by the photodiode?

FWIW, I came across a reference to Multiple Exciton Generation .  I don't know much about it, and after reading the Wikipidia article, it suspect I may not be alone in this.

The aphesa article linked by Mike703 looks interesting, with a discussion of recombination and some of the other loss mechanisms in photodiodes.

Cheers.

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Alan Robinson

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