Photoelectron to electron hole pair doubts

[Manit]

Hi,
I have a couple of questions on photoelectrons.

When a photoelectron of about 3-eV (varies) interacts within 0.2-um depletion region of silicon, what happens?

I know, it will generate an electron-hole pair with an efficiency of 1 for 3.6-eV photoelectron. But what happens if the photoelectron energy is lower than 3.6-eV?

Now, what happens to the current following this generation? Since, this pair is in a reverse biased p-n diode it will be pulled to form a signal. The question is what would be the magnitude and how much time would they be available before being lost or filled up (sorry, I am little rusty with the terminology - have moved away from electronics, but have come across this problem).

I believe, current (I) = # electrons*charge/time

Where, # electrons are the number of electrons; charge- 1.6e-19 C; time - the amount of time the electron is present to generate/affect the signal before they are lost.

So, if 25 photoelectrons (3.6-eV) with 100% efficiency then we see a generation of 25 pairs. Let's say, they each are available for a time of 1-us. So, current will be 25*1.6e-19/1e-6 (A). Am I wrong?

I don't know what happens when the energy is lower than 3.6-eV and the time this electron-hole pair are available.

Please share your valuable thoughts.

Regards,
Manit.

 

[mfb]

I know, it will generate an electron-hole pair with an efficiency of 1 for 3.6-eV photoelectron. But what happens if the photoelectron energy is lower than 3.6-eV?

If there is no suitable transition at the energy of the photon, it will just pass through. The threshold is not 3.6 eV, however, the bandgap of silicon is 1.1 eV. You can create an electron/hole pair with a photon with 1.2 eV. It just does not happen with a 100% efficiency. If you take efficiency into account, particle detectors usually get one pair per 3.6 eV of high-energetic electrons or photons.

So, if 25 photoelectrons (3.6-eV) with 100% efficiency then we see a generation of 25 pairs. Let's say, they each are available for a time of 1-us. So, current will be 25*1.6e-19/1e-6 (A). Am I wrong?

The time is related to the thickness of the sensor and the drift speed of the electrons and holes in the material. It is not as simple as distance/speed, however, as your charges start inducing charges in the electrodes before they arrive there.

 

[Manit]

Thank you very much for your reply, Mr. MFB.

So, to account/estimate the current induced, what is the best way? or any electrical software that can help in measuring the relation between number of photoelectrons (electron-hole pairs) and the current?

Regards,
M

 

[mfb]

There are software packages to simulate that (but I don't know them). Capacitances in the system are important as well.

 

[Manit]

Hi,

Thank you for your prompt reply.

I believe, you are referring to this equation:

I = nevA

I - Current
n = no of electrons per unit volume,
e = charge of an electron,
v = drift speed of an electron,
A = cross sectional area of the wire

However, n will be too large in general (10^18 - semiconductors). # of electrons added from photoelectrons would be marginal. What I am trying to estimate is the current magnitude from the photoelectrons.

Also, I see, v=10 m/s in semiconductors. Is it close or in fact a standard value that I can assume for hand calculation? I mean, is 10 m/s drift speed for an electron in an electron-hole pair?

Thank you for your time and help!

 

[mfb]

I believe, you are referring to this equation:

I = nevA

No.
The current will depend on the electrical environment of the sensor and the readout electronics, it is not as simple as plugging in numbers into a formula.