Main reason why semiconductor is used and not a pure metal for detec

[abotiz]

Main reason why semiconductor is used and not a "pure" metal for detec

Hi,

I have question regarding why semiconductor is preferred in the field of radiation detection.
I have read that they can be doped and their conductivity can be manipulated, but I can't see why this is not possible with "pure" metals, e.g. copper.

Can someone explain this to me?

Also, I have read that the energy to create an electron and hole pair is about 3eV for germanium, the band gap is about 0.7eV, so obviously there is other process that require energy to liberate this pair. Which are they?

Thank you very much!

 

[f95toli]

You can make detectors out of all sorts of materials, not only semiconductors. The most basic detector one can think of would be a bolometer, and they are often made from metals (e.g. gold, tungsten etc)

It really depends of the type of detector you are talking about,

 

[ZapperZ]

Hi,

I have question regarding why semiconductor is preferred in the field of radiation detection.
I have read that they can be doped and their conductivity can be manipulated, but I can't see why this is not possible with "pure" metals, e.g. copper.

Can someone explain this to me?

You need to explain this a bit more. What do you mean by "radiation detection"? EM radiation? And what type of detector are you talking about? Photodetector?

Also, I have read that the energy to create an electron and hole pair is about 3eV for germanium, the band gap is about 0.7eV, so obviously there is other process that require energy to liberate this pair. Which are they?

Thank you very much!

You have to consider not only the band gap, but also the electron affinity, which defines where the vacuum level is from the bottom of the conduction band. The question also isn't clear because are you trying to liberate the electron from the solid, or are you trying to maintain an electron-hole pair in the solid, i.e. create an exciton? One doesn't "liberate" an electron-hole pair. One liberates the electron from the metal, the hole gets "reabsorbed".

Zz.

 

[abotiz]

Hi,

I meant the use of a diode with charged particle radiation (e.g. beta radiation). Its the depleted region inside a P-N type diode (junction) that is used as the active volume of the detector. Is the main reason, why semiconductor are used instad of "pure" metals, due to that the depleted region can not be created in "pure" metals, if so, why?

Yes, sorry for the unclear question. I am not sure what they (my textbooks) mean either, they say it takes about 3eV to "create" an electron and hole pair, in terms of interaction of a charged particle with a covalent electron. The hole is just the vacancy of the electron. At minimum the charged particle needs to transfer the band gap energy, 0.7eV, and Iam guessing an extra energy to liberate the electron from the conduction band, is there something else? Otherwise, the energy to liberate the electron from the conduction band requires approx 2.3eV in Germanium. Is this true?

Thank you

 

[ZapperZ]

Hi,

I meant the use of a diode with charged particle radiation (e.g. beta radiation). Its the depleted region inside a P-N type diode (junction) that is used as the active volume of the detector. Is the main reason, why semiconductor are used instad of "pure" metals, due to that the depleted region can not be created in "pure" metals, if so, why?

You cannot have such a region because (i) there's no such thing as a p-doped or n-doped metals and (ii) the mobile electrons will simply wash out any internal E-field, unlike a semiconductor.

Also note that electrons cannot penetrate very far in metals due to collisions with the conduction electrons, unlike a semiconductor.

Yes, sorry for the unclear question. I am not sure what they (my textbooks) mean either, they say it takes about 3eV to "create" an electron and hole pair, in terms of interaction of a charged particle with a covalent electron. The hole is just the vacancy of the electron. At minimum the charged particle needs to transfer the band gap energy, 0.7eV, and Iam guessing an extra energy to liberate the electron from the conduction band, is there something else? Otherwise, the energy to liberate the electron from the conduction band requires approx 2.3eV in Germanium. Is this true?

Thank you

2.3 eV is the electron affinity, which is the energy from the bottom of the conduction band, to the vacuum level.

0.7 eV is the band gap.

Thus, to get an electron from the top of the valence band to the vacuum level, it is 2.3 + 0.7 = 3.0 eV.

Zz.

 

[abotiz]

You cannot have such a region because (i) there's no such thing as a p-doped or n-doped metals and (ii) the mobile electrons will simply wash out any internal E-field, unlike a semiconductor.

(i) Why? E.g. Silicon has an electron configuration that together with Boron and Phosphorus, can create P and N type, is there no atom with an electron configuration that can create a P or N type with e.g Copper?

(ii) If enough impurity would be added to the "pure" metal, would this not create a sufficient electric field to hold of the (intrinsic) electrons?

2.3 eV is the electron affinity, which is the energy from the bottom of the conduction band, to the vacuum level.

0.7 eV is the band gap.

Thus, to get an electron from the top of the valence band to the vacuum level, it is 2.3 + 0.7 = 3.0 eV.

Zz.

Nice =D Thank you!

 

[nsaspook]

(ii) If enough impurity would be added to the "pure" metal, would this not create a sufficient electric field to hold of the (intrinsic) electrons?

Sure, there are many compound semiconductors like gallium arsenide but pure single-crystal compound substrates are expensive and are harder to scale to smaller structures.

http://en.wikipedia.org/wiki/Gallium_arsenide#Silicon_advantages
http://www.doitpoms.ac.uk/tlplib/semiconductors/compound.php

 

[ZapperZ]

(i) Why? E.g. Silicon has an electron configuration that together with Boron and Phosphorus, can create P and N type, is there no atom with an electron configuration that can create a P or N type with e.g Copper?

(ii) If enough impurity would be added to the "pure" metal, would this not create a sufficient electric field to hold of the (intrinsic) electrons?

There are no band gap in a metal. The Fermi level is right in the conduction band!

The doping that's done in semiconductors introduced the dopant level in the band gap itself.

Zz.

 

[analogdesign]

Also, keep in mind that for a radiation detector to be useful, it must be read out. If it were made of pure metal it would be hard to collect the charge liberated by the impinging radiation. A diode creates a current proportional to the incident radiation, which is what you want.

A semiconductor is so useful for this because it can be depleted. Therefore, liberated e-h pairs are swept to a readout electrode and are unlikely to recombine before collection. This gives a high SNR.

Let me ask you this. If you were to make a radiation detector out of copper, how would you read it out?

 

[abotiz]

There are no band gap in a metal. The Fermi level is right in the conduction band!

The doping that's done in semiconductors introduced the dopant level in the band gap itself.

Zz.

Could you please explain this further? If you can, in simpler terms.

As I understand, its the depletion region that is the main principle (the active volume), the band gap is no longer of importance. So, as long as the crystal can be made to have a depletion region (no mobile carriers, remaining electrons in covalent bonds), I don't see any problems, therefore I don't see why this is not possible with pure metals.

However, if ALL the electrons in a solid metal behaves like valence electrons (loosely bound), then if you apply a voltage, that would sweep all the electrons away and there would be no electrons left bound to the nucleus that the radiation could knock out and ultimately no signal would be produced.

If NOT all the electrons are swept away, then only the tightly bound electrons would remain, creating a similar situation as the P-N junction. The majority carriers are removed. Therefore, metals can also be used due to the fact that an applied voltage to e.g. a copper cube would remove all the valence electrons (majority mobile carriers), leaving behind the tightly bound electrons which the incomming radiation can interact with, the electron is ejected, its motion constitutes to a current that can be measured. Right?

Thank you for your time!

 

[analogdesign]

Could you please explain this further? If you can, in simpler terms.

As I understand, its the depletion region that is the main principle (the active volume), the band gap is no longer of importance. So, as long as the crystal can be made to have a depletion region (no mobile carriers, remaining electrons in covalent bonds), I don't see any problems, therefore I don't see why this is not possible with pure metals.

However, if ALL the electrons in a solid metal behaves like valence electrons (loosely bound), then if you apply a voltage, that would sweep all the electrons away and there would be no electrons left bound to the nucleus that the radiation could knock out and ultimately no signal would be produced.

If NOT all the electrons are swept away, then only the tightly bound electrons would remain, creating a similar situation as the P-N junction. The majority carriers are removed. Therefore, metals can also be used due to the fact that an applied voltage to e.g. a copper cube would remove all the valence electrons (majority mobile carriers), leaving behind the tightly bound electrons which the incomming radiation can interact with, the electron is ejected, its motion constitutes to a current that can be measured. Right?

Thank you for your time!

I think you are a little bit fuzzy on the quantum theory of solids.

The concept of "depletion" really makes no sense in a metal. As ZapperZ said, there is no bandgap in a metal. That means virtually all of the carriers are free already at room temperature. Doping them wouldn't change anything. It would be like adding a glass of water to the Pacific Ocean. Could you measure the difference?

You are also confusing "valence electrons" with "conduction electrons". In a semiconductor, the valence band is the energy levels where electrons are localized to a single lattice site.

In a metal, virtually all electrons are free so all electrons would be swept away. There is no small population of "tightly bound electrons" remaining. if there were, then you would be describing a semiconductor. Remember there is no band gap. What energy levels are these "tightly bound electrons" occupying?

Lastly, your first comment confuses me. How can a metal be a metal and yet have "no mobile carriers"?

 

[M Quack]

The radiation detectors I know mostly work by applying a "bias" voltage across a material with very low conductivity. Radiation produces pairs of positive and negative charges that are collected at electrodes and generate a current there. The current is then amplified electronically.

There are many types of detectors. The earliest ones are gas based (Geiger counter, proportional counter, ionization chamber). Gas based counters are still used extensively to detect neutrons.

Other types are semiconductor-based detectors such as photodiodes (in many variations, avalanche, drift, pin, ...) and scintillation detectors where a high-energy particle is converted into visible photons that are then detected otherwise (e.g. in a photomultiplier, ccd device ...).

Finally, there are photographic plates that have lost some popularity in recent times.

Oh, and we've used a solid block of copper to measure the intensity of an x-ray beam. But that beam had several 100W of power, and we just measured the temperature before and after a few seconds exposure to determine the (average) power in the beam. Through a very thick window of leaded glass you could see the ionization trail in thin air, and the place reeked of ozone.

 

[abotiz]

Thank you for your answers!

virtually all of the carriers are free already at room temperature.

So this means that there is no "electron configuration" in a metal, all electrons are shared with all the atoms? E.g K-shell and L-shell electrons does not exist in a solid metal?

My initial thought was this, radiation interactions or crosssection (photoelectric) increases with atomic number. So, with that being said, why isn't a metal used e.g. lead detectors.

The trouble Iam having is basically understanding the advantages of using a semiconductor. A similar discussion can be held about gaseous detectors, which gas is preferred (ionization energy required, density, recombination etc etc). But this discussion about solid-detectors, as I understand there is three types, metals, semiconductors and insulators.

As analogdesign explained, all the electrons are free in a metal so its not possible to create an electron hole pair. Okay! Just one question, why is it then important to dope the semiconductors, would it no be sufficient just to have remaining electrons in which the radiation can knock out?

What about Insulators?

If I took a piece of glass, applied a high voltage (e.g 4KV), the "loosely bound" electrons would be swept away to the electrode. The remaining electrons is bound to the nucleus. When the radiation knocks out one of these bound electrons, the strong electric field would collect the electron (at the electrode) before it is recombined, and I would measure a current. This current would be proportional to the energy deposited in the glass (insulator). In addition I would have smaller leakage current compared to semiconductor (thermal generated - as the band gap is larger). Is this a poorer choice only due to insulators have lower atomic number (lower crossection)?

Thank you for your time!