Why is P-N junction so important in semiconductor diodes?

AI Thread Summary
The discussion highlights the importance of semiconductor materials, specifically P-N junctions, in constructing photodiodes for radiation detection. Semiconductors are preferred due to their high density, which increases interaction probabilities, and their ability to create a depletion region that mitigates thermal leakage currents. While insulators like glass could theoretically collect electrons under high voltage, they present challenges such as low atomic numbers and difficulty in electron collection, resulting in minimal currents. The P-N junction effectively allows for efficient current generation even without a bias voltage, making it a superior choice for applications requiring precise detection. Overall, the semiconductor's properties are crucial for optimizing performance in various sensing applications.
abotiz
Messages
72
Reaction score
0
Hi,

Iam having a hard time understanding the benefits of using a semiconductor to construct e.g. a photodiode that creates a current in proportion of the energy deposited by radiation.

Text books says that semiconductor is preferred as detectors due to high density (larger chance for interaction with the detector) can be made small etc etc.

Then they start to explain the large leakage current produced due to the thermal generated charge carriers because of the relatively low band gap etc (they contribute to a current of the order 0.1A, and radiation induced current is like 0.00001). So they continue to explain that this needs to be dealt with, so you "fuse" a P and N type crystal so that a depletion region is created near the junction, where the thermally generated charges are depleted. But let's just stop here, all of this just to create a situation where you get rid of these thermal generated charges, well, why do you use a semiconductor in the first place, why not an insulator.

If I took a piece of glass, applied a high voltage (e.g 4KV), then when the radiation knocks out an electron, 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. In addition I would have almost no leakage current (thermal generated). Why is this a bad idea? Am I missing something? Is it ONLY because insulators have low atomic number and therefore not good radiation detectors (you need high atomic number for high interaction probability)

Thank you for your time!
 
Engineering news on Phys.org
You pick the sensor for the application - and the application determines the requirements.

So if you want to detect infrared (simple heat detector which can also detect motion) you will need a sensor sensitive to this wavelength. This is easy to do with a semiconductor, but I don't know how you would do it with glass.

I think the point they are making is that the PN junction (which is required to make a diode) is an easy way to engineer the semiconductor so that its properties are closer to what you want.
 
I think that freed electrons in an insulator are hard to collect (because it is an insulator). This would make the resulting currents very small and the required voltages very high. In a pn junction the depletion region sweeps the freed electrons and produces a current even in absense of a bias voltage. Read the wikipedia pages in photo-electric effect and photo-diode.

Maybe I am missing your point?
 
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I have recently moved into a new (rather ancient) house and had a few trips of my Residual Current breaker. I dug out my old Socket tester which tell me the three pins are correct. But then the Red warning light tells me my socket(s) fail the loop test. I never had this before but my last house had an overhead supply with no Earth from the company. The tester said "get this checked" and the man said the (high but not ridiculous) earth resistance was acceptable. I stuck a new copper earth...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top