Direct Thin Film Transistor Array (Flat Panel) | X-ray Physics | Radiology Physics Course #36
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- Опубликовано: 12 сен 2024
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Direct thin film transistor array detectors have the ability to convert X-ray photon energy directly into an electrical signal. This is due to the amorphous selenium semiconductor layer which feeds ions towards the TFT array. This current is then stored in the capacitor prior to readout. Direct digital radiography systems have excellent spatial resolution but do not handle high energy X-ray photons well. This makes them ideal for mammography which we will look at more closely in a future module.
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If you're here, you're likely studying for a radiology physics exam. I've spent the last few months collating past papers from multiple different countries selecting the most commonly asked questions. You'll be surprised how often questions repeat themselves!
The types of questions asked in FRCR, RANZCR AIT, ARRT, FC Rad Diag (SA), ABR qualifying Core Physics and MICR part 1 are surprisingly similar and the key concepts remain the same throughout. I've taken the most high-yield questions and answered them in video format so that I can take you through why certain answers are correct and others are not.
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Ah, thank you Sukrati. It means a lot. Appreciate the kind words 🙂
amorphous selenium or amorphous silicon.
The purpose of those two materials is to
provide a source of electrons to the TFT
Such a foreign concept when reading it from the book but your explanations help make sense of the material which I truly appreciate thank you so much! 🙏
Yay! That’s exactly what I was hoping for. Reading this stuff is so tedious!
Superb explanation, highly recommended physics video series for radiology residents
🙏
Hello! On a mock exam, I had a question asking which has the potential for the lowest exposure dose and the following answer choices:
Conventional radiography using film screen
CR
DR using a TFT (not specified if direct or indirect)
DR using a CDC.
My gut is leaning towards DR with TFT (vs with CDC as my second choice), but without the specification between indirect or direct, I am not completely confident. You mentioned that with direct TFT arrays, these are more beneficial with handling lower energies which is why I went with that answer. Not sure if I am overthinking (very likely that I am) but could you explain how you would tackle this question if I misunderstood? Thanks in advance. Love your videos by the way they have been life savers!
Ultimate radiology teacher
Thank you Shubham!
Thank you Soo much.. perfectly explained 😊
Always a pleasure!
Thx 4 the vids 😊👍🐈🐾🐾
Aren't the "+" and "-" signals inverted in the diagram? Wouldn't the positive signal be positioned inferiorly in the selenium? This way, the negative charge would be attracted to the closer region of the TFT layer. It's a doubt, if you can explain. Thank you for the content!
Great observation. This orientation of the +/- symbols is due to a certain property of the semiconductor layer (a little too advanced for this course). In fact what it happening here is instead of electrons heading towards the TFT layer - something known as electrons holes are. These holes are formed by the absence of an electron in the valence shell of an atom. The holes are formed by electrons moving away from the TFT layer. The electron holes can also effectively be measured as current at the sensitive area. It’s a technicality which requires much greater explanation than I could give in the video but would encourage you to read further if you’re interested. Hope that makes some sense 😆
@@radiologytutorials Great! Thanks for support, solving my question. Wonderful contents. Stay on these roads, Michael!
Hi Dr Nel,
Still struggle to understand the direction of cathode to anode here.
My understanding is - The semicondutor modelled using the solid band theory can be visualised with its respective valence band close to the TFT array and conduction band close to the patient. The incident X-ray striking a-Se causes excitation of the electron in the valence band to jump to the higher energy state conduction band. This leads to the 'electron hole' and a resultant positively charged Se ion - i.e a positively charged ion closer to the valence band/TFT array resulting in an intrinsic charge differential from conduction band (closer to the patient being negative) to the valence band(closer to the TFT array being positive). The electrodes that sandwich the semiconductor establish a uniform potential difference which is negative near patient and positive near the TFT array in order to establish linear electric field lines and hence allow linear flow of electrons towards the TFT array - this current is then detected by the TFT array and helps localise the exact point of xray interaction/ionpair-electron hole generation resulting in direct radiography's excellent spatial resolution.
This is my current understanding and hence am confused with +electrode being close to patient and -electrode near the TFT array in the video presentation above. Am i missing something or do the signs need to be different on the video slide?
Thank for your help. Your video's are great and I've signed up on Radiopaedia for both your courses in preperation for my RANZCR exam!@@radiologytutorials
PS - Thought about this a bit more. The charge signs on your slides would make sense if the positive charged x-ray receiving surface closer to the patient means that the released electrons flow towards the patient and positive ion (aka electron hole) flows towards the TFT array (in opposite directions), along the electric field lines. The low bias voltage would be low enough the electron flux to the surface elecrode would cancel the surface charge and the variation in charge near the TFT array is what is measured by the TFT array to generate a signal for digital conversion. Resolution would be much precise as the electron hole flow (representing the site of xray interaction) is linear along the electric field lines leading to accurate representation of xray interaction.
Would this be a better way to understand this. I can forsee an exam question asking to describe how direct radiography works.
Thank you very much
amazing
Thank you!