Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

What is the difference between Buried Photodiode and Pinned Photodiode? I understand that the P+/N/P structure where the P+ and P layers have the same potential is the Pinned Photodiode. So what is the buried Photodiode?

share|improve this question
1  
Is this a "pinned" photodiode or a PIN (P/Intrinsic/N) photodiode? –  Brian Drummond Sep 21 '13 at 13:41

1 Answer 1

up vote 0 down vote accepted

This is a commonly misunderstood misused set of terminologies.

First off these are not PIN Photodiodes - which stands for P - Intrinsic- N. These have large depletion regions for higher internal QE (Quantum Efficiency) and faster response. You can't make an array with this design though.

Pinning, refers to fermi-level pinning or pinning to a certain voltage level. Or also the forcing or prevention of the fermi-level/voltage from moving in energy space.

You can get surface state pinning from the dangling Si/SiO2 bonds providing trapping centers. A buried PD (Photodiode) has a shallow implant that forces the charge carriers away from these surface traps. The Si/SiO2 surface contributes to increased leakage (dark current) and noise (particularly 1/f noise from trapping/de-trapping). So confusingly a buried PD avoids pinning of the fermi-level at the surface.

A pinned PD is by necessity a buried PD, but not all buried PD's are pinned. The first Pinned PD was invented by Teranishi at Sony and is used in ILT CCD PD's, these same PD's are used in most CMOS imagers built today.

A pinned PD is designed to have the collection region deplete out when reset. AS the PD depletes it becomes disconnected from the readout circuit and if designed properly will drain all charge out of the collection region (accomplishing complete charge transfer). And interesting side effect is that the capacitance of the PD drops to effectively zero and therefore the KTC noise \$ q_n = sqrt(KTC) \$ also goes to zero. When you design the depletion of the PD to deplete at a certain voltage you are pinning that PD to that voltage. That is where the term comes from.

share|improve this answer
    
"QE"? <padding> –  jippie Sep 21 '13 at 15:10
    
@jippie Explanation added - thanks –  placeholder Sep 21 '13 at 15:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.