Talk:Quantum efficiency

The formula for QE does not look correct. It is stated as being the ratio of the number of electrons produced to the number of photons absorbed. However, at say X-ray wavelengths, the number of electron-hole pairs created by the absorption of one X-ray photon can be greater than one, which leads to a QE>1. For this reason, I think that the formula needs to be revised. Ty8inf (talk) 03:33, 10 June 2008 (UTC)

Yes, in that case, the QE can exceed 1. I'm not sure they call that, though. Do you have any relevant sources? Dicklyon (talk) 03:35, 10 June 2008 (UTC)

The QE curves for the detectors used in X-ray astrophysics (Chandra, XMM/Newton, Suzaku, etc) all have values less than 1. For example here is a URL for the Chandra ACIS CCDs: [1].Ty8inf (talk) 05:23, 10 June 2008 (UTC)

In addition to the QE>1 problem, the current definition makes no distinction between a charge-cloud produced in the active layer and one produced by absorbtion of a photon in a dead layer. For these reasons, I think that the definition should be changed to be something along the lines of "The quantum-efficency of a CCD is the probability that a photon incident upon it will produce a charge-cloud that is recorded by the CCD electronics."Ty8inf (talk) 02:39, 11 June 2008 (UTC)

That's interesting. I wonder what the QE is defined as in the context of an X-Ray detector such as Chandra. Is your proposed definition made up, or sourced? Dicklyon (talk) 02:54, 11 June 2008 (UTC)

My definition is constent with how it is used in X-ray astrophysics and I believe it is used that way in particle physics. It also makes sense to me given the words "quantum" and "efficiency". Here "quantum" refers to the photon and "efficiency" refers to how well the device can detect the photon. Ty8inf (talk) 03:11, 11 June 2008 (UTC)

That's great. Then all we need is a source; and if it differs from the way it's used in light detectors, we need to distinguish the two uses. Dicklyon (talk) 03:20, 11 June 2008 (UTC)

Here is one: [2]. I believe that it is consistent with my definition and appears to be more general. Ty8inf (talk) 03:33, 11 June 2008 (UTC)

That sounds like a good general way to put it. We would definitely want to also keep in the more specialized but common case of electrons per photon for typical light detectors such as silicon image sensors. Is there also a source for the "cloud" concept or terminology? How do these X-Ray detectors work? Dicklyon (talk) 03:41, 11 June 2008 (UTC)

I think that it shows that the notion of QE is much broader than that indicated by the article. As far as how X-ray detectors work, it depends upon what you want to measure. If you want to measure something with milli-second resolution, then a CCD is probably not going to work well. Something like a microchannel plate (MCP) detector would be a better choice. The way an MCP works is very different from the way a CCD works. The article Gamma_spectroscopy provides an overview of various X- and Gamma-Ray detectors and each of them can be characterized in terms of a QE. Ty8inf (talk) 04:01, 11 June 2008 (UTC)

But how much broader? If we can describe a few applications, via sources, we can give a good idea how broad it is, I think. I had only ever used it in silicon light detectors myself, so my perspective was narrow. But since my perspective largely agreed with the cited sources, I didn't have a good way to learn more. WP:SOFIXIT please. Dicklyon (talk) 04:09, 11 June 2008 (UTC)

How about LEDs? When talking about LEDs our output is the photons and our input is electrons? Casey boy (talk) 14:40, 6 March 2010 (UTC)

I think the QE of an LED is quite legitimate to talk about. I wish it was used more often. For the blue LEDs in LED lamps, the power-conversion can be about 25%, and the voltage drop is nearly equal to the photon-energy in eV (electron-volts), so the QE is close to 25% too. For laser diodes operating well above their startup threshold current, it is often 50-80% I think. Such devices are limited to the intuitive 100% QE. — Preceding unsigned comment added by Jim Swenson (talk • contribs) 04:28, 10 February 2016 (UTC)

Quantum efficiency and image noise

A portion of the noise in an image will come from shot noise due to a finite number of photons hitting each photoreceptor during the exposure. If a sensor had low readout noise and high quantum efficiency, I would think that the number of counts recorded for each photoreceptor would be close to the exact number of photons to hit the photoreceptor. On the other hand, if the QE is low but the electronics are otherwise very good, it seems like the the image noise would be dominated by the noise due to QE; that is, if a sensor had a QE of 10%, it would take on average 10 photons to increment the count for that photoreceptor, but there would be a chance that 9 photons would increment the count.

Is this accurate? Can low QE lead to increased noise? All other things being equal, will increasing QE and reducing exposure time lead to reduced noise, up to the shot-noise limit? If so, does that lead to diminishing returns on QE so that more than say 80% QE just wouldn't help any more. Is this what happens in practice or do other factors such as readout noise start to dominate? —Ben FrantzDale (talk) 13:07, 14 October 2008 (UTC)

Yes, sort of. But it's much simpler than that. When QE is low (or high or medium for that matter), the shot noise is the shot noise associated with the number of electrons. If you have QE of 25%, you get a quarter as many electrons, on average, and half as much shot noise variance, so the SNR is a factor of two worse. You don't need to consider the number of photons separately to get the right answer, since the electron events are independent.

At the low end of the dynamic range, readout noise always dominates, and at the high end shot noise always dominates; QE just scales the intensity level linearly, with no other complications. Dicklyon (talk) 16:01, 14 October 2008 (UTC)

I think our "noise" and "efficiency" terms are confusing, non-intuitive, or otherwise lacking in linguistic value. And that's what is impeding resolution of a fairly legitimate concern in this discussion. For example, avalanche photodiodes have lower input-referred shot noise than photomultipliers because the initial photon-to-carrier conversion has a higher Quantum Efficiency. The carrier-multiplication gain which comes after conversion then makes our use of the term "quantum efficiency" ambiguous: does QE refer to output electron-charges per photon, or countable pulse responses per photon? So perhaps it is incumbent on the writer using the term "QE" to specify what ratio it refers to. Or we make a newish term... "quantum response efficiency"? I wish a term for quantum noise was in use, rather than just "shot noise". jimswen (talk) 03:55, 10 February 2016 (UTC)

Quantum efficiency dimensionless ?

In the article, it is stated that "as a ratio, QE is dimensionless", but this isn't a ratio of numbers with the same units! The QE unit is e-/photon that's it. I think when QE is expressed in %, it is the ratio of the real QE and the theoretical maximum QE. If someone knows how to compute the theoretical maximum QE, it could be a nice addition to the article.

What is quantum efficiency in astronomy?

In astronomy, quantum efficiency (QE) refers to the effectiveness of an astronomical detector, such as a CCD camera, in converting incoming photons (light particles) into electrons, which are then recorded as an image. High QE is crucial in astrophotography, as it determines how many of the photons that hit the sensor are transformed into a useful signal, affecting the detector’s ability to capture faint astronomical objects and details. 馬鈴薯不是土豆 (talk) 03:23, 5 February 2024 (UTC)