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I am very new to semiconductors and I'm trying to understand how photodiodes work for imaging sensors.

If I am not wrong, a PN junction is required to detect electron/hole pairs that are photo-generated. The electric field in the depletion zone will move the charge carriers and create a current that can be exploited.

Why not only use a semiconductor without doping? And then apply an electric field to the semiconductor: when light is emitted, it slightly improves the material's conductivity, and a current is detected.

Why is it not that simple?

Thanks

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    $\begingroup$ You can do that. What the pn junction does is place a fairly large field (since you reverse bias the junction) close to the surface (where more photons will be absorbed) so that your get more signal. $\endgroup$
    – Jon Custer
    Commented Jul 10, 2023 at 12:45
  • $\begingroup$ Thanks a lot for your answer ! Would you mean that with a PN junction, the electric field distribution is better for the task than if you polarize the cristal with a generator? How does the electric field look like in the material if you apply an electric field? Is it rather concentrated inside the cristal, where photons are less likely to travel (as most of them are absorbed at the surface)? $\endgroup$
    – bruno
    Commented Jul 10, 2023 at 14:18
  • $\begingroup$ Related: physics.stackexchange.com/questions/770248/… $\endgroup$
    – John Doty
    Commented Jul 10, 2023 at 14:27
  • $\begingroup$ Consider a half-micron thick depletion layer with 100V across it. That said, for transient conductivity measurements you will see increases in current in a lateral pattern from the photocarriers. $\endgroup$
    – Jon Custer
    Commented Jul 10, 2023 at 14:37

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Why not only use a semiconductor without doping? And then apply an electric field to the semiconductor: when light is emitted, it slightly improves the material's conductivity, and a current is detected.

This is commonly done. The device is called a photoresistor or light-dependent resistor (LDR). This technology is used when cost must be kept low and high sensitivity is not required.

It's also possible to make vacuum tube photodetectors that involve no semiconductors at all.

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  • $\begingroup$ Thanks a lot for your answer! $\endgroup$
    – bruno
    Commented Jul 11, 2023 at 7:32
  • $\begingroup$ And could you explain why the sensitivity is higher with a PN junction? Does it have to do with the distribution of the electric field ? $\endgroup$
    – bruno
    Commented Jul 11, 2023 at 8:09
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    $\begingroup$ @bruno, because, like the other answer says, the leakage current is much lower, so if you see a few electrons pass through you know they're caused by a light signal and not just by the basic conductivity of the device. $\endgroup$
    – The Photon
    Commented Jul 11, 2023 at 14:23
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The advantage of using a reverse biased PN junction is that the dark current is zero. Then when we illuminate the junction the light creates minority carriers that cause a small current to flow through the reverse biased junction.

We do this because it is easy to measure a small current when the original current was zero. Without the junction we would have to measure a small change in a large dark current. This can be done of course, but it's much harder to get the same accuracy as when the dark current was zero.

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  • $\begingroup$ Dark current is not zero, it can even be quite high for IR detectors. $\endgroup$
    – Matt
    Commented Jul 10, 2023 at 22:03
  • $\begingroup$ Why would the dark current be minimized in a PN junction? $\endgroup$
    – bruno
    Commented Jul 11, 2023 at 8:08
  • $\begingroup$ @bruno Because if you reverse bias a PN junction no current will flow. That's what a diode does. It allows current to flow in the forward direction but blocks current flow in the reverse direction. So to use a photodiode you reverse bias it so no current flows in the dark. When you illuminate it the light produces minority carriers that will flow across the reverse biased junction so light causes a current. $\endgroup$
    – John Rennie
    Commented Jul 11, 2023 at 8:11
  • $\begingroup$ Thank you! I am a bit confused: I understood that minority carriers can also form in the depletion zone due to heat, right? And these electrons will also flow, so it is impossible to distinguish photo-generated electrons from thermally generated electrons. Do you mean that the reverse bias diode prevents thermally generated electrons to flow apart from those created in the depletion zone? So SNR is better this way ? $\endgroup$
    – bruno
    Commented Jul 11, 2023 at 9:03
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    $\begingroup$ @JohnRennie, you know this, but the dark current is not strictly zero in a photoconductive detector --- and it isn't insignificant either --- it's a critical parameter to determining the device's sensitivity. If you want 0 dark current, operate in photovoltaic mode. $\endgroup$
    – The Photon
    Commented Jul 11, 2023 at 14:25

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