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Beam centering using a photodiode

  1. Sep 27, 2017 #1

    Doc

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    Hi all,

    I am a mechanical engineer so I'm hoping to get a bit of guidance on something I am working on at the moment. I am working on a project which requires we keep actively keep a laser beam centered on a mirror. The mirror transfers a small percentage of beam power and I am hoping to use a quadrant photodiode to detect the position of the beam. The beam has a gaussian distribution, and the beam size will be larger than the active area of the photodiode due to physical constraints. The way I understand that this will function is that the incident beam flux on the photodiode will result in the photodiode generating output current. My question is:

    How can I detect where the bright central core of the beam is (I don't care about the low power outer regions of the beam.)?

    My feeling is that I will be doing this anyway: whichever quadrant the highest amount of flux is incident upon (if the bright core is mostly in one quadrant), then that quadrant will be outputting the most current, see Figure 1 below.

    figure 1.jpeg

    If I then needed to drive another mirror earlier in the system to recenter the beam, I assume that I effectively would want to move the active mirror in such a way that results in i1=i2=i3=i4 (approximately).

    Am I understanding this correctly?

    Thanks,
    Doc
     
  2. jcsd
  3. Sep 27, 2017 #2

    berkeman

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    Can you say more about the application? Can it tolerate a few percent dithering in the incident beam power? If it can, then just wiggle the mirror in the x and y axes by a percent or two, and use the detected beam power with your single photodiode to stay centered on the center of the beam.

    If not, it seems that you would need to use several photodetectors and calibration to try to ensure that the main mirror was staying centered...
     
  4. Sep 28, 2017 #3

    Doc

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    I have added another diagram below to give you an idea of the setup. The beam input at the bottom right will be changing angle back and forth and I want to ensure that, by using the active mirror, the beam is centered on the static mirror. The photodiode will provide feedback to the active mirror.

    figure 2.jpeg

    I want to use a quadrant photodiode (a diode split into a grid of 2x2 active elements).

    Thanks,
    Doc
     
  5. Sep 28, 2017 #4

    Baluncore

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    A Gaussian distribution has an infinite beam width so it must be wider than the photodiode separation. Is the beam so big it is always there, or do you need the mirror to acquire the beam when first turned on ?

    If the diode separation is too small, the top of a Gaussian distribution will appear flat and the mirror will never settle. You can better find and lock to the axis of the beam by separating the diodes so they are always on the slope either side of the peak on the Gaussian distribution.

    If you have independent X and Y control of the mirror, then you can have two independent channels. You do not need to look for equality of i1 = i2 = i3 = i4, all at the same time.

    Take the left and right diode currents and subtract them to provide the direction and magnitude of the mirror X drive. Take the high and low diode currents, subtract them to provide the direction and magnitude of the mirror Y drive.

    With an orthogonal diode array you might sum pairs of currents before subtraction. But if the diode array is rotated by 45° it will spread the diode separation to 140%. You can then subtract diode currents by wiring two outer diodes in series and converting the junction difference current to a voltage with an op-amp and feedback resistor.
     
  6. Oct 4, 2017 #5

    Doc

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    The beam size will always be larger than the photodiode active area so I don't think that initial acquisition will be necessary.


    This is why I want to use a quadrant photodiode (basically four diodes fixed together in a 2x2 grid with a small gap between each.) Unless there is something else that I am not understanding about this part?

    This makes sense thanks.

    The next thing that I am wondering about is the current output from the photodiode. The system will be running at two separate power modes: either at approximately 20 W when operating, or at approximately 1 mW for alignment. I have done some preliminary calculations but will need to check these: I am expecting output currents on the order of uA and nA respectively. I need to feed these currents to...something (to tell the correcting mirror what to do)? I assume that I also may need an amplifier to boost the signal somewhat, nA seems very small? Do I want to be reading out amps, or is it better to run it through a resistor to convert to voltage for some reason? I am pretty fuzzy on this part, so if anybody could provide a fairly simple conceptual explanation I would very much appreciate it.

    Regards,
    Doc
     
  7. Oct 4, 2017 #6

    Baluncore

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    They will be matched, but they may be too close together for the diameter of your beam.

    You have not given the electrical specifications of the mirror positioning drive.

    Attached is a conceptual circuit that will take the difference of two photodiode currents and converts the difference current to a voltage. The photo-diodes operate with 5V of reverse bias. The two Si diodes in the feedback give it a logarithmic response to current. R1 can be used to set a linear response to low currents. The system will need some feedback capacitance to slow the response rate of the control loop.
    Without better specifications for beam width and diode characteristics it is not really possible to go further with the design of a current difference amplifier.
     

    Attached Files:

  8. Oct 9, 2017 #7

    Doc

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    Okay thanks for the feedback. I might leave it at that stage for the time being as that's a little outside of my area of knowledge.

    I would like to check what I have done up until this point if you don't mind commenting further?

    I have included a quick sketch of the general layout of what is happening in Figure 1. The mirror (at left) receives a collimated gaussian laser beam of approximately 20 W. The mirror surface has a reflective coating of about 99.5%. The hashed area is a cylindrical mount (barrel) that the mirror is connected to. There is an aperture of approximately 8 mm behind the mirror leading into the barrel. After the barrel aperture there is an ND3 filter to step the laser power down. After the ND filter there is a quadrant diode with an approximate active area of 5x5 mm^2. It can be seen that the collimated beam will be focused to some point (approximately 100 mm from the curved surface) behind the barrel.

    side.jpeg
    Figure 1: General layout of system.

    Figure 2 below shows power lost from the beam before reaching the photodiode. There will be a portion of the beam mechanically vignetted by the barrel aperture (dashed area). What is remaining of the beam power that is not vignetted will be divided by 1000 (due to the ND3 filter). Of the reduced power area only a portion of it will actually interact with the photodiode active area.

    front.jpeg
    Figure 2: Power lost due to small photodiode active area and mechanical vignetting.

    Any comments on my understanding/errors so far would be much appreciated.

    From here I can look at specific photodiodes and calculate the saturation power, noise equivalent power, etc and make a decision on which one to go with based on those numbers?

    Regards,
    Doc
     
  9. Oct 12, 2017 #8

    Baluncore

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    This is part of a larger system. You have not given it context so it is hard to know what is important.
    You say the beam is Gaussian, but how wide is a standard deviation?
    It is vignetted by the mechanical aperture, what is the aperture diameter?
    Only the nose of the beam falls on a diode array.
    Mirror is 99.5% reflective, 20W * 0.5% = 1W. Then 1W* ND3 = 1mW total. But we do not know the area.

    We need an accurate plot of intensity across the beam, with the aperture shown.
     
  10. Oct 12, 2017 #9

    Doc

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    I will do my best to answer your questions here, Baluncore. Forgive me, laser physics and photonics is not my area, so if something does not make sense (I am probably explaining it badly or using the wrong nomenclature) please let me know.

    If you refer to my previous post and look at Figure 1: General layout of the system, when the beam is collimated and incident upon the curved reflective surface of the mirror at left it has a waist diameter of 10 mm (I understand this to mean the point at which the intensity begins to drop off by 1/e^2, or approximately 86%). Figure 1 below shows a plot of intensity vs beam diameter.

    beam_no_ap.png
    Figure 1: Gaussian beam distribution. Dashed vertical lines represent the active area of the quadrant photodiode. The dashed horizontal line represents the 1/e^2 intensity point.

    I replotted this graph to include the aperture in Figure 2 below.

    beam_with_ap.png
    Figure 2: Gaussian beam distribution with 8 mm aperture included.

    The approximate difference in intensity between the maximum beam intensity incident on the photodiode and the minimum beam intensity incident upon the photodiode is approximately 25%.

    Did you mean: 20*(1-0.995) = 0.1 W. Then 0.1 W*ND3 = 100 uW total?

    Thanks,
    Doc
     
  11. Oct 12, 2017 #10

    Baluncore

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    Wording is important, how can vertical lines represent an area of a quadrant photodiode? Are the diodes each 5x5 mm, making a square of four diodes a 10x10 mm square?

    During the beam acquisition phase, the skirt on one side of the beam will pass through the aperture to reach the photodiodes. The amplified diode difference currents then drive the beam towards the axis. The beam locks on axis when the currents are balanced.

    Both the acquisition and the axial locking benefit from widely spaced diodes. There is no reason why the diodes must be contiguous, they can be separated if the beam diameter is known. When locked, each diode should be centred on the steepest part of the beam skirts.

    If the diode area is greater than the aperture, any misalignment of the diodes with the aperture will bias the beam off-axis. That is not the case with a bigger aperture. Why such a small aperture?

    You are aligning the beam based on the profile of the beam part way to the focal point. I don't know if that might pose a problem.

    Laser wavelength may be important. I have not considered diffraction effects at the aperture, or the thickness of the ND filter measured in quarter wavelengths. Will it have wavelength dependent deep notches in transmission or does the attenuation separate the faces and so prevent cancellation.

    You are right. I was asleep.
     
  12. Oct 12, 2017 #11

    Doc

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    My apologies, bad phrasing. I should have said that the vertical lines mark the boundaries of the photodiode active area. So yes, the diodes are 5x5 mm each, making a square of four diodes of 10x10 mm^2 as you said.

    This makes sense, but I don't believe that the barrel diameter can be made much larger than 10 mm. I can see why it would be advantageous to make it so, but for now I'd like to assume that this cannot be changed.

    Yes this is a good point, I don't think that I would have thought of this. Thank you for bringing it up. The size of the aperture can probably be increased, it is simply a limitation of the current mechanical design of the physical mounting of the mirror.

    I'm sorry I do not understand the question. The laser power and modes will be as follows:
    - 20 W at 589 nm
    - 1 mW at either 532 nm or 589 nm

    Thanks,
    Doc
     
  13. Oct 17, 2017 #12

    Doc

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    Hi all,

    I'm still looking for a bit of feedback here if anybody has time: I'm still fairly unsure of how to proceed.

    Regards,
    Doc
     
  14. Oct 17, 2017 #13

    Tom.G

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    Hmmm... with an ND of three, wouldn't interference from the second surface reflection be reduced by a factor 1 000 000? Doesn't sound like much ripple.

    Or were you referring to ND filter-to-sensor cancellation?
     
  15. Oct 18, 2017 #14

    Baluncore

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    Exactly. When using a narrow-band laser source it is necessary to be aware of possible wavelength related problems.
    The varying thickness of the convex 99.5% mirror should not be a problem, but the construction and placement of the ND filter needs to be considered. While ND filter-to-sensor cancellation is possible, I expect the plane surfaces at the back of the mirror and the front of the NDF also need to be considered. A tuned thickness coating might be considered if there is a problem.

    We have insufficient detail information and as advocates to the Devil we really can only guess at what might later become a problem.
     
  16. Oct 21, 2017 #15
  17. Oct 22, 2017 #16

    Doc

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    Hi Baluncore,

    What further information do you require? I am happy to try to provide it.

    Regards,
    Doc
     
  18. Oct 22, 2017 #17

    Baluncore

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    What sort of laser? If the laser light is very narrow band then you may well have interference problems.

    You need to avoid parallel surfaces where there is an RI mismatch such as an air to glass contact. You could eliminate two air-glass interfaces by attaching the ND plate to the flat back of the mirror with an optical grease.
     
  19. Oct 23, 2017 #18

    Doc

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    It is continuous wave if that is what you are asking? Attaching the ND filter to the back of the mirror may be possible. Is an RI mismatch a huge problem considering the application?

    Regards,
    Doc
     
  20. Oct 23, 2017 #19

    Baluncore

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    No, but if it casts a partial interference pattern of fringes onto the photodiodes, it might not centre the beam correctly. It is hard to tell what will happen when the beam position detectors are part way between the mirror and the focal point.
    You need to set up your proposed optical path components on an optical bench and see what happens. Manually adjust the positions and find the sensitivity to beam position detection and look for the presence of fringes.
     
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