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I like to make a simple experiment to characterize the angular response of a particular photodiode. (SFH206K). Current thinking is to use a flat table, use a red laser, turn off the lights, point the laser to the photodiode at various angles and look at the current passing through with a multimeter.

I have done this test and find out my results correlate with datasheet nicely if the orientation of the photo diode is 0 degrees. However, if I change the axis of the photodiode, results do not match datasheet. (for ex. 45 degree, the photodiode's angular response is spherical, so this angular change should not have impacted the measurements).

I think my test set up is too rudimentary. Any ideas to measure what I am trying to measure more precisely? I am open to buying profesional equipment as well (needs to be very low cost though, say <1K)

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  • \$\begingroup\$ Please let us see photodiode you are talking about ) \$\endgroup\$ Commented Aug 13, 2011 at 7:00
  • \$\begingroup\$ You doubt your test when it doesn't agree with the datasheet, because you trust the datasheet? Why do you want the measurement yourself then? \$\endgroup\$
    – stevenvh
    Commented Aug 13, 2011 at 7:02
  • \$\begingroup\$ I am generally a curious guy also datasheet doesn't specify how those measurements are done. I take everything with a grain of salt. \$\endgroup\$
    – Frank
    Commented Aug 13, 2011 at 7:07
  • \$\begingroup\$ @barsmonster added the part number \$\endgroup\$
    – Frank
    Commented Aug 13, 2011 at 7:12
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    \$\begingroup\$ I added a link to the part's datasheet. We're trying to make users aware of the importance of this, especially for less common parts, so that others don't have to go searching for it and that everybody is sure to be talking about the same thing. Just trying to cultivate good habits. \$\endgroup\$
    – stevenvh
    Commented Aug 13, 2011 at 7:20

5 Answers 5

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NEW:

OK - 3rd time lucky - I'll edit all this "after the event" to remove material that has got too far off course:

The problem [tm]

  • My total argument is based around the understanding that you are tilting the photodiode off axis and NOT rotating it. If you are tilting it off axis relative to the centre of a half sphere then the following applies. If you are rotating it, then read the rest of the material :-).

The problem is that when you tilt the sensor off axis and then sweep the beam across the surface of the sensors "half sphere of sensitivity response" he beam is not crossing the centre of the sensors sensitivity sphere, except when tilt angle is zero. ie the variable geometry of your method is not representing the sensor response sphere properly. All you have to do to get a proper 3D characterisation is to rotate the sensor and then sweep the beam through 180 degrees.

The 3D graph below and partial table show you what is happening.

The graph is a plot of 15 degree steps across a half sphere from 0 degrees to 180 degrees in X and Y axis.

  • X axis = sweep of sensor 0 - 180 degrees.

  • Y axis = tilt of sensor 0 - 180 degree, with 90 degrees being normal to mounting surface.

  • Z axis is the product of sine (x angle) * sine(y angle)

The graph represents the mapping of the intensity of light falling normally on your sensor due to variations in angle due to sweeping of your beam in one axis AND tilting of the senor on another. The resultant SinA x SinB 3D surface forms a series of "hills" with varying height. The only legitimate curves are those which pass through the centre of the half sphere. The only legitimate curve shown here is the dark green one with Y axis = 90 degrees. All the other curves represent a tilted sensor with the beam never sweeping across the centre of the sensor. QED.

enter image description here

I've left the rest of the material in here as it all has some relevance - but I think the above (corresponding to the "cut to the chase" section at the end probably contains your answer. You may have to work through the spherical model section preceding it to visualise what is happening. If so, get a plastic beach ball and some dry erase markers before you start :-). I can visualise it now but initially it wouldn't gel mentally.


OLD:

  • Method is suspect

  • My experience with LED characterisation discussed - inverse of you problem and much in common

  • Sources of error discussed.

  • A "certain to work" method of checking half-spherical response is proposed. but - You may consider "cutting to the chase" and reading "sure fire evaluation rig" at the end :-).

  • Arising from the "surefire method" is the indication that the "true" angle of pointing when tilt and rotation are combined may not be what it seems to be. Tilt and rotate appears to be varying two angles at once with the sensor response being the product of the sensitivities for the two angles involved - which is probably what you are seeing.

I have used similar setups to characterise LEDs and arrays of LEDs. It only occurred to me as I wrote this that your assumption about half spherical symmetry may be incorrect. As this is uncertain and not wholly intuitive I have dealt with it last under "symmetry"

Items to look for are light paths other than the intended one and light sources other than intended. I found that beam reflections off unintended surfaces made a significant difference to my measurements.

You say "change the orientation of the photodiode" but you may be changing the orientation of the LASER. Results can be different.

  • If you are tilting the photodiode then it's most sensitive 0 degrees axis is "looking" at off beam objects while you illuminate it.

  • If you have a horizontal table or desk surface in your setup - even if it is notionally non reflective and even if there should be little beam energy there it can have an effect.

  • If there are off axis sources the sensor will respond differently if angled towards or away from them.

I used a simple aperture plate (aka mask or guide or shield or ...) to mask out most alternative light paths.

A good test is to take a sheet of opaque material and "wander it around" in the workspace while the beam is being measured without obstructing the intended beam bath. Clearly it should have no effect. If it does you have alternative paths or sources.


The diagram below shows the datasheet response curve for your device. You say the response will be spherical but it would of course be "half spherical" at best (and I assume that that is what you meant). I assume that you meant that the response curve with angle from axis would be the same in any radial direction. eg Response at 45 degrees is shown in the graph below as being about 0.71 of the 0 degree response. In the absence of information to the contrary, this should be the same result at 45 degrees immediately up from center, 45 degree down from centre, 45 degrees left of centre or 45 degrees right of centre.

If this is not true then you can try a test which seeks to demonstrate that the response of the device rather than the illumination from the measuring setup varies.

enter image description here

Below are two representations of the device as viewed from the top and from the side with major sensitivity axis to the right of the diagram in each case.

In the top view, a beam from H45 (horizontal, 45 degrees off axis) should elicit about 70% of the response of the same beam from H0 (Horizontal, 0 degrees off axis). A beam from h-45 (not shown but obviously 45 degrees down from h0 in this diagram or in practice 45 degrees radially off centre in a horizontal plane through the centre of the sensor).

In the side view (sensitive window facing towards right of diagram) V0 is the exact same axis as Ho in the top view, V45 is a beam from 45 degrees above the center of the sensor window and v-45 would be a similar beam from 45 degrees below the centre. V45 response should be about 70% of V0 response and V-45 should equal v45. Also properly V45 = v-45 = H45 = H-45

The same arrangements apply for beams from other angels such as H30, H-30 etc, with the "proper" response compared to H0/V0 being read from the sensitivity graph.

Possible happenstances:

From what you said, H45 = H-45, H30 = H-30 etc.

From what you said H45 <> V45 etc. So

  • Does V45 = V-45 - ie is their symmetry in the above and below plane at various angles. If not, why not.

    • If you vary spot size on the sensor does the effect on response shape vary? ie is perhaps the sensor being under illuminated areally when illuminated off axis? Or is it being over illuminated with excess beam being reflected in the housing or elsewhere.

    • If you are tilting the sensor to get off horizontal axis rotational results, does adding a series of horizontal slits so that all light enters only horizontally make a difference.

    • And more ...

enter image description here

Symmetry:

Your stated assumption about spherical symmetry (or half spherical) seemed to make sense initially. For emitters such as LEDs they tend to be point sources with the output level curve (the equivalent of your response curve) having "rotational symmetry" around the centre axis.

I would "expect" the device to be spherically symmetrical but it is conceivable that light from above and below axis exhibits an additional sine of off-axis-vertical-angle decrease - or some other law. Plotting results achieved at off axis angles of tilt would show visually if there was some smooth "law" in action. (eg linear decrease wuth tilt angle or sine, square, square-root etc).

Sure-fire evaluation method.

Best last :-).

You say you are willing to spend up to $1000. This costs much less than that in material but takes a degree of effort.

(1) Construct a half sphere of material of your choice of as large a size as you are reasonably able to do. This could eg by done with papier mache on an exercise ball. Newsprint and some paste and a happy few hours playing should produce a workable result. There are probably easy ways of getting larger half spheres than that.

(2) Provide a horizontal surface with sensor facing vertically upwards at centre (or vertical surface with sensor horizontal, but horizontal arrangement makes next step easier.)

(3) Coat flat surface plus inside of sphere with matt black surface - matt black polyurethane enamel should suffice.

(4) Provide small entry holes for light source spherically evenly distributed across half sphere. Probably every 30 degrees in arcs of the same separation would suffice to start. eg if spaced at 30 degree steps there are 7 holes at 0 30 60 90 120 150 180 degrees. As the 0 and 180 are about useless you could be happier with 20 or 15 degree steps.
Holes = (180/step_angle) + 1 - See: "number of holes" below.

(5) Consider masking unused holes with black mask - tape or whatever. this is annoying in operation and you could start with minimal external light compared with beam and see what happens.

(6) Insert light at each hole in turn, normal to sphere surface and impinging radially on sensor. If beam can be focused (which seems a very very good idea) try to reduce spot size to be about sensor surface size at distance involved.

Cut to the chase: Prior to building and trialling the above carry out some thoughts experiments (pen and paper and spreadsheeting may also be used if less capable than Isaac Newton at visualisation of effect of 3D motions involved.) Think about what happens when you tilt the beam to say 60 degrees off centre and traverse an arc. As ends are approached the beam angle is dropping towrds 90 degrees in two directions at once !. Response is (probably) the product of the sensitivities for the two angles involved !!!

To get true 1D variation in angle you need to traverse a 1/4 circular path on the sphere surface but tilt and rotate is tracing out an arc with angle varying in 2 dimensions at once. I think :-) (Not being Sir Isaac class myself).

Number of holes:

For 30 degree steps as above with 7 holes per arc, there are also 7 "arcs" The 0 degree arc is flat on the horizontal surface and all holes are at 90 degrees to the sensor. The next arc is at 30 degrees off the surface and has 7 holes but the first and last are common to all other arcs so there are 5 unique holes per arc.

Total holes = ((180/step + 1) arcs x (180/step -1) unique holes per arc ) + 2

= (180/step)^2 +1 holes

eg
for 30 degree steps, holes needed = (180/30)^2 +1 = 37
for 20 degree steps, holes needed = (180/20)^2 +1 = 82
for 15 degree steps, holes needed = (180/15)^2 +1 = 145

Starting with 30 degree steps seems wise :-)

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  • \$\begingroup\$ @Russell_McMahon Wow.. These types of answers are why I love this forum. I need to think about what you said however very detailed answer with considerable time spent. \$\endgroup\$
    – Frank
    Commented Aug 14, 2011 at 4:20
  • \$\begingroup\$ @Frank - are you TILTING the sensor rather than rotating it. If so I'm about 99% sure that the mechanical description at the top is your issue. If you are "r0tating" the sensor then this is not your issue. The rest is useful but probably not your main issue. \$\endgroup\$
    – Russell McMahon
    Commented Aug 14, 2011 at 9:29
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Reflections and other light sources are going to be your big pain. Most "black" colors are actually up to 30% reflective.

I think your best bet, setup wise, is to use a fiber optic, preferably single-mode but multi-mode might work as well, to illuminate the diode. Make it so it rotates the diode, not the source, and put the thing in a sealed box.

Good luck.

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  • \$\begingroup\$ @Mike_desimone I didn't know black was reflecting up to 30%. Good data.. Thx \$\endgroup\$
    – Frank
    Commented Aug 14, 2011 at 4:21
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Keep in mind that a laser diode is polarized. Polarized light reflects from a surface at different intensities based on the angle. See the Wikipedia article on Fresnel equations. For measuring this sort of thing in our lab, we use a semi-colimated white light source (I think a flashlight would work for your purposes). Fix the light source and then rotate the diode. Make sure you're rotating it about the point you want to measure - do you care about the total performance or the actual PIN diode? If it's the total performance, rotate about the diode about the front face.

Once you have the measured data, you'll need to normalize by \$1/cos(\theta)\$ because what you're really measuring is \$P_{optical}/A_{surface}\$. The surface area is reduced by the \$cos\$ of the angle that it makes with the incident beam. So, in order to measure just the power at each angle, you'll need to remove the affect of the reduced surface area that the light falls on.

To help understand this, think about holding a square card in front of you, now rotate the card and the surface area you see begins to decrease. Eventually, when you're looking directly at the card edge, you won't see any surface area. So, even if the photodiode had a perfectly flat response to angle, you'll still see the measured light decrease as \$cos(\theta)\$ because the light is falling on a smaller and smaller surface area.

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  • \$\begingroup\$ interesting point. I didn't know polarized light will have a different intensity based on the angle. This is a good point. I will change the light and try again. \$\endgroup\$
    – Frank
    Commented Aug 14, 2011 at 4:25
  • \$\begingroup\$ The different reflectivity of polarized light, based on the incident angle, is the principle by which polarized sunglasses cut down on glare. \$\endgroup\$
    – gallamine
    Commented Aug 15, 2011 at 19:19
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I think you may be misinterpreting the datasheet? Looking at the datasheet, the half angle is 60 degrees. At 45 degrees you should be seeing about 75% of the output compared to 0 degrees.

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  • \$\begingroup\$ think of it this way. The diode looks like "|" than I change the orientation to "/" and with same position of the laser, I expected to read the same thing but I didn't and that is why I was looking to understand what is going on. \$\endgroup\$
    – Frank
    Commented Aug 14, 2011 at 4:23
  • \$\begingroup\$ Frank, this is unclear. I can see what the diode looks like from the datasheet, but I think you should make a drawing of your setup. \$\endgroup\$
    – morten
    Commented Aug 15, 2011 at 20:19
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The datasheet presents the angular response with a tacit assumption that it is radially symmetric, but it just isn't.

You can see this with the naked eye. Look at the die of your photodetector from 45deg off normal at a few different angles around the perimeter of the device. For some angles the die is replicated in a reflection off the top face of the resin.

You can bet that effect shows up in the responsivity at those angles.

enter image description here

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