GB2232550A - Optical sensor - Google Patents

Description

OPTICAL SENSOR This invention relates to an optical sensor capable of

determining the direction of arrival of a beam of light. Such devices are known as direction-of-arrival sensors (DOAS).

DOAS are known. One known device is termed a quadrant detector and consists of four planar photodetectors, one each disposed in the four quadrants of a plane about imaginary X-Y axes. Any beam of light striking the detectors is first focussed by a lens. The actual direction of arrival of the light beam determines precisely where, upon the four quadrants, the focussed beam strikes. The relative conduction of the four photodetectors can then be translated into a geometric direction of arrival. A further known device replaces the four quadrant photodetectors with a CCD array. Again, depending upon where the focussed beam strikes the array, the received signal can be translated into a direction of arrival. DOAS have use in tracking optical objects or acting as an indicator to warn when a collimated light beam (e.g. a laser beam) is illuminating the detector, and from which direction the beam is arriving.

The present invention provides an improved DOAS which not only has very high accuracy in terms of angular resolution, but also can be fabricated in miniaturised form. It is thus capable of providing high resolution at short working distances.

According to the invention there is provided a direction-of-arrival optical sensor comprising a planar base and a three-dimensional body extending therefrom, said three-dimensional body being constituted by at least four planar contiguous reflective surfaces disposed such that the join lines of adjacent surfaces lie parallel to one another and perpendicular to the base, at least four optical sensors being disposed on the surface on said base, one for each reflective surface, each sensor extending outwardly on said base surface from the edge where its reflective surface contacts the base surface.

Four reflective surfaces form a cuboidal three-dimensional body. For simplicity of signal processing four reflective surfaces forming a rectangular parallelepiped (preferable a cube) are desirable.

A preferred embodiment of the invention will now be described# by way of example, with reference to the accompanying drawings, in which:- Figure 1 is a perspective view of a DOAS according to the invention.

Figure 2 is a 2-dimension-reduced side view of the DOAS of Figure 1, and Appendix.

Figure 3 is a plan view of the DOAS of Figure 1 to explain how the direction of arrival of a beam of light is computed.

Figure 4 is used for explanation purposes in the Referring to Figure 1, a DOAS according to the invention is shown which comprises a semiconductive substrate 2 from which extends a cube 4 having its four upstanding surfaces 6 mirrored to provide as perfect reflectors as possible for the incoming light whose direction of arrival it is desired to sense. Extending from the base edge of each of these four reflective surfaces, supported on substrate 2, is a photodetector 8 having a square active area. The conductive leads from the four photodetectors are lead to detector electronics (shown schematically at 10). The latter determines, by means of voltage ratios, the ratio of conduction, and hence incident light falling on the two opposed pairs of photodetectors. By processing these ratios, as explained below, it is possible to determine the angle of incidence of a beam of light striking the DOAS. A focussing lens disposed between the cube and photodetectors, and the source of light, is not required. A flat protective 1 window may, however, be disposed.

The manner in which the angle of incidence is determined will now be explained. For the purposes of the explanation, it will be assumed that the size of the DOAS relative to the distance to the light source is such that the light waves striking the DOAS are planar.

Referring to Figure 2, a 2-dimensional reduced version of the DOAS is shown. In essence, this consists just of one pair of the opposing pairs of photodetectors 8 and associated mirrors 6.

The incident light striking the arrangement from a source, S, is shown with an angle of incidence G where G = tan-1 h/w where h is the height of the mirrors 6 and w is the length of the active area of the photodetectors.

The useful detecting range 6 is also shown. This is defined by the limiting angle E?aR at which light incident on mirrors 6 ceases to be reflected back onto photodetectors 8 and is reflected back into space. In such circumstance, the theoretical useful range is:t (90- E?aR). For h=w, aR=450 and the useful range is 900. In practice a useful range less than this is likely as, at the limits, one or other of the photodetectors would be barely illuminated and the device least accurate. A more practical useful range is probably about 800.

Consider now the 3-dimensional version, in Figure 3, as represented by the DOAS of the invention. The light source whose angle of incidence is being detected can be considered to lie on a section of a hemisphere, radius R, directly above the DOAS. The projection of this hemisphere onto the ground plane of photodetectors 8 (i.e. onto substrate 2) is shown in Figure 3 as P, as is the boundary of a section of the hemisphere (PS) at which the angle of the beam of light to the normal to the ground plane is (90- G aR). An imaginary x-y axis in the ground plane whose origin is at the centre of the cube 4 and whose axes are parallel to the two pairs of photodetectors is also shown.

As the projection of the hemisphere section is symmetrical about the normal, only one quadrant of the hemisphere section need be considered, as illustrated. The solution for the position of the projection of the light source on this quadrant section is identical for other 3 quadrants. It is within one of these four quadrants that the projection of light source, S, must lie if it is to be detected in position by the DOAS.

The problem of detecting the position of S within the quadrant, and hence its direction-of-arrival, can be solved from the ratios of the voltages on the two opposing pairs of photodetectors. The comprehensive solution to this problem is given in Appendix 1 hereto, together with an assessment of accuracy of the device. The solution will be summarised here. In order to practice the invention, only the final results to be quoted are necessary. The solution given in Appendix 1 is simply for additional guidance.

Let the photodetectors be identified as Dxpr Dxn; Dyp, Dyn and the ratio of the voltages on these pairs be expressed as P(Dxp) P(Dxn) and P (Dyp) P (Dyn) The problem of determining the position of S is solved by considering the quadrant section to be divided into 3 areas: Area 1 The projection of S is at S1, where the coordinates of S1 are x < w/2, y <,,, w/2 (w being the length of the 1 - R - side of the cube).

Sl is then almost directly above the cube and there will be no light incident on mirrors 6, and thus no reflected light falling on the photodetectors. Each will receive approximately the same amount of incident light. Thus, to a first approximation, if S is at Sl:

P(Dxp) P(DX2) = 1.

P(Dxn) P(Dyn) If the 3-dimension Cartesian coordinates of S are XS, YS, ZS, then XS=YS=O; ZS=l. Area 2 The projection of S is within one of two narrow bands at the edges of the quadrant, extending parallel to the x and y-axes respectively. These are shown as S2a and S2b.

For S2a. x w/2 y > w/2 S2b y w/2 x > w/2 Only one of these bands needs to be considered, as the solution for the other is identical (but orthogonal).

If the projection is within S2a, there will be direct illumination of each of the four photodetectors and additionally reflection off the mirror for Dyp. There will also be an area in shade on at least one of Dxn and Dxp, but to a first approximation this can be ignored (see Appendix 1). Thus if S is within S2a:

Dxp = 1 and a12 ', 1 - Dxn Dyn (If PI]2 = 1 and Dxp \ 1, then S is within S2b).

Dyn Dxn If the Cartesian coordinates of S on its hemisphere are again XS, YS, ZS, then: XS = 0 YS = tan Oa, where a is the angle of the ZS 6 - line OS, from the origin to S, to the Z-axis in the Y plane.

If RY = P(Dyn) P (Dyp) it can be shown (Appendix 1) that Oa = tan-l( 1-Ry) 1+RY Area 3 The projection of S is shown as S3, where the coordinates of S3 are: x > w/2 Y > w/2.

It can easily be seen that:

P(Dxp) P(DYP) 1.

P(Dxn) P(Dyn) Again, let the coordinates of S be XS, YS, ZS.

Let P(DX2) R P(Dyn) P (Dxp) = R n P (Dxn) C) A = XS ZS B = YS R nj-1 R (+1 R ^_1 (d) = X5 = K6 It can then be shown (Appendix 1) that:

B = [-(K6-K5-2)-((K6-K5-2)2 -8K5)1/2]/2 A = [-K5-K6-2)-((K5-K6-2)2-8K6)1/21/2 giving XS and YS, and herice the position of S.

ZS ZS It is more useful to convert A and B into direction cosines for the unit vector, U, in the direction of S.

zs = cos '6 R YS = cos R = cos P Again, it can be shown:

(A2 + B2 + l)-1/2 = cos B = cos 3 (A2+B2+1)1/2 - A = co S (A2+B2+1)1/2 Thus the direction of S in Area (cos CK ' cos p ' cos) - 3 is given by In a working embodiment of the invention, a mirror cube of 1 cm3 was formed of metal electroplated with gold to provide a highly reflective surface for a 633 nm He/Ne laser light source. The cube was adhered to a substrate carrying four silicon photodetectors of 1 cm2 in area. This arrangement proved to have a repeatability and sensitivity accurate to better than 0.20 and an absolute accuracy of -11' 10 at the extremes of a 300 cone within which the light source was positioned.

8 APPENDIX 1 The same symbols and references as used in the preceding description will continue to be employed.

Assume in a working example that w = h = 250 pm and Rmin -"' 12 cm (where Rmin is the minimum working distance of the source, S, from the DOAS).

S emits spherical waves, but at Rmin these will be locally plane over a dimension-tw. Assume that the accuracy required is Ck. 0.20 (this is 1 part in 450 over a range of 90). Since the problem is symmetrical, an accuracy of 0.20 is required over a range of 0 < 0 < 450. Thus 1 part in 225 is a broad requirement for the ratio measurement of the powers falling on the photodetectors, assuming they have reasonably flat sensitivities. Thus, if a factor in the position calculation has a value substantially less than 1/225, it can be neglected.

1. INCIDENT POWER ON A PLANE SURFACE Consider the incident power falling on a plane area, W2, from the source S. Let the intensity on the plane area, positioned at a distance R from the source S and with the normal to the plane directed at the source, be ILR-2. Then 2 2 the incident power on this area 'S ILR- W. If the plane is tilted relative to the source by an angle 0, then the incident power is reduced by cos 0. The power, P, is then:

p = 1LR-2W2 COS 0 Unless source S is directly over the DOAS (x=O, y=O), then for any two opposed photodetectors (i.e. Dxp, Dxn; Dyp, Dyn) distance R is not quite the same.

Assume that the distance to one photodetector (e.g.

Dxp) is R, and the distance to the opposing photodetector (Dxn) is R+8.

Then P (Dxp) ILR7 2W2 COS 0 P (Dxn) L (R+S) -2W2 COS 0 and the ratio of these two power is:

( _.1) 2 R+8 1 9 As & < w < R ( R '2 1 R+6) The conclusion from this is that the small difference in distance of source S from any two opposed pairs of photodetectors can be ignored when considering power ratios. 2. SHADOW CALCULATION

For various of the calculations given below, it is necessary to determine the position of any shadows cast by the cube 4 onto the photodetectors.

One such shadow is shown in Figure 4, in relation to Dxn. Figure 4 is a plan view looking down onto the DOAS.

Let the top corners of the cube be A, B, C, D. The edge of the shadow on Dxn is the intersection of the plane containing S and cube edge CD with the plane of Dxn (the x,y plane). This intersection is shown as line L, and the actual extent of the shadow line is RQ. Such shadow lines are always parallel to the sides of the cube and therefore one needs to determine only one point on the shadow line to define its position. For Dxn (or Dxp) one needs only to calculate the x-coordinate of the shadow line on the x,y plane. 25 Let S be denoted by S(Xs, Ys, Zs). Let any point P in the plane SCD be P(Xp, Yp, Zp). SP is given by (Xp-Xs, YpYS, ZP-Zs). Let 1Stl = M then the line containing S and P is given by X-Xs Y-Ys Z-Zs t UX UY Uz and unit vector U XP-Xs' YP-ys, ZP-U M M M identical to U = (Ux, Uy, Uz).

When Z = 0, t -Zs -ZS.M Uz Zp-Zs then X = tUx + Xs -ZS.M.(XP-XS) + Xs (ZP-ZS). M X = ZpXs-ZSXp xi (1) Zp-Zs Y = zpYS-zSYP Zp-Zs Yi (2) Xi, Yi are the coordinates of the intersection of the line SP with the xy plane and def ine the position of the shadow line.

For example, when P is taken to be the corner, C, of the cube, then point Xi, Yi is point R on the shadow line. 3. AREA 1 When the source S is directly over the centre of the cube (Xs=o, Ys=0), all four photodetectors are symmetrically illuminated and clearly P(Dxp) P(Dyp) exactly 1.

P(Dxn) P(Dyn) The worst error arising from differences in shadows occurs when the source is directly above one edge of the cube (i.e. at the edge of Area 1). Referring to Figure 4, assume that S is directly above cube edge AB. Dxp will be fully illuminated whereas the edge of Dxn closest to the cube will be in shadow. Under such circumstances, one can calculate the position of the shadow line using equations (1) and (2) above with S = (w/2, w/2, R) and point P = point C = (-w/2, w/2, 0).

It can be shown by simple trigonometry using (1) and (2) that whereas the width of photodetector Dxn is w, the width of the shadow line equals W2 R-w 11 The fractional width covered by the shadow on D= is thus:

W R-w Using the numerical examples for R and w given above, this gives a fractional width of 1/479. This is within the margin of error quoted at the beginning of this Appendix.

Therefore, to a first approximation, within Area 1, any shadow differences can be ignored when considering the photocurrent ratios of the photodetectors.

4. AREA 2 Consider Area 2a. When the projection of S is within 2a - not at its edge - we have four areas of direct illumination (on Dxp, Dxn, Dyp, Dyn), one area of reflected illumination (on Dyp), and three areas in shadow (on Dxp, Dxn, Dyn).

Dxp, Dxn shadows On each of Dxp, Dxn there are two shadow lines. For Dxn, for example, the first shadow line is in direction L, and the extent of this falling on Dxn is RQ1 (Figure 4). The second shadow line is the intersection of the plane containing S and the cube vertical edge to corner C with the x, y plane. This is in direction LO. These two shadow lines intersect on Dxn at R, from which the shadow area thereon can be deduced. This is depicted by the shaded area, on Figure 4, Co (directly below C), R, Q', Do (directly below D). A similar shadow area occurs on Dxp.

The worst difference in the shadow areas on Dxp, Dxn occurs when source S is at the edge of Area Sa, e.g. Xs=w12. In this case Dxp has no shadow at all and is totally illuminated. It can be shown using the numerical examples for R and w quoted at the beginning of this Appendix that, at this position, the fractional area of Dxn in shadow is 1/456. This is insignificant and can be discounted.

Therefore, within Area 2a, any differences in the shadows on Dxp, Dxn can be ignored.

Thus 12 P (Dxp) =1. P (Dxn) Dyn shadow The position of the ' edge of the shadow on Dyn can be calculated from the projection of the line SA (from source S to cube corner A) onto the x,y plane on Dyn.

A = (w/2, -w/2, w) S (0 k-.-, w/ 2, w/ 2 -+ R, R4-7 R) -2 J 72 The y coordinate of the intersection, Yi, is independent 15 of Xs (equation (2), above). To a first approximation, y 2 + z 2 = R 2, since X is small.

Thus Yi = W \ - S+ Zs 1 - (Y ( w-Z s 2) The illuminated area on Dyn, as a fraction of the total area (W2), is (Yi + 312w)1w Ys-Zs+3/2w W-Zs, (3) Dyr) Dyp receives full direct illumination and a reflection off its mirror face. As the projection of S is within the narrow area S2a, and assuming 100% reflectance, and R>>w, it can be assumed that Dyp captures all of the reflected light.

Within Area 2a, source S is at an angle to the detectors and to the mirror adjacent Dyp (angle 0 quoted in this Appendix, Section 1 above).

To a first approximation, again as R>>w, the angle, of the plane Dyp (and Dyn) to S is Zs/R. The angle, P, for 13 the mirror adjacent Dyp is Ys/R. Now:

P(Dyp) = P(direct illumination)+P(reflected illumination) = I LR-Mcos 7 + ILR-2w 2 cos J6 ILR -2 Zs W 2 + I L R -2 Ys W 2 R R (4) P(Dyn) = P(direct illumination)(fraction illuminated) where (fraction illuminated) is expression (3) above.

P(Dyn) = R -2ZS W2) Ys-Zs+312w) (5) CL - 1 - R (1 W-Zs Dividing (5) by (4) and neglecting terms in w P(Dyn) 1-(Ys/Zs) P(Dyp) 1+(YS/ZS) From the specification description above,

Ys/Zs = tan Oa P (Dyn) = Ry P (Dyp) then Oa = tan-' (I-Ry) 1+Ry This result provides a single valued graph Oa against Ry which, to a first approximation, can be considered linear.

5. AREA 3 If S is at the position illustrated in Figure 4, then Dxn and Dyn receive direct illumination but are each partially in shadow. Dyp and Dxp receive direct illumination and light reflected off its respective adjacent cube mirror. consider detectors Dxn and Dxp in turn, 14 Detector Dxn The illuminated area Axn 5. consists of a rectangular portion 5 portion. z.

and a triangular Axn = (Xc 1 + 312w)w + (-w/2 - Xcl)(w/2-Yc 1).112 is:

(w 2 - 1 S-Zs+3/2w - (w+2Xs)(W-2YS) 7 zs 8(w-Zs) P (Dxn) = L R-2ZS.Axn R = I lzS 2 R-2(_ (M: S) (W-2Ys) (6) L1 - _) (Xs-Zs+3/2w-(w+2X -Zs 8(W-Zs) R W A similar result for Dyn is:

P (Dyn) = Detector DxD The illuminated area Axp consists of the whole area of Dxp receiving direct illumination plus reflected illumination from its adjacent mirror. The reflected illumination only partially overlaps Dxp I R -2(LS)(Ys-Zs+3/2w-(W-2Xs)(W+2YS (7) R 8(w-Zs) It can be shown that the proportion of Dxp receiving reflected illumination, i.e.

area illuminated by relected light total area of Dxp 3 2w-2 Zs+Ys 2 (w-Zs) (8) Thus, the power captured by Dxp by direct illumination is ILR -2 ZS W2, and R the power captured by its adjacent mirror is ILR -2 XS W2 R (9) (10) The amount of power reflected onto Dxp is thus (10) multiplied by (8). Thus, P(Dxp) = (9) + (10).(8) 15 which, simplifying terms, gives P (Dxp) = IL R-2 W2 z s+Xs(312w+Ys-2Zs? R 2(w-Zs) J A similar result can be obtained for Dyp (11) P (Dyp) = L R-2 W2 s+YS(3/2W-2ZS+Xs (12) R 2(w-Zs) There are six possible power ratios from the powers of the four photodetectors, but as only two pieces of information, Xs and Ys, are required, only two of these Fs- PS ratios need be considered Consider P(Dyp) P(Dyn) P (Dyp) = (12) = P@ P (Dyn) (7) If terms in w are neglected then 2-- 1+(YSJZS-(1/2)(XS/ZS)(YS/ZS)) 1-(YSIZS-(112)(Xs/Zs)(ys/Zs)) 16 rearranging Let A = Xs/Zs; B = Ys/Zs then B - AB = Kc Similarly for P(Dxp) Pn, P(Dxn) pie 1+(Xs/ZS-(1/2)(XS/ZS)(ys/ZS)) 1-(Xs/Zs-(1/2)(Xs/Zs)(Ys/Zs)) (XSJZS-(112)(XS,ZS)(ys/ZS)) = C, Ybi A - AB = K6 where K,, K6 are constants From (13), (14) (Ys/Zs-(112) (Xs1Zs) (ys/Zs)) K5 B = [-(K6 -K5-2) I(K6 -K5-2)2_8K5]12 A = [-(K5-K6-2) 61/2 (13) (14) The range of valid roots within a 450 capture for the DOAS is 0 < A, B < 1 1 (15) is 1 7 Consider the roots which will satisfy (15).

For B, a = 1 since 1<R2<mo, O<K6<1 b = (K6-KS-2) -3<b<-1 1<Rj<(0, O<K5<i c = 2K5 0<c<2 Consider (-b)+(b 2 -4ac) 1/2 2 and (-b) is positive, hence _b+ (b2 -4ac) 1/2 < 1 b 1 2 4ac is positive but when (4ac) is large, K. is large and K6 is small, hence 1 b 1 ---bjp 3.

When (4ac) is small, K. is small, and K 6 is large, hence Ibl->l.

It is reasonable to use the negative root since (-b) - (b 2 -4ac) 1/2 is positive and <1 is desired. A similar 2 result holds for A.

Thus B = [-(K CKS-2)-((K6-KS-2) 2 -8K 5) 1/2]12 A = [-(K 5 -K 6 -2) - ( (K5-K 6 -2)2 -SK 6) 1/2] 12 giving the formulae quoted in the patent specification above.

Claims (9)

1. CLAIMS:1. A direction of arrival optical sensor which comprises a planar

   base and a three-dimensional body extending therefrom, said threedimensional body being constituted by at least four planar contiguous reflective surfaces disposed such that the join lines of adjacent surfaces lie parallel to one another and perpendicular to the base, at least four optical sensors being disposed on the surface on said base, one for each reflective surface, each sensor extending outwardly on said base surface from the edge where its reflective surface contacts the base surface.
2. 2. A sensor according to claim 1 wherein said three-dimensional body is a rectangular parallelepiped.
3. 3. A sensor according to claim 2 wherein the parallelepiped is a cube.
4. 4. A sensor according to any of claims 1 to 3 wherein each optical sensor is a semiconductive photosensor.
5. 5. A sensor according to any of claims 1 to 4 wherein each sensor has an optically-sensitive area of rectangular shape.
6. 6. A sensor according to claim 5 wherein each rectangle is a square.
7. 7. A sensor according to any of claims 1 to 6 wherein said base comprises a semiconductive substrate.
8. 8. A sensor according to any of claims 1 to 7 which additionally comprises means for comparing the outputs received from said optical sensors and for computing therefrom the direction of arrival of an incident beam of light.
9. 9. A sensor according to claim 8 wherein said comparing means comprises means for comparing ratios of voltages of respective pairs of said optical sensors.

   Published 1990rt The patent Office, State House,6671 High Holborn, LondonWCIR4TP.Purtbe.r copies MaYbeobtainedfrom The Patent Office. Sales Branch. St Mary Cray, Orpington, Kent BR5 3RD. Printed by Multiplex techniclues ltd, St M8.17 Cray, Kent, Con. 1/87