Optical Sensing for theHBCS-1100

Application Note 1008

IntroductionThe rapid growth of the digitalprocessing used in commercial,industrial and consumerproducts, has created the needfor sensors that convert physicalparameters into electricalsignals which may directlyinterface to a digital system.Optical sensors have utility ineach of these areas, in that theyprovide a quick, non-contactresponse to the parameterssought as a data source.Commercial applications includebar code scanning, paper edgesensing, end of tape sensing,and position and magnetic tapeloop stabilizing. Industrial usesmay consist of opticaltachometry, assembly linemonitoring, and safetyinterlocks, while the consumeruses of the optical sensor maybe found in audio products,entertainment products, andvideo games.

This application note describesthe electrical and optical designconsiderations for using discreteoptoelectronic devices, or theHBCS-1100 High ResolutionOptical Reflective Sensor. Theapplication areas addressed in-clude non-contact transmittive,or reflective sensor systems.

Each of these application areasincludes an optical emitter, atransmission path, and adetector to perform the sensingfunction. The sensing may occurby having the object obstructthe transmission path, orcomplete it by reflecting theemitter beam to the detector. Ineither the transmissive orreflective sensing configuration,there are optical, electrical, andmechanical considerations thatmust be addressed in order toinsure optimum performance ofthe emitter/detector system.

System ElementsEvery optical sensor system in-cludes a source of optical flux, atransmission path, and a receiv-ing detector. In most sensorsystem analysis, the availableflux and the responsivity of thereceiver are considered to beconstant and consequently thedynamic incidance change foundat the receiver results frommodifications of the transmis-sion path. Figure 1 shows apaper tape reader and densitom-eter examples of transmissivesensor systems. The presence orabsence of the paper hole resultsin a binary ON-OFF transmis-

Figure 1. Different Techniques of Flux Path Modulation.

REFLECTIVE TRANSMISSIVE

100%MODULATIONEDGE/HOLE

SENSING

PARTIALDIGITAL

MODULATION

ANALOGMODULATION

(DENSITOMETRY)

2

sion path attenuation. The den-sitometer functions by causingan analog modulation of thetransmission path. Bar codescanning and paper edge sensingare applications of reflective sen-sor techniques shown in Figure1. The bar code provides a reflec-tive contrast between thebackground and the bars. It isthe reflective differential thatmodulates the transmissionpath. In paper edge sensing ap-plications, the paper edgecreates a digital respondingtransmission system. When thepaper edge is not present in thereflecting field of the sensor, theflux transmission will be zero,and the transmittance will in-crease when the edge is in thefield.

Optical Transfer FunctionThe characteristics of the trans-mission path can be estimatedthrough the use of an opticaltransfer function, OTF. Thefunction is the ratio of the totaloptical flux available, φe, to theincident flux arriving at the re-ceiver, φe(R). This functionallows the designer to calculatethe amount of photocurrentavailable for the detector ampli-fier.

a transmittance, T, of the mate-rial for the wavelength of thesource. The transmittance isequal to:

T = (1 – τ) (2)

Another cause of attenuation isthe incomplete coupling of theflux from the source to the re-ceiver caused by the mismatchof the receiver relative apertureto the source relative aperture.

Coupling FundamentalsThe general case of source fluxcoupled to a receiver is depen-dent upon the source radiationpattern, the source receiverspacing, and the receiver area.

The following analyses show howthese parameters affect the OTF.Figure 2 shows a source which islocated at the center of a hemi-sphere with a receiver of anarea, A, located at a distance, d,from the apex. The ratio of thereceiver area to the distancesquared defines the solid angle,ω, subtended by the area, A. Thetotal flux, φe, being radiated isthe integral of the flux incidentwithin the hemisphere. The ra-diation pattern of a Lambertiansource is shown in Figure 3. Thepattern describes the ratio of theradiant intensity at an off-axisangle, Ie(θ) to the axial radiantintensity, Ie(0). The radiationpattern for a Lambertian source(or the reception pattern for aLambertian receiver) is de-scribed by the cosine of theoff-axis angle. The Lambertianradiation pattern function is:

Ie (θ) = Ie (0) cos θ (3)

When the radiation pattern,Ie (θ) / Ie (0), and the axial radi-ant intensity, Ie (0) are known,Equation (4) can be used to de-termine the total flux into thehemisphere, (θ = 90°) or the fluxinto a cone created by the area, A.

Figure 3. Radiation Pattern of a Lambertian Source.

OTF = (1)φe(RECElVED)

φe(AVAILABLE)

The optical flux attenuationcomes from a number of causes.One cause is the transmissionloss as the flux is incident uponand passes through the trans-mission media. These lossescome about because of reflectionat the surface of the material,and scattering and absorptionwithin the material. For the cal-culation of the optical transferfunction, it is customary to de-scribe these losses, τ, in terms of

Figure 2. Definition of the SolidAngle, ω.

Ie (0)

Ie (θ)≈ d sin θ

θ

d

ω = Ad2

90°

80°

70°

60°

50°

40°

30°

20°10° 0°

10° 20° 30° 40° 50° 60° 70° 80° 90° 100°

.2

.4

.6

.8

1.0 Ie (θ) = Io cos θ

3

When a Lambertian radiator isused as a source, the flux intothe cone specified by the off-axisangle, θ, is calculated fromEquations (3) and (4).

φe (θ) = Ie (0)o∫θ Ie (θ)

Ie (0)2π sin θ d θ

(4)

φe (θ) = Ie (0)o∫θ

φe (θ) = Ie (0) π sin2 θ

2π cos θ sin θ d θ

(5)

The total flux, φe, of theLambertian source is obtainedfrom Equation (5) when θ = 90° .Thus, the total flux φe is π timesthe axial radiant intensity, Ie(0)

The amount of flux coupled intothe receiver area A relates to therelative aperture of the receiver.The relative aperture, alsoknown as the numerical aper-ture, N.A., defines the ability ofthe receiver to accept flux arriv-ing at off-axis angles. When asource with a specific radiationpattern is considered, a receiverwith a large numerical aperturewill capture more flux than onewith a smaller N.A. The numeri-cal aperture is defined as thesine of one half the includedangle of the receiver cone. Thus,

N.A. = sin θ (6)

where θ = 1/2 cone angle.

As seen in Figure 4, when smallangles of θ are considered, thenumerical aperture can be calcu-lated as:

N.A. = sin θ ≈ D / 2d

N.A. = sin θ ≈ (A/π)1/2 / d

N.A.2 = sin2 θ ≈ A/πd2 (7)

Figure 4. Illustration of Numerical Aperture Relationships.

RECEIVER

d

SOURCE

θ1

θ2

AR

AS

D

In Figure 4 are shown a sourceand a receiver separated by adistance, d. At that distance, thecone (θ1) of radiation from thesource irradiates an area, AS.Clearly, the receiver having anarea, AR, much smaller than AS,describes a smaller cone (θ2) andwill receive only a fraction of theflux radiated by the source. Thisfraction describes the OpticalTransfer Function (OTF) for thissimple situation, and a good es-timate of the value of thisfraction is the ratio of the areasAR/AS. Then, applying the rela-tionship derived in Equation (7),the OTF can be defined in termsof the cones for source and re-ceiver which are described bynumerical apertures:

OTF =φR

φS

AR

AS

πd2N.A.R2

πd2N.A.S2=

=

=

N.A.RN.A.S( =((

2

(2

sin θ2

sin θ1

(8)

In estimating OTF, it is neces-sary only to recognize andevaluate such cones of coupling— exit cone (or N.A.) and accep-tance cones. As a general rule,the cone angle, θ, is defined asthe angle at which coupling is0.1 (10%) of the axial value(θ = 0). A case of special interestis that of a Lambertian emitterhaving a radiation pattern vary-ing as cos θ. The angle at whichcos θ = 0.1 is 84.26°, at which sinθ = N.A. = 0.995; thus, for practi-cal purposes, a Lambertiansource is regarded as havingN.A. = 1. If, in Figure 4, aLambertian source were used,the relationship in Equation (8)reduces:

OTF = N.A.RN.A.S( =(

2N.A.R

1( (2

(9)= (N.A.R)2

This simplification is importantas it is used extensively in thesections to follow. Notice, how-

4

ever, that OTF cannot exceedunity, so that if N.A.R > N.A.S,then OTF = 1. This occurs whenthe acceptance cone of the re-ceiver is larger than the exitcone of the source.

Lens FundamentalsIn practical sensor applications,the source-receiver distancemight be ten or more times thediameter of the receiver. If thereceiver area is small relative tod, then by Equation (7), the re-ceiver numerical aperture willbe small. If the source isLambertian, the total flux cou-pling will then relate to thesquare of this small numericalaperture.

Through the use of lens, moreefficient coupling will result. Inorder to understand how this isaccomplished, it is helpful topresent a few rules of optics. Thefirst is the basic lens equation.Referring to Figure 5, a doubleconvex condensing lens is shownrelaying an image from thesource to the receiver. Thesource is located a distance, dS,from the lens and the real imageis located at a distance, dR,which is related to the focallength of the lens. Thus, the re-lationship of dS, dR, f is calledthe basic lens equation.

Where f = Focal Length

dS= Source Lens Distance

dR= Received ImageDistance

=1

f1

dS+

1

dR(10)

This relay lens system is focusedwhen the receiver is placed atthe distance, dR, [from Equation(10) when f and dS are known]from the lens.

Under the focused condition, theimage of the source (at the planeof the receiver) is magnified. Thedegree of magnification, m, isequal to the ratio of dR to dS.

m =

Where AS,M = Source Image(Source Magnified)

AS = Source Area

m2 =

(11)

(12)

dS

dR

AS

AS,M

Using Equations (10) and (11),the magnification can be repre-sented in terms of the focallength and source to lens dis-tance, dS.

Figure 5. Focused Emitter Detector System Using a Double Convex Lens.

ds

DS

dR

N.A. = SIN θ

OTF = N.A.L2 • T, FOR A LAMBERTIAN SOURCE WITH IMAGE AREA SMALLER THAN RECEIVER AREA

θ

ds

1+

dR

1=

f

1

(13)

AS,M = m2 AS = dS – f

f( (2

• AS

When the source-lens distance,dS, is twice the focal length, f,the image will appear at dR = 2fon the other side of the lens. Un-der a 2f focused condition, thearea of the image AS,M, is equalto the area of the source AS. Thisconfiguration is termed a 1:1magnification or 2f system.

Lenses also have relative aper-ture qualities. The relative

aperture may be specified as ei-ther a numerical aperture, N.A.,or as an f-number, f/. The f-num-ber is equal to the ratio of thefocal length to the diameter ofthe lens. The f-number and thenumerical aperture are inverselyrelated, such that a smaller f-number will result in aproportionately larger N.A.

In Figure 5, the effective diam-eter of the lens is DL, andrelative to this, the numericalaperture is approximately DL2dSas in Equation (7). Comparingthe definitions of f/no and N.A.:

When dS = f, the source is fo-cused at infinity, makingN.A. = 1/2(f/ no); when dS < f,there is no real image, but a vir-tual image of the source lying onthe same side of the lens, andthe lens is less effective in im-proving the coupling.

Lens CouplingThe amount of flux coupled intothe lens is dependent upon thenumerical aperture of the lensand the exit aperture of thesource. When a focused lens sys-tem is considered such as that

( ( (14)

f/no =

f/no =

N.A. ≈DL

f DL

2dS

f

dS

1

2N.A.

5

shown in Figure 5, the flux thatarrives at the received imagepoint is equal to the ratio of thesquare of the numerical apertureof the lens divided by the squareof the exit numerical aperture ofthe source, this total times thetransmittance, T, of the lens.Thus, the lensed OTF is equalto:

OTF =

When the source is Lambertian, the equation simplifies to:

φR

φS= N.A.L

N.A.S[ [ (15)• T

2

OTF =φR

φS= N.A.L2 • T (16)

The following will illustrate apractical example:

Given:• Detector Area AD = 0.1 mm2

• Lambertian Source AS = 0.1mm2, φS = 100, µW

• Lens Numerical ApertureN.A.L = 0.5

• Lens Focal Length, f = 5 mm• Lens Transmittance, T = 0.95• Lens to Source Distance,

dS = 20 mm• Lens to Receiver Distance,

dR = 6.67 mm

Using Equation (15), the flux atthe receiver will be:

φR = φS N.A.L2 • T (17)

φR = 100 µW (0.5)2 • 0.95

φR = 23.75 µW

The received area is determinedby Equation ( 12).

Thus, into an area of 0.011 mm2,a flux of 23.75 µW is concen-trated. Using the basic lensequation, (10), the received im-age lies 6.7 mm from the lens. Ifa photodiode with an area, AD,were placed at this distance, dR,the fraction of this flux thatcouples the detector is the ratioof AD to AS,M if the detector arealies entirely within the sourceimage area. In general, this frac-tion is the ratio of that portion ofthe source image that overlapsthe receiver, divided by the totalsource image area, and thereforecannot exceed unity, even whenthe detector area AD is muchlarger than the source imagearea, AS, M.

When a lens coupling is com-pared with non-lens coupling,the improvement is describedby:

Detector Coupling = KD =

(19)

≤ 1AD

AR

;N.A.L2 • T

N.A.R2=

N.A.R2 = AD/[π(dS + dR)2]

Where N.A.R is found from Equation (7) with d = dS + dR

φL

φn-L

(.5)2•0.95

4.47 × 10-5= 5.3 × 103=

φL

φn-L

Thus, this simple example usinga lens improves the couplinggain by 37 dB.

Reflector FundamentalsSensor applications, such as barcode scanning, paper edge detec-tion, and optical tachometry, usethe reflective properties of theobject or element that theysense.

The reflection of the incidentflux may either be specular ordiffuse. A specular reflector isone where the angle of the re-flected ray of flux is equal to theincident ray of flux. Thus, if aray of flux was incident to thereflector at 20° from the normal,the reflected ray would also be20° from the normal and 40°from the incident. A first surfacemirror or a highly polished codewheel are examples of specularreflectors. They are character-ized as having reflectioncoefficients, ρ, of almost unity. Ithas a numerical aperture,N.A.R -S, equal to the N.A. of thesource that is incident upon it.The reflected flux would be equalto incident flux times the reflec-tion coefficient.

The reflection coefficient is as-sumed to be constant over thewavelengths of interest. If this isnot true, then a correction coeffi-cient, kρ(λ) must be introducedto correct for this spectral prop-erty.

A perfectly diffuse reflector ischaracterized as having aLambertian radiation pattern.

(20)

• ρ;N.A.L2

N.A.R-S2

φOUT = φIN

where N.A.R-S = N.A.L

φOUT = φINρ

(18)

AS,M =

AS,M =

dS – f

f( (2

• AS

20 mm – 5 mm

5 mm( (2

• 0.1 mm2 = 0.011 mm2

6

The flux that is coupled from thereflector is equal to:

(21)

• ρN.A.RECEIVER

2

N.A.REFLECTOR2

φOUT

φIN=

In typical applications, the nu-merical aperture of the receiverwill be that of the receiving lens.Also, the N.A. of the diffuse sur-face is unity; therefore, Equation(21) can be rewritten as:

(22)N.A.L2 • ρφOUT

φIN=

Most reflectors are neither per-fectly specular nor diffuse, but acombination of the two. A typicaldiffuse reflector may appearmore specular at one wavelengthand more diffuse at another.Also, the angle of incidence maymodify the reflection properties.

It may be desirable in reflectivesensor applications to computethe relative ratio of reflectionbetween a specular and a diffusereflector. Assuming both appli-cations have the same φIN andthe same receiving lens, and us-ing Equations (20) and (21), thisratio becomes:

N.A.L2 ρD

(23)

ρS=φOUT (SPECULAR)

φOUT (DIFFUSE)

Thus, if the receiving lens had aN.A.L = 0.3, the effective gain ofusing a specular over a diffusereflector would be 10.45 dB.

Confocal CouplingIt was shown in the lens cou-pling section that the area of thesource could be reduced or mag-nified through the use of a lens.The technique of reducing theimage of the receiver and the

source, such that they lie on thesame plane located a distancebetween the source and sensorcan be accomplished by usingtwo confocally spaced pair oflenses.

Figure 6 shows two plano-convexlenses positioned so that theyare confocally focused. Using thecoupling concept developed inthe coupling fundamentals theoverall OTF can be developed inthe following steps.

Figure 6. Confocal Spaced Emitter Detector System Employing Two PlanoConvex Lenses.

The convenient fact about such atransfer function is that each el-ement of the linear system canbe evaluated and then the totalfunction is the product of the in-dividual terms.

In a practical application, theexit N.A. of lens 1 and the en-trance N.A. of lens 2 will beequal. If a Lambertian sourcewere used in Equation (24), itwould reduce to N.A.L12 TL1.These two conditions allow thefollowing.

ds dSL dST dT

TL1 TL2

L1 L2

OTF = N.A.L12 • TL1 • TL2

WHERE = N.A.L1 = N.A.L2 FOR A LAMBERTIAN SOURCE

ds

1+

dSL

1=

f

1

CONFOCAL SPACINGWITH LENSES

m =ds

dSL

D

The image size appearing equi-distant between the two lenssurfaces is determined by themagnification factor, m = dS1 /dS.Thus, a large source and receivercan be focused down to a point inspace, and when confocallycoupled, the imaged source-receiver area can become the in-terruption point for a hole edge,or paper edge sensor.

Step 1. Flux into Lens 1.

OTF =

Step 3. Total OTF.

φL2

φS

(24)

(25)

(26)

entrance N.A.L12

N.A.S2•TLI

•TL1 •TL2

=φLI

φS

Step 2. Flux into Lens 2 from Lens 1.

entrance N.A.L22

exit N.A.L12

=φL2

φL1

entrance N.A.L22

exit N.A.L12

•

entrance N.A.L12

exit N.A.S2=

Confocal OTF =φL2

φS

(27)

= N.A.L12 •TL1 •TL2

7

Lens Reflective CouplingIf one of the lenses of a confo-cally coupled lens system wereplaced adjacent to the other lenselement and skewed so that thetwo images in front of the lensesintersected, a lensed reflectivesensor would result. This isshown in Figure 7.

OTF = N.A.L12 • TL1 • TL2 • N.A.L22 • ρ REFLECTOR • O.F.

θ1

φS

dS

(f/no)SSOURCE

RECEIVER (f/no)R

L1

dT1

θ2 φR

dR L2

dT2

LAMBERTIAN RADIATOR

TL1 , TL2 – TRANSMITTANCES OF LENS 1 AND LENS 2N.A.L1 = SIN θ1

N.A.L2 = SIN θ2

Figure 7. Reflective Coupling Employing Confocally Spaced Lenses.

Every mechanical system hasalignment tolerances which mayallow the two images at the fo-cused point to move with respectto one another so that the areaof the source and receiver im-ages do not totally overlap. The

(28)

O.F. =

≤1

AREA OF SOURCE IMAGEWHICH IS OVERLAPPED

BY RECEIVER IMAGE

TOTAL AREA OFSOURCE IMAGE

ratio of the image overlap istermed the overlap fraction, OF.

the receiver image overlaps theentire source image, the O.F.will have its maximum value ofunity. Having a larger receiverimage provides a more consis-tent O. F. by reducing variabilitydue to alignment difficulty.

The amount of flux couplingwould be dependent upon thetype and reflectance of the re-flector placed at the imageintersection, as well as on theN.A.’s of the lenses.

Using Equations (26) or (27),along with (22), and (28), the op-tical transfer function for adiffuse reflector can be deter-mined.

where ρD = reflection coefficientof a diffuse reflector and otherterms are as defined in Figure28.

In a similar manner, the OTFfor the specular reflector may bedetermined.

(29)

OTF = N.A.L12 •TL1•TL2

•N.A.L22 •ρD•O.F.

(30)

OTF = N.A.L12 •TL1•TL2

•ρS•O.F.

where ρS = reflectance coefficientof a specular reflector

The conclusions to be drawnhere are that a specular reflectorwill provide a much larger re-ceived flux. However, it suffersfrom a coupling problem where amovement of the normal of thereflecting plane, with respect tothe normal of the confocallyspaced lens system, causes theincident flux upon the reflectorto be reflected outside of the ap-erture of the receiving lens. Thisis shown in Figure 8.Figure 8. Positioning Sensitivity Caused by the Use of a Specular

Reflecting Surface.

RECEIVER

SOURCE

SPECULARREFLECTOR

SOLID = REFLECTOR SKEWED TO SENSORDASHED = REFLECTOR NORMAL TO SENSOR

α

γ

β

γ = 2 (β + α)

This fraction can vary from zeroto unity. When it is zero no cou-pling occurs and at unity,maximum coupling results. If

8

HBCS-1100 ReflectiveCouplingMany of the alignment problemsfound in discrete confocallyspaced reflective sensor systemscan be eliminated with the use ofthe HBCS-1100 High ResolutionOptical Reflective Sensor. Thissensor includes a source and re-ceiver focused with 2f optics. Theoptics system is a bifurcated as-pheric lens with an effectivenumerical aperture of 0.3. Theseelements are housed in a TO-5package with a glass windowwhich is shown in Figure 9.

In the data sheet, radiant flux,φe, is specified as the fluxcoupled by the source lens to theimage. This means that to de-velop the optical transferfunction, only the receiving lensOTF need be described. The OTFfor the HBCS-1100 and a diffusereflector will appear very similarto that of a single lens systemwith the addition of the trans-mittance of the glass window,TG, and the OTF of the reflector.

EMITTER5

6

7

8

1

2

3

4

DETECTOR

M.S.P.*

EPOXY SEAL

GLASSWINDOW

OPTICAXIS

REFLECTEDLIGHT

CANAPERTURE

TO-5HEADER

*MAXIMUM SIGNAL POINT

LENS

RTV

STRAY LIGHTSPOT WELD

The HBCS-1100 receiver areaAR is 0.160 mm2, and the sourcearea AS is 0.023 mm2. The ratioof AR/AS is greater than onewhich means that for focusedoperation, the overlap fractionO.F. is equal to one.

The following example showsthe expected flux at the receiverphotodiode from a diffuse reflector.

Figure 9. Elements of the HBCS-1100 High Resolution Optical Reflective Sensor.

(31)

φRECEIVER

φREFLECTOR

= TL•TG • O.F. • OTFREFLECTOR

OTF =

(32)

Step 1. Diffuse Reflector OTF.

OTF = N.A.L2 ρD

Step 2. Total OTF for HBCS-1100 Using Step1 and Equation (31).

Step 3. Using the following:

OTF = TL•TG • O.F. • N.A.L2 ρD

N.A.L = 0.3 ρD = 98%,

Step 4. φREFLECTOR

= φe (Data Sheet)

φRECEIVER = φREFLECTOR(OTF)

φRECEIVER = 576 nW

φe (Data Sheet) = 9µW

TG = 0.9 TL = 0.8 O.F. = 1

OTF = (.8) (.9) (1) (.3)2 (.98) = 0.064

If a specular reflector were used,the flux at the receiver would bethe product of the transmittanceof the glass and lens, the overlapfraction, O.F., and the reflec-tance of the specular reflector ρS.This is shown in Equation (33).

(33)

φRECEIVER = φREFLECTOR(TL)

(TG)(O.F.)( ρS)

ρS = 0.95

φRECEIVER = 9 µW (0.8) (0.9) (1)(0.95) = 6.16 µW

9

Through the use of the numeri-cal aperture of the receiver lens,it is possible to determine theOptical Transfer Function, andusing this OTF, the radiant fluxthat appears at the receiver sur-face. From the responsivity ofthe receiver diode, the photocur-rent can be estimated.

Optical SensorParameters

IntroductionBar code scanning, paper edgesensing, and optical tachometryapplications place specific re-quirements on optical resolutionand electrical performance of areflective sensor system.

This section will describe the ex-pected optical resolution andelectrical performance of theHBCS-1100 as the object beingsensed is placed at a locationother than the optical focus point.

Modulation TransferFunctionThe optical resolution of a reflec-tive sensor system is determinedby the overlap area of the im-ages focused at the reflectorsurface. This may be limited byeither source or receiver imagesize, whichever is the smaller.Optical resolution of a reflectivesensor system is defined as theability to discriminate the reflec-tion of closely spaced lines withunequal reflectances. Figure 10shows a series of reflectors andbars, and the response to thispattern as the imaged sourceand receiver, are moved laterallyacross the surface. The assump-tion is made that the reflectorreflectance, ρREFLECTOR , ismuch greater than the reflec-tance of the bar between thereflectors. The conclusion that

W(TYP) RECEIVER

BEAMSOURCE

BEAM

S1

S2

S3

SENSORMOVEMENT

d%

RE

FLE

CT

ED

% R

EF

LEC

TE

D

% R

EF

LEC

TE

D

ALL REFLECTOR WIDTHS, W, ARE EQUAL

W S WS1 = 1/2d

W S WS2 = 1/4d

W S WS3 = 1/8d

ρREFLECTOR >> ρBAR

can be drawn from this illustra-tion is that the ability of thesensor to discriminate betweenreflectors spaced a distance, s,apart increases as the space, s,increases.

Figure 10. Resolution of an Optical Reflective Sensor System.

(34)Resolution αs

d

Figure 11a, b shows a reflectorobject with a reflecting surfacecomposed of equal width reflect-ing (white), and non-reflecting(black) lines. When a receiverwith a circular image with a di-ameter, d, is positioned over thereflector so that the image areatotally intersects the white line,a maximum or 100% peak re-flected response will result. In asimilar manner, when the imageis positioned over the black line,a minimum or 0% peak reflectedresponse will occur. The differ-

ence between maximum andminimum response specifies thepeak amplitude response underthese line width-image diameterconditions.

If this scanning spot werescanned laterally across theblack-white transition, the re-sponse would be a ramp with aslope (% RESPONSE/lateral dis-tance) determined by the imagediameter. This is shown in Fig-ure 12a. The 50% response pointshown in Figure 12b indicatesthat the image area is equallyintersecting the black-white re-flecting areas. If the lateralscanning were continued acrossthe surface, the reflected re-sponse for Figure 11a would be atrapezoidal waveform. This isshown in Figure 11 b.

If an image is used to scan ablack-white pattern where the

10

line width is much smaller thanthe image diameter, the mini-mum to maximum response willbe reduced from the responseobtained when the line width ismuch larger than the image size.

An example of this is shown inFigure 13. A total 0 – 100% re-sponse occurs for a W2 = 3condition, while only a 33%minimum to maximum responseis obtained when W1 = 1. Thus,as the line width gets smallerwith respect to the image, thedifference between the minimumand maximum is also reduced.When the performance atsmaller line widths is comparedto the performance for linewidths resulting in 100% re-sponse, a modulation ratio isobtained. This ratio is shown inEquation (35) using data fromFigure 13.

LATERALMOVEMENT

IMAGE

W

100%

0%RE

FLE

CT

ED

RE

SP

ON

SE

PEAK WHITE

PEAK BLACK

LATERAL MOVEMENT

Figure 11. a,b. Image Response for Equally Spaced White and Black Lines.

100%

50%

0%

RE

FLE

CT

ED

RE

SP

ON

SE

WHITEAREA

BLACKAREA

10% - 90% RESPONSE = IMAGE DIAMETER

d

TRANSITIONLATERAL IMAGE MOVEMENT

a) EDGE RESPONSE

WHITE BLACK

IMAGE

LATERALMOVEMENT

TRANSITIONEDGE

b) 50% REFLECTED RESPONSE LOCATION

Figure 12. Image Transition Response.a) Edge Responseb) 50% Reflected Response Location

W

0 1 2 3 4LATERAL DISTANCE

5 6 7

RESPONSEFOR W2

RESPONSEFOR W1

100%

50%

0

W1 = 1

MINMAX

W2 = 3

c)

b)

a)

Figure 13. Reflection Modulation forDifferent Line Widths.

(35)Modulation =MAX–MIN

MAX+MIN

The modulation performance fordifferent line pair widths for afixed image size is called theModulation Transfer Function,MTF, of the optical image sys-tem. The MTF is specified as apercent response at a particularspatial frequency. The spatialfrequency, F, is defined as thenumber of equal width white-black line pairs per lateraldistance. The common units forspatial frequency is line pairsper millimeter, ln pr/mm. Thespatial frequency is determinedby Equation (36).

(36)

F = ln pr/mm =1

2 line width (mm)

11

Figure 14. Modulation TransferFunction of the HBCS-1100.

110

100

90

80

70

60

50

40

30

20

10

0

% A

MP

LIT

UD

E M

OD

UL

AT

ION

(P

-P)

0 1 2 3 4 5 6SPATIAL FREQUENCY (LINE PAIR/mm)

Figure 14 shows the modulationtransfer function, MTF, for theHBCS-1100. In applicationssuch as bar code scanning andoptical tachometry, the spatialfrequency can be calculated withEquation 36. If a black-whiteline pattern with a line width of0.254 mm (0.01 in.) were to bescanned by the HBCS-1100, theperformance can be determinedthrough the use of Figure 14 andEquation (36). The spatial fre-quency is determined to be 1.97ln pr/mm which results in anMTF response of 70%.

The MTF performance indicatesthat when a line pattern isscanned by the HBCS-1100, thereflected flux is degraded fromthe amount of flux obtained froma non-patterned reflector. Thus,the MTF response becomes an-other element to be added to theoptical transfer function, OTF,as given in Equations (32) and(33).

Depth of FieldThe optical transfer function,OTF, of a reflective sensing sys-tem was presented in the LensReflective Coupling, and theHBCS-1100 Reflective CouplingSections. In each case, the as-

FOCUS POINT

A

BLURRED IMAGE SIZE

ds

∆ INSIDE

∆ OUTSIDE

β = ANGULAR BLUR

β

sumption was that the sourceand the received image were fo-cused on the same plane. Inmost applications, the mechani-cal alignment of the sensor tothe reflecting element will notbe at the fixed focused point.

As the reflecting object movesaway from the focus point, theimage will become defocused re-sulting in a blurred image. In areflective sensor system, thedefocusing will occur for boththe source and received image.The ratio of the intersection ofthe two image areas determinesan overlap fraction, O.F.. As thesystem is defocused, the overlap,O.F., decreases.

The defocused coupling responseversus reflector distance is re-ferred to as the depth of field, ∆l.Figure 18 shows the relative re-sponse versus reflector distance.One will note an assymetricalresponse on either side of themaximum signal point. There isa much sharper % Ip (∆l) re-sponse roll-off for the distancebetween the reference plane tothe maximum signal point,MSP, than from the MSP, to dis-tances further away. This is dueto the fact that the amount of

Figure 15. Defocusing Versus Depth of Field.

blurring on the near side of theMSP is less than that on the farside. This is shown in Figure 15.The blurred image at ∆l inside issmaller than the blurred imageat ∆l outside. When a reflectivetype of lensed system, as shownin Figure 9, is considered, theoverlap fraction, O.F., is smallerat a ∆l inside than that for thesame ∆l outside.

The defocusing of the optical sys-tem also impacts the modulationtransfer function. As the systemis defocused, the image size willincrease causing a reduction inthe MTF for a specified line pairper millimeter.

HBCS-1100 Total TransferFunctionThe optical transfer function forthe HBCS-1100 was developed inthe HBCS-1100 Reflective Cou-pling Section. This developmentspecified the performance of thereflective sensor as a ratio of fluxincident at the receiver φR to theflux φe incident at the reflector,OTF= φR / φe. The electrical de-signer needs to know therelationship of the current sup-plied to the emitter, IF, whichproduces a photocurrent, IPR , inthe detector. This relationship

12

will be referred to as the sensorelectrical transfer function ortotal transfer function, TTF.

The TTF is the product of theoptical transfer function, theflux from the emitter, φe. and theflux responsivity, Rφ, of the pho-todiode. This is shown inEquation (37).

Figure 16. K-factor of φe vs. LED DCForward Current.

1.6

1.4

1.2

1.0(k)

.8

.6

.4

.2

0

RA

DIA

NT

FL

UX

φE

k-F

AC

TO

R(N

OR

MA

LIZ

ED

AT

I F =

35m

A, T

A =

25°

C)

0 10 3020 40 50 60 70 80IF – DC FORWARD CURRENT (mA)

-20° C0° C

25° C

50° C70° C

= Rφ • OTF • φe TTF =

(37)

IPR

IF

(IF) • K

The flux responsivity, Rφ , of thephotodiode at 700 nm is speci-fied as 0.22A/ W. The fluxresponsivity, Rφ , will be consid-ered a constant throughout thecalculations. The radiant flux,φe, from the source is dependentupon the current through theLED emitter. The K-factor rela-tionship of the output flux, φe, toLED forward current, IF, isshown in Figure 16. This graphis normalized at 35 mA and25°C. Thus, the data sheet typi-cal of 9 µW occurs at 35 mA and25° C.

The following example will illus-trate how the TTF is used for abar code scanner.

Given:

The total transfer function, TTF, is:

IF = 45 mA; φe (35 mA) = 9µW; TA = 25°C;Reflector = Lambertian,ρD = 0.85 @ 700 nmBar Width = 0.01", = 0.254 nmDepth of Field = ∆ = 0.6 mmN.A.L = 0.3; TG = 0.9;TL = 0.8; Rφ = 0.22 A/W

= Rφ • TL • TGTTF =

× MTF • φe • K

(38)IPR

IF

• N.A.L2 • ρD • O.F. ( ∆ )

The first step is to evaluate theoverlap fraction, O.F. (∆l) = %IPR (∆l), a function of the depthof field, ∆l. From Figure 19, the%IPR(∆l) = 50% = 0.5. Thus, theO.F. is equal to 0.5.

The second step is to determinethe MTF response for a barwidth of 0.254 mm, for a depthof field of 0.6 mm. Equation (36)is used to arrive at an F(APP) =1.97 lnpr/mm. Figure 14 is usedto determine the MTF (APP) of70%. Thus, MTF (1.97 ln pr/ mm)is equal to 70%.

The third step is to determinethe K-factor. This can be foundfrom Figure 16. The K-factor foran IF = 45 mA is equal to 1.3 at25°C.

These values of O.F. (∆l),MTF(APP), and K are substi-tuted into Equation (38) todetermine the reflected photo-current, IPR.

The conclusion that can bedrawn from this example is thatthere are many factors contrib-uting to the total transferfunction.

HBCS-1100 Logic InterfacingOptical sensing applications maybe accomplished with the HBCS-1100 High Resolution ReflectiveOptical Sensor. This device in-cludes a 700 nm emitter, abifurcated aspheric lens, and aphotodetector. The cathode ofthe LED emitter and the sub-strate of the photodetector areelectrically connected to the me-chanical package. The photo-detector may be interconnectedas a discrete photodiode or aphoto diode-transistor amplifier.

Photodiode InterconnectionThe photodiode, within the inte-grated photodetector, is isolatedfrom the substrate-case by sub-strate diodes. These diodesappear from the common sub-strate-case to the transistorcollector, and to the cathode ofthe photodiode.

Figure 17 shows recommendedinterconnection of the unusedterminals of the sensor whendiscrete photodiode operation isdesired. Care should be taken toensure that these substrate di-odes are always reverse biasedso that they do not create a con-ductive path that may damagethe substrate or other circuit ele-ments.

The photodiode behaves like acurrent source, such that whenoptical flux falls on the device, it

IPR = 0.22 A/W • 0.8 • 0.9 • (.3)2

• 0.85 • 0.5 • 0.7 • 9µW • 1.3

IPR = 49.6 nA

13

will generate a photocurrent inrelationship to its responsivity,Rφ, of approximately 0.22 µA/µWat 700 nm. The total photocur-rent, IP, generated by thisphotodiode is the summation oftwo currents, the reflected pho-tocurrent, IPR, and a strayphotocurrent, IPS. Thus,IP = IPR + IPS.

Stray Photocurrent—IPSThe stray photocurrent, IPS, re-sults from flux falling on thedetector from sources other thanthe reflector surface. The prin-ciple source of strayphotocurrent is from the scat-tered flux of the LED emitterthat is reflected within the me-chanical package. Ambient lightcan also be a source of the strayphotocurrent, but this sourcehas been greatly minimized bythe use of an optical long wavefilter. This filter action is pro-vided by the red coloration foundin the bifurcated lens.

IPR—IPS RatioThe magnitude of the stray pho-tocurrent resulting from internalscattering is directly propor-tional to the forward current, IF,through the LED and the emit-ter relative efficiency. DCoperation of the emitter will re-

VD

RF

VO

VO = -IPRFIP

–

+

1

Figure 17. Photodiode Transresistance Amplifier.

sult in a steady state stray pho-tocurrent that will range indirect relationship to the specifi-cations and the worst case ortypical value of IP (MIN) andIP (MAX) related to the stray pho-tocurrent ratio, IPR/IPS. Thephotocurrent specified for theHBCS-1100 is the total photo-current, IP, which is equal to thesum of IPR and IPS. The ratio ofIPR to IPS is designated a qualityor Q-factor of the sensor. Thus,as Q increases for a given IP, thevalue of stray photocurrent, IPS,decreases. A worst case analysisfor IPS under the condition ofminimum Q = 4, and an LEDcurrent of 35 mA results in anIPS (MIN) = 20 nA for IP(MIN) =100 nA, and IPS(MAX) = 50 nA forIP(MAX) = 250 nA. A typicalvalue of Q = 6.5 would cause IPSto range from 13 nA to 33 nA.

The quality factor, Q, relation-ship to IP, IPS, and IPR is shownin Equation (39).

Depth of Field With Respectto Maximum Signal PointFigure 18 shows that the 100%maximum reflected photocur-rent, IPR , response occurs at thelocation from the reference planedefined as the Maximum SignalPoint, MSP. It also shows thatthe value of the reflected photo-current IPR is reduced as thereflector is moved in either direc-tion away from the maximumsignal point. The HBCS-1100has a relatively symetrical re-sponse of IPR versus thedistance, l, from the MSP. Thedepth of field of this optical sys-tem is defined as the distance,∆l, between two equal percent-age response points on eitherside of the MSP. The 50% IPRresponse is referred to as thedepth of field full width halfmaximum, FWHM. The depth offield, ∆l, FWHM shown in Figure18 is found to be 1.2 mm. Thus,if the reflector were moved onehalf of the total FWHM distance,∆l, from the MSP, the 50% IPRpoint would occur approximately0.6 mm on either side of theMSP location. The reflected pho-tocurrent, IPR, response at a

(39)IP = IPS + IPRQ =IPR

IPS

IPR = IP [Q/(Q + 1)]

IPS = IP [1/(Q + 1)]

Where

Q = Quality Factor

IP = Total PhotocurrentIPR = Reflected PhotocurrentIPS = Stray Photocurrent

Figure 18. Depth of Field vs. Maxi-mum Signal Point.

110

100

90

80

70

60

50

40

30

20

10

0

% R

EF

LE

CT

ED

PH

OT

OC

UR

RE

NT

0 1 2 3 4 5 6 – REFLECTOR DISTANCE (mm)

MAXIMUM SIGNALPOINT

% IP (∆ ) FWHM

% IP (∆ ) HWHM

14

specific depth of field is referredto as the %IPR (∆l). This value isalways less than or equal to100%.

The specific value of the IPR isdependent upon the flux fromthe emitter and the type and re-flection coefficient, ρ, of thereflector. The reflector funda-mentals section presented thecharacteristics of the reflectors,and demonstrated that a specu-lar reflecting surface offers anorder of magnitude improvementin the reflecting photocurrentwhen compared to a diffuse sur-face.

When a diffuse reflector is used,the expected value of the re-flected photocurrent, at aspecific depth of field, IPR (∆l), isthe product of the percent re-sponse of IPR at the depth offield %IPR (∆l), the reflection co-efficient, ρ, of the reflector, thetotal photocurrent measured atthe MSP from a diffuse reflectorat a specific LED current Ip (IF),and the quality ratio Q/Q + 1.This relationship is shown inEquation (40).

When a specular reflector isused, an additional coefficient isintroduced. The HBCS-1100 IPRperformance is specified for adiffuse reflector; thus, when aspecular reflector is used, an im-provement factor dictated byEquation (23) is obtained. Thisfactor indicates the IPR improve-ment, for PS = PD, is inverse ofthe lens numerical aperturesquared. The IPR(∆l) responsefor a specular reflector is shownin Equation (41).

(40)

IPR ( ∆ ) = %IPR ( ∆ ) • ρ • Iρ

(IF) • Q/(Q + 1)

(41)

IPR ( ∆ ) = %IPR ( ∆ ) • ρ • Iρ

(IF) • Q/(Q + 1) • 1/N.A.L2

The expected value of IPR (∆l)using a diffuse reflector with areflector having a reflectance of75%, a total depth of field of 1.2mm (0.6 mm each way), andLED emitter current of 35 mA,and quality factor Q = 6.5, canbe determined from Equa-tion(40). The depth of field of 1.2mm is equal to a 51PR (∆l) of50%, and typical IP (35 mA =140 nA. The IPR (∆l) under theseconditions is equal to 45.5 nA. Ifa specular reflector were used,an IPR(∆l) = 506 NA would befound from Equation (41), forN.A.L = 0.3.

These two equations are usefulin determining the expectedrange of IPR for a given reflectorand a depth of field. These twosystem elements are very impor-tant in bar code scanning andpaper edge sensing where thetype of reflector and specificdepth of field are variables.

Amplifier ConsiderationsEach sensor application willgenerally specify the electricalinterface required and the rangeof the types of reflectors whichwill be utilized. The magnitudeof the photocurrent generated bythe photodiode is normally toosmall to interface directly to alogic gate. This condition indi-cates that an amplifier isneeded. The amplifier electricalperformance parameters, suchas current and voltage gain, andthe type of coupling are deter-mined by the logic family to beinterfaced, the application, andthe magnitude of the reflectedphotocurrent.

Applications using specular re-flectors include tachometry andoptical limit sensing, while dif-fuse reflectors are moregenerally found in paper edgesensing and bar code reading. Anac coupled amplifier is accept-able in tachometry and bar codereading, while a dc coupled am-plifier is required for steadystate applications such as paperedge sensing and optical limitsensing.

The relationship of the reflectedphotocurrent, IPR, to stray pho-tocurrent, IPS, has a large effecton the type of dc amplifier de-sign selected. The initial step inamplifier design is to determinethe worst case magnitude of thestray photocurrent, IPS. It is thisworst case value of IPS that be-comes the input quiescent biascurrent, which sets the thresholdfor the dc amplifier output voltage.

Transresistance—TTL InterfaceA very common dc amplifier usedwith photodiodes is thetransresistance type of amplifier.The simplest form is shown inFigure 15.4.1-1. The circuit con-figuration described by theelectrical transfer function,VO = – IPRF, is often called acurrent to voltage converter. Asingle power supply trans-resistance amplifier is shown inFigure 20. Here the photodiodeis connected to the inverting in-put, and an offset voltagederived from VCC is determinedby a resistive voltage divider, 1 +R2/R1 and is applied to the non-inverting input. The electricaltransfer function is:

Where IP = IPR + IPS

VCC

1 + R2/R1

–IPRF

(42)

Vo =

15

Equation (43) indicates that un-der the condition of zeroreflected photocurrent, IPR = 0,the output voltage, VO, will bethe offset voltage less the volt-age developed by the strayphotocurrent, IPS, times thetransresistance, RF. Thus, therelationship for IPR = 0 is:

Figure 19. Photodiode Transresistance Amplifier with Offset Voltage.Figure 20. Graphical Solution ofTransresistance Amplifier Design.

REFLECTOR

REFERENCEPLANE

ANODE

3 1

VF 6

CATHODE 4

2SUBSTRATE, CASE

DS

DS

IP

VCC

8

R2

R1

RF

VOUT

= – IPRF1 + R2/R1

VOUTVCC

+

–

5VOFFSET = 4.4V

[IPS (MAX)' VIH]

[IPS (MIN)' VIL]

IP (MIN)

IPS (MAN)

4

20 40 60 80 100 120

3

2

1

5

VO

na

1/RF

When the transresistance ampli-fier shown in Figure 19 is usedto interface the photodiode to aTTL logic device, the output volt-age, VO, of the amplifier mustchange from a logic high, VIH, of2.0 V to a logic low, VIL, of 0.8 V.To improve the noise immunityof the interface, it is desirable tobroaden the range of VIH to VILto 2.4 V and 0.4 V. The offsetvoltage and the value of thetransresistance resistor, RF, isselected to insure that the maxi-mum value of strayphotocurrent, IPS (MAX) = 50 nA,does not cause the output volt-age, VO, to fall below VIH of 2.4 V,and the minimum total photo-current, IP(MIN) = 100 nA, willcause the VO to be equal to aVIL = 0.4V.

(43)VCC

1 + R2/R1

–IPSRFVo =

It is very unlikely that anIPS (MAX) = 50 nA and IP(MIN) =100 nA will occur simulta-neously for a single device. Adevice that has an IPS (MAX) of50 nA would also have anIP(MIN) of 250 nA. In a similarmanner, a device that has anIP(MIN) of 100 nA would mostlikely have an IPS (MAX) of 20 nA.

Figure 20 shows the graphicconstruction of the electricaltransfer function for Equation(42). The interface conditions of[IPS (MAX), VIH] and [IP(MIN), VIL]describe a line whose y interceptdictates the offset voltage,Voffset, and whose slope deter-mines the transresistance, RF.Using Equations (44) and (45),the offset voltage, Voffset, andthe transresistance can be deter-mined.

The value of Voffset and RF whichsatisfy the TTL interface condi-tions can be determined fromEquations (44) and (45). For theexample, Voffset = 4.4V, and RF =40 MΩ.

This example of photodiode-logicinterfaces places parametric de-mands on the instrumentationoperational amplifier selected.This amplifier should have avery low input offset current,thus allowing sensing of IP atvery low levels. It should have again greater than that requiredfor the interface. For example,the required current gain for thephotodiode-TTL amplifier is ap-proximately 85 dB; thus, anamplifier with an open loop gainof 100 dB would be desirable.The slew rate of a trans-resis-tance amplifier may be slowerthan that required for TTL logicinterconnection; thus, it may benecessary to specify a Schmitttrigger gate as the interconnect-ing logic element.

CMOS InterfaceThe internal transistor of theHBCS-1100 may also be used asa gain element in a single ormultiple stage amplifier. Figure21 shows an example of intercon-

(44)

(45)RF = –

Voffset =

VIL IPS (MAX) – VIH IP (MIN)

IPS (MAX) – IP (MIN)

VIH – VIL

IPS (MAX) – IP (MIN)

16

necting the photodiode to aCMOS buffer gate, CD4049, us-ing the internal transistor as again element.

It was shown previously that thevalue of IPS and IP can vary fromunit to unit under similar condi-tions of IF, reflector type anddistance, l . There will also bevariations of the hFE of the tran-sistor from unit to unit. Thedesign of the photodiode-transis-tor amplifier to CMOS logic gatemust take into consideration thevariations of IPS, IP, and hFEwhen a direct coupled intercon-nection is desired. Figure 21presents a design for an HBCS-1100 CMOS interface. The firststep is to calculate the worstcase stray photocurrent, IPS(MAX).The IPS(MAX) becomes transistorbase current which, when multi-plied by the hFE (MAX), willdetermine the maximum collec-tor current resulting from strayphotocurrent. It is a requirementof this circuit that the collectorcurrent resulting from IP(MAX)does not cause the collector out-put voltage to fall below theinput logic high level, VIH, of 4volts. Thus, the maximum valueof the load resistor is selectedbased on the IPS(MAX) and

IP

IF

VCC

VO

RL

CD4049

CONDITIONSIF = 35 mAhFE = MIN = 100, MAX = 300IP = MIN = 100 nA, MAX = 250 nA

IPR

IPS(MIN)

= Q(MIN) = 4 ∴ IPS =20 nA MIN

50 nA MAX

hFE (MAX) of the sensor.

It is desirable in this type of in-terface to have at least a two toone difference between theIPS(MAX) and the IP(MIN). Such astipulation will place constraintson the type of reflector, thedepth of field, and the allowablespread between hFE(MIN) tohFE (MAX). Equations (40) and(41) are used to calculate theIP (∆l) for worst case conditionsof IP(MIN) when a depth of fieldof 1.2 mm is desired. Thus, adiffuse reflector would cause37.5 nA IP(MIN) and a specularreflector would result in anIP(MIN) of 416 nA. Using the cri-teria of IP (∆l)/IPS(MAX) ≥ 2[where IP (∆l) is evaluated forIP(MIN)] indicates that a reflectorwith specular properties shouldbe used.

The next step is to determinethe minimum value of RL for theminimum value of IP (∆l) andhFE (MIN) which causes the tran-sistor collector voltage to fallbelow the VIL of the CMOS gate.

This worst case analysis of Table1 indicates that there is a verynarrow range between theRL(MAX) and RL(MIN). If a

smaller depth of field ∆l were se-lected, the range would belarger, thus giving a greater de-sign margin.

Current Feedback AmplifierAnother common design problemis to interface the photo diode-transistor amplifier to adifferential comparator such asthe LM311 family. Here the de-sign goals are similar to thosespecified in Figure 21 but aneven greater degree of outputvoltage, VO, stability is desired.By using a simple current feed-back amplifier such as the oneshown in Figure 22, the varia-tions of VO caused by ∆hFE and∆IPS will be minimized. In thisamplifier design, the trade offsare between voltage/currentgain, stability, and amplifierspeed. As the ratio of RF to RLapproaches unity, the VO stabil-ity improves, but there is a lossin signal gain. The difference be-tween the IPS and IP(∆l) willspecify a current that will causea change in the output voltage,VO. When a larger swing in VO isdesired, the value of RL is in-creased, such that ∆VOα(IP—IPS) RL. However, as RLincreases, the speed of the cir-cuit decreases. Tables 2 and 3show a design example using anRL = 100k, RF = 10 MΩ . In Fig-ure 22, the differentialcomparator threshold is set bythe resistor ratio R1, R2 andshould be equal to 1.25 V. This isbelow the minimum quiescent,VO, of 1.3 V caused by the varia-tions of hFE and IPS. The changeof the quiescent voltage causedby the variation of hFE can becalculated through the use of thestability factor defined as s".This factor is the incrementalchange of IC caused by an incre-mental change in hFE.

Figure 21. HBCS-1100 Interface to a CMOS Gate.

17

Specifically,

The VO stability is improved asthe value of s" is reduced. Theincremental VO change of thecircuit shown in Figure 22 canbe calculated from the followingrelationship:

IP = 0(46)s"= ( (∆IC

∆hFE

(47)

s"RL + ∆VO = -∆hFEδ2VO

δhFEδIP [ [

= 50 nA=

QMIN =

IPS (MAX) =

Step1. Maximum Stray Photocurrent

Step 2. RL (MAX) for VIH

IPS (MAX)

QMIN + 1

IPR (MIN)

IPS (MAX)

250 nA

4 + 1

= 66.7k=RL (MAX) = VCC – VIH

hFE (MAX) • IPS (MAX)

5.0 – 4.0

300 • 50 × 10-9

Table 1. HBCS-1100 to CMOS Interface Design Procedure

Step 3. Minimum Photocurrent from Specular Reflector at∆ = 1.2 mm

From Equation (41):

1

0.3= 416 nA

ρ = 75% IP (MIN) @ IF = 35 mA = 100 nA

=

IP (∆ ) = %IP (∆ ) • ρ • Iρ

IP (∆ ) = 0.5 • 0.75 • 100 nA

N.A.(SURFACE)

N.A. (LENS)

RL =

Step 5. Select RL = 66.2k 1%

VCC – VIL

hFE (MIN) • IP( ∆ )

[[

[2

[2

N.A.(SURFACE) = 1

N.A.(LENS) = 0.3

%IP ( ∆ ), ∆ 1.2 mm = 0.5

Step 4. RL (MIN) for VIL at ∆

= 66.1k5.0 – 2.25

100 × 416 nA

The s" parameter is important indc coupled amplifier circuits be-cause the change in VO causedby the ∆hFE may result in thesucceeding gain stages beingdriven into saturation.

Current – Voltage FeedbackAmplifierWhen even greater output volt-age stability is desired, amodified current-voltage feed-back amplifier may benecessary. This bias approachuses RF and RN as shown in Fig-ure 23, to force a VB which setsIC to a level determined by VB/RE.This circuit can offer an s" tentimes better than the currentfeedback amplifier. The design issuch that an hFE variation of100—300 will cause a 0.3 Vchange in VO in the current-volt-age feedback configuration,while this same ∆hFE for Figure22 will cause a 1.2 V change.

18

Figure 22. Current Feedback Amplifier Interface to an Analog Comparator.

IF

VCC

RL100k

R1

R2

RF

10M

1k

–

+LM311

Table 2. Current Feedback Amplifier Design Procedure.

OUTPUT VOLTAGE, VO

VO =

VCC +VBERF

hFERL

1

hFE( (( (–RFIP

+ 1 +

1 +

RF

hFERL

1

hFE( (( (

STABILITY FACTOR, s"= ∆IC

∆hFE

s" ≈ (RL + RF) (VCC – VBE)

(RF + RL + RL hFE)2

hFE MIN = 100 MAX = 300 IPS(MAX) = 41 nA

RL = 100kΩ RF = 10MΩ VCC = 5.0V, VBE = 0.6V

hFE = 100 hFE = 300

VO 2.60V 1.39Vs" 1.1 × 10-7

Table 3. Practical Example of a Current Feedback Amplifier Design.

19

Figure 23. Current-Voltage Feedback Amplifier.

IF

VCC = 5V

VO

VE

RL500k

RE100k

RN

RF

12M

15M10µF

The design example given inTable 4 sets the VO at 2.5 V andoffers an s" of 7.5 x 10-9. The out-put voltage VO will be reducedfrom a design center of 2.5 V to aworst case (VO – s" ∆hFERL) =2.25 V. This implies that thethreshold of the comparatorshould be set to a level of 1.55 V.This amplifier offers atransresistance of 8 MΩ whichmeans for a 100 nA IP ( ∆l ), theoutput voltage VO will fall to1.5 V which is a sufficient differ-ential to cause the LM311output to change logic states.

LSTTL InterfaceThe previous circuits dealt withCMOS and comparator type in-terfaces. Figure 24 shows a twotransistor amplifier to LSTTLinterface. This circuit can eitherbe ac or dc coupled, only the di-rect coupled configuration will bepresented. The design approachis similar to that of Figure 21with the additional analysis ofthe second stage.

The first transistor, Q1, is biasedby the photodiode in a commonemitter configuration. Under theconditions of IPS(MAX), the collec-tor of Q1 is pulled up to within0.5 V of VCC, thus insuring thatQ2 is not conducting. This condi-tion sets the maximum value ofR1. When a reflected photocur-rent is present, the resulting ICof Q1 is the combination of cur-rent through R1 and the IB of Q2.Thus, R1 must be large enoughthat the current sinking capabil-ity of Q1 (dictated by hFE and IP)will result in sufficient IB in Q2to cause Q2 to saturate.

In the absence of reflected photo-current, both Q1 and Q2 arenormally off. Under this condi-tion, the load resistor, R2, mustbe able to sink the IIL of the

Step 1. Select RL Given: VO and IC

= 500k=RL = (VCC – VO)

IC

2.5V5µA

≈RN = (VE + VBE)

IN

0.5 + 0.675 nA

= 101k ≈ 100k

= 14.6MΩ ≈ 15MΩ

=RE = VEIE

0.5

4.91 × 10-6

(g + 1)( (

Step 2. Select RE Given: VE and IE

Step 3. Select RN Given: IN

=RF =

1 +

VO – VE – VBE

IN + IB

2.5 – 0.5 – 0.675 nA + 50 nA

= 11.2MΩ ≈ 12MΩ

Step 4. Select RF Given: IN and IB

Step 5. Stability Factor, s"

s" =

RF – RLRN

VCC – VBE

RF [1 + (1 + hFE)g]2

[ [+

s" = 7.5 × 10-9

Where g = RE + RLRF

RERN ( (1 +

RLRF

Table 4. Design Equations for the Current-Voltage FeedbackAmplifier.

20

Figure 24. DC Coupled HBCS-1100 to LSTTL Interface.

IF

VCC = 5V

R143k

R21.8k

IB

IPIILQ2

Q1

2N3906

74LS14

LSTTL gate at the desired VIL.To satisfy the desired logic con-dition, R2 must be less than VIL/IIL. The minimum value of R2 isdetermined by the current sourc-ing capability of Q2 produced byIB. The collector current of Q2must generate a voltage dropacross R2 greater than the VIH ofthe gate. It is recommended thata high gain, low leakage PNPtransistor, such as a 2N3906, beselected for Q2.

The rate at which the outputvoltage of Q2 changes is directlyrelated to the speed at which thereflecting surface is moving intothe reflection plane of the sen-sor. In many applications, therate of change of VO through theswitching region of the LS gateis so slow that it may cause logiclevel chatter at the output of thegate. If this chatter is observed,it is recommended that aSchmitt trigger gate, such as the74LS14, be used.

Reflective SensorApplications

Rotary TachometryA reflective sensor can be usedas the transducer to determinethe rotary speed of a motorshaft. This can be accomplishedby utilizing a disc with equallyspaced reflective and non-reflec-

tive lines placed around the cir-cumference of the disc. Thenumber of line pairs per revolu-tion will then give a specificpulse count per revolution. Inmany applications, it is desiredto have a very high density ofline pairs around the perimeterof a small diameter disc. Theperformance of the reflectivesensor is determined by theMTF for the spatial frequency,F, of line pairs on the disc. Thespatial frequency for a disc isdetermined by Equation (48).

Using Figure 25, the spatial fre-quency can be determinedassuming the radius is to thecenter point of the line pattern.Given the radius, r = 10 mm,and 220 lines/ revolution, a spa-tial frequency of 1.75 ln pr/mmis calculated. When an HBCS-1100 is used as the sensor forthis code wheel, an MTF re-sponse of 75% is obtained fromFigure 14.

The code wheel is affixed to ahub which is placed on the rotat-ing shaft. The reflective sensoris positioned perpendicular tothe disc and at a distance suchthat the maximum signal point,MSP, is at the plane of the codepattern. The highest reflected

(48)F = lines/rev.

4πr

Figure 25. Spatial Frequency of aCode Wheel.

photocurrent, IPR , is obtainedfrom a specular reflecting codepattern. This can be imple-mented by photolithographing apattern of opaque bars on ashiny metallic wheel-hub assem-bly. A diffuse code wheelassembly should be used whenthe mechanical tolerance of theaxial alignment of the HBCS-1100 to the normal of the codewheel exceeds 10°.

Tachometry applications allowan ac coupled amplifier to beused, such as the current feed-back type in Figure 22. ACcoupling the output of theHBCS-1100 eliminates the dcoutput offset voltage variationscaused by the stray photocur-rent.

HBCS-1100 AnalogTachometerThe HBCS-1100 can be used asthe transducer in high speed ro-tary tachometry applications.Figure 26 shows a circuit dia-gram that uses the reflectivesensor as a pulse source input toa frequency to voltage converter.The HBCS-1100 is configured asa current feedback amplifier andac coupled to an LM2907 fre-quency to voltage converter. The

r

LINES/REV

4πrF =

21

transistor Q1 is used as a cur-rent source to supply the IF tothe LED emitter.

The reflective sensor generates npulses per revolution, where n isthe number of line pairs perrevolution. The magnitude of thefrequency, f, applied to the F-Vconverter is n times the numberof revolutions per minute. Thisis the relationship shown inEquation (49).

VCC = 9.0V

VOUT

9 10

8

5

100k

3-42

11 LM2907

1M

100k

62 pFC

47kR

.01µF

1µF

50Ω

330Ω

10Ω1

123

5082-4487

8

4

HBCS-1100

62N2904

10M –+

–+

Figure 26. Analog Tachometer Circuit.

The capacitor, C, is the domi-nant factor in determining themaximum output voltage for arequired full scale frequency in-dication. The capacitor requiredto determine a full scale outputvoltage for a specific full scalefrequency is shown in Equation(50).

minsec

160

(49)

n =

f = n rev/min

where f = Hz

πr

line width

2πr

In pr width=

•×

(50)C =VOUT FULL SCALE

R • VCC • fFULL SCALE

A VOUT FULL SCALE = 1 V, 0 —25,000 rev/min tachometer canbe designed using a code wheelwith a radius of 20 mm and aline width of 0.63 mm. The ap-proach is to determinemaximum full scale frequencyfrom Equation (49), and then us-ing Equation (50), the full scalefrequency is 41.5 kHz and C iscalculated to be 57 pF. A 62 pFcapacitor is used for this ex-ample.

The F-V converter will respondto a minimum input signalswing of 250 mV. This inputlevel can be insured through theuse of a specular reflector. In theHBCS-1100 Total TransferFunction Section it was shownthat the IPR increases by 10.45dB when a specular reflector isused over a diffuse reflector. Thelimitation of a specular reflect-ing code wheel is that theHBCS-1100 alignment to the

code wheel must not be greaterthan 10° from the normal. If thedeviation is greater than 10° theimage of the source will not bereflected to the detector.

Paper Edge SensorThe accurate detection of theedge of a piece of paper can beaccomplished with an HBCS-1100 reflective sensor. If therange of reflectivity of the paperis known, either a paper reflec-tive or an obscuration systemcan be selected.

When a paper type which ishighly reflective is considered, itis desirable to utilize a reflectivesystem of the type that positionsthe sensor so that the maximumsignal point lies at the surface ofthe paper platten. This approachis shown in Figure 27. When alow reflectance paper type is be-ing sensed, the obscuration typesystem may be more suitable.Such a system is shown in Fig-ure 28.

The edge position sensing accu-racy is dependent on the spot

22

location as referenced to the me-chanical system. The HBCS-1100 offers a reflective sensingspot location of ±0.51 mm withrespect to the package centerline.

When an obscuration sensor sys-tem is used, the two transistoramplifiers shown in Figure 24provide a convenient dc couplingto a 74LS logic family. When thereflective system is applied, atransresistance amplifier of thetype shown in Figure 19 shouldbe considered.

Figure 27. Reflective Type Paper Edge Sensor.

PLATTEN

PAPER

HBCS-1100

SOCKET

PLATTEN

PAPER

MIRRORIMAGE SPOT

HBCS-1100

PRINTED CIRCUIT BOARD

Bar Code ScannerA reflective optical sensor can beused as the transducer in a barcode scanner application. Thebar code is an encoded form ofbinary data storage. The relativewidth difference bar to bar, andspace to space, describe the typi-cal encoding scheme ofDifferential Width encoding.This is the data format used inthe Universal Product Code,UPC.

The sensor provides an electricaloutput signal with a pulse width

Figure 28. Obscuration Type Paper Edge Sensor.

determined by the bar and spacewidths, and signal amplitude de-pendent upon the bar and spacereflection coefficients.

The Differential Width encodingscheme requires that outputpulse width, bar to bar, or spaceto space, be an accurate repre-sentation of the distance per unittime. The accuracy of the scan-ning output improves when thereflecting spot size is smallerthan the minimum bar or spacewidth. The smaller the scanningimage, the more abrupt the tran-sition from bar to space.

The output signal amplitude isdetermined by the difference be-tween the bar reflectance andspace reflectance. The minimumoutput signal to maximum out-put signal ratio is directlyproportional to the bar to spacereflectance.

The signal amplifier that is in-terconnected to the sensor musthave a large dynamic operatingrange to accommodate the varia-tions of reflector types, and alsoprovide adequate signal differen-tial for the bar to spacereflectivity difference.

Figure 13 shows the trapezoidalpulse train that is obtained fromscanning a bar code of equalwidth bars and spaces. As theimage size increases due todefocusing, the pulse train am-plitude is reduced and thewaveform becomes triangular. Itis desirable that the amplifierprovides the signal amplitudechange at the same scanning lo-cation as the bar to spacetransition.

23

Figure 29 shows a schematic foran amplifier system that willconvert the bar and space widthsinto TTL compatible logic sig-nals. The circuit uses theCA3130 as a transresistance am-plifier for the HBCS-1100photodiode. The output of theamplifier is applied to positivepeak (LM124-1) and negativepeak (LM124-2) detectors. Theresistors R1 and R2 set the refer-ence (negative-going) input tothe code comparator (LM124-3)

at a voltage which is halfway be-tween the positive peak and thenegative peak, so the switchingthreshold is therefore at 50% ofthe peak-to-peak modulation.The noise gate (LM124-4) com-pares the negative peak to avoltage which is two diode volt-age drops (D1 and D2) below thepositive peak, so unless thepeak-to-peak amplitude exceedstwo diode drops, the G input ofthe 74LS75 remains low and Qcannot change. This ensures

Figure 29. Bar Code Scanner Circuit.

+VCC

240Ω

2-25pF

HBCS-110022M

CA3130

LM124-1–

+VCC = VCC = 10V

++

–

1N914

1N914

1N914

D11N914

D21N914

15µF

+

–

LM124-247k

47k

4.7k

4.7k

74LS75

15µF 10M

R2

R1

+

–

LM124-4

+

–

LM124-3

D

G Q–

OUTPUT

+VCC

that the Q output of the 74LS75will remain fixed unless the ex-cursions at the output of theCA3130 are of adequate ampli-tude (two diode drops) that noisewill not interfere.

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Data subject to change.Copyright © 1999 Agilent Technologies, Inc.Obsoletes 5953-04605091-7363E (11/99)