attenuation due to connectors and splices

18
ATTENUATION DUE TO CONNECTORS AND SPLICES Tyler Martin Zebedee Smith Phillip Chukwu

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Tyler Martin Zebedee Smith Phillip Chukwu. Attenuation due to Connectors and Splices. Overlap of fiber cores. The fibers may be parallel, but ate not lined up vertically. Light leaks out of the core of the first fiber into the cladding of the second fiber. Diameter Overlapping. - PowerPoint PPT Presentation

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Page 1: Attenuation due to Connectors and Splices

ATTENUATION DUE TO CONNECTORS AND SPLICES

Tyler MartinZebedee SmithPhillip Chukwu

Page 2: Attenuation due to Connectors and Splices

Overlap of fiber cores

The fibers may be parallel, but ate not lined up vertically.

Light leaks out of the core of the first fiber into the cladding of the second fiber.

Page 3: Attenuation due to Connectors and Splices
Page 4: Attenuation due to Connectors and Splices

Diameter Overlapping

Differences in the diameter of the cores will also create overlapping.

The loss due to this can be found as: Loss = (d1

2 – d22 / d1

2

Where d1 is the diameter of the core of the sending fiber, and d2 is the diameter of the core of the receiving fiber.

Page 5: Attenuation due to Connectors and Splices

Example

Example: Calculate the loss due to overlapping if the diameter of the receiving fiber core is 9µm and the diameter of the sending fiber core is 15 µm.

Page 6: Attenuation due to Connectors and Splices

Others

Problems also occur from differences in fiber types, and when there are elliptical and off0centered cores.

Page 7: Attenuation due to Connectors and Splices

Alignment of Fiber Axes

This simply refers to the angle of difference.

Fibers with a higher Numerical Aperture can collect light over a large range of angles.

Page 8: Attenuation due to Connectors and Splices

Spacing between fibers

Page 9: Attenuation due to Connectors and Splices

SPACING BETWEEN FIBERS• There are two ways that light can be

lost if there is any spacing between the fibers.1.There is the spreading of the light

when it leaves a fiber.2.There is the reflection of light

passing between materials.

Page 10: Attenuation due to Connectors and Splices

SPACING BETWEEN FIBERS(END-SEPERATION LOSS)

• Light leaves fibers in a cone shape• The spreading angle of the light is

dependent on the numerical aperture.

• The loss increases as the NA of the input fiber increases because a higher NA causes the light to spread faster.

• The more the light spreads the less light the receiving fiber will collect.

Page 11: Attenuation due to Connectors and Splices
Page 12: Attenuation due to Connectors and Splices

SPACING BETWEEN FIBERS(END-SEPERATION LOSS)

The equation for loss is

𝐿𝑜𝑠𝑠 (𝑑𝐵 )=10 log 10( 𝑑/2

𝑑 /2+(𝑆 tan(sin−1(𝑁𝐴𝑛0 ))))• d is the core diameter• S is the fiber spacing• NA is the numerical aperture

• n0 is the refractive index of the material between the fibers

Page 13: Attenuation due to Connectors and Splices

End-reflection loss

Page 14: Attenuation due to Connectors and Splices

END-REFLECTION LOSS• Another way light can be lost

through spacing between the fibers is through end-reflection loss.

• This occurs when light passes between two materials with different refractive indexes. This process is called Fresnel Reflection.

Page 15: Attenuation due to Connectors and Splices
Page 16: Attenuation due to Connectors and Splices

END-REFLECTION LOSSThe equation for the fraction of light reflected is

• R is fraction of light reflected• nfiber is the refractive index of

the fiber where light is coming

from• n is the refractive index of the

object where the light is heading to

𝑅=(𝑛 𝑓𝑖𝑏𝑒𝑟−𝑛𝑛𝑓𝑖𝑏𝑒𝑟+𝑛 )

2

Page 17: Attenuation due to Connectors and Splices

END-REFLECTION LOSSThe equation for loss in dB is

𝑙𝑜𝑠𝑠 (𝑑𝐵 )/ 𝑠𝑢𝑟𝑓𝑎𝑐𝑒=−10 log(1−(𝑛𝑓𝑖𝑏𝑒𝑟−𝑛𝑛 𝑓𝑖𝑏𝑒𝑟+𝑛 )

2

)• This gives the loss per glass-air

interface.

Page 18: Attenuation due to Connectors and Splices

Example• Find the End-reflection loss in dB if the fiber core is made

of glass and the gap is air.

𝑑𝐵=−10 log(1−(𝑛𝑓𝑖𝑏𝑒𝑟−𝑛𝑠𝑝𝑎𝑐𝑒

𝑛 𝑓𝑖𝑏𝑒𝑟+𝑛𝑠𝑝𝑎𝑐𝑒 )2

)𝑑𝐵=−10 log (1−( 1.5−11.5+1 )

2)𝑑𝐵=−10 log(1−( .52.5 )

2)

𝑑𝐵=−10 log (1− .04 )

𝑑𝐵=.177 𝑑𝐵