Abstract

The design of a mechanical nanosecond light pulsing system which tests the optical response time of a semiconductor film is discussed. The system consists of a high speed rotating mirror which is used to scan a laser beam across a film target. The system is capable of producing variable duration light pulses as fast as 3.5 nsec. A discussion of how shorter exposure times may be realized is also presented.

© 1983 Optical Society of America

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References

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  1. S. A. Tretter, Introduction to Discrete Time Signal Processing (Wiley, New York, 1976), pp. 39–63.

Tretter, S. A.

S. A. Tretter, Introduction to Discrete Time Signal Processing (Wiley, New York, 1976), pp. 39–63.

Other (1)

S. A. Tretter, Introduction to Discrete Time Signal Processing (Wiley, New York, 1976), pp. 39–63.

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Figures (8)

Fig. 1
Fig. 1

Double mirror arrangement with slight rotation of one mirror.

Fig. 2
Fig. 2

Design parameters for a 2-nsec exposure time.

Fig. 3
Fig. 3

Variable time duration light pulsing system.

Fig. 4
Fig. 4

Laser beam path: 1, laser beam entering the system; 2, first and last reflections at spinning mirror; 3, first and last reflections at parallel mirror; 4, reflections at folding mirror; 5, intermediate reflections at spinning mirror.

Fig. 5
Fig. 5

Variables for spot progression determination in an off-parallel mirror system.

Fig. 6
Fig. 6

Skewed mirror surface.

Fig. 7
Fig. 7

Superposition of 5000 photodetector output pulses.

Fig. 8
Fig. 8

Experimental and theoretical exposure times vs number of bounces.

Equations (27)

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ϕ [ n ] = ϕ [ n 1 ] + 2 θ .
x [ n ] = x [ n 1 ] + 2 L ϕ [ n 1 ] ,
z [ Φ ( z ) ϕ [ 0 ] ] = Φ ( z ) + 2 θ 1 z 1 ,
z [ X ( z ) x [ 0 ] ] = X ( z ) + 2 L Φ ( z ) ,
X ( z ) = x [ 0 ] 1 z 1 + 2 L ϕ [ 0 ] ( 1 z 1 ) 2 + 4 L θ z 2 ( 1 z 1 ) 3 .
x [ n ] = x [ 0 ] + 2 L ϕ [ 0 ] + 2 L θ ( n 2 n ) .
x [ n ] = 2 L θ ( n 2 n ) .
υ [ n ] = 2 L ω ( n 2 n ) ,
t = D υ [ n + 1 ] ,
υ [ n + 1 ] = υ [ n + 1 ] + 2 ω n L ,
α = ( 2 λ ) / s ,
D = A s + 1 A α h .
A = ( α h s ) 1 / 2 .
t = ( 2 α s ) 1 / 2 ω L 1 / 2 [ n 3 / 2 + n 1 / 2 ( 1 + L L ) ] .
t = A s + 2 A α n L 2 L ω [ n 2 + n ( 1 + L L ) ] .
t min = α ω A ( n + 1 + L L ) .
n = 25 , ω = 25 , 000 rpm , L = 3 m , L = 1.5 m .
Y ( z ) = y [ 0 ] 1 z 1 + 2 L b z ( 1 z 1 ) 2 + 4 L a z 2 ( 1 z 1 ) 3 .
y [ n ] = y [ 0 ] + 2 n L b + 2 L a ( n 2 n ) .
b = D / L ,
a = 0.02 ( D / L ) . ( 20 )
y [ 13 ] = 6.8 D .
e [ n ] = y [ n ] L m e ,
E ( z ) = e L m Y ( z ) . ( 23 )
X ( z ) = x [ 0 ] 1 z 1 + 2 L ϕ [ 0 ] z ( 1 z 1 ) 2 + 4 e L L m z 2 ( 1 z 1 ) 2 Y ( z ) .
x [ n ] = x [ 0 ] + 2 L ϕ [ 0 ] + 4 e L L m { y [ 0 ] 2 ( n 2 n ) + L ( 2.5 × 10 5 n 4 + 1.6 × 10 4 n 3 8.7 × 10 3 n 2 + 5.7 × 10 3 n ) } .
e = x [ n ] L m 4 L 2 ( 2.5 × 10 5 n 4 + 1.6 × 10 4 n 3 8.7 × 10 3 n 2 + 5.7 × 10 3 n ) . ( 26 )

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