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Optical MEMS pressure sensor based on Fabry-Perot interferometry

Ming Li, Ming Wang, and Hongpu Li

Open Access Open Access

Abstract

By employing the surface and bulk micro-electro-mechanical system (MEMS) techniques, we design and demonstrate a simple and miniature optical Fabry-Perot interferometric pressure sensor, where the loaded pressure is gauged by measuring the spectrum shift of the reflected optical signal. From the simulation results based on a multiple cavities interference model, we find that the response range and sensitivity of this pressure sensor can be simply altered by adjusting the size of sensing area. The experimental results show that high linear response in the range of 0.2–1.0 Mpa and a reasonable sensitivity of 10.07 nm/MPa (spectrum shift/pressure) have been obtained for this sensor.

©2006 Optical Society of America

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Figures (8)
Fig. 1.
Fig. 1. Configuration of the optical MEMS pressure sensor

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Fig. 2.
Fig. 2. Multiple cavities interference model of the optical MEMS pressure sensor

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Fig. 3.
Fig. 3. Influence of loaded pressures on the reflected spectrum.

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Fig. 4.
Fig. 4. Influence of the glass thickness on the reflected spectrum.

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Fig. 5.
Fig. 5. Processing steps for fabrication of the optical MEMS pressure sensor

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Fig. 6.
Fig. 6. Experimental setup for the optical MEMS Pressure sensor

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Fig. 7.
Fig. 7. Optical signals obtained from OSA. (a) Spectrum of the ASE light source, (b) reflection spectrum of the sensor; and (c) the filtered reflection spectrum.

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Fig. 8.
Fig. 8. Results of the pressure measurements and its linear fit.

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Equations (13)

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ω ( r ) = 3 P R 0 4 ( 1 − υ 2 ) 16 E h 3 ( 1 − r 2 R 0 2 ) 2 = ω 0 ( 1 − r 2 R 0 2 ) 2
L = L 0 − ω
r 1 ′ = r 1 + κ 1 exp ( − iϕ 1 ) + κ 2 exp ( − iϕ 2 ) + κ 3 exp ( − iϕ 3 )
κ 1 = η 1 ( 1 − r 1 2 ) r 2
κ 2 = η 1 η 2 ( 1 − r 1 2 ) ( 1 − r 2 2 ) r 3
κ 3 = η 1 η 2 η 3 ( 1 − r 1 2 ) ( 1 − r 2 2 ) ( 1 − r 3 2 ) r 4
∣ r 1 ′ ∣ ≈ Re [ r 1 ′ ] = r 1 + κ 1 cos ( ϕ 1 ) + κ 2 cos ( ϕ 2 ) + κ 3 cos ( ϕ 3 )
φ 2 = 4 πn air L λ
Δ φ 2 Δ λ ∣ λ = λ m = 0 = 4 πn air ( Δ L λ m − Δλ m λ m 2 L )
Δ φ 2 = 4 πn air ( L 1 L 2 − L 2 λ 1 ) λ 1 λ 2 ≈ 4 πn air Δ L λ
Δ φ 2 Δ P = 4 πn air λ L λ m ∙ Δλ m Δ P ≈ 4 πn air L λ m ( P 0 ) [ 1 λ m ( P 0 ) ∙ Δ λ m Δ P ]
Se = 1 λ m ( P 0 ) ∙ Δ λ m Δ P = Δ λ m λ m ( P 0 ) Δ L ∙ 3 R 0 4 ( 1 − υ 2 ) 16 Eh 3
Δ P max = 16 Eh 3 3 R 0 4 ( 1 − υ 2 ) Δ L = 8 Eh 3 3 R 0 4 ( 1 − υ 2 ) λ n air
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