Instrumentation & Measurement Magazine 24-5 - 14

Fig. 3. The curvature of deflected diaphragm versus pressure (for a Silicon-Nitride diaphragm, E=300 GPa; v = 0.24): (a) for various diaphragm thicknesses t(mm)
when its radius is b=15mm; (b) for various diaphragm radiuses b(mm) when its thickness is t=0.2 mm.
Fig. 2a and Fig. 2b show the deflection versus the radial
coordinate of an under uniform pressure diaphragm: larger
diaphragm radius and less diaphragm thickness leads to
more deflections. Also, Fig. 2c and Fig. 2d show the central/
maximum deflection versus pressure for various amounts of
diaphragm thicknesses and diameters, respectively.
Considering ,Yr for a definite  as Y(r), one can calculate
the diaphragm curvature using the concept of the
osculating circle by substituting in [28]:
 

      
T r
 10r
r
        (3.a)
Tr r


Yr YY


Yr r  3 122 DD 
1
  

22
r
16
b  
; Y r br
16 bb
 
 
R Rr 

where 

, and T are the ancillary vectors for calculating the curvature
of the diaphragm R. So, we work in a desirable radial
coordinate r. The prime symbols show the derivatives relative
to it.
We solved such equations with a self-programmed
Mathematica module to find R. Ultimately the curvature is
considered at the basic spot-size on the diaphragm, namely
 10
w rr . Always, wb
b
/2
is much smaller than the diaphragm
radius. Fig. 3 shows that a very tiny deflection leads
to a large curvature of the diaphragm. Such curvatures are in
a range between infinity for a flat diaphragm, down to about
2 m, corresponded to a central deflection which is less than 2
mm. Fig. 3a and Fig. 3b show that the near-center curvature of
the diaphragm changes more smoothly in thicker and more
extensive diaphragms in comparison with thinner and less extensive
ones. So, we have chosen a rather extensive and thick
diaphragm in a way that its maximum deflection, as well as
14
 

1 Y 2
Y
3/2
(3.c)
22 
  
  
(3.b)
its sensitivity, lays in a suitable range for the desired pressure
domain.
Optical Ray Analysis
If Pin
would be the total power of the light guided into the fiber,
then the ejected light intensity from the fiber considered has a
Gaussian envelope like:
Iz   ww
,
Pin

2
22 ;
exp



(4)
where w = w(z) is the effective radius of the light spot at an arbitrary
z-distance from the fiber endpoint. For example, z can
identify the fiber-to-reflector distance 0
h hh in Fig. 4,
b ,
when we want to calculate the intensity of light on the diaphragm;
also z = 2h for the receiving plane, namely returning
again to the fiber.
Matrix ray analysis [29] helps us to derive the w at the lightreceiving
point: light ray travels an h-distance from the end tip
of the fiber, then the ray reflects from the diaphragm considered
as a convex mirror with radius R, then it returns again from an hdistance
to the fiber endpoint. Every step can be described by its
specified matrix. So, if we show the emitting ray with a vector as
00 
y , , and the receiving one as f

y , f
, a triplet set of matrix
productsM will describe these three steps of light ray traveling.
Fig. 4. Displacement diagram for a double-clad reflective-diaphragm fiberoptic
pressure sensor (DFOPS).
IEEE Instrumentation & Measurement Magazine
August 2021

Instrumentation & Measurement Magazine 24-5

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