CHAPTER II.5.11
Cochlear Prostheses
PROBLEMS
Problem 1
Sound at the level of the tympanic membrane demonstrates a pressure peak at a frequency of 2.5 kHz in the
human. Consider the boundary conditions of the ear
canal and explain why that resonance occurs.
Problem 2
Considering the areas of the stapes and the tympanic
membrane, as well as the properties of the ossicular
chain, compute the approximate ratio of pressures that
is found when the tympanic membrane is driven at low
frequency. Note that this is an approximate calculation
that does not consider the dynamics of the system.
Problem 3
Focusing stimuli has been proposed by several people
(Suesserman and Spelman, 1993; Jolly et al., 1996) as
a solution to the problem of field interference between
monopole sources. Consider using dipole and quadrupole sources. Assume that the sources are 50 μm
hemispheres located on the surface of an insulating
boundary. Let the hemispheres be separated by 200 μm
in both cases. Plot the potential fields along two lines:
one that is 100 μm above and parallel to the sources, the
other that is 200 μm above and parallel to the sources.
Describe the properties of the fields that are produced.
Look particularly at the peak potentials and the halfamplitude widths of the fields. Hints: A dipole consists
of two sources, one carrying current I and the other
carrying current −I. A quadrupole is a special case of
a tripole. Three sources are used. The central source
carries current I, while the two flanking sources carry
current −I/2. For ease of calculation, place one of the
sources at the origin, and let the other sources lie on,
e.g., the x-axis.
Problem 4
Two electrode array designs are considered. One uses a
silicone substrate, and the other a liquid crystal polymer
substrate. If a circular cross-section is used, with an outside diameter of 200 μm, find the force that would be
exerted on a free end whose length is 1 mm and whose
deflection is 20 μm. The modulus of elasticity of silicone is
2.76 Mpa, and that of liquid crystal polymer is 158 MPa.
The polar moment of inertia is:
4
Ip =
πd
32
.
Problem 5
Izzo and Richter address the intensity of optical stimuli to excite the inner ear in terms of the energy density
required to stimulate neurons. They state that the energy
density well above threshold is 3 mJ/cm2 at a pulse width
of 5 μsec. Assume that the optical source is a Vertical
Cavity Surface Emitting Laser (VCSEL) whose emitter
has a 20 μm diameter. Assume further that pulses repeat
every 100 μsec (Izzo et al., 2008). How much DC power
is required per channel if the VCSEL is 15% efficient?
ANSWERS
Problem 1
The ear canal is bounded at one end by the tympanic
membrane (a closed end), at the other by the opening in
the pinna (an open end). The canal can be approximated
by a cylinder about 1 cm in diameter with a length of
about 2–3.5 cm. The first resonance will occur at a frequency whose wavelength is 8–14 cm (a quarter-wave
resonance, since the proximal end is closed and the distal
end of the canal is opened). The velocity of sound in air
is 331.6 m/sec. The resonant frequency is:
f=
c
λ
Hz
If λ, the wavelength, is 1.33 cm, then f = 2.5 kHz.
Problem 2
The area of the tympanic membrane is 20 times that of
the footplate of the stapes. The pressure at the footplate
of the stapes will be 20 times that applied to the tympanic membrane. However, the motion of the footplate
of the stapes is about 75% of that of the tympanic membrane. There is an additional mechanical advantage of
4/3 for force. Hence, the pressure at the footplate of the
stapes will be about 27 times that applied to the tympanic membrane.
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Problem 3
field. Note also that the quadrupole field decreases more
rapidly with distance than does the dipole field.
The fields can be computed by summing the potentials
produced by the sources. The potential produced by a
spherical source is identical to that produced by a point
source and is:
Problem 4
V(r) =
Iρ
4π
volts
The maximum deflection of a cantilever beam occurs at
its loaded end and is:
ymax =
where I is the applied current in Amperes, assumed to
be 10 μA in this case; ρ is the resistivity in ohm-cm, here,
let the resistivity be 53 ohm-cm (the resistivity of perilymph); r is the radial distance from the source. For a
source on an insulator, the potential field is twice that
produced by the same source in a conductive medium.
Since the medium is linear, it is possible to superpose the
sources that produce the dipole and quadrupole fields.
Using Mathcad® 2000 (MathSoft, Cambridge, MA) to
compute the fields, it is possible to get the following
results.
Figure II.5.11.3 shows that the peak potential of the
dipole source is greater than that of the quadrupole (about
twice as much), and that the half amplitude width is
greater as well. The quadrupole produces a more focused
Pl3
3EI
solving for P, the force necessary to deflect the beam:
P=
3EIymax
l3
For this beam, d = 200 μm and I = 1.57 × 10−16 m4. Thus:
P = 9.425 × 10 –12 E N
E for silicone is 2.76 Mpa, and 158 MPa for silicone and
liquid crystal polymer, respectively. The force required
to deflect the silicone is 2.6 × 10−5 N, or about 2.6 mg.
The liquid crystal polymer requires a force of 1.5 ×
10−3 N or 150 mg to deflect.
This is the program Question3, that determines the potential fields 100 and 200 microns above a
dipole and quadrupole source. The sources are separated by 200 microns. In the case of the
dipole, the sources are at x= -100 and x= +100 microns. In the case of the quadrupole, the
sources are at x= -200, x= 0 and x= +200 microns. Current is 10 A.
Compute the potential fields:
rh := 53
−5
I := 10
ohmcm
x1 := −0.01 cm
i := 0 .. 800
x2 := 0.01
k := rh⋅
I
2⋅π
x3 := −0.02 cm
cm
 i − 1

 400 
xi := 0.04 ⋅ 
A
cm
j := 1 .. 2
Vcm
x4 := 0.02 cm
y j := 0.01 ⋅ j
1
1


−
0.5
0.5
2
2
2
2
  (xi − x1) + (y j) 
 (xi − x2) + (y j)  


Vdi , j := ( k) ⋅ 
cm
volts
1
−0.5
0.5


+
−
0.5
0.5
0.5
2
2
2
2
2
2
  (xi − x1) + (y j) 
 (xi) + (y j) 
 (xi − x2) + (y j)  


Vqi , j := ( k) ⋅ 
e2
volts
Potential vs Distance for Dipole and Quadrupole Sources
0.00
0.00
0.00
0.00
0
FIGURE II.5.11.3 Potential pro-
0.00
0.00
0.00
0.00
0.04
0.0
0.0
0.0
0
0.0
Problem 5
The energy density of the optical pulse is given as
3 mJ/cm2. The width of the pulse applied to the cochlea
is 5 μsec. Assume that the applied energy doesn’t vary
during its application. In that case, the power density is
given by:
papp =
wapp
tpulse
And
papp =
Or,
0.003
5 · 10 − 6
W / cm2
papp = 600 W / cm2
The light source is a VCSEL with a small area of emission,
Svcsel = π(10
−3 2
2
) cm
Svcsel = 3.14159 × 10 − 6 cm2
0 .0
0 .0
0 .04
duced by the dipole and quadrupole sources of problem 3. The
dipole field is shown in orange for
a distance of 100 microns and in
blue for 200 microns. The quadrupole field is shown in green for
a distance of 100 microns and in
cyan for 200 microns. This shows
that while the quadrupole source
produces a more focused potential
field, the field is dramatically smaller
than that produced by dipole.
The applied power is the product of the power density
and the area of emission:
Papp = papp (Svcsel )W
Papp = 1.884 × 10 − 3 W
Now the VCSEL is 15% efficient, and the applied
power per pulse is the optical power divided by the
efficiency of the VCSEL. However, the pulsatile power
is applied for only 5% of the repetition rate, so that
the total required DC power will be 5% of the power
required per pulse. Thus,
PDC = Papp / 0.03 W
PDC = 1.884 × 10 − 3 / 0.03 W
PDC = 0.628 mW
The power per stimulus channel is quite reasonable,
indeed, ten channels could be driven simultaneously with
less than ten milliwatts of power.
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