Biomedical Engineering Reference
In-Depth Information
of the figure,
Z
R
¼
Z
W
(impedance of water). The acous-
tic loss for this case at resonance can be found from (note all these impedances include the
area factor
Z
L
¼
Z
B
(backing impedance) and
)
A
AL
ð
f
Þ¼
Z
R
=ð
Z
L
þ
Z
R
Þ¼
Z
W
=ð
Z
B
þ
Z
W
Þ
ð
16
:
32
Þ
0
The physical meaning of this equation is that the electrical power reaching the radiation
resistance is converted into acoustical power radiating from the two electroded faces of
the piezoelectric; the proportion of power in the forward (right) direction is represented
by the ratio in Eq. (16.32).
Construction of a single crystal transducer is shown in Figure 16.11. The key parts are the
backing, the piezoelectric crystal coated with thin electrodes for electrical contact, matching
layers, and a lens material, with an acoustic impedance close to water, for focusing (dis-
cussed in more detail in Section 16.2.6). Matching layers will be discussed shortly; in prac-
tice, more than one matching layer is used to broaden the bandwidth of the transducer.
In the first simple model, the crystal was loaded by acoustic impedances equal to that of
the crystal, or
Z
C
¼
Z
L
¼
Z
R
so that
AL
(
f
0
)
¼
0.5, as expected. Another example is that in
which the left side is loaded by air or
Z
L
¼
0
A
MRayls, and the right by water,
Z
R
¼
1.5
A
MRayls. In this case,
1.0 but at the expense of extremely narrow bandwidth and
a correspondingly long pulse. One case of interest is where the backing material is matched
to that of the crystal so that
AL
(
f
0
)
¼
Z
L
¼
Z
c
¼
30
A
MRayls and
Z
R
¼
1.5
A
MRayls, and
AL
(
f
0
)
¼
0.048, an inefficient transfer.
To improve the forward transfer efficiency, as described by the acoustic loss factor,
matching layers are employed. At the resonant frequency, a matching layer is designed to
be a quarter of a wavelength thick and to be the mean value of the two impedances to be
matched, or
p
Z
1
Z
2
Z
ml
¼
ð
16
:
33
Þ
EXAMPLE PROBLEM 16.5
Match a crystal of impedance
Z
c
¼
30
A
MRayls to water,
Z
w
¼
1.5
A
MRayls.
Solution
From Eq. (16.33), the matching layer impedance is
p
Z
W
Z
C
p
1
Z
ml
¼
¼
:
5
A
MRayls
30
A
MRayls
¼
6
:
7
A
MRayls
The acoustic input impedance at resonance is
2
ml
=
Z
2
Z
1
¼
Z
therefore, the impedance looking to the right is
2
2
ml
=
Z
W
¼ð
Z
R
¼
Z
6
:
7
A
Þ
=
1
:
5
A
¼
30
A
MRayls
leading to A(
f
0
)
¼
30/(30
þ
30)
¼
0.5 for a water load with an intervening matching layer or an
improvement of over 10 dB.
Continued
