Biomedical Engineering Reference
In-Depth Information
where current is
is charge. The total power radiated from both sides
of the transducer into a surrounding medium of a modified specific acoustic impedance
(including an area factor),
I
¼
i
o
Q
¼
i
o
C
0
V
and
Q
Z
C
¼ r
cA
, equal to that of the crystal, is
2
2
2
W
A
¼
ATT
*
=ð
2
Z
C
=
A
Þ¼
A
j
F
ð
f
Þ=
A
j
=
2
Z
C
¼j
hC
V
sin
ðp
f
=
2
f
Þj
=
2
Z
C
:
ð
16
:
26b
Þ
0
0
Setting the powers of Eqs. (16.26a) and (16.26b) equal leads to a solution for
R
A
sin c
2
R
A
ð
f
Þ¼
R
AC
ð
f
¼
2
f
0
Þ
ð
16
:
27a
Þ
where sin c(
x
)
¼
sin (
p
x
)/(
p
x
) and
2
T
2
2
T
k
0
¼
d
k
R
AC
¼
ð
16
:
27b
Þ
4
f
C
2
A
e
S
0
p
C
D
=
in which the electroacoustic coupling constant is
is
inversely proportional to the capacitance and area of the transducer and directly dependent
on the square of the thickness,
k
T
,and
k
T
¼
h
=
e
S
:
Note that
R
AC
d
. Also, note that at resonance,
2
T
Þ¼
k
R
A
ð
f
ð
16
:
27c
Þ
0
p
2
f
0
C
0
Network theory requires that the imaginary part of an impedance be related to the real
part through a Hilbert transform, so the radiation reactance can be found as
X
A
ð
f
Þ¼
`
Hi
½
R
A
ð
f
Þ ¼
R
AC
½
sin
ðp
f
=
f
0
Þp
f
=
f
0
2
ð
16
:
28
Þ
2
ðp
f
=
2
f
Þ
0
The transducer impedance is plotted as a function of frequency in Figure 16.9b. Here
R
A
is
maximum at the resonant frequency
X
A
is zero.
This simple model describes the essential characteristics of a piezoelectric transducer.
The electrical impedance has a maximum of real radiation resistance at the resonant fre-
quency. The force spectrum is also peaked at the resonant frequency and has a certain
shape.
f
0
and
EXAMPLE PROBLEM 16.4
Find the transducer impedance at resonance for a square transducer 2.5 mm on a side for a res-
onant frequency of 3 MHz for the ceramic PZT5A with
e
s
/
e
0
¼
830,
c
¼
4.35 km/s,
k
T
¼
0.49, and
e
0
¼
8.85 pF/m.
Solution
First find the thickness needed to achieve a resonant frequency of 3 MHz from
f
0
¼
c
/2
d
,
or
d
¼
c
/2
f
0
¼
4.35 mm/
m
s
/2
3 MHz
¼
0.725 mm. Then the capacitance is
10
12
10
3
m
2
830
8
:
85
F
=
m
ð
2
:
5
Þ
C
0
¼
¼
63
:
3pF
10
4
m
7
:
25
