Transducers
[<Lat. Trans, across + ducere, to lead]
• Devices to transform signals between
different physical domains.
• Taxonomy
– Active: input signal modulates output energy;
e.g. liquid crystal display.
– Passive: input signal transformed to output energy;
• Bidirectional, e.g. motor/generator;
• Unidirectional, e.g. photodiode
Signal domains
• Signals may occur in many physical forms
• Examples
– Electrical signals on wires
– Sound in free air
– Light fields in free space
• Transducers connect signal domains
– input signals of one physical form
– produce signals in a different physical form.
Signals, power and energy
• Physical signals are carried by energy
– Energy = force x distance (mechanical) [Joules]
• May take many equivalent forms.
– Within a domain: potential vs kinetic;
– In different domains: electrical, nuclear, heat, etc.
• Is conserved: can transform but not create or destroy.
– Power = time derivative of energy [Watts]
• Signal strength is a measure of signal power.
• Signal to noise ratio is a ratio of powers.
• A transducer receives signals in one domain and
generates them in another.
Potential and kinetic energy
• Potential energy is stored in reactive tension:
– Voltage on a capacitor
– Force compressing a spring
• Kinetic energy is stored in reactive motion:
– current in an inductor
– velocity of a mass
• Reactance types
– Potential: voltage, force, electric field, etc.;
– Kinetic: current, velocity, magnetic field, etc.
Voltage, current, force, &cet.
• Power is expressible by pairs of variables:
– Voltage and current,
– Force and velocity,
– Pressure and flow, etc.
• These are potential-kinetic variable pairs in
particular physical domains.
• Each domain has an associated measure of
impedance and its reciprocal, admittance,
although nomenclature varies.
An electrostatic microphone
Passive bidirectional transducer example
• We first explore the physics relating
voltage to pressure and current to volumevelocity. The result is non-linear.
• We use small-signal assumptions to
linearize the model.
• We then represent these results in terms of
two-port matrices which linearly relate
variables between domains.
Microphone construction
Microphone is a conducting membrane
stretched over an insulated ring above a
conducting back plate
Cross section of unbiased microphone
Biased membrane under tension
Insulated annular support rings
Parallel plate capacitor physics
Consider a capacitor with fixed lower plate and movable
upper plate, both of area A, separated by distance x. An
amount Q of charge has been moved from one plate to the
other, causing an electric field between them to develop a
voltage V. The field of the lower plate acts on the charge
of the upper plate to
produce a force F on it.
These effects provide the
A
V
coupling between the
-Q
electrical and mechanical
Q
domains.
x
Electric field of the lower plate
Coulomb’s law in differential and integral forms:
 

  D      E  nˆda    ( x)d 3 x,
S
V
where D = E, E is electric field,  is charge density, and
8.85x10-12 for air dielectric. The first integral is of the
field normal to a closed surface, and the second is of the
charge in the volume enclosed by the surface. Thus the
field near a flat plate of area A with charge Q is,



  E  nˆda    E  nˆda    E  nˆda  Q
S1
S2
S edge

Q
ˆ
Q  2  E  nda  2EA  E 
2A
A
S1
S2
Capacitance between the plates
The definition of capacitance is Q=CV. The total field
between the plates is the superposition of fields from both
plates, or E=Q/A. By the definition of voltage we have,
  Qx
V   E  dx 
x
A
The latter because the field between the plates is nearly
uniform. From this we can obtain the formula for a parallel
plate capacitor:
x
x
A
V Q
 CV
 C
A
A
x
Force on the upper plate
The field E from the charge on the lower plate exerts force
on the charge Q on the upper plate which is negative:

 3
F   Ed x  F   Ed 3 x  QE ,
V
V
the latter because the field is nearly uniform over the plate
area. Substituting for the electric field, we get a nonlinear
force as a function of Q, which is always attractive.
Q
 Q2
F
.
2A
F
E
Force - voltage relationship
Let the total voltage VT = V0 + V, the sum of the bias and
signal voltages respectively. Similarly, FT = F0 + F, where
F0 is force due to the bias voltage alone and F is the signal
force. Also, xT = x0 + x, where x0 is the plate separation at
rest, and x is the signal displacement. The total force is,
 QT
 (CTVT ) 2  (A / xT ) 2 VT2  AVT2
FT 



.
2
2A
2A
2A
2 xT
2
If the signal displacement x is held to zero,
 AVT2
FT 
.
2
2 x0
This is a non-linear relationship which is always attractive.
Linearizing the relationship
This behavior can be “linearized” by assuming that
V  V0 , F  F0 , and x  x0 .
Expanding FT and VT we make the approximation,
V02  2V0V  V 2
V02  2V0V
F0  F  A
 A
.
2
2
2 x0
2 x0
Subtractin g out F0  AV02 / 2 x02 leaves the linear form,
F  (AV0 / x02 )V .
Substituti ng C0  A / x0 , and defining pressure p  F / A,
 V0C0
p
V.
Ax0
Velocity - current relationship
If now the signal voltage is held to zero, the current and
displaceme nt must vary as follows :
dQ d
dCT
d  A   AV0 dx
 
I
 (CT V0 )  V0
 V0 
dt dt
dt
dt  x0  x  ( x0  x) 2 dt
 AV0
 AV0 
dx
 2

x.
2
2
x0  2 x0 x  x dt
x0

Let us define volume velocity as u   A x . Then we have,
V0
V0C0
I 2 u
u.
x0
Ax0
Two-port transducer matrices
• Relate two input variables to two output variables.
– The product of the input variables must be power
– The product of the output variables must be power
– Input and output variables may be in any domains
• A microphone has inputs p and u, outputs V and I.
• The hybrid, or h-parameter matrix H is written,
u
p
I
H
V
 p   h11 h12  u 
 
   
 I   h21 h22 V 
* (See Hanspeter Schmid, “Tables: Two-Port Matrices,” people.ee.ethz.ch/~hps/publications/twoport.pdf )
The microphone’s H matrix
We have shown that
 V0C0
V0C0
p
V and I 
u.
Ax0
Ax0
Holding u  0  x  0, I  C0
dV
.
dt
Holding V  0  p  kAx  k  udt (k  spring force).
In frequency domain, these are, I  sC0V and p 
Thus, we can write,
 k

 p  s
  
 I   V0C0
 Ax
0

 V0C0 

Ax0  u 
 

sC0 V 

k
u.
s
The microphone’s T matrix
It was convenient to use the H matrix because we could derive the
behaviors of p and I when u and V were held fixed, respectively.
It is also useful to have a
u
I
transmission matrix T providing
p
electrical outputs, given acoustic
V
T
inputs. Conversion of the H
matrix to the T matrix is a
V   t11 t12  p 
problem in linear algebra, the
 
   
solution of which is,
t
t
I
u
  
1  1

T
h12  h22
h11   Ax0  1

 
det H  V0C0  sC0
21
22




2
(V0C0 / Ax0 )  kC0 
k/s
The microphone’s Z matrix
It is also sometimes useful to
represent this as an impedance
matrix Z facilitating impedance
matching calculations at the
electrical and acoustic ports. In
this case, the solution is,
u
p
I
Z
 p   z11
   
V   z21
2
1  det H h12 
1  (V0C0 / Ax0 )  kC0


 
Z
h22   h21 1  sC0 
 V0C0 / Ax0
V
z12  u 
 
z22  I 
 V0C0 / Ax0 


1
