For CMUTs that are designed to work in immersion, the membranes need to be fixed on all sides to prevent the coupling medium from getting into the cavity. However, for air-coupled CMUTs, that requirement can be lifted. In order to generate maximum pressure, the volume displaced by the membrane during vibration should be as large as possible.
For a rectangular membrane, that condition is achieved when only two opposite sides of the membrane are fixed, and the other two sides are free. A comparison between having two sides and all four sides fixed was presented in [23], and it was proved by simulations that more ultrasonic energy would be emitted when only two opposite sides were fixed.
Figure 6.2: A fixed-fixed beam that is used to model an air-coupled CMUT.
This structure essentially becomes a fixed-fixed beam, as illustrated in Figure 6.2, and the length, width, and thickness of the beam are denoted by L, b, and h respectively.
The static deflection of an electrically actuated fixed-fixed beam can be described using the following equation found in [86]:
where w(x) is the beam displacement in the transverse direction, q(x) is the distributed load per unit length, L, N , and I are the length, the axial load and the second moment of area of the beam respectively. The axial load arises from the residual stress and is proportional to the cross-sectional area of the beam. If the width of the beam is large compared to its length (L≤W ), the effective Youngs modulus is ˆE = E/(1 − v2) where E is the Youngs modulus and v is the Poisson’s ratio of the material [87]. For a narrow beam (L≥10W ), the effective Young’s modulus is equal to the material’s Young’s modulus. If the beam width falls in between the two extremes, a linear interpolation can be done to estimate the effective Young’s modulus1.
The non-linear term in equation (6.1) is the contribution from mid-plane stretching, which is caused by the lengthening of the beam when it deflects. Mid-plane stretching
1Email communication with Prof. Eihab Abdel-Rahman.
increases the resonant frequency of the beam because of the increase in axial tension [86].
However, in most multi-user MEMS processes, the gap height, defined by the sacrificial layer thickness, is not much larger than the beam thickness. For example, in PolyMUMPs, the first sacrificial layer thickness is 2µm and the polysilicon layer is also 2µm thick.
Therefore, the mid-plane stretching term can be removed as suggested by Lee [88], who reasoned that the non-linear term can be neglected when the gap to beam thickness ratio is much less than five. Equation (6.1) can be rewritten as
EIˆ d4w
If equation (6.2) is integrated to get the form kw = F , the spring constant, k, can be found using the mode shape of the beam, as presented in [88].
k = 1024
The fundamental resonant frequency of the beam is f0 = 2π1 qk
m, where the effective mass, m, of the beam can be estimated as m = 128ρbhL/315 [89], and ρ is the density of the beam material. Therefore, the resonant frequency is
f0 = 1
Two observations can be made from equation (6.5). First, a tensile axial stress (N >
0) increases the resonant frequency by making the beam stiffer. Secondly, the resonant frequency, to the first order, is independent of the width of the beam because the second moment of area (I) and the axial load (N ) are directly proportional to the width.
The derivation of the resonant frequency so far did not take into account the bias volt-age, which reduces the resonant frequency because of the spring softening effect. Consider
the pull-in voltage of a fixed-fixed beam given by [88]:
Vpi∼= 0.6036 s
kh03
εA (6.6)
where h0 and A are the unactuated height and the area of the beam, respectively. Recall the softened spring constant given in (3.14):
ksof t = k − εSV2
d3 (6.7)
If the bias voltage is set at 80% of the pull-in voltage, which is a typical bias voltage choice for CMUTs, then by substituting 0.8Vpi into (6.7) and assuming that the beam area A is equivalent to the piston area S, ksof t can be calculated to be 0.767k, and the resonant frequency is decreased by 12.4%.
The beam thickness, h, is determined by the process and there is nothing that a user of a standard process can do except choosing a different structural layer, if available. The length, L, is chosen according to the desired resonant frequency. The only dimensional parameter left, then, is the width, b. As it turns out, the beam width affects mainly the torsional vibration frequency. Torsional vibration of a fixed-fixed beam means that one free side of the beam goes up while the other free side goes down, as opposed to lateral vibration where both free sides go up and down together. For a well designed beam, the fundamental vibrational mode is lateral because the impulse response of the beam should mainly consist of lateral vibration. Figure 6.3 shows the two vibrational mode shapes of a fixed-fixed beam.
The torsional equation of motion of a fixed-fixed beam with axial load is given in [90]:
ρJd2τ
dt2 − (Gkt+ N ζ)d2τ
dx2 = 0 (6.8)
Where τ is torsional displacement; J = hb(h2 + b2)/12 is the polar second moment of area;
G = E/2(1 + v) is the shear modulus of the beam, ζ = (h2+ b2)/8, and kt is the torsional stiffness coefficient [91], which can be estimated as
kt= bh3 1
Figure 6.3: Lateral (left) and torsional (right) vibration mode shapes of a fixed-fixed beam.
Solving equation (6.8) with the boundary conditions of a fixed-fixed beam will lead to the torsional resonant frequency ft [90]:
The width of the beam should be picked such that the torsional resonant frequency is not an integer multiple of the fundamental lateral resonant frequency, in order to avoid any coupling of energy to the second mode. However, spring softening is not taken into account in (6.10). An analytical expression for such a case is not available, thus numerical modelling will be used to find the torsional vibration frequency when a bias voltage is applied. Nevertheless, for initial design purpose, the torsional vibration frequency can be compared with the unactuated lateral vibration frequency, assuming that spring softening will cause both to decease by a similar percentage.