Sensor Systems

Sidebar

theory:sensor_technology:st8_accelerometer_model

The Accelerometer Model

In three sections, the mathematic background behind some typical electrostatic transducer phenomena is described. First we will see the coupling of resonance and sensitivity. Next, we will construct the non-linear and linear transduction equations and the related coupling coefficient. Finally, the pull-in voltage is discussed.

An accelerometer and its second-order response

The most common method to implement an accelerometer is to use a proof mass on a spring. In that case, the excursion of the proof mass is proportional to the acceleration. Consider two simple mechanical formulas, the first being Newton’s second law of motion

$$F = M \cdot a \label{eq:NewtonSecond}$$

with $F$ the force on the mass, $M$ the mass and $a$ the acceleration. The second equation is Hooke’s law

$$F = k \cdot x \label{eq:HookeLaw}$$

for an extension $x$ of the spring due to a force $F$ with $k$ the spring constant. When a mass is connected to a spring, these two forces are equal and we find

$$F = k \cdot x = M \cdot a \rightarrow x= \frac {M}{k} a \label{eq:Acceleration}$$

stating that the excursion of the mass (or the extension of the spring) is proportional to the applied acceleration $a$. The proportionality is given by the spring constant $k$ and the proof mass $M$. Later, we will see that the $M/k$ ratio determines the resonance frequency as well, resulting into a system where sensitivity and bandwidth can not be decoupled.

A mass-spring combination is a second order system characterised by a low-pass behaviour and a resonance frequency. The differential equation is given by

$$F_{ext} = M\frac{\mathrm{d}^{2}x }{\mathrm{d} t^{2}} + D \frac{\mathrm{d} x }{\mathrm{d} t} + k \cdot x$$

With $M$ the proof mass, $D$ a damping coefficient and $k$ the spring constant. If the external force $F_{ext}$ is initiated from an external acceleration $a$, as given by equation \eqref{eq:NewtonSecond}, then we find

$$F_{ext} = M \cdot a = M\frac{\mathrm{d}^{2}x }{\mathrm{d} t^{2}} + D \frac{\mathrm{d} x }{\mathrm{d} t} + k \cdot x.$$

By a Fourier transform with respect to the displacement $x$, we find the relation for the frequency domain

$$F_{ext} = M \cdot a = M \left ( j \omega \right )^{2} x + D j \omega x + kx$$

which can be rewritten as

$$x \left ( j \omega \right )=\frac{Mk^{-1}}{\left ( j \omega \right )^{2}Mk^{-1} + j \omega Dk^{-1} + 1}a \label{eq:TransferOmega}$$

For a fixed acceleration, the displacement is constant:

$$x = \frac{M}{k}a$$

which is equal to equation \eqref{eq:Acceleration}. The mechanical spectrum has a low-pass behaviour up to the undamped resonance frequency

$$\omega _{0}=2 \pi f_{0}=\sqrt{\frac{k}{M}}. \label{eq:Omega0}$$

Equation \eqref{eq:TransferOmega} can be written in terms of the resonance frequency and quality factor $Q$

$$x \left ( j \omega \right )=\frac{a}{\omega_{0}^{2}}\left ( \left ( j \frac{\omega}{\omega_{0}} \right )^{2} + j \frac{\omega}{\omega_{0}} \frac{1}{Q}+ 1 \right )^{-1} \label{eq:TransferOmega0}$$

with

$$Q=\frac{\sqrt{Mk}}{D}$$

The excursion $x$ of the mass due to an acceleration a has a sensitivity $S = x/a$. From equation \eqref{eq:Acceleration} we can see this sensitivity is given by $S = M/k$ below the resonance frequency. In addition, we see the same ratio $M/k$ appearing in the resonance frequency according to equation \eqref{eq:Omega0}. Therefore, we may conclude that the sensitivity $S = x/a$ and the resonance frequency are coupled inherently by means of

$$S=\frac{M}{k} \leftrightarrow f_{0}=\frac{1}{2 \pi \sqrt{S}}.$$

The frequency $f_{0}$ is the undamped resonance (in vacuum). Due to damping, the real resonance peak will be slightly lower than $f_{0}$, especially for low $Q$’s. For $Q> 1/\sqrt{2} \approx 0.7$, we have a visible peak which is located at1)

$$f_{res}=f_{0} \sqrt{ 1 - \frac{1}{2Q^{2}}}.$$

For lower $Q$’s, the system is overdamped and the bandwidth is determined by the damping and the spring constant only. In that case, the cut-off frequency becomes

$$\omega_{-3dB}=\omega_{0}Q=\frac{k}{D}.$$

For electrical engineers known as the RC-time.

In figure 1, four curves are drawn for the transfer function given by equation \eqref{eq:TransferOmega0}. This graph shows the modulus of $x \left (j\omega \right )/a \left ( j \omega \right )$, being the square root of $\left ( Re^{2}+Im^{2} \right )$. The physical meaning is how the amplitude of a sinusoidal acceleration transfers into an amplitude of the sinusoidal excursion $x$. The central black solid curve is for $Q=1$ and an undamped resonance frequency of $f_{0}$. The effect of double and half the value of $Q$ is shown by the other two black solid curves. When we decrease the resonance frequency to $f_{0}/2$, the de sensitivity increases by a factor of four. This can be seen by the grey dotted line. Note that the two axes are logarithmic.

Fig. 1: Typical responses of second-order systems

Note that we assume a second-order system comprising a single (effective) mass and a single (effective) spring constant. A practical microphone will not have a single resonance: there will be multiple modes. The theory above will still apply, but is only meaningful for the basic resonance.

Electrostatic transduction

Consider a parallel plate set-up as drawn in figure 2. This configuration acts as a parallel plate capacitor with variable gap. First, consider the situation without friction and without spring suspension.

Fig. 2: Parallel plate set-up

According to the Maxwell equations, the electrical displacement field $D$ equals the free surface charge $\sigma$. Note that the displacement field $D$ should not be confused with the damping constant $D$ as used in the previous section. Therefore, the electric field strength is

$$E = \frac{\sigma}{\epsilon_{0}} \label{eq:ElectricFieldStrength}$$

with $\epsilon_{0}$ the permittivity of air. Subsequently, the potential over the capacitor can be calculated

$$u = E \cdot d = \frac{\sigma d}{\epsilon_{0}} = \frac{q d}{\epsilon_{0}A} \label{eq:PotentialCapacitor}$$

in which the surface charge $\sigma$ is expressed as total charge $q$ divided by the size $A$ of the capacitor. In fact, the last equation is the capacitor equation from which we can easily see the capacitance $C$ of the parallel plate capacitor

$$u = \frac{q}{C} \rightarrow C= \frac{\epsilon_{0}A}{d}. \label{eq:ParallelCapacitor}$$

On the bottom plate there is a charge $–q$, on the upper plate a charge $+q$. In fact, the upper plate is placed in the electric field of the bottom plate. The result is that the upper plate experiences an electrostatic force caused by the electric field of the magnitude $E/2$. This force is

$$F_{el} = q \cdot \frac{E}{2} = \frac{q^{2}}{2 \epsilon_{0}A}$$

using equation \eqref{eq:ElectricFieldStrength} and the fact that $\sigma = q/A$.

Next, imagine a mechanical spring attached to the upper plate. The upper plate can move and the bottom plate is fixed as shown in figure 3. In this graph, the air gap at rest is $d$ and the displacement due to an electrostatic force is $x$.

Fig. 3: A parallel plate capacitor suspended from a spring

The force needed to extend the spring by a distance $x$ is given by

$$F_{spring} = -k \cdot x$$

with $k$ the spring constant. The force balance equation is

\begin{aligned} F_{ext} \quad & = \quad F_{el} +F_{spring} \\ F_{ext} \quad & = \quad \frac{q^{2}}{2 \epsilon_{0}A} - k \cdot x \end{aligned} \label{eq:ForceBalance}

with $F_{ext}$ an external force due to pressure, sound, acceleration etc. From this equation, we can already feel that for a certain applied voltage, the $q^{2}$ term becomes that large, that there might not be a sufficiently high spring force to balance the equation. This is the moment of pull-in and so the system will become instable for a certain bias voltage.

Equation \eqref{eq:ForceBalance} relates phenomena in the electrical domain to phenomena in the mechanical domain. A systematic approach for mathematically characterising a transducer is by exploring the relation between the efforts and the state variables in each domain. The word effort is used for the physical variable which is the cause of a change in the state variable. In the mechanical domain, we have the state variable $x$ and the effort $F$. In the electrical domain, the state variable is the charge $q$ which is related to an effort, being the potential difference $u$. The transduction equations can be described by the series expansion of \eqref{eq:PotentialCapacitor} and \eqref{eq:ForceBalance}

\begin{aligned} \mathrm{d}u \quad & = \quad \frac{d_{0}}{\epsilon_{0}A}\mathrm{d}q + \frac{q_{0}}{\epsilon_{0}A}\mathrm{d}x + \frac{1}{\epsilon_{0}A}\mathrm{d}q\mathrm{d}x \\ \mathrm{d}F \quad & = \quad \frac{q_{0}}{\epsilon_{0}A}\mathrm{d}q+k\mathrm{d}x+\frac{1}{2\epsilon_{0}A}\mathrm{d}q^{2}\end{aligned}

with $d_{0}$ and $q_{0}$ the gap size and capacitor charge at the operational point at which the series expansion is taken. This is the full expansion, which can be reduced to the linearised transduction equations for small variations on top of a bias ($x<<d_{0}$):

\begin{aligned} \mathrm{d}u \quad & = \quad \frac{d_{0}}{\epsilon_{0}A}\mathrm{d}q + \frac{q_{0}}{\epsilon_{0}A}\mathrm{d}x \\ \mathrm{d}F \quad & = \quad \frac{q_{0}}{\epsilon_{0}A}\mathrm{d}q+k\mathrm{d}x \label{eq_electrostatic_transduction_equations} \end{aligned}

This set of equations is known as the characteristic equations and we can learn a lot from them using generic methods from the mathematics behind the transduction technology. First, we write them in a matrix form

$$\begin{bmatrix} \mathrm{d}u\\ \mathrm{d}F \end{bmatrix} = \begin{bmatrix} \frac{d_{0}}{\epsilon_{0}A} & \frac{q_{0}}{\epsilon_{0}A}\\ \frac{q_{0}}{\epsilon_{0}A} & k \end{bmatrix} \begin{bmatrix} \mathrm{d}q\\ \mathrm{d}x \end{bmatrix} =M \cdot \begin{bmatrix} \mathrm{d}q\\ \mathrm{d}x \end{bmatrix}$$

using the transduction matrix $M$, not to be confused with the mass $M$. In practical microphones and accelerometers, we know all elements of the matrix $M$. The $d_{0}$ does not deviate so much from the gap size, $A$ is the plate area, $k$ the (effective) spring constant of the suspension and $q_{0}$ a direct consequence of the biasing voltage by means of

$$q_{0}=C \cdot V_{bias} = \frac{\epsilon _{0}A}{d_{0}}V_{bias}.$$

The cross entries $M_{12}$ and $M_{21}$ in the transduction matrix indicate there is a coupling between the two do-mains when they are non-zero. These terms are equal and referred to as the transduction coefficient. When they are non-zero, a displacement $dx$ translates into a potential difference $du$ and a charge variation dq into a force variation $dF$. This means we can relate the mechanical domain to the electrical domain and back.

Pull-in voltage

The common read-out mechanisms for capacitive microphones are using a biasing technique. In this case, a voltage is applied to the microphone plates with a very high imped-ance. The result is that the change in capacitance, due to sound, is converted into a charge flow (= current). In terms of the previous section, biasing makes the transduction coefficients $M_{12}=M_{21}$ unequal to zero. Note that also accelerometer read-out using an analog to digital converter of the sigma-delta type uses a DC offset on the capacitive plates of the accelerometer.

In a biased capacitor with moveable plates, we have the phenomenon of pull-in. Consider a DC biasing voltage applied to the configuration of figure 3. Equation \eqref{eq:ForceBalance} can be used for the situation where $F_{ext} = 0$:

$$\frac{q^{2}}{2 \epsilon_{0} A} - kx=0 \\ \frac{V_{bias}^{2}C_{el}^{2}}{2 \epsilon_{0} A} - kx=0 \\ \frac{V_{bias}^{2} \epsilon_{0} A}{2 \left ( d-x \right )^{2}} - kx=0$$

where the capacitor relation $q = C \cdot u$ is substituted first, followed by substitution of \eqref{eq:ParallelCapacitor}. Note that we are now using a gap size of $d$ again, instead of $d_{0}$. This is because we are using the total analytic equations instead of linearised small signal variations around a gap size $d_{0}$. In the large signal model, the instantaneous gap size equals $d-x$. Manipulation yields

$$V_{bias}=\sqrt{\frac{2k}{\epsilon_{0}A} \left ( d-x \right )^{2}x}$$

which has a maximum for $x = d/3$. This means we can increase the bias voltage up to the level where the top-plate is displaced up to $d/3$, above that voltage the system becomes unstable and the top-plate is pulled down to the bottom plate. After substitution of $x = d/3$ we can find this maximum voltage which is referred to as the pull-in voltage:

$$u_{pull-in}=\sqrt{\frac{8k}{27 \epsilon_{0}A} d_{0}^{3}}.$$

Note once more that this is for a parallel-plate set-up: in our case we have a flexible membrane in the microphone. In the calculations above, the deflection profile is not taken into account.

Sensor Technology TOC

1)
J. Millman and A. Grabel, Microelectronics, McGraw-Hill, Inc., Singapore, 1988