# Electrostatics - Electrostatically actuated tunable MEMS capacitor

## Analytical calculation of electrotatic force between two parallel plates

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This page proposes an analytical calculation of a plate electrostatically actuated toward a second parallel plate and shows the influence of the different parameters on the electrostatic force, and its evolution with gap.

This kind of architecture could be used as an electrostatic tunable MEMS capacitor: the distance between the plate controls the capacity value.

## Static capacitor

### Basics

First of all, to understand electrostatic actuation, let's remember the principle of a capacity:
A capacity is made of two conductive solid elements (the *electrodes*) separated by a dielectric non-conductive environment(void, gas, solid, liquid, whatever!!). If you apply a voltage between the conductive part, you create an electric field accross the dielectric material. Electrical charges will agglomerate on the conductive parts, corresponding to an equilibrium made with the electric field.

The capacity is the ability to store a quantity of electrical charges, depending on the voltage difference between the electrodes forming the capacity.

Behaviour of a static capacity

Once you have let the charges reach the equilibrium point, the capacity is loaded: the agglomeration of charges is an energy the capacity is able to give back! So, if you cut off the voltage supplier, and let the capacity in an open circuit, it keeps the voltage difference between the electrodes, and if you branch a closed circuit, it will give back its electrical charges, generating current, until it is completely unloaded.

Of course, loading and unloading imply electrical charges displacement, so electrical current! Capacity consume current only when loading, the unloading process being just a restitution of the accumulated energy.

Now, let's see this in a more physicist way!

From here and to the end of this page, you will need a few electromagnetics basics. Even if the capacitor device physics is explained, it would be better to have a minimal knowledge of the electrostatic equations.

### Hypothesis

Let's consider two rectangular plate parallel to the plane defined by $(\mathrm{x},\mathrm{y})$ axes, of size $L$ by $W$, separated by a thickness $d$ along $z$ axis, of air having an electrical permittivity of ${\epsilon}_{0}$. To simplify notations, we use $S=L.W$ to note the surface of the plates.

Two rectangular plates forming a capacitor

We consider the two plates perfectly conductive. They are under an electric potential difference(a voltage) ${V}_{0}$.

We neglect, in this calculation, the edge effect (electrical field path not straight between the two plates around the plates edges). So the electrical field between the two plates has the form:

$$\overrightarrow{{E}_{z}}={E}_{\mathrm{z}}\left(z\right)\overrightarrow{{u}_{z}}$$

To calculate the capacitive effect, we must establish a relation between the voltage across the device and the electrical charges on the plates. For both theses elements, we will use the electric field to get the values.

### Relation between electric field and voltage

According electric field definition, we have:

$$\overrightarrow{E}=-\overrightarrow{gradV}$$

But we also know that air doesn't contain electric charges, so we can use *Maxwell-Gauss* equation to find:

$$div\overrightarrow{E}=\frac{\rho}{{\epsilon}_{0}}=0$$

So we can conclude that $\overrightarrow{E}$ is a constant vector ${E}_{0}\overrightarrow{{u}_{z}}$.

This implies for the electrical potential the following solution:

$${V}_{0}\left(z\right)=-{E}_{0}.d+V\left(0\right)$$

Considering that the bottom plate is at a potential $V=0$, we have a simple solution to the problem.

$$\Delta V=-{E}_{0}.d$$

### Relation between electrical field and charges

We've said earlier that charges were accumulating on the conductive plates. So we need to calculate them using *Maxwell-Gauss* equation.

Let's consider a box containing one of the plates. Inside the box, we have

$$div\overrightarrow{E}=\frac{\rho}{{\epsilon}_{0}}$$

that we integrate on the volume, and then we use the *Ostrogradski* theorem to get

$${\oint}_{S}\overrightarrow{{E}_{0}}.\overrightarrow{\mathrm{dS}}={\oint}_{V}\frac{\rho}{{\epsilon}_{0}}$$

The integrations lead to:

$$ES=\frac{Q}{{\epsilon}_{0}}$$

replacing $E$ with $\frac{\Delta V}{d}$, we get:

$$Q=\Delta V\frac{{\epsilon}_{0}S}{d}$$

and finally, we can say:

$$C=\frac{{\epsilon}_{0}S}{d}$$

that is the capacity for parallel plates.

## Tunable capacitor

Now that we know the capacity for static plates, let's imagine that the upper plate is attached to a spring but free to move along the $z$ axis while the lower plate is fixed.

In this case, the charges and the electric field will give a force applied on the upper plate.

Schematics of the Coulomb's force applied on the upper plate.

Our objective is to predict the position of the mobile plate depending on the voltage applied between the electrodes. So, we need an equilibrium equation:

### Equilibrium equation

There is two ways to get the equilibrium equation allowing to calculate the position of the mobile plate along the $z$ axis. The first one is based on the first principle of dynamics, and the second is based on energy variations.

The first principle of dynamics tells us:

$$\overrightarrow{\sum \overrightarrow{F}}=m\overrightarrow{a}$$

Considering that we're looking for the equilibrium point, the acceleration is null. The detail of the forces in presence are:

$$\overrightarrow{{F}_{\mathrm{elecstq}}}=Q.\overrightarrow{E}$$

that we can detail saying:

$$\overrightarrow{E}=\overrightarrow{-\mathrm{grad}\left(V\right)}$$ and $$Q=C.V$$

that leads to:

$$\overrightarrow{{F}_{\mathrm{elecstq}}}=-\frac{\epsilon S}{2}\frac{{V}^{2}}{{z}^{2}}\overrightarrow{{u}_{z}}$$

$$\overrightarrow{{F}_{\mathrm{spring}}}=-k(z-{z}_{0})\overrightarrow{{u}_{z}}$$

At equilibrium position, these forces are equal.

The energy calculation considers the principle that total energy of a system is a constant. So, the energy accumulated by the capacitor (increasing with the plate moving to the fixed plate) is equal to the energy transfered to the spring.

The energy accumulated in the capacitor and in the spring are:

$${E}_{\mathrm{elecstq}}=1/2C{V}^{2}$$

$${E}_{\mathrm{spring}}=1/2k{(z-{z}_{0})}^{2}$$

If we develop and say these energies are equal, we find the same equation for both methods:

$${z}^{3}-{z}_{0}.{z}^{2}+\frac{\epsilon S{V}^{2}}{k}=0$$

### Pull-in, pull-out and hysteresis

Now we go back to the forces expression. We can calculate with a numeric equation the position of the mobile plate for each voltage. Drawing the curve of forces amplitudes in function of the displacement $z$ show an interesting information when reaching a particular configuration.

For a normal case, the two curves will cross in a location corresponding to the equilibrium point, giving the position of the plate. But beyond a critical voltage, it is impossible to find an equilibrium point, resulting in an electrostatic force with no counter force. The plate is accelerated until it is stopped by the fixed plate. This phenomenon is called *pull-in*. Most of time, the lower electrode is protected with an isolating layer so that there is no short-cut between the electrodes.

We can calculate the position of the plate where the *pull-in* effect occurs:

Consider the global energy ${E}_{\mathrm{elecstq}}-{E}_{\mathrm{spring}}$:
Most of time, it is null at the equilibrium point. The *pull-in* effect corresponds to an instability, meaning a changing in the sign of the energy function along $z$ axis, so that energy get higher and higher until it is mechanically controlled by the contact of the mobile plate. This can be retrieved by nullifying the derivation of the energy functions:

$$3{{z}_{\mathrm{pull-in}}}^{2}-2{z}_{0}{z}_{\mathrm{pull-in}}=0$$

so we can easily calculate ${z}_{\mathrm{pull-in}}$, and by reinjecting it in the equilibrium formula, we even get ${V}_{\mathrm{pull-in}}$:
$${z}_{\mathrm{pull-in}}=\frac{2}{3}{z}_{0}$$

$${V}_{\mathrm{pull-in}}=\sqrt{\frac{8}{27}\frac{k}{\epsilon S}{{z}_{0}}^{3}}$$

Our plate is now almost sticked to the bottom electrode, only an isolation layer help avoiding a shortcut. This gives a new configuration for our calculation:

Now, the initial distance between the two plates does not correspond to an equilibrium, but to a tiny distance (let's call it ${t}_{\mathrm{isol}}$). So even if you redude voltage, this weak distance will help electrostatic force being far stronger thant spring force! You will have to reduce far more the voltage to make the mobile plate get back to an equilibrium position. This is called the *pull-out* phenomenon. Exactly like the pull-in, you can find the corresponding voltage thanks to the energies derivation:

First of all, let's calculate the energy considering the configuration:

$${E}_{\mathrm{elecstq}}=\frac{1}{2}\frac{{\epsilon}_{\mathrm{isol}}S}{{t}_{\mathrm{isol}}^{2}}{V}^{2}$$

$${E}_{\mathrm{spring}}=\frac{1}{2}k{({z}_{0}-{t}_{\mathrm{isol}})}^{2}$$

The voltage corresponding to the nullification of the sum of these energy is the *pull-out* voltage.

so, we finally find:

$${V}_{\mathrm{pull-out}}={t}_{\mathrm{isol}}\sqrt{2{z}_{0}\frac{k}{{\epsilon}_{\mathrm{isol}}S}}$$

in which we consider that ${t}_{\mathrm{isol}}{z}_{0}$ to simplify the notation.

Schematics and graphics showing the electrostatic actuation cycle with pull-in, pull-out, and the hysteresis behaviour

The *pull-in*/*pull-out* phenomenons make the relation between $z$ position and $V$ voltage being a *hysteresis* cycle. This is shown on the animation.