# Polarization of electrostatic actuator

Introduction
DC polarization
AC polarization
Introduction
Applying a pure sinusoidal voltage
What happen if we add an offset voltage to the sinusoidal voltage?
And what about a square, triangular, or patatoidal voltage?
Conclusion
A ±30V voltage amplifier

## Introduction

Polarizing an electrostatic actuator seems quite simple but may be hindered by quite a few mistakes that only experience can learn... (Put in a simpler way, I've done all of these mistakes but I probably overlooked many things, thus if my explanations suit you, take them number , but otherwise tell me what you think is the truth :-)
First of all, electrosatic actuators dislike stray charge that may come from airborne particule, or by improper grounding. To lessen these effects you must ground everything than can be grounded and leave a ground plane just below the actuator. If you forget this you may expect unreliable behaviour and, for example, actuator working for a while and then stopping (build-up of charge in insulating layer that stick the actuator), or actuator operation varying with humidity level (a film of water helps to reduce the static charge by increasing the conductivity) or even actuator not working at all... Actually even with well grounded structure you may already be affected by these problems...

## DC polarization

This is the simplest one: you apply voltage between the two electrodes of the actuator and you increase it slowly while watching the structure through your microscope. If it moves you're quite a happy man! An interesting point with the electrostatic actuator is that the force generateed is proportional to the square of the voltage, ie:
F ~ V²
Hence when the voltage is multiplied by 3 the force increase almost by an order of magnitude.
For a comb-drive actuator, where the force does not change with the displacement, and with a linearly increasing restoring force (ie, a spring) it means that the displacement also increases with the square of the voltage. Thus take care not to increase to rapidly your voltage:, when you go from 1 to 10 V, you increase the force by a factor of 100 and at 31.5V it is a factor of 1000!
Moreover, for a gap-closing actuator, where the force increase quadratically with the displacement, associated with a linearly increasing restoring force (ie, a spring), after the actuator has moved 1/3 of the initial gap width, it will snap through and 'close' the gap. This may cause a short-circuit if no insulator or landing point has been designed. However this problem does not always appear and I've noted that for doped polysilicon actuator short-circuit do not appear for moderate voltage, presumably because of the native oxide layer and the large rouhghness of the side that yields a very small contact area.
And finally, if the structure does not move? Hum, hum. Actually this is the most common case :-( Then either your structure is stuck (check your release process, and why not try SAM coating?) or more probably the voltage is not high enough (Actually, and to be honest, it means that your structure is too stiff or the electrodes surface too small :-). Then, you need a DC power supply which can deliver higher voltage or you may try to excite your structure by applying AC signal, which will give you more information on your structure.

## AC polarization

### Introduction

The main interest of applying AC voltage to an electrostatic actuator is the measurement of its resonnant frequency. But it may also be simply used to check if a stiff structure is moving, because at the resonnant frequency the displacement is increased by the Q factor of the structure, often more than ten, and more than enough to see something with the optical microscope without investing in laser vibrometer...

### Applying a pure sinusoidal voltage

As we have already pointed out the force delivered by an electrostatic actuator vary with the square of the voltage. Thus what happen if we apply a sinusoidal voltage V = V0 sin(wt + Ø ) to it? We have:
F ~ V² = (V0 sin(wt + Ø ))² = V0²(½ - ½cos(2wt + 2Ø)) = ½V0²(1 + sin(-2wt - 2Ø + ¶/2))
Thus we see that the force has a constant component of magnitude proportional to ½V0² and a varying component with frequency twice the frequency of the applied voltage and a complex phase factor. Thus under the microscope we will see the actuator oscillates at a frequency twice that of the exciting voltage around a deflected position different from the rest position. It should be noted that the existence of a deflected position will certainly result in a different mode shape than the one corresponding to the tested mode, and will add to the innacuracy of this method to check the resonant frequency (actually even if the structure is driven by the actuator around its rest position the deformation will not correspond exactly to the shape of the corresponding mode :-).

### What happen if we add an offset voltage to the sinusoidal voltage?

We return back to the expression of the force, which gives in this case:
F ~ V² = (V0 sin(wt + Ø ) + V1)² = ½V0² + V1² + 2V0V1sin(wt + Ø ) - ½V0²cos(2wt + 2Ø) =
F ~ ½V0² + V1² + 2V0V1sin(wt + Ø ) + ½V0²sin(-2wt - 2Ø + ¶/2)
As you see there is a larger DC component and now there is a signal driving the structure at a pulsation equal to the pulsation of the voltage. The interest of such arrangement is that it provides a mean to control the offset displacement (ie, the DC component) and that it will provide a displacement synchronized with the driving signal. Actually, if we choose V0 « V1 then the expression of the force may be simplified as:
F ~  V1² + 2V0V1sin(wt + Ø )
then the driving force is at the same pulsation than the applied voltage which may be very helpful if we want to use synchronous detection of the movement. Once more the start from a deflected position will result in a different mode shape than the one corresponding to the tested mode. Moreover, we should not forget the existence of the second order harmonic (2wt) which may create spurious excitation if its amplitude (½V0²) is too large (or simply if the Q-factor is large).

### And what about a square, triangular, or patatoidal voltage?

Actually the whatever periodic signal you use may be decomposed in Fourier series and, for example, you will find that a square signal is the sum of sinusoidal signals whose pulsations ar an odd multiple of the original pulsation (ie, the odd harmonics) and whose amplitude decrease as the inverse of the rank of the harmonic. Generally, signals will have a more complex decomposition, including odd and even harmonics with amplitude varying as a complex function of the harmonic rank (it is similar to the signal with offset case, where we had only the two first harmonics).
Then the force is once more proportional to the square of the signal and thus it will have frequency components which will be a multiple of the original signal frequency, and including this frequency. The amplitude of the different harmonics will depend on the signal and may be computed using Fourier analysis. However it is not a simple calculus, as not only square of the signal will appear but also cross-product between harmonics which result in the generation of signal having a frequency corresponding to the difference of the two signal mixed.
Practically you will see your structure being excited in resonance with many different frequencies that you may mistakenly take for higher order mode. However, if you find that they are an integer multiple of the lowest resonnance frequency (which will be equal to the signal frequency and not twice its value as with a pure sinusoidal signal)... then it is most probably just the harmonics of the AC signal that excite one after the other the resonance of your structure.

### Conclusion

Generally speaking, except for special case, you should avoid signal other than sinusoidal when you want to measure resonnant frequency of your structures. If you want to use synchronous detection of displacement you need to add an offset voltage, preferably with an amplitude much larger than the amplitude of the sinusoidal component, and the structure will be driven at the same frequency than the AC signal. If you use a pure sinusoidal signal (without offset) the structure will be driven at a frequency twice that of the AC signal.