effects of bias voltage polarity on differential capacitive sensitive devices

Sensors and Actuators A 112 (2004) 253–261

Effects of bias voltage polarity on differential capacitive sensitive devicesLufeng Che∗, Bin Xiong, Linxi Dong, Yuelin Wang

State Key Laboratory of Transducer Technology, Shanghai Institute of Microsystem and Information Technology,Chinese Academy of Science, 865 Changning Road, Shanghai 200050, PR China

Received 21 April 2003; received in revised form 7 January 2004; accepted 13 January 2004

Abstract

For capacitive sensing, a driving signal with a dc bias voltage and an ac voltage is usually necessary for the measurement of capacitance.The electrostatic forces caused by the driving signal on the mechanical capacitor may interfere significantly with the measurement and thenormal operation of the devices. In this paper, quantitative analyses on the effects of driving signal with different bias voltage polarities aremade for double-sided driving and double-sided driving with voltage feedback. For double-sided driving, the effects of the dc bias voltagewith different polarities on devices are the same. For double-sided driving with voltage feedback, positive–positive and positive–negativebias voltage configurations may improve the measurement linearity of the device and mostly eliminate the pull-in effect caused by thedriving signal, while negative–negative and negative–positive bias voltage configurations maximize the side effects of the driving signaland reduce largely the safe operation range of the device.© 2004 Elsevier B.V. All rights reserved.

Keywords: Electrostatic forces; Capacitive sensors; Bias voltage polarity

1. Introduction

Capacitive sensors have been increasing in popularity inrecent years for many MEMS devices due to its process com-patibility with most mechanical structures, high sensitivityand low temperature drift.

As it is well-known, for capacitive sensing, a certain formof voltage driving signal is usually necessary for the mea-surement of capacitance[1–4]. The voltage driving signalcauses electrostatic forces on the movable electrode and in-terferes significantly with the measurement and the normaloperation of the devices. In operation, the movable elec-trode in a capacitive sensor is subjected to four differentforces: the force caused by the measurand, the elastic forceof the suspension structure, the damping force and the elec-trostatic force generated by the driving signal. In this paper,the damping effect is not considered.

Puers and Lapadatu[5] noticed that the electrostatic forceby the single-sided driving signal may cause offset errorsin the measuring process and the collapse of the sensingstructure if critical voltages/charges are exceeded. The volt-age for pull-in to occur was found for single-sided driving

∗ Corresponding author. Tel.:+86-21-62511070×5975;fax: +86-21-62513510.E-mail address: [email protected] (L. Che).

scheme and a double-sidedQ-mode driving for the symmet-rical structure was only suggested.

Li et al. [6] analyzed the electrostatic force in open-loopoperating mode of the capacitive accelerometer and men-tioned the sensitivity and operating range of the accelerome-ter may be adjusted by changing the bias voltages. But thereis no detailed report to analyze the effects of the bias voltagepolarity for the open-loop operating mode of the capacitiveaccelerometer.

Bao et al.[7] analyzed quantitatively the effects of theelectrostatic force on the measurement and the operation ofthe sensors for three commonly used configurations for ca-pacitive sensors: single-sided driving, double-sided drivingand double-sided driving with an electromechanical feed-back and discussed the effects of driving signal on the off-set of output, on sensitivity and on the pull-in effect ofthe movable electrode. However he did not consider effectsof bias voltage polarity on differential capacitive sensitivedevices.

In this paper, the effects of the electrostatic force on themeasurement and the operation of the differential capaci-tive sensitive devices are analyzed quantitatively by chang-ing polarity of bias voltage. For the sake of simplicity, wetake capacitive accelerometers as examples in the discus-sion. Analyses will be made for double-sided driving anddouble-sided driving with voltage feedback using two biasvoltages with different polarity respectively. Based on the

0924-4247/$ – see front matter © 2004 Elsevier B.V. All rights reserved.doi:10.1016/j.sna.2004.01.005

254 L. Che et al. / Sensors and Actuators A 112 (2004) 253–261

analyses, the effects of the driving signal with different biasvoltage polarities on the linearity and on the pull-in effectof the movable electrode will be discussed. The bias voltagepolarity of the driving signal can be selected among the re-quirements on the linearity and the reliability of the doubledriving devices with voltage feedback.

2. Double-sided driving

For a double-sided driving structure as shown inFig. 1(feedback voltageVr = 0), the equation to find the displace-ment of the movable platex is:

Fe + ma − kx = 0 (1)

whereFe is the electrostatic driving force,ma is the inertialforce inx-direction,k is the stiffness coefficient.

Generally, the driving voltage consists of a dc componentand an ac component. A commonly used form of the drivingsignal is (±V0±V1 sinωt). For positive–positive bias voltageconfiguration, i.e. top bias voltage polarity and bottom biasvoltage polarity are all positive,Vt = V0 + V1 sinωt, Vb =V0 − V1 sinωt, the electrostatic force on the mass is:

Fe = Aε0

2d02

[Vt

2

(1 − x)2− Vb

2

(1 + x)2

](2)

whereA is the area of the electrodes,d0 is the original gapsof the two capacitors,ε0 is the free space permittivity,Vtand Vb are the driving voltage,x = x/d0. Eq. (2) can berearranged as

Fe = 2Aε0

d20(1 − x2)2

[(V 20 + V1

2 sin2 ωt)x

+ V0V1 sinωt + x2V0V1 sinωt] (3)

For positive–negative bias voltage configuration, i.e. topbias voltage polarity is positive and bottom bias voltagepolarity is negative,Vt = V0 + V1 sinωt, Vb = −V0 −V1 sinωt, the electrostatic force on the mass is:

Fe = 2Aε0x

d20(1 − x2)2

[V 20 + V1

2 sin2 ωt + 2V0V1 sinωt] (4)

Fig. 1. Schematic structure for double-sided driving accelerometer.

For negative–negative bias voltage configuration,Vt =−V0 + V1 sinωt, Vb = −V0 − V1 sinωt, the electrostaticforce on the mass is:

Fe = 2Aε0

d20(1 − x2)2

[(V 20 + V1

2 sin2 ωt)x

− V0V1 sinωt − x2V0V1 sinωt] (5)

For negative–positive bias voltage configuration,Vt =−V0 +V1 sinωt, Vb = V0 −V1 sinωt, the electrostatic forceon the mass is:

Fe = 2Aε0x

d20(1 − x2)2

[V 20 + V1

2 sin2 ωt − 2V0V1 sinωt] (6)

As the driving frequency,ω, is much larger than thesignal frequency and the natural vibration frequency ofthe mechanical structure, the force applied on the movableelectrode is the average of the electrostatic force of thedriving voltage. For the four configurations, the averageforce applied on the mass is the same:

Fe = 2Aε0x

d20(1 − x2)2

[V 20 + 1

2V12] (7)

So for double-sided driving, it is not necessary to considereffects of bias voltage polarity when analyzing quantitativelythe effects of the electrostatic force on the measurement andthe operation of the devices.

3. Double-sided driving with feedback voltage

For a force-balanced accelerometer with feedback volt-age,Vr, the simplified model is shown inFig. 1. For a smalldisplacement, the feedback voltage is proportional to thedisplacement of the mass:Vr = βV1x, whereβ is a constantdesignating the strength of the feedback operation.

1. For positive–positive bias voltage configuration with thefeedback voltage,Vt = V0 + V1 sinωt, Vb = V0 −V1 sinωt, the electrostatic force on the mass is:

Fe = Aε0

2d02

[(Vt − Vr)

2

(1 − x)2− (Vb − Vr)

2

(1 + x)2

](8)

which can be rearranged as

Fe = 2Aε0

d20(1 − x2)2

[V1(V0 − Vr)(1 + x2) sinωt

+ ((V0 − Vr)2 + V1

2 sin2 ωt)x] (9)

As the driving frequency,ω, is much larger than thesignal frequency and the natural vibration frequency ofthe mechanical structure, the average force applied onthe mass is:

Fe = 2Aε0

d20(1 − x2)2

[(V 20 + 1

2V12 + V 2

r − 2V0Vr)x] (10)

L. Che et al. / Sensors and Actuators A 112 (2004) 253–261 255

Using the notation ofV1 = αV0 andFe0 = (Aε0/2d20)V 2

0 ,one gets:

Fe = 4Fe0

(1 − x2)2[(1 + 1

2α2 + (αβx)2 − 2(αβx))x] (11)

Therefore, the force-balanced equation for the proofmass with an accelerationa becomes:

ma − kx + 4Fe0

(1 − x2)2[1 + 1

2α2 + (αβx)2 − 2(αβx)]x = 0

(12)

If p = Fe0/kd0 and q = ma/kd0 are defined, thefollowing equation is obtained:

q = x − 4p

(1 − x2)2[1 + 1

2α2 + (αβx)2 − 2(αβx)]x (13)

For β = 0, Eq. (13)becomesEq. (14):

q = x − 4p

(1 − x2)2(1 + 1

2α2)x (14)

Obviously, the maximum value ofVr(αβxV0) is re-stricted by the supply voltageV0 of the electronic sys-tem, i.e.Vr ≤ V0 or αβx ≤ 1, for positive–positive biasvoltage configuration, the relation betweenq and x canbe found by defining the function off(x, β, p, α):

f(x, β, p, α) =

x − 4p

(1 − x2)2[1 + 1

2α2 + (αβx)2 − 2(αβx)]x, x ≤ 1

αβ

x − 4p

(1 − x2)2[ 1

2α2]x, x ≥ 1

αβ

(15)

2. For positive–negative bias voltage configuration,Vt =V0 + V1sinωt, Vb = −V0 − V1sinωt. According toEq. (10)and similar derivation, the relation betweenq andx can be found by defining the function ofs(x, β, p, α):

s(x, β, p, α) =

x − 4p

(1 − x2)2[(1 + 1

2α2)x + (αβx)2x − (αβx) − (αβx)x2], x ≤ 1

αβ

x − 4p

(1 − x2)2[(2 + 1

2α2)x + 1 − x2], x ≥ 1

αβ

(16)

3. For negative–negative bias voltage configuration,Vt =−V0 + V1 sinωt, Vb = −V0 − V1 sinωt, the relationbetweenq and x can be found by defining the functionof h(x, β, p, α):

h(x, β, p, α) =

x − 4p

(1 − x2)2[1 + 1

2α2 + (αβx)2 + 2(αβx)]x, x ≤ 1

αβ

x − 4p

(1 − x2)2[4 + 1

2α2]x, x ≥ 1

αβ

(17)

4. For negative–positive bias voltage configuration,Vt =−V0 + V1 sinωt, Vb = V0 − V1 sinωt, the relation be-tweenq and x can be found by defining the function ofr(x, β, p, α):

r(x, β, p, α) =

x − 4p

(1 − x2)2[(1 + 1

2α2)x + (αβx)2x + (αβx) + (αβx)x2], x ≤ 1

αβ

x − 4p

(1 − x2)2[(1 + 1

2α2)x + x + 1 + x2], x ≥ 1

αβ

(18)

As a special case, forβ = 0, Eqs. (15)–(18)becomeEq. (14), i.e. double-sided driving with voltage feedbackmode becomes double-sided driving mode.

4. Results and discussions

We draw the curves forf(x, β, p, α), s(x, β, p, α),h(x, β, p, α)and r(x, β, p, α) with different β-values forspecificp andα values, the curves off(x), s(x), h(x) andr(x) can be obtained with graphic solutions. Two curves areshown in figure for two sets ofp andα.

(1) According toFigs. 2a,b and 3a,b, for smallβ the max-ima (fmax) of the curves off(x)and the maxima (smax) of thecurves ofs(x), are much smaller than unity due to the effectof the electrostatic force of the driving signal, respectively,this means that the critical acceleration that causes pull-ineffect, i.e.,ac = fmaxkd0/m (or smaxkd0/m), is small. Whenβ = 0 (without feedback voltage), the critical accelerationcauses pull-in effect is the smallest. With increasedβ, fmaxand smax increase withβ, respectively, untilβ is about 5when the curves have the highest linearity (for the two setsof p andα considered), this means the effect of positive feed-back caused by the driving signal is mostly compensated


Fig. 2. For positive–positive bias voltage configuration, graphic solutions for double-sided driving with electromechanical feedback: (a) forp = 0.05,α = 0.2 andVr ≤ V0; (b) for p = 0.1, α = 0.2 andVr ≤ V0; (c) the curve offmax vs. β.


Fig. 3. For positive–negative bias voltage configuration, graphic solutions for double-sided driving with electromechanical feedback: (a) forp = 0.05,α = 0.2 andVr ≤ V0; (b) for p = 0.1, α = 0.2 andVr ≤ V0; (c) the curve ofsmax vs. β.


Fig. 4. For negative–negative bias voltage configuration, graphic solutions for double-sided driving with electromechanical feedback: (a) forp = 0.05,α = 0.2 andVr ≤ V0; (b) for p = 0.1, α = 0.2 andVr ≤ V0; (c) the curve ofhmax vs. β.


Fig. 5. For negative–positive bias voltage configuration, graphic solutions for double-sided driving with electromechanical feedback: (a) forp = 0.05,α = 0.2 andVr ≤ V0; (b) for p = 0.1, α = 0.2 andVr ≤ V0; (c) the curve ofrmax vs. β.


for by the negative electromechanical feedback. With fur-ther increasedβ, all the curves off(x) with β-values largerthan 5 will merge together and the effect of positive feed-back caused by the driving signal will be fully compensatedby the negative electromechanical feedback, while for allthe curves ofs(x), the displacement of the mass decreasesfor the same acceleration due to the effect of electromechan-ical feedback, meanwhile the direction of the electrostaticforce is changed because of positive feedback caused by thedriving signal smaller than the negative electromechanicalfeedback. All the curves ofs(x) with β-values larger than5 merge together for largex, this is because the feedbackvoltages are restricted to a maximum value ofV0.

It can be seen fromFigs. 2c and 3cthat for smallβ(β ≤ 5), the larger thep-value the smaller the maxima ofthe curves due to the larger electrostatic force of the drivingsignal, this means that the larger thep-value, the smallerthe critical acceleration,ac, that causes the pull-in problem.However, if the feedback is large enough (say,β > 5, forthe two sets ofp andα considered),fmax can be quite closeto 0.8 andsmax can be quite close to 0.9, this means thatthe pull-in effect caused by the driving signal can be mostlyeliminated by the feedback configuration.

(2) According toFig. 4a,b and 5a,b, when β = 0, thecurves ofh(x) and the curves ofr(x) have the maximum, re-spectively. The largerβ-value the smaller the maxima (hmax)of the curves ofh(x) and the maxima (rmax)of the curvesof r(x), this means that the critical acceleration that causespull-in effect is smaller. With increasedβ, hmax andrmax de-crease withβ until the curve is close to 0, respectively, thismeans the effect of positive feedback caused by the drivingsignal is strengthened by the electromechanical feedback. Itcan be seen fromFig. 4c and 5cthat the larger thep-valuethe smaller the maxima of the curves due to the larger elec-trostatic force of the driving signal. However, if the feed-back is large enough (say,β > 5), hmax and rmax can bequite close to 0, respectively, this means that the pull-in ef-fect caused by the driving signal can occur easily with anacceleration signal.

Now let us consider the accelerometer described in Ref.[8] as an example. As the mass of the seismic mass ism =1.6 × 10−10 kg and the resonant frequency of the structureis 17.7 kHz, the elastic constant of the flexure is found to beabout 1.98 N/m. As the original distances between the elec-trodes ared0 = 1.3�m, the original capacitances areC0 =0.1 pF and the driving signals have a dc voltage ofV0 =1.6 V and a high frequency (1 MHz) square wave with anamplitude ofV1 = 0.3 V, Fe0 is found to be 9.85× 10−8 N.According to the definition forp, α andβ, the parameters arefound to bep = 0.038,α = 0.1875 andβ ∼= 8.5. Therefore,for positive–positive, positive–negative, negative–negativeand negative–positive bias voltage configurations,fmax, smax,hmax, rmax are found byEqs. (15)–(18)to be about 0.8350,0.8775, 0.1818, 0.1368, respectively. However, if the feed-back is taken away, thesmax, fmax, hmax, rmax are only about0.3766.

Normally, the level of the acceleration signal for the de-vices is usually not larger than its nominal operation range of50 g. For different bias voltage configurations, the maximumaccelerations,ac, that causes a permanent pull-in effect areabout 1343, 1412, 292 and 221 g, respectively. However, ifthe feedback is taken away, the correspondingac is reducedto about 600 g. For positive–positive and positive–negativebias voltage configurations, this is considered to be adequatefor most applications, However, for negative–negative andnegative–positive bias voltage configurations, This would betoo small for many applications so that the accelerometerwould easily be pulled into contact with one of the fixedelectrodes.

5. Conclusion

For double-sided driving with voltage feedback, positive–positive and positive–negative bias voltage configurations,the electromechanical feedback is negative in nature. Whenthe feedback is large enough, the pull-in effect caused bythe driving signal can be mostly eliminated. The effect ofpositive feedback caused by the driving signal may be mostlycompensated for by the negative electromechanical feedbackand the device may have relatively good linearity.

For negative–negative and negative–positive bias voltageconfigurations, the electromechanical feedback is positivein nature. The two configurations reduce the safe operationrange of the device with increase of the strength of the feed-back operation. Whenβ is larger than 30, the mass is not sta-ble even without an acceleration signal and easily be pulledinto contact with one of the fixed electrodes.

The overall conclusion is: for positive–positive andpositive–negative bias voltage configurations, double-sideddriving with electromechanical feedback has the best per-formance and is least affected by the side effects of thedriving signal.

Acknowledgements

We would like to thank Prof. Minhang Bao of FudanUniversity for his helpful discussions and suggestions. Thiswork is supported by the Major State Basic Research Devel-opment Program of China ‘Integrated micro-optical-electro-mechanical system’ (no. G1999033101).

References

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[2] R.P. van Kampen, M.J. Vellekoop, P.M. Sarro, R.F. Wolffenbuttel,Application of electrostatic feedback to critical damping of an inte-grated silicon capacitive accelerometer, Sens. Actuators A43 (1994)100–106.


[3] O. Kromer, O. Fromhein, H. Gemmeke, T. Kiihner, J. Mohr, M.Strohrmann, High-precision readout circuit for LIGA acceleration sen-sors, Sens. Actuators A 46–47 (1995) 196–200.

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Biographies

Lufeng Che obtained his PhD degree in mechanical engineering fromZhejiang University in 1999. From 1999 to 2001, he was a post-doc atState Key Laboratory of Transducer Technology, Shanghai Institute ofMicrosystem and Information Technology, Chinese Academy of Sciences(CAS). Now he is working here. His research interests include MEMSinertial sensor and its system integration technology.

Bin Xiong received his BS degree from Southeast University in 1984,MS degree and PhD degree from Shanghai Institute of Microsystem andInformation Technology (SIMIT), Chinese Academy of Sciences (CAS),in 1997 and 2001, respectively. He has been working at the ShanghaiInstitute of Microsystem and Information Technology, Shanghai, China,since 1984. He was a visiting scholar at Hong Kong University of Scienceand Technology in 1997. Now he is a professor in SIMIT. His currentresearch interests include silicon micro gyroscopes and micromachiningtechnology.

Linxi Dong received his MS degree from the college of mechanical andenergy engineering of Zhejiang University in 2001. He is now pursuinghis PhD degree in the college of information science and engineering ofZhejiang University. His current research interest is in micro-mechanicaltransducers and technologies.

Yuelin Wang received his BS, MS and PhD degrees from Zhejiang Uni-versity, Harbin Institute of Technology and Tsinghua University, China,in 1982, 1985 and 1989, respectively. From 1991 to 1993, he was anassociate professor at the Department of Information and Electron Engi-neering, Zhejiang University. Since 1993 he was been a professor at thesame department. He was a visiting scholar in Hong Kong Universityof Science and Technology and Tohoku University in 1995 and 1996,respectively. Since 1998 he is with Shanghai Institute of Microsystemand Information Technology, CAS. His current research interests includemicromachining technologies, sensors and micromechanical optics. He isa senior member of the IEEE. As a chief scientist, now he is presidingover the Major State Basic Research Development Program of China‘Integrated micro-optical-electro-mechanical system’.

effects of bias voltage polarity on differential capacitive sensitive devices

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