ShreyasSenISSS

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Proceedings of ISSS 2005International Conference on Smart Materials Structures and Systems

July 28-30, 2005, Bangalore, India

ISSS-2005/PS-04

Design of the optimum microheater for smart MEMS gas

sensor

Partha Bhattacharyya, Shreyash Sen ,Avhishek Chatterjee and Hiranmay Saha

IC Design and Fabrication center , Dept. of Electronics and Telecommunication Engg.

Jadavpur University, Kolkata- 700032, India

Phone: (033)24146217

Fax : (033)24146217Email:[email protected]

ABSTRACT

Conventional Metal Oxide gas sensors commonly used for sensing inflammable hydrocarbon gases and

other toxic gases. However, they suffer from the two limitations, viz. (a) their relatively high operating

temperature (300 c) and (b) large power dissipation (1 Watt). Micromachined silicon based metaloxide gas sensors are being developed to overcome these limitations.The main part of power

consumption in a micro-machined gas sensor consists of various thermal losses like conduction through

bulk silicon substrate, convection in air from all exposed surfaces and radiation. The thermal

characteristics of micro-machined metal oxide based gas sensors have to be optimized with respect to

low power consumption, well controlled temperature distribution over the sensing layer and fasttransient response. However microheater for the MEMS metal oxide gas sensors have not yet been

optimized. In this paper we have developed a methodology (software) for designing and optimizingmicroheater for MEMS based gas sensor. Using this software we can estimate power requirement forachieving a particular temperature as well as the temperature distribution over the active layer.

Keywords: MEMS metal oxide gas sensor, microheater, thermal analysis, power optimization

1 . I NTRO DUCTI O N

Semiconducting Metal oxides like SnO2, ZnO, TiO2 have long been used for detecting poisonous (CO) and

inflammable gases(CH4) by their change in conductivity. For sensing gases the temperature of the sensing

layer should be raised to a particular value (typically from 300-450 c) thereby requiring a large amount ofpower. [1-3]. Before the evolution of micro electromechanical structure or MEMS the gas sensors(mainly

ceramic based) has very high power consumption(of the order of 500mW -2W) [3]due to their excessive

thermal mass. Moreover response time was also very high. Microheater for them have been designed andoptimized. But with MEMS and proper thermal isolation between sensor element and substrate, this powerconsumption has been scaled down to about only 30-150 mW [4-6]. An embedded heater is incorporatedto maintain a particular temperature (at a particular temperature the sensitivity of the metal oxide is

maximum) over the sensing metal oxide layer, which eventually leads to high power consumption. For

determining the thermal characteristics, there are softwares like ANSYS, Coventorware that employ

finite element method (FEM). But these softwares are very costly and hence are not in the reach of all.In this paper a new method is devised for analyzing the power consumption and thermal characteristics.

In this paper wedevelopedthe theory of heat transfer with the basis of basic thermodynamic equations.

The results of some sample run of programs build based on the theory discussed to show its validity and has been


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justified with those obtained using coventorware. Using this easy and cheap methodology anyone can optimize

power consumption with respect to a particular sensing temperature (to reach a particular temperature how much

power is required).

2. STRUCTURAL DESCRIPTION

Fig 2.1: Side view of the MEMS metal oxide gas sensor

The SiO2 layer is a square with dimension 1cm x 1cm on the top surface and the layer is 1.5-2 micron thick.

In the middle of this SiO2 layer a metal oxide (SnO2/ZnO) layer is deposited for sensing. This is also a

square with dimensions 5mm x 5mm and thickness 200 nm. The heater is embedded beneath the SnO2 layerin the SiO2 layer 0.1 micron down from the top surface of SiO2. The heater keeps the active layer at a

temperature of 350c (typically) for maximum sensitivity to reducing gases. The bulk Si is 350 m thick.The Si is etched in the middle section with KOH solution so that the membrane (Si) thickness is 50m .The

etched portion covers the portion of the SiO2 layer above which the SnO2/ZnO layer is placed. All the

dimensions given here specifies a particular MEMS sensor, these may be varied within certain limits.

Fig 2.2:Top view of the sensor (SnO2 /ZnO and SiO2 layer)

3. THERMAL ANALYSIS

In this section we devise a simple and basic theory for calculation of total heat loss.

3.1 Heat source approximation

In top view the heater area (embedded in the SiO2 layer) looks like a square. For simplicity of calculation

we approximate the square by a circle whose radius is taken such that the error in approximation becomesminimum. This helps us to approximate the heat source as a cylinder whose radius is calculated as follows

and height is the thickness of the SiO2 layer.To find out the right r0 we take r0= a Where a is the side of the square and is the factor to be found.


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Fig 3.1.1:Heat source area approximated as a circle

To minimize the error in approximation we have to minimize the areas where the square and the circle do

not overlap. For this it can be readily seen that we need to minimize the area ABD + area BCE = area e (for

error).

We can write; Area e = r2/8 a2/8 + 2* area ABDWhich could be written in terms of a and asArea e = a2/8 [1- 2 -2(42 1) + 8 2 tan-1 ((42 1))]Putting de/d = 0 we get = 0.5412

So r0becomes 0 0.5412r a a= = The temperature of the lateral wall of the cylinder is taken as the heater temperature and this approximationis valid as the temperature gradient along the thickness of the SiO2 layer is very low (it is also proved later

by calculation in sec 3. 3 .4).Hence the cylindrical heat source looks like

Fig: 3.1.2:Cylindrical heat source

3.2 Pathways of heat transfer

As shown in the figure 3.2.1 there are 3 components of the heat loss Qa, Qb, Qc. So total heat loss Q=Qa

+Qb+Qc. Heat loss from the surface is due to radiation, conduction through air and convection. The last two lossesare considered as a whole and we call them loss through air .Qa: heat loss from the top surface of the SnO2 layer by radiation and through air.Qb: heat loss from the etched surface of the Si bulk by radiation and through air.Qc: heat coming out from the lateral surface of the cylindrical source.Qc: heat loss from the top surface of the SiO2 layer

Qc: heat loss from the bottommost surface of the Si layer.


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Qc=Qc+QcFig 3.2.1:Pathways of heat flow

So the total heat coming out from the lateral surface of the cylindrical source is bypassed as Qc and Qc and

hence no heat goes out through the farthest vertical surfaces of the SiO2 layer where the temperature dies down

to ambient temperature (T0).Electric current chooses its path such that (Vsource Vsink)*I is minimum.

Drawing the analogy for heat conduction, heat should flow in such a way so that (Tsource-Tsink)* (dQ/dt)is minimum. Now as Th-T*>>Th-T1, (as shown later in sec 3. 3. 4) heat will avoid the bypass path

(marked by cross in Fig 3.2.1) and heat bypassing is negligible compared to Qb.

3. 3: Heat loss calculations

In this section Qa, Qb, Qc are separately calculated.

3. 3. 1: Calculation for Qa

Qa is the heat loss from the upper surface (ie. SnO2 surface). It includes heat loss by radiation and throughair. If Th is the temperature of the heater and To is the ambient temperature then considering the SnO2

surface temperature =Th (as the thickness of the SiO2 layer over the heater is very small (0.1 micron) and

SnO2 is conductive and also very thin) we can writeQa=Qradiation + Qair

= 44 0 02 ( ) 4 ( )air oSnOh h h A T T k r T T + ------(1)

Here Ah =area of the heater surface, = emissivity constant, =Stephans constant, airk = thermalconductivity of air.

3. 3. 2: Calculation for Qb

As in the case of upper surface heat is also lost from the lower surface (i.e. The etched surface). The temp.

of the lower surface be T1 , then Qb has two parts:

Qb= Qradiation + Qair =44

0 1 01( ) 4 ( )air oA T T k r T T + ----(2)

Where A= area of the lower surface.Again this Qb is supplied by conduction .So we can write

Qb= A (Th-T1)/ (h/Ksi +h/KSiO2) ----(3)

Where h is the distance between the heater and the Si layer along the SiO2 layer.

3. 3. 3: Calculation for Qc

Heat loss as Qc requires more attention than Qa or Qb. Here heat is flowing horizontally outwards from the

cylindrical heat source. Now M and N edges are far away compared to P, Q, R, and S edges. Due to

relatively high thermal resistance in reaching M or N we approximate that at M and N edges temperature is


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ambient temperature. Now as the source is cylindrical heat flows in cylindrical isothermal surfaces. Hence

we can analyze the heat flow considering some concentric cylinders.Here the heat from the heater goes

radially outwards through the concentric cylinders. Let T and T+T be the temperatures of the isothermal

surfaces (A & B) at a distances x and x+x from the center respectively. Some part of the heat going into

surface A is dissipated from the piece and some lesser amount of heat comes out at surface B. thisdissipated heat has two parts,

1) One part goes upwards and is lost by radiation and through air.( 1rQ , 1airQ ).2) Another part goes to the lower part of the Si structure through conduction (Qcd) and then is lost by

radiation and through air. (2r

Q ,2air

Q ).

During conduction the amount of heat which is dissipated per unit volume at a point where the temperature

is T is given as2

2

d TK

dx. So in our case heat dissipated in the region between surfaces= Heat loss from

upper and lower surfaces.( as shown in Figure 3.3.1)

Or, KA (d2T/dx2) *x =Qr1+Qair1+Qr2+Qair2 ---(4 )Where the temperature of the lower surface is T* So equation (4) can be written as

24 4

0022 2

4( ) ( )

d TairSiSiO SiOdx f

Ak A h A T T T T k r = + +

4 40 0

2

4( ) ( ) airSiOf

A A T T T T k r +

Where A= 2 x x .

Figure 3.3.1:Heat loss components

.Or,

24 4

002

41 ( ) ( )

d TairSi

dx f K T T T T k r = + +

4 40 0

2

4( ) ( ) airSiO

fT T T T k r

+ ----(5)

Where K1 =2 2

SiO SiOk h

Assuming T-T*= (T-To)/10

Where T* = Temp. at the bottom surface and To = ambient temp. =300kSolving we get T*=0.9T+30

Expanding both sides as 4th power of T, we get

T*4=810000+97200T+4374T2+87.48 T3+0.6561T4

With the help of this from equation 5 we have to find


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0

dT

rdxand then we have ,Qc= Ksio2 hsio2[dT/dx] ro * 2. ro -----(6)

With the typical dimensions specified in section 2.

hSiO2=61.5 10 m, h=1.4*10 -6m ( height of the SiO2 layer under heater),

hsi=50 *10-6

(height of the Si layer under heater)

A= 25*10- 6 m2, Ksio2 = 1.4 W m-1K-1, Ksi= 150 W m-1K-1, airk = 0.044 W m-1 k-1,

2SnO =0.65, Si =0.2, 2Si O =0.7

We calculate all the loss components for heater temperature Th= 3500 C =623 K

Qa=Ah 2SnO (Th4-To

4) + 04 airr k (Th-To) as per ---(1)

=25*10^ -6*0.65*5.7*10^ -8*[6234-3004] +4*0.5412*5*10^-3[623-300]=0.615 W

Qb=A Si (T14-To

4) + 04 airr k (T1-To)as per ---(2)

Now for this we have to calculate T1.

T1 can be calculated by equating equation 2 and 3, i.e.

A Si (T14-To

4) + 04 airr k (T1-To) = A (Th-T1)/ (hsi/Ksi +h/KSiO2)

25*10 -6*5.7*10^-8*0.2*(T14-3004) +59.84(T1-300)

= 25*10-6(

623-T1)/ (50*10^-6/150 + 1.4*10-6

/1.4)Solving this equation we get T1=622.972

Now solving the same equation for different Th we get

Th 623 400 800 1200

T1 622.972 399.992 799.954 1199.9

Table 3. 3. 4. 1 : T1 for different Th

So we can confer that for all practical values of Th we can assumeT1=Th. (this is the reference talked aboutpreviously).So ,

Qb=A Si (Th4-To

4) + 04 airr k (Th-To) as per ---(1)

=25*10^ -6*0.2*5.7*10^ -8*[6234-3004] +4*0.5412*5*10^-3[623-300]=0.5239 W

Qc= Ksio2 hsio2 [dT/dx] ro * 2. ro as per ----(6)

For this we have to calculate [dT/dx] roAfter putting the values into the equation(5)Or, (2.1*10-6)(d2T/dx2) = 3.75*10-8 T4+3.49* 10 -6*T3 +1.74*10-4 T2 + 61.79T-18952.142(d2T/dx2) = 0.0178 T4 + 1.662*T3+92.857 T2 + 29.42*10 6 T-9.02*109

Now we have eqn. in the formd2T/dx2 =f (T)

or, 2 (dT/dx )( d2T/dx2)=2 f(T) (dT/dx )

or,d[(dT/dx )2]/dx=2 f(T) (dT/dx )

or, d[(dT/dx )2]=2f(T) . dT

d[(dT/dx )

2

]=

2f(T) . dT

0.5

2[( / ) ]

2

0

[( / ) ]

dT dx ro

d dT dx = 623

300

)( dTTf

Where f(T)= 0.0178 T4 + 1.662*T3 +92.857 T2 + 29.42*10 6 T-9.02*109

[dT/dx]r0 = -1.93 X106 Kelvin/m(-ve sign comes as T decreases with increasing x, we would only take the

magnitude).Putting this value in equation 7 we get


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Qc=0.0689 W

So the total heat power consumed is Q=(0.615+0.5239+0.0689)=1.2078 W4. RESULTS

A matlab program has been developed on the basis of above-mentioned theory, which calculates the total

heat loss (in Watt), which comprises of three components as discussed above, for a specified temperatureand also for different temperatures.

The plot of total power and individual power consumption components vs. heater temperature curve is shown inthe fig below.

Fig 4.1: Heat loss (different components and total power) vs. temperature of heater

The sensitivity varies with temperature as shown below (for buried Pt [5] )

Fig: 4.2: Sensor sensitivity vs. heater temperature

Comparing the power consumption and the sensor response (sensitivity) it can be concluded

which is the optimum temperature for the operation of the sensor and the power consumed atthat particular temperature.


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Fig 4.3:Heat (temp.) distribution along the tin oxide layer (top view) from our calculation (left) and

verification from Coventorware (right)

5. CONCLUSION

The method of heat analysis employed in this paper is simple enough and matlab programs on the basis ofthis theory have been written which calculate the heat loss (i.e. power consumption) at different

temperatures. We have also found the function for dT/dx from which T vs. x curve has been plotted (as

shown in Fig 4.3 where x is the running horizontal distance from the center of the sensing layer. The

equation for this plot isdT/dx=(2*(1.3692*10^12+0.0178*x^5/5+1.662*x^4/4+92.857*x^3/3+29.42*10 6*x^2/2-

9.02*10^9*x))^.5;

And the plot matches the plot of the same in Coventorware. Plot of heat loss vs. temperature for differentdimensions of sensing layer (from table 4.2) is also possible as shown in example [1, 2, 3, 4]. ANSYS or

Coventorware gives the all the thermal characteristics but these softwares are not available to all due to

their excessive cost and moreover they are not user friendly. Apart from that these softwares offers no

insight to the nature of heat flow within and from the structure. The methods employed in the paper offersinsight as well as gives almost correct results with much less calculation hazards. The program build on this

theory gives results readily as shown in the results section. The MEMS structure has been discussed and

then the theory has been derived, based on which program has also been made and results obtained from

those have also been included in this paper.

References

1. Isolde Simon, Nicole Barsan, Michael Bauer, U. Weimer, Micromachined metal oxide sensors:

opportunities to improve sensor performance, Sensors and Actuators, B73, ELSEVIER, 9-12

2. P.Hille, H.Strack,A heated membrane for a capacitive gas sensor, Sensors and Actuators,A32 (1992)321-325

3. U.Grigull, H.Sandner, Warmeleitung, Springer, Berlin, 1979

4. S. Astie, A. Gue, E. Schield, L. Lescouzeres. A. Cassagnes. Optimization of an integrated SnO 2 gassensor using FEM simulator, Sens. Actuators, A69 (1998) 205-211

5.G.G.MandayoE. Castano, Carbon monoxide detector fabricated on the basis of a Tin oxide novel dopingmethod, IEEE Sensors Journal Vol 2 No 4 Aug 2002, 326.

6. G. Sberveglieri, W. Hellmich, G. Muller, Silicon hotplates for metal oxide gas sensor elements,Microsyst. Technol. 3 (1997) 183-190.

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