Holding device

Abstract

To suppress damage to a joining unit by improving cooling performance in a holding device even when heat input to a plate-like unit for holding an object becomes large.SOLUTION: A holding device for holding an object includes: a plate-like unit formed in a plate-like shape; a base unit that supports the plate-like unit, is mainly composed of a material having a thermal expansion coefficient different from the thermal expansion coefficient of a material mainly forming the plate-like unit, has a cooling function, and is formed in a plate-like shape; and a joining unit that is arranged between the plate-like unit and the base unit, includes an adhesive, and joins the plate-like unit and the base unit to each other. The thermal resistance of the joining unit is 5.0×10-4 (m2K / W) or less, and the strain amount at the time of maximum shearing stress of the joining unit is 0.5 mm or more.SELECTED DRAWING: Figure 1

Description

[Technical Field]

[0001] This disclosure relates to a holding device. [Background technology]

[0002] Conventionally, as a holding device for holding objects, for example, when manufacturing semiconductors, a holding device for wafers etc. Electrostatic chucks are known for holding objects. Generally, electrostatic chucks are used when the object is placed on them. A ceramic part, a base part in which a refrigerant flow path is formed, and a connection between the ceramic part and the base part. It comprises a joint that fits together. For example, Patent Document 1 describes the material of the adhesive layer (joint) as , silicone resin, or heat-treated silicone resin such as aluminum oxide or aluminum nitride The document states that an adhesive consisting of a composite resin with conductive fillers added may be used. [Prior art documents] [Patent Documents]

[0003] [Patent Document 1] Japanese Patent Publication No. 2008-300491 [Overview of the Initiative] [Problems that the invention aims to solve]

[0004] According to the technology described in Patent Document 1 above, silicone resin is used as the material for the joint. Therefore, even under relatively high temperature conditions, the stress between the ceramic part and the base part is relieved. This is possible. However, if, for example, an electrostatic chuck is exposed to a higher-power plasma... When the heat input to the ceramic part becomes large, as in the case of using it as shown, the ceramic part There was a possibility that the temperature of the ceramic part would rise excessively, reducing the processing accuracy of the wafer. Due to the stress generated in the joint and other parts caused by the excessive temperature rise, the joint might be damaged. . Therefore, even when the heat input to the ceramic part increases, a technology that can enhance the cooling performance in the holding device and suppress the occurrence of inconveniences such as damage to the joint has been desired.

Means for Solving the Problem

[0005] The present disclosure can be realized in the following forms. (1) According to one aspect of the present disclosure, a holding device is provided. This holding device includes a plate-shaped part formed in a plate shape, a base part that supports the plate-shaped part and mainly made of a material having a coefficient of thermal expansion different from that of the material mainly forming the plate-shaped part, has a cooling function, and is formed in a plate shape, and a joint part that is disposed between the plate-shaped part and the base part, contains an adhesive, and joins the plate-shaped part and the base part. The thermal resistance of the joint part is 5.0×10 (m K / W) or less, and the strain amount at the maximum shear stress of the joint part is 0.5 mm or more. According to the holding device of this aspect, even when the heat input to the plate-shaped part increases, the cooling performance in the holding device can be enhanced, and the inconveniences caused by the stress generated in the joint part can be suppressed. , , , 2 , , , , , , , -4 , , , , (m 2 K / W) or less, and the strain amount at the maximum shear stress of the joint part is 0.5 mm or more. According to the holding device of this aspect, even when the heat input to the plate-shaped part increases, the cooling performance in the holding device can be enhanced, and the inconveniences caused by the stress generated in the joint part can be suppressed. (2) In the holding device of the above aspect, for the joint part, when a tensile test is performed on the joint part, the shear stress when the deformation length due to the shear force reaches 0.5 mm may be 0.1 MPa or more and 3.5 MPa or less. With such a configuration, the shear stress generated in the joint part can be suppressed, so that the effect of suppressing the inconveniences caused by the stress generated in the joint part can be further enhanced. (3) In the above-described holding device, the joint portion includes the adhesive and an inorganic filler. It may also be made of composite materials. In such a configuration, it can be joined with an adhesive. The flexibility of the part is ensured to increase the amount of strain at maximum shear stress, and inorganic fillers are used This ensures heat transfer at the joint and reduces the thermal resistance at the joint. (4) In the above-described holding device, the radius of the inscribed circle when the inorganic filler is viewed in cross-section. Let R1 be the value of the circumscribed circle, and let R2 be the radius of the circumscribed circle. If the average value of R2 / R1 is 1.10 or greater, This configuration may also be used. With this configuration, more contact points are formed between the inorganic fillers. This facilitates the formation of heat conduction paths within the joint, thereby reducing the thermal resistance of the joint. ru. (5) In the holding device of the above form, the rigidity ratio of the joint is 0.03 MPa or more, 2.0 It may also be less than MPa. With such a configuration, flexibility of the joint is ensured. Therefore, to improve the stress relaxation performance at the joint, when a shear force is applied to the joint... Damage to the joint can be reduced. Furthermore, the shape retention of the joint can be improved. (6) In the holding device of the above form, the joint portion comprises a resin layer containing the adhesive and the resin Formed to spread in the planar direction of the joint within the lipid layer, and composed of an inorganic material. The system may also include an inorganic sheet. With such a configuration, the inorganic sheet By providing this feature, the thermal resistance and the strain at maximum shear stress can be made uniform within the plane of the joint. This allows for the creation of a uniform temperature distribution within the plane of the joint. (7) In the above-described holding device, the plate-shaped portion is mainly composed of ceramic, and the target It includes an adsorption electrode for holding an object, but does not include a heater electrode for heating the object. This may also be done. With this configuration, the heater electrode heats the object and the plate-like part. Even when not heated, the heat input to the plate-like part is large, and the temperature difference between the plate-like part and the base is In usage conditions where the size tends to increase, the cooling performance of the holding device is improved, and the joints This method significantly reduces the problems caused by the resulting stress. This disclosure can be implemented in various forms other than those described above, for example, a semiconductor-made device including a holding device. This can be realized in the form of a manufacturing device, a method for manufacturing a holding device, or a method for forming a joint. [Brief explanation of the drawing]

[0006] [Figure 1] A perspective view showing a schematic representation of the appearance of the electrostatic chuck according to the first embodiment. [Figure 2] A schematic cross-sectional view showing the configuration of an electrostatic chuck. [Figure 3] An explanatory diagram showing how to determine the "R2 / R1" ratio of inorganic filler particles. [Figure 4] An explanatory diagram showing how heat transfer paths are formed between inorganic filler particles. [Figure 5] An explanatory diagram showing the relationship between the inorganic filler content and the thermal conductivity of the joint. [Figure 6] An explanatory diagram regarding the stress generated at the joint. [Figure 7] A schematic cross-sectional view showing the configuration of the electrostatic chuck of the second embodiment. [Figure 8] An explanatory diagram summarizing the measurement and evaluation results for each sample. [Figure 9] An explanatory diagram summarizing the measurement and evaluation results for each sample. [Figure 10] An explanatory diagram summarizing the measurement and evaluation results for each sample. [Figure 11] An explanatory diagram summarizing the measurement and evaluation results for each sample. [Figure 12]A schematic diagram illustrating the calculation method for maximum shear stress and strain. [Figure 13] An explanatory diagram showing an example of measured values ​​obtained from a tensile test (sample S25). [Figure 14] An explanatory diagram showing an example of measured values ​​obtained from a tensile test (sample S25). [Modes for carrying out the invention]

[0007] A. First Embodiment: (A-1) Structure of the electrostatic chuck: Figure 1 is a schematic perspective view showing the appearance of the electrostatic chuck 10 in the first embodiment. Figure 2 is a schematic cross-sectional view showing the configuration of the electrostatic chuck 10. In Figure 1, the electrostatic chuck 1 A portion of the 0 is shown broken. Also, Figures 1, 2, and Figures 6 and 7 (described later) show the method To specify the direction, mutually orthogonal X, Y, and Z axes are shown. The X and Y axes are shown in each figure. The Z-axis and X-axis each represent the same direction. In this specification, the Z-axis indicates the vertical direction, and X The axis and Y-axis indicate the horizontal direction. Note that the above diagrams schematically represent the arrangement of each part. Therefore, it does not accurately represent the ratio of the dimensions of each part.

[0008] The electrostatic chuck 10 is a device that attracts and holds an object by electrostatic attraction, for example, a semi-electrostatic chuck. It is used to fix the wafer W, which is the object to be manufactured, inside the vacuum chamber of a conductor manufacturing apparatus. The electrostatic chuck 10 comprises a ceramic part 20, a base part 30, and a joint part 40. These are the ceramic part 20, the joint part 40, and the base, oriented in the direction of the -Z axis (vertically downward). The parts are stacked in the order of section 30. The electrostatic chuck 10 in this embodiment is referred to as a "holding device". They also call them that.

[0009] The ceramic part 20 is a roughly circular plate-shaped member, and is made of ceramic (for example, aluminum oxide). It is formed mainly from aluminum (such as aluminum nitride). In this specification, the specific The phrase "it is the main component" or "it is the main forming material" refers to the content of the specific component in question. However, this means that it is 50% by volume or more. The diameter of the ceramic part 20 is, for example, 50 mm. A length of around 500mm is sufficient, and typically it is around 200mm to 350mm. The thickness of the part 20 should be, for example, about 1 mm to 10 mm. The ceramic part 20 is It is also called the "plate-like part".

[0010] As shown in Figure 2, an adsorption electrode 22 is placed inside the ceramic part 20. The electrode 22 is formed from a conductive material such as tungsten or molybdenum. When a voltage is applied to the adsorption electrode 22 from a power source (not shown), an electrostatic attraction force is generated. This electrostatic attraction causes the wafer W to be adsorbed and fixed to the mounting surface 24 of the ceramic part 20. The electrode 22 may be bipolar or unipolar. Also, the ceramic part 20 Inside, there is a resistive generator formed by a conductive material (e.g., tungsten or molybdenum). A heating element, not shown, is used to heat the wafer W which is adsorbed and fixed to the mounting surface 24. A heater electrode may be provided.

[0011] The base portion 30 is a plate-like member made of metal and formed in a substantially circular shape. For example, aluminum, magnesium, molybdenum, titanium, tungsten, nickel It may contain at least one of our metals: molybdenum, titanium, tan. Gusten has a relatively small coefficient of thermal expansion among the metals mentioned above, therefore, among these, When the base portion 30 is constructed using one type of metal, the base portion 30 and the ceramic portion It is desirable to be able to suppress the difference in thermal expansion coefficient between 20 and 20. The term "coefficient of thermal expansion" refers to the "coefficient of linear expansion." Also, magnesium has a relatively small Young's modulus. Therefore, when magnesium is used to construct the base portion 30, the thermal response generated in the base portion 30 It is desirable that the force can be reduced. Also, aluminum has a relatively high thermal conductivity, It is easy to manufacture and low-cost. Therefore, when aluminum is used to construct the base part 30 In addition, the cooling efficiency of the ceramic part 20 and the wafer W by the base part 30 is increased. This is desirable as it reduces the manufacturing cost of the electrostatic chuck 10. From the perspective of improving cooling efficiency while keeping manufacturing costs down, the metal content in the base part 30 A higher proportion is desirable, and it is desirable that the base portion 30 be mainly composed of metal. , containing 90% or more by mass of highly versatile aluminum (for example, A6061, A5 It is preferable that the base portion 30 be made of an aluminum alloy such as 052. It may contain non-metallic components such as ceramics. The diameter of the base part 30 is, for example, A range of approximately 220mm to 550mm is sufficient, and typically it is between 220mm and 350mm. The thickness of the base portion 30 can be, for example, about 20mm to 40mm.

[0012] Multiple refrigerant flow paths 32 are formed inside the base portion 30 so as to follow the XY plane. By flowing a refrigerant such as a fluorine-based inert liquid, water, or liquid nitrogen through the refrigerant flow path 32, The base portion 30 is cooled. Then, the base portion 30 and the ceramic portion are connected via the joint portion 40. The ceramic part 20 is cooled by heat transfer between it and the 20, and the mounting surface 24 of the ceramic part 20 The held wafer W is cooled. This enables temperature control of the wafer W. In addition to a configuration in which the refrigerant flow path 32 is located inside the s section 30, the base section 30 is located outside the base section 30. The base portion 30 may be given a cooling function by cooling it.

[0013] The joint 40 is positioned between the ceramic part 20 and the base part 30, and the ceramic part 2 The 0 and the base part 30 are joined together. The joining part 40 contains an adhesive formed from a resin material. The joint 40 further includes the properties of the joint 40 and the paste used to form the joint 40. It may contain various fillers (inorganic fillers) to adjust its properties. The joint 40 may be composed of a composite material containing an adhesive and an inorganic filler. However, If the joint 40 satisfies the properties described later, the joint 40 does not contain inorganic fillers. This may also be done. The thickness of the joint 40 is, from the viewpoint of reducing the thermal resistance of the joint 40, for example, It is acceptable if it is 1.00 mm or less, preferably 0.45 mm or less, and even better if it is 0.40 mm or less. Preferably, it is 0.35 mm or less. The thickness of the joint 40 is, for example, the joint 4 From the perspective of ensuring flexibility and strength, a thickness of 0.05 mm or more is sufficient.

[0014] The electrostatic chuck 10 is further provided with multiple gas supply passages 50. 50 is provided penetrating the ceramic part 20, the joint part 40, and the base part 30 in the Z direction. It is located at a gas outlet 52 formed on the mounting surface 24 (see Figure 1). The gas supply line 50 supplies an inert gas, such as helium gas, from a gas supply device (not shown). The gas is supplied and an inert gas is discharged from the gas outlet 52 into the space between the mounting surface 24 and the wafer W. It supplies heat. This improves the heat transfer between the ceramic part 20 and the wafer W, The controllability of the temperature distribution of HW is further improved. Note that the gas supply path 50 is not essential, and static The electric chuck 10 may be left without a gas supply passage 50.

[0015] (A-2) Composition of the joint: The following describes the configuration of the joint 40. As previously mentioned, the joint 40 is made of resin It comprises an adhesive composed of and further includes inorganic fillers. Furthermore, the joint portion 40 of the electrostatic chuck 10 of this embodiment has a thermal resistance of 5.0 × 10 -4 (m 2 The stress ratio (k / w) is less than or equal to 0.5 mm, and the strain at maximum shear stress is 0.5 mm or more. The strain amount under heavy shear stress is a numerical value that serves as an indicator of the flexibility and stress relaxation performance of the joint 40. The amount of strain at maximum shear stress will be explained in detail later.

[0016] Examples of adhesives that make up the joint 40 include silicone resin, acrylic resin, and Epoxy resins and the like can be used. Silicone resins, in particular, have relatively good heat resistance and It is desirable because of its excellent flexibility. Among these resins, silicone resin has a relatively high elastic modulus. Because it is low, it has a high function of mitigating thermal stress generated at the joint 40, and the heat resistance temperature is relatively low. Because it is expensive, it is desirable.

[0017] Inorganic fillers include ceramics, metal oxides, metals, or other inorganic compounds. Various inorganic materials can be used, in granular or powder form, etc. Examples of inorganic fillers include aluminum nitride (AlN) and aluminum oxide. (Alumina: Al2O3), Zirconium oxide (Zirconia: ZrO2), Yttrium oxide Yttrium (Y2O3), yttrium fluoride (YF3), silicon carbide (SiC) Silicon nitride (Si3N4), silicon dioxide (silica: SiO2), iron oxide, valucleated sulfate Inorganic fillers can be made from the above-mentioned inorganic materials. Because the material generally has a higher thermal conductivity than the resin used as an adhesive, inorganic filler is used in the joint 40. By adding -, the thermal conductivity at the joint 40 can be increased. In particular, heat From the perspective of having relatively high conductivity and making it easier to suppress the thermal resistance of the joint 40, inorganic fillers Preferred materials for this structure include aluminum nitride, aluminum oxide, and silicon carbide. Aluminum nitride and aluminum oxide are particularly preferred.

[0018] Furthermore, the joint 40 is further equipped with a catalyst to promote the curing reaction and to promote curing and adhesion. Silane coupling agents, crosslinking agents, and reaction agents for adjusting the curing speed of adhesives to impart properties. It may contain a reaction inhibitor or viscosity modifier, etc. The catalyst contained in the joint 40 is: Various conventionally known catalysts can be used, such as platinum catalysts, rhodium catalysts, and titanium catalysts. A catalyst such as a bismuth catalyst can be used. In particular, a highly reactive platinum catalyst can be used. This is desirable. There are no particular restrictions on the silane coupling agent contained in the joint 40, for example, As organic reactive groups, vinyl groups, epoxy groups, methacrylic groups, amino groups, mercapto groups, Among conventionally known silane coupling agents, those having any of the isocyanate groups. can be appropriately selected as needed. Further, instead of the above silane coupling agent, a titanium -based coupling agent or an aluminate-based coupling agent may be used. As the crosslinking agent contained in the joint portion 40, an organohydro genpolysiloxane having at least three hydrosilyl groups in one molecule can be used. More specifically, for example, polymethylha ydrogensiloxane and / or at least one of poly(dimethylsiloxane-methylhydrogensilo xane) can be used.

[0019] As the reaction inhibitor contained in the joint portion 40, various conventionally known reaction inhibitors can be used, for example, 1,3,5,7-tetramethyl-1,3,5,7-tetravinylcyclotet rasiloxane, 1,1,3,3-tetramethyl-1,3-divinyldisiloxane, tria lyl isocyanurate, etc. can be used. As the viscosity modifier contained in the joint portion 40, various conventionally known viscosity modifiers can be used, for example, fumed silica, pyrogenic silica, colloidal silica, fumed alumina, pyrogenic alumina, colloidal alumina, etc. can be used. The types and amounts of the above-mentioned catalyst, silane coupling agent, crosslinking agent, reaction inhibitor, or viscosity modifier, etc. may be appropriately selected according to, for example, the type of the resin constituting the joint portion 40.

[0020] As described above, in this embodiment, the thermal resistance of the joint portion 40 is 5.0×10 -4 (m 2 K / W) or less. It is desirable that the thermal resistance of the joint portion 40 be 4.5×10 -4 (m 2 K / W) or less, and 4.0×10 -4 (m 2It is more desirable to keep it below kW. 3.5 × 10 -4 (m 2 It is even more desirable to keep it below (K / W). Note that the thermal conductivity λ is Here, the thermal resistance R refers to the value at room temperature (22°C). The thermal resistance of the joint 40 is R (m 2 (K / W), thickness of joint 40 is t (m), thermal conductivity of joint 40 is λ (W / mK) Therefore, the thermal resistance R of the joint 40 can be calculated by the following equation (1).

[0021] R(m 2 K / W) = t(m) ÷ λ(W / mK) …(1)

[0022] By setting the thermal resistance of the joint 40 to the above value, the base portion 30 and the joint 40 are connected. Heat transfer between the ceramic part 20 and the base part 30, that is, heat transfer from the ceramic part 20 to the base part 30. This makes it easier to perform the operation and improves the cooling efficiency of the electrostatic chuck 10.

[0023] The thermal resistance of the joint 40 can be reduced by reducing the thickness of the joint 40, according to equation (1). This can be done. For example, by making the thickness of the joint 40 less than 0.5 mm, the joint 4 The thermal resistance of 0 is 5.0 × 10 -4 (m 2 It becomes easier to keep it below kW. However, By increasing the thermal conductivity of the joint 40 to, for example, 1.0 W / mK or more, the thickness of the joint 40 Even if the thickness is 0.5 mm or more, the thermal resistance of the joint 40 is 5.0 × 10 -4 (m 2 K / W) It becomes relatively easy to do this downwards. To increase the thermal conductivity of the joint 40, for example, as will be described later... To that end, inorganic fillers made of materials with higher thermal conductivity are used, or the inorganic fillers are included You can increase the percentage of people who have it.

[0024] Furthermore, the thermal resistance of the joint 40 is, for example, due to the material of the inorganic filler contained in the joint 40, The amount of inorganic filler in the joint 40 and the type of resin contained in the joint 40 will vary. It can be improved by using inorganic fillers made of materials with higher thermal conductivity. By doing so, and by increasing the inorganic filler content, and by improving the thermal conductivity By using a high-performance resin, the thermal conductivity of the joint 40 is increased, thereby reducing the thermal resistance of the joint 40. It can be made smaller.

[0025] Furthermore, the thermal resistance of the joint 40 depends on the shape of the inorganic filler particles contained in the joint 40. It can be changed. In order to reduce the thermal resistance of the joint 40, the inorganic filler is cut off. When viewed from the surface, if the radius of the inscribed circle is R1 and the radius of the circumscribed circle is R2, then "R2 / R It is desirable that the average value of "1" be 1.05 or higher, and even better if it is 1.10 or higher. Desirable. The closer "R2 / R1" is to 1.0, the closer the particle shape is to a perfect sphere (in the following, true This indicates a high degree of accuracy (also known as a high degree of ball-scoring).

[0026] The methods for identifying the inscribed and circumscribed circles, and for finding "R2 / R1", are as follows: That is, in the image obtained by observing the cross-section of the joint 40 with a scanning electron microscope (SEM), Focus on one observed particle. Enclose the particle and touch the three vertices within the particle. Draw a circle and make it the circumscribed circle. Also, it is enclosed within the particle and touches the outer circumference of the particle at three points. Draw a circle as shown, and make it the inscribed circle. If multiple circles can be drawn, the radius of the circle that is smallest will be drawn. Let the circle be defined as the inscribed circle and the circumscribed circle. Similarly, identify the inscribed and circumscribed circles for 10 particles. Then, measure the radius R1 of the inscribed circle and the radius R2 of the circumscribed circle, and calculate "R2 / R1" from the average value. Calculate ".

[0027] Figure 3 shows how to determine the "R2 / R1" ratio when viewing the inorganic filler 42 particles in cross-section. Figure 4 is an explanatory diagram showing how heat transfer paths are formed between the inorganic fillers 42. In Figure 3, the inscribed circle of the inorganic filler 42 when viewed in cross-section is shown by a dashed line, and the circumscribed circle is shown by It is shown by a dashed line. Furthermore, the center of the inscribed circle is shown as O1, and the center of the circumscribed circle is shown as O2. This is shown in Figure 4. In Figure 4, heat conduction paths are formed at the points where the inorganic fillers 42 come into contact with each other. The appearance is represented by double-headed arrows. The "R2 / R" when viewing the inorganic filler 42 in cross-section. The value of "1" is greater than or equal to the lower limit mentioned above, and the shape of the inorganic filler 42 is different from a perfect sphere on the surface. The more uneven the surface, the more contact points there are between the inorganic fillers 42, as shown in Figure 4, resulting in bonding. Heat conduction paths are more easily formed within part 40, and the thermal resistance of the joint part 40 is reduced.

[0028] Figure 5 shows the inorganic filler at the joint 40 for two types of inorganic fillers with different particle shapes. This is an explanatory diagram illustrating an example of the relationship between the film content and the thermal conductivity of the joint 40. It uses aluminum nitride (AlN) and aluminum oxide (alumina) as inorganic fillers. (Al2O3) is used, and among the inorganic fillers used here, aluminum nitride is The "R2 / R1" value is larger than that of aluminum oxide. Specifically, as shown in Figure 5, aluminum nitride... The R2 / R1 ratio of aluminum is 1.15, while the R2 / R1 ratio of aluminum oxide is 1. It is 05.

[0029] As shown in Figure 5, even if the inorganic filler content in the joint 40 is the same, Aluminum nitride with a larger R2 / R1 value has a higher thermal conductivity at the joint 40. As shown in Figure 5, the higher the proportion of inorganic filler, the greater the aluminum nitride. The difference in the magnitude of thermal conductivity relative to the content ratio between um and aluminum oxide is significant. This is because the higher the proportion of inorganic fillers, the greater the gaps between the inorganic filler particles. The reduced distance makes contact easier, and the increased number of contact points between inorganic fillers leads to heat transfer. This is thought to be because the effect on conductivity becomes more apparent. In other words, the inorganic filler content is If the amount is somewhat large, and the inorganic filler is close to a perfect sphere, it only slightly reduces the content ratio. Therefore, the contact points between inorganic fillers tend to decrease significantly, and if there are many irregularities on the surface of the inorganic filler, the content This is because reducing the proportion has a relatively small impact on the reduction of contact points between inorganic fillers. This is possible when the inorganic filler content is relatively small (for example, 20 vol%). In the following cases, the influence of the shape of the inorganic filler on the thermal conductivity is small, and the inorganic filler Due to differences in the thermal conductivity of the materials themselves, a slight difference in the thermal conductivity of the joint 40 occurs. When the inorganic filler content is relatively high and inorganic fillers are likely to come into contact with each other, the inorganic fillers Because the shape of the film has a significant impact on the thermal conductivity, the thermal conductivity of the joint 40 is greatly affected. It is thought that differences will arise. Note that the inorganic filler used in Figure 5 is aluminum nitride. The value of "R2 / R1" is larger for this than for aluminum oxide, but the value of "R2 / R1" is Regardless of the material of the inorganic filler, various modifications can be made by adjusting the manufacturing method, etc. ru.

[0030] In addition to being formed from homogeneous materials, inorganic fillers can also be constructed, for example, by structuring the particle body of the inorganic filler. A coating layer made of a different material from the constituent material may be provided on the surface. By appropriately selecting the appropriate components, the water resistance of the inorganic filler can be enhanced, and the interaction between the inorganic filler and the resin can be improved. This makes it possible to improve the fit and increase the flexibility of the joint 40. For example, inorganic When using aluminum nitride powder as a filler, the aluminum nitride that constitutes the inorganic filler On the surface of the aluminum particles, silicon dioxide (SiO2), aluminum oxide (Al2O3), Alternatively, a coating layer containing aluminum phosphate (AlPO4) may be provided.

[0031] The lower limit of the thermal resistance of the joint 40 is, for example, 0.6 × 10 -4 (m 2 (K / W) This is possible. In order to reduce the thermal resistance of the joint 40, an inorganic filler can be used as described above. One possible approach is to appropriately select the constituent materials and increase the proportion of inorganic fillers. However, if the proportion of inorganic filler is increased excessively, the flexibility of the joint 40 will be reduced. This is impaired, making it difficult to ensure a sufficient amount of strain at the maximum shear stress of the joint 40, as described later. This may increase the risk of heat loss. Also, in order to reduce the thermal resistance of the joint 40, the joint 40 can be made thinner. One possible method is to make the joint 40 excessively thin. As the value decreases, it becomes difficult to ensure the flexibility of the joint 40, and the amount of strain at maximum shear stress decreases. It may become difficult to achieve the desired numerical range. Therefore, the thermal resistance of the joint 40 is , 0.6 × 10 as described above -4 (m 2 It is desirable to set it to kW or higher.

[0032] Furthermore, as previously described, in this embodiment, the amount of strain at maximum shear stress of the joint 40 is The minimum required strain is 0.5 mm. The maximum strain at joint 40 under maximum shear stress is 0.6 mm or more. It is preferable to have a thickness of 1.0 mm or more, and more preferably 1.1 mm or more. It is even more desirable to do so. Note that the strain amount at maximum shear stress of the joint 40 is usually 5 It will be 0.0 mm or less. The amount of strain at maximum shear stress refers to the flexibility and stress relaxation properties of the joint 40. This is an index value that represents the performance, and it is the value generated at the joint 40 when a shear force is applied to the joint 40. When the shear stress is at its maximum, that is, when the maximum shear stress occurs at the joint 40 This refers to the magnitude of the strain (displacement in the direction of the shear force) that occurs at the joint 40. A larger strain under force indicates higher flexibility of the joint 40. The specific measurement method using a tensile testing machine to measure the amount of seized material will be explained in detail later. To clarify, if the strain amount at maximum shear stress of the joint 40 is set to the above value, the flexibility of the joint 40 will be This makes it possible to ensure sufficient durability and stress relaxation performance.

[0033] Figure 6 is an explanatory diagram regarding the stress generated at the joint 40. When using the electrostatic chuck 10. Due to heat input to the ceramic part 20, such as exposure to plasma, the ceramic Part 20 is heated and expands. The base part 30 is cooled by the refrigerant flow path 32. It then shrinks. Figure 6 shows how the ceramic part 20 is heated and expands, and the base The white arrows indicate how part 30 cools and shrinks. At this time, the joint part 40 In this case, a tensile force acts on the ceramic part 20 and the base part 30. Figure 6 shows the joint. The force with which part 40 pulls on the ceramic part 20, and the force with which part 30 pulls on the base part, respectively The arrows indicate this. Furthermore, in Figure 6, the amount of deformation in the X-axis direction that occurs in the joint 40 is shown as strain. It is shown as the quantity δ.

[0034] Furthermore, in the electrostatic chuck 10, the base portion 30 is generally larger than the ceramic portion 20. It has a high coefficient of thermal expansion and expands and contracts significantly with temperature changes. Therefore, depending on the temperature conditions... Therefore, the amount of thermal expansion of the base portion 30 is greater than the amount of thermal expansion of the ceramic portion 20. It may also become larger. In this way, expansion and contraction may occur between the ceramic part 20 and the base part 30. Due to the difference in degree of contraction, a shear force in the X-axis direction is applied to the joint 40, and shear Stress is generated. Therefore, as described above, the amount of strain at the maximum shear stress of the joint 40 is large. By increasing the width, even when a large shear force is applied to the joint 40, This reduces the shear stress generated at the joint 40 and minimizes damage to the joint 40.

[0035] The amount of strain at maximum shear stress in the joint 40 is determined by the adhesive (resin) contained in the joint 40. It can be changed. For example, the maximum shear resistance can be changed depending on the type of resin that makes up the joint 40. In addition to being able to change the amount of strain under force, even when using the same type of resin, the polymer material is resin By controlling the distance between crosslinking points in the resin, the amount of strain at maximum shear stress in the joint 40 can be controlled. It can be changed. Specifically, the content of functional groups that form crosslinking points in the above resin (functional group equivalent). Reducing the crosslinking point increases the distance between crosslinking points, improving the flexibility of the resin, thus reducing the maximum shear The amount of strain under stress can be increased. Also, the content of reactive functional groups in the resin Even in equivalent cases, the distance between crosslinking points can vary depending on curing conditions such as the curing temperature and curing time of the resin. This allows for control over curing temperature, or lengthening the curing time. In this case, the resin hardening progresses further, the crosslinking density increases, and the distance between crosslinking points shortens. Yes.

[0036] Furthermore, the amount of strain at maximum shear stress in the joint 40 can be increased by increasing the thickness of the joint 40. Therefore, it can be made larger.

[0037] Furthermore, the higher the proportion of inorganic filler in the joint 40, the greater the maximum capacity of the joint 40. The amount of strain under sectional stress tends to be small. This is because the inorganic filler content is high. The more the inorganic filler restrains the surrounding resin (resin composition), the greater the degree to which the joint 4 This is thought to be because the flexibility of 0 decreases, making the joint 40 less susceptible to strain.

[0038] According to the electrostatic chuck 10 of this embodiment configured as described above, the electrostatic chuck 10 The thermal resistance of the joint 40 is 5.0 × 10 -4 (m 2 (K / W) or less, and the maximum shear load Because the strain under force is 0.5 mm or more, the heat input to the ceramic part 20 is greater. Even in such cases, the cooling performance of the electrostatic chuck 10 is improved, and the joint portion 40 This can suppress problems caused by the resulting stress. Specifically, as previously described, the joint The lower the thermal resistance of part 40, the higher the cooling efficiency in the electrostatic chuck 10 can be. Therefore, by setting the thermal resistance of the joint 40 to the above value, for example, excess of the ceramic part 20 This suppresses the temperature rise. Therefore, it suppresses the decrease in wafer processing accuracy caused by the temperature rise. Furthermore, the shear stress generated at the joint 40 can be suppressed. The greater the strain amount at maximum shear stress, the higher the stress relaxation performance of the joint 40. By making the strain amount at the maximum shear stress of the joint 40 0.5 mm or more, the joint 40 Even if shear stress occurs, damage to the joint 40 caused by shear stress is prevented. This can be suppressed. Therefore, the thermal resistance of the joint 40 and the amount of strain at maximum shear stress are By setting the above values, the overall performance of the electrostatic chuck 10 can be improved.

[0039] Such effects are particularly pronounced when the electrostatic chuck 10 is exposed to higher-power plasma. As shown above, this effect is particularly noticeable when the heat input to the ceramic part is large. Even without providing heater electrodes for heating the buck section 20, high plasma power When using the electrostatic chuck 10 together with the ceramic part 20, the mounting surface of the ceramic part 20 and the base part 30 Because the temperature difference between the two points tends to become large, the effects of this embodiment are significantly enhanced.

[0040] The stress relaxation performance at the joint 40 is improved, so that when a shear force is applied to the joint 40... In order to minimize damage to the joint 40, when a shear force is applied to the joint 40, It is desirable that the shear stress generated at the joint 40 be small. From the perspective of suppressing this, a tensile test was performed on the joint 40, and the length of deformation due to shear force was 0 The shear stress generated at the joint 40 when it becomes 0.5 mm (hereinafter referred to as "shear strain amount 0.") The shear stress at 5 mm (also called the shear stress at 5 mm) should preferably be 3.5 MPa or less. It is more desirable that it be less than or equal to 2.0. A specific method using a tensile testing machine to measure the shear stress at a tensile strain of 0.5 mm. The measurement method will be explained in detail later.

[0041] Furthermore, the shear stress at a shear strain of 0.5 mm must be 0.1 MPa or higher. This is desirable. For example, as an example of how the electrostatic chuck 10 is used, the electrostatic chuck 10 is used in a vacuum chuck There is a structure that forms part of the outer wall of the rump, and in this case, the electrostatic chuck 10 contains ceramic The Mick section 20 side may be exposed to a vacuum, while the base section 30 side may be exposed to the atmosphere. If the "shear stress at a shear strain of 0.5 mm" is set to 0.1 MPa or higher, then Even with the usage described above, it is easy to ensure sufficient strength at the joint 40. Yes. However, if the usage is different from the above, for example, "when shear strain is 0.5 mm" The shear stress may be less than 0.1 MPa.

[0042] The stress relaxation performance at the joint 40 is improved, so that when a shear force is applied to the joint 40... To minimize damage to the joint 40, as previously described, flexibility of the joint 40 must be ensured. This is important. Another indicator of the flexibility of the joint 40 is the stiffness modulus of the joint 40. This can be achieved. The rigidity ratio of the joint 40 is preferably 2.5 MPa or less. Furthermore, it is more desirable for the pressure to be 2.0 MPa or less, and even more desirable for it to be 1.5 MPa or less. This is desirable. However, if the rigidity ratio of the joint 40 is too low, the shape of the joint 40 may be maintained. When the performance deteriorates, it becomes difficult to attach and detach wafers from the electrostatic chuck 10, or to transport the electrostatic chuck 10. When this happens, the ceramic part 20 may move to an undesirable degree. From the perspective of minimizing such inconveniences, the rigidity ratio of the joint 40 should be set to 0.03 MPa or higher. The method for determining the stiffness modulus will be explained in detail later.

[0043] As described above, the joint 40 requires a shear strain of 0.5 mm or more, and this must be met. To achieve this, it is desirable that the tensile elongation of the adhesive used in the joint 40 be 40% or more. A percentage of 50% or more is more desirable, and 60% or more is even more desirable. While a higher tensile elongation is preferable, it is usually less than 500%. (Tensile elongation of adhesive) This means that when a tensile test is performed using a tensile testing machine, the hardened adhesive sample breaks. This is the value obtained by dividing the length of elongation at that time by the initial length of the sample. For adhesives exceeding a certain quantity, the conditions for bonding and the flexibility index of the bonded portion 40 are considered. Measurements such as maximum shear stress, strain at maximum shear stress, and shear strain of 0.5 By measuring the shear stress and stiffness modulus at mm intervals, the efficiency of adhesive selection can be improved. Cut.

[0044] As described above, the joint 40 has sufficient flexibility, so the ceramic of the electrostatic chuck 10 The flatness of the mounting surface 24 of the Mick unit 20 can be increased. As previously described, electrostatic chat Even when a shear force is applied to the joint 40 in the case of 10 due to temperature changes, By increasing the flexibility of the joint 40, the shear stress generated at the joint 40 can be suppressed. In both cases, the force exerted by the joint 40 on the ceramic part 20 is suppressed, and the deformation of the ceramic part 20 is reduced. In other words, the decrease in the flatness of the ceramic part 20 is suppressed. Note that flatness refers to JIS standards. As defined in B0621 "Definition and Representation of Geometric Deviations", "Geometry of Planar Forms" This indicates the magnitude of the deviation from the perfectly correct plane.

[0045] B. Second Embodiment: Figure 7 is a schematic cross-sectional view showing the configuration of the electrostatic chuck 110 of the second embodiment. The electrostatic chuck 110 of this embodiment is equipped with a multilayer joint 140 instead of a single-layer joint 40. Except for the above, it has the same configuration as the first embodiment. Electrostatic chuck 1 of the second embodiment In 10, the same reference numeral is used for parts that are common to the electrostatic chuck 10 of the first embodiment. Note that Figure 7 omits the description of the components within the base section 30, such as the refrigerant flow path 32.

[0046] The joint 140 consists of a first resin layer 142, an inorganic sheet 144, and a second resin layer 146. In the joint 140, the first resin layer 14 is positioned in the direction of the -Z axis (vertically downward). 2. The inorganic sheet 144 and the second resin layer 146 are laminated in that order. In this embodiment, The first resin layer 142 and the second resin layer 146 together are also simply called the "resin layer." In terms of form, the inorganic sheet 144 is in the planar direction (X-axis direction) of the joint portion 140 within the resin layer. It is formed to spread out. And the heat of the joint 140 which has a three-layer structure as described above The resistance and strain at maximum shear stress are within the same numerical range as the joint 40 of the first embodiment. This means that the thermal resistance of the entire joint 140 is 5.0 × 10 -4 (m 2 K / The stress is less than or equal to W, and the strain at maximum shear stress is 0.5 mm or more.

[0047] The first resin layer 142 and the second resin layer 146 are equipped with an adhesive made of resin, As the adhesive, the same adhesive described above as in the joint 40 of the first embodiment can be used. In particular, it is desirable to use a silicone resin with a relatively low modulus of elasticity and a relatively high heat resistance temperature. Furthermore, the first resin layer 142 and the second resin layer 146 are further connected to the first embodiment. It may also contain the same inorganic fillers as described above as in the joint 40. Examples of inorganic fillers include: Aluminum nitride, aluminum oxide, and silicon carbide are desirable because they have relatively high thermal conductivity. The first resin layer 142 and the second resin layer 146 are made of a type of adhesive and an inorganic filler. If inorganic fillers are present, the amount of inorganic fillers may differ from that of the other.

[0048] The inorganic sheet 144 is made of an inorganic material, for example, the ceramic part 20 and Similarly, the ceramics mentioned above (for example, aluminum oxide and aluminum nitride) are the main components. It can be formed as a component. However, the inorganic sheet 144 and the ceramic part 20 are different. It may also be formed using a type of ceramic. In addition, the inorganic sheet 144 is a ceramic In addition, or instead of ceramics, other inorganic materials such as carbon fiber can be used. It may also be constructed using graphite or metal. Furthermore, the inorganic sheet 144 may be, for example, made of... It may constitute a data set.

[0049] In the second embodiment, the joint 140 as a whole has a maximum shear stress strain and thermal resistance. The thickness of the first resin layer 142 and the second resin layer 146 is set so that the value of is within the numerical range described above. The thickness of the inorganic sheet 144, and the resins constituting the first resin layer 142 and the second resin layer 146 Type, content of inorganic filler in the first resin layer 142 and the second resin layer 146 (inorganic filler The presence or absence of a film can be set as appropriate. For example, the maximum shear load of the joint 140 The amount of strain under force increases as the thickness of the first resin layer 142 and the second resin layer 146 increases. Furthermore, the amount of strain at maximum shear stress in the first resin layer 142 and the second resin layer 146 is increased. The more you do it, the larger it becomes. And the thermal resistance of the joint 140 is the first resin layer 142, inorganic sheet 1 The value decreases as the thermal resistance of 44 and the second resin layer 146 are reduced.

[0050] Here, the thermal resistance of each of the first resin layer 142, the inorganic sheet 144, and the second resin layer 146 The resistance can be calculated using the following equation (2). In equation (2), R n This represents the thermal resistance of each layer. , t n λ represents the thickness of each layer. n represents the thermal conductivity of each layer. However, in equation (2), The first resin layer 142 has n=1, the inorganic sheet 144 has n=2, and the second resin layer 146 n=3.

[0051] R n (m 2 (K / W) = t n (m) ÷ λ n (W / mK) …(2)

[0052] The thermal resistance R of the entire joint 140 can be calculated using the following equation (3). (3) In this, N represents the sum of the number of resin layers and the number of inorganic sheets, and the joint 1 of the second embodiment For 40, N=3.

[0053] R(m 2 (k / w) = R1 + R2 + ... + R N …(3)

[0054] With this configuration, the thermal resistance of the entire joint 140 and the strain at maximum shear stress are By setting the value of to the numerical range described above, the same effect as in the first embodiment can be obtained. Furthermore, Since the joint 140 includes an inorganic sheet 144 with a higher thermal conductivity than the resin layer, the first resin The sum of the thicknesses of layer 142 and the second resin layer 146 is greater than the thickness of the joint 40 in the first embodiment. It can also be made thinner. This is because when an inorganic sheet 144 is provided, the inorganic sheet 14 Since 4 also contributes to thermal resistance, the thermal resistance of the entire joint 140 is kept within the numerical range described above, and the resin Compared to obtaining the same thermal resistance with only the layers, the thermal resistance of the first resin layer 142 and the second resin layer 146 This allows for a smaller resistance, and as a result, the first resin layer 142 and the second resin layer 146 This is because the total thickness can be made thinner. Therefore, the side surface of the joint 140 (parallel to the Z-axis) By reducing the surface area of ​​the adhesive layer exposed on the surface, the plasma resistance of the joint 140 is improved. This is possible. Furthermore, according to the second embodiment, the joint portion 140 comprises an inorganic sheet 144. This makes the thermal resistance and the strain at maximum shear stress uniform within the plane of the joint 140. This allows for the temperature distribution within the plane of the joint 140 to be made uniform.

[0055] In the second embodiment, the joint portion 140 is provided with a single layer of inorganic sheet 144. However, it is also possible to have multiple inorganic sheets. In this case, there is a gap between each inorganic sheet. The same resin layers as the first resin layer 142 and the second resin layer 146 can be arranged. Even if the part comprises multiple inorganic sheets, the ceramic part 20 in contact with the joint The layer including the surface and the layer including the surface in contact with the base portion 30 may be made of resin. If an inorganic sheet is provided, at least one inorganic sheet must be different from the other inorganic sheets. It may be composed of such materials. Even if it comprises multiple inorganic sheets, the bonding If the values ​​of thermal resistance and strain at maximum shear stress are within the aforementioned numerical range for the entire section, , a similar effect can be obtained. When the joint comprises multiple inorganic sheets, each inorganic The thermal resistance of the sheet and resin layer can be determined by the above equation (2), and the thermal resistance of the entire joint The resistance can be calculated using equation (3) above.

[0056] C. Other embodiments: This disclosure applies to holding devices other than electrostatic chucks that hold wafers W using electrostatic attraction. It is also possible to have a plate-like portion, a base portion, and a joint portion that connects the plate-like portion and the base portion. , and other holding devices for holding an object on the surface of the ceramic part, for example, CVD, P This technology can also be applied to heater devices for vacuum equipment such as VDs and PLDs, as well as vacuum chucks.

[0057] Furthermore, in each of the embodiments described above, the plate-like portion having a mounting surface is made mainly of ceramic. Although the ceramic part 20 is used, the plate-like part will mainly consist of materials other than ceramic. This is also fine. Furthermore, the base portion 30 has a thermal expansion coefficient different from that of the material that mainly forms the plate-like portion. It is sufficient if the main component is a material with tension. Such a plate-like part, a base part, and a plate-like part If the holding device includes a joint that connects the base portion to the main body, then the main body exemplified in each embodiment is as follows: By applying the disclosed configuration, the cooling performance in the retaining device is improved, and the joints are affected. In addition to suppressing stress, it also has the same effect of suppressing damage to the joint even if shear stress occurs at the joint. You can obtain this. [Examples]

[0058] The holding device of this disclosure will be described below based on examples. Here, thermal resistance As a sample to correspond to various joints with different strain amounts at maximum shear stress, Sheet-like samples from S1 to sample S33 were prepared. A sample of an electrostatic chuck with a joint having the same composition as S1-S33 was prepared. Below, even for samples in the form of an electrostatic chuck, a sheet-like sample with the same composition as the joint is used. It is called by the same sample number as Ru.

[0059] Figures 8-11 show the thermal resistance, strain at maximum shear stress, thickness, and heat of the joint of each sample. Along with the conductivity value, the cooling performance of the ceramic part through the joint, and the ease of damage to the joint are also important. This is an explanatory diagram summarizing the results of the evaluation of the fibers (peeling). Figures 10 and 11 further show... Tensile tests were then performed on the joints of samples S24-S33 to determine the deformation due to shear force. Shear stress when the length is 0.5 mm (shear stress when shear strain is 0.5 mm) The values ​​and the results of evaluating the flatness of the ceramic part 20 in the electrostatic chuck are shown together. Figure 10 further shows the stiffness moduli of samples S24 to S29. Figure 11 further shows the inorganic filler "R2 / R1" present in the joints of samples S30 and S31. The value of " is also shown.

[0060] <Preparation of each sample> The electrostatic chucks of samples S1 to S32 are all electrostatic chucks of the first embodiment shown in Figure 2. Having a similar configuration to chuck 10, the electrostatic chuck of sample S33 is as shown in Figure 7. It has the same configuration as the electrostatic chuck 110 of the second embodiment. Below, first, each sump This section describes the methods for preparing sheet-like samples and electrostatic chuck-type samples common to all types of samples. Next, the materials and preparation conditions for each sample will be explained.

[0061] [Preparation of sheet-like samples] The method for preparing the sheet-like sample is as follows: The resin material is the adhesive before curing. (100 parts by weight) and inorganic filler, plus platinum catalyst (0.003 parts by weight of platinum content). The constituent material includes (parts), silane coupling agent (2 parts by weight), and crosslinking agent (3 parts by weight). After stirring under vacuum using a vacuum degassing agitator, the mixture is kneaded using a three-roll mill. A paste-like resin composition (adhesive paste) was prepared. Subsequently, the prepared adhesive paste was used Then, using a doctor blade, apply it onto the polyethylene terephthalate (PET) film. It was spread out. Next, the PET film with the adhesive paste spread on it was cut, and then cut. The adhesive paste with the attached PET film is dried in a dryer at an appropriately set temperature and time. By heating under intermediate conditions, each sample is partially cured or fully cured. I made adhesive sheets.

[0062] In each sample, silicone resin was used as the adhesive. Specifically, the bonding was performed before curing. Vinyl-terminated polydimethylsiloxane was used as the agent material (resin material). To obtain this, a difunctional D-unit having two organic substituents is used as a starting material: dimethyl dichloro Rosilane ((CH3)2SiCl2) was used. Furthermore, a monofunctional compound with three organic substituents was used. As raw materials for M units, trimethylchlorosilane ((CH3)3SiCl) and hexameth Ludisiloxane ((CH3)3SiOSi(CH3)3) was used. Furthermore, functionalities were added to the terminals. In order to introduce the group, dimethylvinylchlorosilane, a raw material of M units having a functional group, ( (CH3)2(CH2=CH)SiCl) is used to introduce a reactive functional group into the molecular chain. Therefore, CH3(CH2=CH)SiCl2, which is a D-unit raw material having a functional group, and (CH3)2(CH2=CH)SiCl was used. Samples S1~S25, S30~S Sample 33 uses the same resin material, and samples S26 to S29 are each made from raw materials in D units. A resin material was used in which the mixing ratio of the raw material and the M-unit raw material was different from that of the above-mentioned sample S1, etc. By varying the mixing ratio of D-unit raw materials and M-unit raw materials, the average molecular weight of the resin material can be reduced. This increased the content (density) of reactive functional groups in the resin material. Specifically, the sample The average molecular weight of resin materials S1-S25 and S30-S33 is 63,000, and the sample The average molecular weight of the resin material of sample S26 is 33,000, and the average molecular weight of the resin material of sample S27 is 33,000. The average molecular weight is 30,000, and the average molecular weight of the resin material in sample S28 is 18,000. The average molecular weight of the resin material in sample S29 is 15,000, and the reactivity is in this order. It has a high content of functional groups.

[0063] [Preparation of electrostatic chuck-type samples] A sample of the electrostatic chuck 10 is made by applying the semi-cured adhesive paste described above to the ceramic It is placed between the lock part 20 and the base part 30, and then the adhesive paste is allowed to harden. It was made from the above-mentioned semi-cured adhesive paper. The sample of the electrostatic chuck 110 was made from the above-mentioned semi-cured adhesive paper. Two sheets of material sandwich an inorganic sheet 144, which is mainly composed of aluminum oxide, and these It is placed between the ceramic part 20 and the base part 30, and then the adhesive paste is cured. It was manufactured by the following method. The ceramic part 20 is made of aluminum oxide. The ceramic part (thermal expansion coefficient of 7 ppm / K) is used, and the base part 30 is made of aluminum. A base portion (with a thermal expansion coefficient of 23 ppm / K) was used. Ceramic portion 20 The diameter of the joint was set at 350 mm.

[0064] [Samples S1-S5] Samples S1 to S5 all use the same resin material described above, and as an inorganic filler Aluminum nitride (average particle size 10 μm) was used. The resin in samples S1 to S5 The curing temperature was set to 150°C in all cases. Furthermore, the curing time of the resin in samples S1 to S5 was also measured. The interval was set so that the strain amount at the same maximum shear stress was the same, with 10 hours for sample S1 and sample Sample S2 was 11 hours, sample S3 was 15 hours, sample S4 was 16 hours, sample S In sample 5, the time was set to 28 hours. Samples S1 to S5 differed in the amount (percentage) of inorganic filler added. Specifically, the inorganic filler content is 32% by volume in sample S1, and Sample S2 contained 28.5% by volume, Sample S3 contained 21% by volume, and Sample S4 contained 20 units. The volume percentage was set to 11.5% for sample S5. As shown in Figure 8, the inorganic filler content The higher the proportion of the sample, the greater the thermal conductivity of the junction and the smaller the thermal resistance.

[0065] [Samples S6-S8] Samples S6 to S8 all use the same resin material as described above, but the curing temperature is different. By making them different, the distance between crosslinking points in the resin is made different. That is, the sample Sample S6 was cured at 155°C, sample S7 at 165°C, and sample S8 at 170°C. The curing time was set to 10 hours in all cases. The higher the curing temperature, the greater the crosslink density and the greater the crosslinking between crosslinking points. It is thought that the distance becomes shorter, and as shown in Figure 8, the higher the curing temperature, the greater the maximum shear stress. The amount of strain over time decreased. Note that samples S6-S8 were used as inorganic fillers. Contains aluminum nitride similar to S1-S5. The amount of inorganic filler added is the same as in sample S6- In all cases of S8, the same 32% volume percentage as in sample S1 was used.

[0066] [Samples S9-S20] Samples S9 to S20 all use the same resin material as described above, but the curing temperature By making the following different, the distance between crosslinking points in the resin is made different. That is, the sample The curing temperature for samples S9, S12, S15, and S18 is 155°C, while for samples S10, S13, and S1 6. The curing temperature of sample S19 is 165°C, and the curing temperatures of samples S11, S14, S17, and S20 are as follows: The curing temperature was set to 170°C. As shown in Figures 8 and 9, the higher the curing temperature, the greater the maximum shear stress. The amount of strain over time decreased. The curing time was 11 hours for samples S9-S11, and for Sun... Pull S12-S14: 15 hours, Sample S15-S17: 16 hours, Sample S18- S20 was set to 28 hours.

[0067] Furthermore, samples S9-S20 were nitrided as inorganic fillers, similar to samples S1-S5. Contains aluminum. The amount of inorganic filler added is 28.5% by volume for samples S9-S11. Samples S12-S14 are 21% by volume, samples S15-S17 are 21% by volume, sample The amount of S18-S20 was set to 11.5 volume%. As shown in Figures 8 and 9, the inorganic filler The greater the amount of additive, the higher the thermal conductivity of the joint and the lower the thermal resistance.

[0068] [Samples S21-S23] Sample S21 is the same as sample S1. Samples S22 and S23 are the joints. Only the thickness was different from sample S21(S1). The thickness of the joint is the same as sample S21(S1). (S1) was 0.3 mm, sample S22 was 0.4 mm, and sample S23 So we set it to 0.5 mm. As shown in Figure 9, the thicker the joint, the greater the strain at maximum shear stress. The quantity has increased.

[0069] [Samples S24-S29] Sample S24 is the same as sample S1. Samples S25~S29 are inorganic phi As a filler, it contains aluminum nitride similar to sample S24(S1), but with inorganic fillers. The amount added was 44.4% by volume, which is higher than that of sample S24 (S1). As shown in Figure 10, samples S25 to S29 are compared to sample S24 (S1). The thermal conductivity of the joint increased, and the thermal resistance decreased.

[0070] Furthermore, samples S25 to S29 differ in the type of adhesive used to form the joint. Sample S25 uses the same resin material as Sample S24 (S1), as previously mentioned. The curing conditions are 140°C for 10 hours, which is a low curing temperature. Therefore, the crosslinking density decreases. It is thought that the distance between the bridge points has increased, and as shown in Figure 10, sample S24(S1 Compared to the previous example, the "shear stress at a shear strain of 0.5 mm" was greater. Sample S 26~S29 are curing conditions the same as sample S25, but as previously mentioned, in the resin The density of functional groups that form crosslinking points has been changed, and as a result, crosslinking in silicone resin The distances between the points are different. As shown in Figure 10, the higher the density of reactive functional groups, In other words, in the order of samples S25 to S29, the shear stress at a shear strain of 0.5 mm is measured. The values ​​of the stiffness modulus and other factors increased.

[0071] [Sample S30~S32] Samples S30 to S32 all use the same resin material as described above, but inorganic The types of fillers are different from each other. As an inorganic filler, sample S30 is aluminum nitride. For sample S31, aluminum oxide was used, and for sample S32, boron nitride was used. I used it.

[0072] The inorganic filler contained in sample S30 is the same as that of aluminum nitride shown in Figure 5. The inorganic filler contained in sample S31 is the same as that of aluminum oxide shown in Figure 5. In both sample S30 and sample S31, the average particle size of the inorganic filler was 10 μm. The inorganic filler content in the joint 40 was set to 44% by volume in all cases.

[0073] In sample S32, the average particle size of the plate-like boron nitride inorganic filler is 2 μm. When using boron nitride as an inorganic filler, aluminum nitride or aluminum oxide is used. Compared to when using aluminum, the viscosity of the adhesive paste tends to be higher, and in the adhesive paste... It was difficult to increase the content of inorganic fillers that could form joints. By limiting the inorganic filler content within the range, the junction of sample S32 is made of inorganic filler. The boron nitride content was set to 16% by volume.

[0074] [Sample S33] The electrostatic chuck of sample S33 is, as previously described, the electrostatic chuck of the second embodiment shown in Figure 7. It has the same configuration as jack 110. In sample S33, the first resin layer 142 and The second resin layer 146 uses the same resin material as described above as in sample S25, with a curing time of no time. The proportions of mechanical and inorganic fillers were the same as in sample S25. The first resin layer 142 and the second resin layer 146 each have a thickness of 0.15 mm, and heat Its conductivity is 1.44 W / mK and its thermal resistance is 1.04 × 10⁻⁶ -4 m 2 It was set to kW. In sample S33, the inorganic sheet 144 is aluminum oxide (alumina: Al2 It is made of O3, has a thickness of 1.0 mm, and has a thermal conductivity of 30 W / mK, and has a thermal resistance Resistance is 0.33 × 10 -4 m 2 The value was set to kW.

[0075] <Measurement of thermal conductivity> Thermal conductivity is measured using a known thermal conductivity meter (Kyoto Electronics Manufacturing Co., Ltd.) on a sheet-like sample. The measurement was performed using a rapid thermal conductivity meter (QTM-710). Note that instead of a sheet-like sample, When measuring thermal conductivity at a joint incorporated into an electrostatic chuck, After grinding away the ceramic part of the jack with a surface grinder or similar machine to expose the joint, use a knife or similar tool. You can then use this method to peel off the joint and measure the thermal conductivity.

[0076] <Calculation of thermal resistance> The thermal resistance R is calculated using the measured thermal conductivity values ​​of each sample as described above, as follows: This was obtained using equation (1). In equation (1), t is the thickness of the joint and λ is the heat of the joint. Conductivity, that is, the thermal conductivity of each sample measured as described above.

[0077] R(m 2 K / W) = t(m) ÷ λ(W / mK) …(1)

[0078] <Measurement of maximum shear stress> The maximum shear stress and the strain at the maximum shear stress, as described later, are determined using a known tensile testing machine. The measurements were taken using a Shimadzu Autograph AGS-5kNX (manufactured by Shimadzu Corporation) via tensile testing.

[0079] Figure 12 is a schematic diagram illustrating the method for calculating the maximum shear stress and strain. Figure 12(A) shows the tensile test viewed from the front, and Figures 12(B) and 12(C) are Figures 12(A) and 12(B) show the view from the side. Figure 12(C) shows the child, and the situation after the start of the test. Test specimens 7 of samples S1 to S33. 0 is a semi-cured adhesive sheet of each sample, measuring 25mm wide x 100mm long x 1mm thick. Two aluminum plates 72, 25mm x 12.5mm from the edge, 12.5mm from the edge. By attaching them to the respective parts and pulling the two aluminum plates 72 in opposite directions, After bonding the parts together in the correct orientation, the aforementioned semi-cured adhesive sheet is cured to create the product. The thickness t of the test specimen 70 at the start of the test was set to the thickness of the joint shown in Figures 8 to 11. Next, the two aluminum plates 72 are moved relative to each other so that a shear force acts on the above test specimen. Here, a tensile testing machine was used to test one of the aluminum plates in one direction parallel to the bonding surface. While moving the object at a tensile speed of 2 mm / min, the load and the strain amount δ (as the distance moved) were measured. (See Figure 12(C)). In Figure 12(B), the relative movement of the two aluminum plates The direction is indicated by a white arrow. The adhesive area of ​​the test specimen before load application (25mm x 12mm) is shown. The shear stress was calculated by dividing by 5 mm. The relative movement was continued until the specimen 70 fractured, and the shear stress at the time of maximum shear stress was measured. The force was taken as the maximum shear stress (unit: MPa).

[0080] <Measurement of the strain amount at the maximum shear stress> The strain amount at the maximum shear stress (unit: mm) was defined as the strain amount δ when the shear stress reached its maximum in the tensile test shown in Fig. 12. 断応力が最大になったときのひずみ量δとした。

[0081] Fig. 13 is an explanatory diagram showing the relationship between the strain amount δ and the shear stress in sample S25 as an example of the measured values obtained from the tensile test. In Fig. 13, the horizontal axis represents the strain amount (mm), and the vertical axis represents the shear stress (MPa). For each sample, as shown in Fig. 13, based on the results of the tensile test, the strain amount δa when the shear stress reached the maximum shear stress τ になるときのひずみ量 縦軸にせん断応力(MPa)を示している。各サンプルについて図13に示すように、 引張試験の結果に基づいて、せん断応力が最大せん断応力τ max になるときのひずみ量 δaを求めた。

[0082] <Measurement of the shear stress at a shear strain of 0.5 mm> The shear stress at a shear strain of 0.5 mm refers to the shear stress when the strain amount δ is 0.5 mm in the tensile test shown in Fig. 12. That is, for each sample, as shown in Fig. 13, based on the results of the tensile test, the shear stress τ 量δが0.5mmのときのせん断応力を指す。すなわち、各サンプルについて、図13に 示すように、引張試験の結果に基づいて、ひずみ量δが0.5mmになるときのせん断応 力τ 0.5 を求めた。

[0083] <Shear modulus> The shear modulus G is the value obtained by dividing the shear stress τ by the shear strain γ, as shown in the following equation (4). The shear strain γ is the value obtained by dividing the strain amount δ by the thickness t0 of the sample before the test, as shown in the following equation (5). である。せん断ひずみγとは以下の(5)式に示すように、ひずみ量δを、試験前のサ ンプルの厚みt0で除した値である。

[0084] G (MPa) = τ (MPa) ÷ γ (-) … (4) γ (-) = δ (mm) ÷ t0 (mm) … (5)

[0085] Figure 14, like Figure 13, shows the results for sample S25 as an example of measured values ​​obtained from a tensile test. This is an explanatory diagram illustrating the point. In Figure 14, the horizontal axis shows the strain instead of the amount of strain (mm). The strain shown on the horizontal axis of Figure 14 is calculated by multiplying the value of the shear strain γ in equation (5) above by 1. This is expressed as a percentage after multiplying by 00. The stiffness modulus G is the strain-shear response shown in Figure 14. This value is calculated from the inclination of the force lines. As an example of deriving the stiffness modulus G for each sample, For sample S25, the calculation results are shown using values ​​when the strain amount δ is 0.5 mm. Since the thickness t0 of sample S25 before testing is 0.3 mm (see Figure 10), the amount of strain Shear strain γ when δ is 0.5 mm 0.5 From equation (5), we get "0.5 ÷ 0.3 ≈ 1." This results in 667 (166.7%). Furthermore, the shear stress when the strain δ is 0.5 mm is also calculated. τ 0.5 is, τ 0.5 Since = 0.90 (see Figure 10), the stiffness modulus G is given by equation (4) "0.90 ÷ 1.667 ≈ 0.54".

[0086] Furthermore, the maximum shear stress and maximum shear stress of the joint already incorporated into the electrostatic chuck When measuring the strain under force, shear stress at a shear strain of 0.5 mm, and stiffness modulus. For example, this can be done as follows: First, the joint is cut using a processing method such as laser cutting. The specimen is cut out together with the adherend (ceramic part and base part). The shape of the cut-out test piece is determined by the tensile test. Figure 12 shows two adherends that can be held by the jig of the test machine and are joined together. Any shape that can be pulled in opposite directions is acceptable. Before performing the tensile test... Then, measure the area of ​​the joint and the thickness of the joint in the cut-out test piece. After that, Perform a tensile test in the same manner as the method described above, and measure the maximum shear stress, the amount of strain at the maximum shear stress, the shear stress at a shear strain amount of 0.5 mm, and the modulus of rigidity.

[0087] <Evaluation of Cooling Property of Ceramic Part> For each sample S1 to S33 of the electrostatic chuck, the cooling property of the ceramic part through the joint was evaluated. Specifically, the initial temperature of the ceramic part of each sample was set to 150 °C and heated until the ceramic part reached 150 °C. Also, refrigerant at -10 °C was supplied to the refrigerant flow path of the base part to cool the base part. Then, after the ceramic part reached 150 °C, the heating of the ce ramic part was stopped, and the time until the ceramic part was cooled to 20 °C was measured. If it takes 30 seconds or more for the ceramic part to be cooled to 20 °C, the cooling performance is low ( ×), and if it takes 25 seconds or less, the cooling performance is very excellent (◎). If the time until the ceramic part is cooled to 20 °C is longer than 25 seconds and less than 30 seconds, it was evaluated that the cooling performance is good (○).

[0088] <Evaluation of Peeling of Joint> A peeling test was performed on each sample S1 to S33 in the form of an electrostatic chuck to evaluate the peeling of the joint. Specifically, each sample was placed in a commercially available thermal cycle tester and subjected to a thermal cycle test at a maximum temperature of 1 50 °C and a minimum temperature of 0 °C for 100 cycles, then returned to room temperature, and the presence or absence of peeling of the joint between the ce ramic part and the base part was evaluated. The presence or absence of peeling of the joint was determined using a known ultrasonic flaw detector. Ultrasonic waves were irradiated from the ceramic part side, and a reflection echo (flaw echo) was detected from the depth corresponding to the joint interface between the ceramic part and the joint. If there was peeling at the bonding interface between the ceramic part and the bonding part, it was determined that there was peeling. Ultrasonic waves were irradiated from the base part side, and if a reflected echo (defect echo) was detected from the depth corresponding to the bonding interface between the base and the bonding part, it was determined that there was peeling at the bonding interface between the base part and the bonding part. For at least one of the two interfaces where peeling was observed, peeling was evaluated as "present", and for those where no peeling was detected by ultrasonic flaw detection, peeling was evaluated as "absent".

[0089] <Calculation of R2 / R1> For each sample in the form of the electrostatic chuck 10, in order to calculate the value of "R2 / R1", an arbitrary location was cut out in a size of 50 mm × 50 mm in the X - Y direction shown in FIG. 1 and penetrated in the Z direction. The part of the bonding part 40 was taken out from the obtained test piece with a cutter. After mirror - polishing the cross - section of the cut - out bonding part, it was observed using a scanning electron microscope (SEM). For 10 particles of inorganic fillers observed in a field of view of 200 μm × 200 μm, their shapes were observed. Using image analysis software WinROOF, the radius R1 of the inscribed circle and the radius R2 of the circumscribed circle were measured to obtain the value of "R2 / R1" (see FIG. 3), and the value of "R2 / R1" for each sample was calculated as the average value of the 10 particles.

[0090] <Flatness> The definition of flatness is as described in JIS B0621 "Definition and Representation of Geometric Deviations", which is the magnitude of the deviation from the geometrically correct plane of a flat - shaped body. When the use surface is sandwiched between two geometrically correct parallel planes, the dimension of the interval when the interval between the two parallel planes is minimized is defined as the flatness. Specifically, the measurement method of flatness may be measured using a known three - dimensional measuring machine. Here, a one - shot 3D shape measuring machine "VR series" manufactured by Keyence was used. For the two parallel planes described above mentioned above [[ID=4le]]<0D<D94>​​​​​​​ When the minimum spacing dimension is less than 50 μm, the flatness is excellent. If the thickness is 200 μm or more, it is evaluated as having low flatness (×). For thicknesses less than 200 μm, the flatness was evaluated as good (○).

[0091] As shown in Figures 8 to 11, the thermal resistance of the joint is 5.0 × 10⁻⁶. -4 (m 2 (kW) or less Sometimes, the cooling performance for the ceramic part is evaluated as good (○), and the thermal resistance of the joint The resistance is 3.5 × 10⁻⁴ (m 2 When the wattage (K / W) is below this value, the cooling performance is excellent (◎). It was confirmed that this evaluation was correct. Furthermore, the amount of strain at the maximum shear stress of the joint was 0.5 m When m or more, delamination does not occur at the joint, and strain at the joint is suppressed, and the joint It was confirmed that the effect of suppressing joint damage caused by strain was sufficiently obtained. Safety factor If we set this to 2, it is particularly desirable that the strain at maximum shear stress be 1.0 mm or more. It's possible.

[0092] As an indicator for ensuring the flexibility of the joint in order to improve the stress relaxation performance at the joint, In addition to the strain amount at maximum shear stress, the "shear stress at a shear strain of 0.5 mm" and the joints The stiffness modulus can be cited. In each sample shown in Figure 10, as previously described, the joint is composed of The curing temperature of the resin, the proportion of inorganic fillers, and the density of reactive functional groups in the resin are all different. By doing so, the values ​​of the indicators mentioned above are changed. As shown in Figure 10, The shear stress at a cross-sectional strain of 0.5 mm must be 3.5 MPa or less, or the rigidity of the joint must be less than or equal to 3.5 MPa. It was confirmed that the flatness of the ceramic part improves by setting the pressure to 2.0 MPa or less. In particular, the shear stress at a shear strain of 0.5 mm should be 2.0 MPa or less, or By reducing the rigidity modulus of the joint to 1.2 MPa or less, the flatness of the ceramic part is further improved. It was confirmed that this would happen.

[0093] Furthermore, as can be seen by comparing sample S30 and sample S31 in Figure 11, inorganic particles If the lar content is the same, the R2 / R1 value is 1.10 or higher, and the sphericity is lower. Sample S30 has a higher thermal conductivity and a lower thermal resistance at the joint. This was confirmed. However, as shown in Figure 11 as sample S32, the inorganic filler was used. When using plate-like boron nitride particles, the sphericity of the inorganic filler is considered to be lower. However, as mentioned above, the viscosity of the adhesive paste increases, and the proportion of inorganic filler increases. Because it is suppressed, the thermal conductivity becomes relatively high. As a result, the thermal resistance is also 5.0 × 10 -4 ( m 2 It exceeded kW.

[0094] Although data is not shown in Figure 11, in addition to samples S30-S32, carbon nano We also prepared samples using tubes as inorganic fillers. Carbon nanotubes themselves are This is because it is known to have a very high thermal conductivity. However, in this case the joint The thermal conductivity value was further reduced (0.27 W / mK). This result is due to carbon nanotubes. When using an inorganic filler as the tube, the inorganic filler has a fibrous shape, so the adhesive base It is thought that the amount of inorganic filler that could be incorporated into the material was limited, which prevented the thermal conductivity from increasing. In other words, fibrous inorganic fillers are different from granular or spherical inorganic fillers. This is likely because the low fluidity of the substance itself limited the amount that could be added.

[0095] This disclosure is not limited to the embodiments described above, but may extend to the extent that it does not depart from the spirit thereof. It can be realized in various configurations. For example, in each of the embodiments described in the section on the summary of the invention... The technical features in the embodiments corresponding to the technical features solve some or all of the above-mentioned problems. For this purpose, or to achieve some or all of the effects described above, substitutions or combinations may be made as appropriate. It is possible to combine them. Furthermore, the technical features of this specification are not considered essential in this specification. If not explained, it can be deleted as appropriate. [Explanation of Symbols]

[0096] 10,110... Electrostatic chuck 20…Ceramic part 22...Adsorption electrode 24… Mounting surface 30...Base section 32… Refrigerant flow path 40,140...Joint part 42… Inorganic fillers 50…Gas supply lines 52...Gas outlet 70…Test piece 72…Aluminum plate 142...First resin layer 144... Inorganic Sheet 146…Second resin layer

Claims

1. A holding device for holding an object, A plate-like portion formed in a plate shape, A base portion that supports the plate-shaped portion, is mainly made of a material having a different coefficient of thermal expansion than the material that mainly forms the plate-shaped portion, has a cooling function, and is formed in a plate shape, A joint portion is placed between the plate-like portion and the base portion, contains an adhesive, and joins the plate-like portion and the base portion. Equipped with, The joint is composed of a composite material including the adhesive and an inorganic filler, which is a general term for fillers consisting of inorganic materials contained in the joint. When the inorganic filler is viewed in cross-section, let R1 be the radius of the inscribed circle and R2 be the radius of the circumscribed circle. If the average value of R2 / R1 is 1.10 or greater, The holding device is characterized in that, when a tensile test is performed on the joint, the shear stress at which the length of deformation due to shear force becomes 0.5 mm is 0.1 MPa or more and 3.5 MPa or less.

2. A holding device for holding an object, A plate-like portion formed in a plate shape, A base portion that supports the plate-shaped portion, is mainly made of a material having a different coefficient of thermal expansion than the material that mainly forms the plate-shaped portion, has a cooling function, and is formed in a plate shape, A joint portion is placed between the plate-like portion and the base portion, contains an adhesive, and joins the plate-like portion and the base portion. Equipped with, The joint is composed of a composite material including the adhesive and an inorganic filler, which is a general term for fillers consisting of inorganic materials contained in the joint. When the inorganic filler is viewed in cross-section, let R1 be the radius of the inscribed circle and R2 be the radius of the circumscribed circle. If the average value of R2 / R1 is 1.10 or greater, A holding device characterized in that the rigidity modulus of the joint is 0.03 MPa or more and 2.0 MPa or less.

3. A holding device for holding an object, A plate-like portion formed in a plate shape, A base portion that supports the plate-shaped portion, is mainly made of a material having a different coefficient of thermal expansion than the material that mainly forms the plate-shaped portion, has a cooling function, and is formed in a plate shape, A joint portion is placed between the plate-like portion and the base portion, contains an adhesive, and joins the plate-like portion and the base portion. Equipped with, The joint is composed of a composite material including the adhesive and an inorganic filler, which is a general term for fillers consisting of inorganic materials contained in the joint. When the inorganic filler is viewed in cross-section, let R1 be the radius of the inscribed circle and R2 be the radius of the circumscribed circle. If the average value of R2 / R1 is 1.10 or greater, The thickness of the joint is 0.05 mm or more and 1.00 mm or less. A holding device characterized in that the average particle size of the inorganic filler is 10 μm or less.

4. A holding device for holding an object, A plate-like portion formed in a plate shape, A base portion that supports the plate-shaped portion, is mainly made of a material having a different coefficient of thermal expansion than the material that mainly forms the plate-shaped portion, has a cooling function, and is formed in a plate shape, A joint portion is disposed between the plate-like portion and the base portion, contains an adhesive, and consists of a single layer that joins the plate-like portion and the base portion. Equipped with, The joint is composed of a composite material including the adhesive and an inorganic filler, which is a general term for fillers consisting of inorganic materials contained in the joint. When the inorganic filler is viewed in cross-section, let R1 be the radius of the inscribed circle and R2 be the radius of the circumscribed circle. If the average value of R2 / R1 is 1.10 or greater, A holding device characterized in that the content of the inorganic filler in the joint exceeds 20 vol%.

5. A holding device according to any one of claims 1 to 4, The thermal conductivity of the aforementioned joint is 0.90 W / mK or higher. holding device.

6. A holding device according to any one of claims 1 to 5, The thermal resistance of the aforementioned joint is 5.0 × 10 －４ (m ２ It is less than or equal to kW, The maximum shear stress strain at the joint is characterized by being 0.5 mm or more. holding device.

7. A holding device according to any one of claims 1 to 6, The adhesive is characterized by being a crosslinked vinyl-terminated polydimethylsiloxane consisting of a bifunctional D unit having two organic substituents and a monofunctional M unit having three organic substituents. holding device.

8. A holding device according to any one of claims 1 to 7, The plate-like portion is characterized in that it is mainly composed of ceramic, includes an adsorption electrode for holding the object, and does not include a heater electrode for heating the object. holding device.

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