TWI220294B - A triple gate oxide process with high-k gate dielectric - Google Patents

A triple gate oxide process with high-k gate dielectric Download PDF

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TWI220294B
TWI220294B TW092120331A TW92120331A TWI220294B TW I220294 B TWI220294 B TW I220294B TW 092120331 A TW092120331 A TW 092120331A TW 92120331 A TW92120331 A TW 92120331A TW I220294 B TWI220294 B TW I220294B
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Chew-Hoe Ang
Wenhe Lin
Jia-Zhen Zheng
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Chartered Semiconductor Mfg
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    • H01L21/823857Complementary field-effect transistors, e.g. CMOS with a particular manufacturing method of the gate insulating layers, e.g. different gate insulating layer thicknesses, particular gate insulator materials or particular gate insulator implants
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    • H01L29/40Electrodes ; Multistep manufacturing processes therefor
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Description

1220294
導體元件製造,特 厚度及組成的閘極 五、發明說明(1) (一)、【發明所屬之技術領域】 本發明係有關於一種積體電路中半 別是關於一種需要形成具有多種程度的 氧化層之電路。 (二)、【先前技術】 積體電路的製造通常需要形成具有多 組成的介電層,諸如問極氧化層。多種厚; 包含有操作不同電壓上的半導體元件的例+,例::古: 壓功率電晶體且使用於程式化EEPR0M元件 = 比使用於DRAM中記體儲存阿的柄雷厭+ 电日曰體 需要 介展u :: 電壓電晶體厚的閘極氧 化層。再者,隨者元件尺寸持續減少,需要排除如 MOSFETs中短通道效應問題的超薄閘極 / :有=電常f高介電常數介電質)的介電曰質/料最所好形以 因此’不僅須使製程能夠形成具有多種厚度的介電 = 供形成具有不同成份的介電質。顯然地,此 類製程亦應有效且產生具有完整性及品f佳的介電層。 ,Hattangady等(美國專利第5, 97〇, 345號)係教 用於,體電路中兩種不同厚度的氧化層之方法,其係 二氧:壓刪電晶體,此方法包含有形成一第 氧化:’ -犧牲層係覆蓋於第一氧化層上且局部地移 :’以谷許將第一氧化層選擇性餘刻掉’然後一第-氧化 牲層上,藉以犧牲層藉由ί 一 第一氧化層區上的犧 猎由第一氧化層所消耗掉,且殘留的 1220294 五、發明說明(2) _ 第一氧化層及第二氧化屉脔 度。 $ &化層覆盍於其上’以形成一層雙厚 有-===(厂美國專利第6’147’ _號)係教導 ?二種不同厚度的閘極氧化層之方法 種形成具 :,卜個希望的最大厚度,想要留下該層的沈” 阻所覆蓋’-氮離子植人係進人到基板的小-光 化層未被覆蓋的部#,離子植入 二::: =二且Ϊ變第二氧化層的厚度。然後移 層?: 第一氧化層,然後成長-第二較薄 第m有-個㈣部份覆蓋於氮離子植入區。 Song(美國專利第6, 191,〇49β1號)亦教導一種 有夕種厚度的閘極氧化層之方法,其係、 :基板,以修改在植入區上基板的氧化物成子 方法教導使用兩種不同離子種類,氮 度及濃度,藉此方法而獲致隨後不同氧化層成長
Zhong等(美國專利第6,268,25 1Β1號)係教導一種製造 多種厚度閘極氧化層之方法,根據此方法,一第一戶 氧化層係沈積達到一第一厚度,其係為氧化層的厚^。一 層多晶矽層係沈積覆蓋於第一閘極氧化層上且將其$坦 化,罩幕多晶矽層,以定義一必要的第二氧化層厚度‘, 然後蝕刻掉基板表面的多晶矽及在其下的氧化層,然後成 長一第二氧化層於基板的暴露表面上,以達到一第:厚 度,其係為中間厚度,一第二多晶矽層沈積於第二氧化層 上且平坦化之。在此之前,罩幕多晶矽,以定義出第三氧
第7頁 1220294 五、發明說明(3) 化層厚度的區,其係最薄的。然後重覆先前的步驟。 隨著元件尺寸變得愈來愈小,使用選擇性離子植人以 控制閘極氧化層成長(如上述所教導的習用技藝一樣),將 會變得不切實際的。此外,希望儘可能的形成多種厚度 層,其係需要將製造步驟減到最小。本發明達到無須離子 植入的目的,且具有比那些習用技藝較少數量及較簡單製 程步驟。 發明内容 第一目的,係在於提供 介電層簡單且有效之方 第二目的,係在於提供 的介電層簡單且有效之 第三目的,係在於提供 的介電層簡單且有效之 "電常數材料,且中間 常數材料形成於一層氧 第四目的,係在於提供 及組成的M0S/CM0S元件 方法, 及最大 化矽上 簡單且 :二,供一種在半導體基板上形成 且有效之方法,其係包括 二 '曰覆蓋於一半導體基板上;2)移除 一部份,以暴露出其下半導體基板表、 本發明之一 形成多種厚度的 本發明之一 上形成多種組成 本發明之一 上形成多種組成 層係為單層的高 括有一層高介電 本發明之一 閘極氧化層厚度 法。 根據本發明 多種厚度及組成 有·· 1 )形成一第 其下的該層的
一種半導體基板上 法 一種在半導體基板 方法。 一種在半導體基板 其中最薄的 的層係為包 的雙層。 一種形成需要多種 有效之方
第8頁 1220294 五、發明說明(4) :品3)形成一第二、較薄的介電層覆蓋藉以 表面上,· 4)移除第一介電声的一笫一 恭路出的基板 的半導體矣而.^電曰的第一°卩伤,以暴露出其下 第二介電i當:成一第三全面性地介電層覆蓋於 丨H第一介電層的殘留部份、及暴露的半導體某 f上。在本發明的較佳實施例中,第一 ^ 氧化石夕声ς 1· η、 Ό咕一 禾一"電層係為 數居。=$ 一2i,第二全面性地介電層係為一高介電常 m 一^的指出,一層閘極電極材料(如摻雜夕曰 ⑽S元件的閘極介電f。電層且將其圖案化’以形顧0S或 (四)、【實施方式】 方法本ϊ:ΐ:較佳實施例係提供一種形成閘極介電層之 八係匕括有一單獨形成高介電常數材料的最薄Α
Li:成高介電常數材料覆蓋於-層氧化石夕上ίΐ:及 最大厚度的部份。 』τ間及 2參閱第1圖’係顯示一半導體基板1〇之橫 圖’其中在此實施例中係為—Si基板,其上覆蓋、面 介電層12,在此實施例中係為一層氧 一 的厚度係獅侧埃之㈤。 減州叫,氧切 :著,參閱第2圖’係顯示第一氧化層12的殘 展:係由一層抗蝕刻材料14所覆蓋’在此實施例中传A :層光阻,已圖案化光阻(未顯示於圖中)進人 1為 J,以使-部份的第一氧化層移除掉,且暴露出其2 板10的表面16。氧化層的蝕刻可藉由一 、 基 刻或已知習用技藝的其他方法而達成。 虱氟酸)蝕 1220294 五、發明說明(5) 接著,參閱第3圖,係顯示第2圖的费、▲ 層介電質材料1 8 (在此實施例中第二声A 其中一第二 成覆蓋於基板10表面16上’其係藉由〜多除二氧,石”係形 *來。氧化層係形成達到一個厚度在10到:‘而f露 的光阻“(第2圖中)藉由化學光阻 殘留 同習用技藝的方法。 溉已移除掉,如 接著,參閱第4圖,係顯示第3圖的製 層抗钱刻材料(#此實施例亦為一層光阻2〇)已开;^第二 第二氧化層18及第-氧化層12上 阻覆盍於 ^的第-氧化層12移除掉,且暴露 光阻2〇 (第4圖中)第藉Ξ化ΐ i: : f造件其中殘留的 4圖相同)而被已移除掉 去除(所有兀件符號均與第 接著,#閱第6圖’係顯示第5圖的製造其
Dn第三介電層24已形成覆蓋;-板10表面上22,第三介電層係為 及基 以,屬氧化物者為較佳,諸如,、ΖΓ〇2、;η〇係 kO3、或其矽酸鹽或鋁酸鹽’此層可^ = (A^VD)或⑽金屬的熱氧化處❹沈積。此層氧^目 ε〇χ/ )T-k)c ^ t ε〇χ^ ;h.^w :氧化物及同"電常數介電質的電容率,且了係 》;, 最好介於4到15埃之間。從圖式可知,已形成介電層;产 的三個區:“ ’《具有中間介電層厚度;區B,其具有度最 im 第10頁 1220294 五、發明說明(6) =電層厚度;及區C,具有最小介電層厚度。區c僅包含 有同介電常數介電質24,而區A及6具有高介 24分別覆蓋於氧化層18及12上。 电申数’丨罨質 於?ίιί閱第7圖,係為三個區Α、β、及C的圖案化部 伤,母個具有一閘極電極(係以多晶矽為較佳)形成於1上 30,其預期形成MOS或CMOS元件覆蓋於適當介電質厚度的 區上若夕B曰矽使用作為閘極電極材料時,可藉由低壓化 學氣相沈積(LPCVD)而形成一層覆蓋於介電層上,適當的 圖案化及蝕刻將會導致第7圖的最後配置。若閘極電極係 為金屬時,則如TiN、TaN、或Mo的金屬將會藉由物理氣相 沈積而沈積,且在類比方式蝕刻處圖案化。 以上所述係藉由貫施例說明本發明之特點,其目的在 使熟習該技術者能暸解本發明之内容並據以實施,、而非限 定本發明之專利範圍,故,凡其他未脫離本發明所揭示之 精神所完成之等效修飾或修改,仍應包含在形成製造M〇s 及CMOS兀件的本發明方法中,且同時提供一種製造M〇s及 CMOS元件之方法。 圖號簡單說明 10 半導體基板 12 第一介電層 層 第11頁 1220294 五、發明說明(7) 20 光阻 22 表面 24 第三介電層 30 閘極電極 1220294 圖式簡單說明 、________
第13頁 底下藉由具體實施例配合所附的圖 =易瞭解本發明…、技術内容、當更 效。 行點及其所達成之功 圖式說明: 同厚Ϊ 2到第6圖係為橫剖面圖,係說明形成具有=葙 t及:俜成::成電二二^ ^ … 閑極電極形成於其成上,•電層之橫剖面圖,係圖案化且具有 _

Claims (1)

1220294 六、申請專利範圍 形成一具有多種厚度及組成的介 種在一半導體基板上 電層之方法,係包括有·· 提供一半導體基板; 具電層覆蓋於該基板上,其該第-介電層係 一第一層抗蝕刻材料覆蓋於該第一介電 ; 電層的第一區刻材料,以暴露出該第-介 藉由一银刻步驟移除該第一介 露其下的-第-基板表面L 的第—區,且糟以暴 ,移除抗蝕刻材料的殘留第一層; 第二介電層’其係在該第二基板表面區上具有一第 第!材料的-全面性第二層覆蓋於第二介電層及 移除一部份的抗姓刻材料的第 電層的第二區; 从暴路出該第一介 區,且藉以暴 藉由一蝕刻步驟移除該第一介 露其下的一第二基板表面區; ㈢的第 移除抗蝕刻材料第二層的殘留部份;及 厚度 形成一全面性第三層的介電材料,1 覆蓋於第二介電層、第-介電層的殘留in… 表面區上。 p份、及第二基板 2.如申請專利範圍第1項所述之方法,其中該基板係為〜 Ml 第14頁 ^^0294 六、申請專利範圍 - 矽基板。 3·如申請專利範圍第2項所述之方法,其中第一及第二介 電層係為S i 〇2層。 4·如申請專利範圍第3項所述之方法,其中該抗蝕刻材料 為光阻。 5·如申請專利範圍第3項所述之方法,其中該蝕刻為一HF 溼式蝕刻。 6·如申請專利範圍第3項所述之方法,其中該第一厚度係 介於30到60埃之間。
7·如申請專利範圍第3項所述之方法,其中該第二厚度係 介於10到30埃之間。 8·如申請專利範圍第3項所述之方法,其中該第三介電層 係為一層高介電常數介電質材料。 9·如申請專利範圍第8項所述之方法,其中該高介電常數 介電質材料係選自於包含有下列:Hf02、Zr02、Ti02、 1〇1 2 3、及LaO3、及其矽酸鹽及鋁酸鹽所組成的組群之一 〇
1 0·如申請專利範圍第8項所述之方法,其中該第三厚度係 2 為’該高介電常數介電質材料具有在4到15埃間的氧當 量厚度的厚度。 3 11·如申請專利範圍第10項所述之方法,其中該高介電常 數介電質係藉由原子層化學氣相沈積(ALCVD)而沈積。 12·如申請專利範圍第iq項所述之方法,其中該高介電常 數介電質係藉由濺鍍金屬的熱氧化處理而形成。 U20294 六、申請專利範圍 13· —種在一半導體基板上形成具有多種厚户 極介電層MOS或CMOS元件之方法,係包括f .,’成的閉 層 提供一半導體基板,其係、具有多種厚度及組成的—介電 形成一層閘極電極材料於該介電層上; 圖案化及蝕刻該閘極電極材料及介電層之下,以製造 依閘極電極厚度所需的元件。 W出 14,如申請專利範圍第13項所述之方法,其中該基板 一矽基板。 、馬 15·如申請專利範圍第14項所述之方法,其中第一及第二 介電層係為Si〇2層。 一 16·如申請專利範圍第15項所述之方法,其中第一厚度係 介於30到60埃之間。 a、 1 7·如申請專利範圍第1 5項所述之方法,其中第二厚度係 介於10到30埃之間。 、 18·如申請專利範圍第15項所述之方法,其中該第三介電 層係為一層高介電常數介電質材料。
六、申請專利範圍 22如^申^/專係^错^•由@原@子層化學氣相沈積ULCVD)而沈積。 23如二直係猎由錢鑛金屬的熱氧化處理而形成、 極材二a利範圍第1 3項所述之方法,其中該層閘極電 極材枓係為一層多晶矽。 24.如中請專利範圍第23項所述之m中該多晶石夕係 籍由低壓化學氣相沈積(LPCVD)而沈積。 25枭如申:青/專利範圍第23項所述之方法,其中該層閘極電 f材料係為一層選自於包含有TiN、TaN、及M〇所組成的 金屬之一。 26·如申請專利範圍第25項所述之方法、其中該金屬係藉 由物理氣相沈積(PVD)而沈積。
第17頁
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