CN102138233A - 相变记忆材料 - Google Patents
相变记忆材料 Download PDFInfo
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- 238000000034 method Methods 0.000 claims description 24
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- 229910052782 aluminium Inorganic materials 0.000 claims description 20
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- 229910052733 gallium Inorganic materials 0.000 claims description 20
- 229910052738 indium Inorganic materials 0.000 claims description 20
- 229910052745 lead Inorganic materials 0.000 claims description 20
- 229910052711 selenium Inorganic materials 0.000 claims description 20
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Abstract
本文描述了相变记忆材料,更具体描述了适用于相变记忆应用,如光学和电子数据存储的碲化GeAs材料。
Description
优先权声明
本专利申请要求2008年8月29日提交的美国临时专利申请61/092868和2009年7月15日提交的美国专利申请第12/503156号的优先权。
发明领域
本发明的实施方式涉及相变记忆材料,更具体涉及适用于相变记忆应用如光学和电子数据存储的碲化GeAs材料。
技术背景
传统的相变记忆装置利用能在具有不同性质的两相之间变化的材料。所述材料通常从无定型相变化到晶相,所述相可具有显著不同的性质,例如不同的电阻率、电导率和/或反射率。
从无定型相到晶相的相变可以通过将无定型材料加热到促进成核、晶体形成、和然后晶化的温度来实现。回到无定型的相变可以通过将晶相加热到高于熔化温度的温度来实现。
目前将硫属化物材料如Ge、Sb和Te合金用于相变记忆应用中,例如用于在可重复写入盘(over writable disk)中存储信息。
目前Matsushita/Panasonic和IBM的工作人员已经开发了一些相变记忆材料。代表性的材料包括基于GeTe-Sb2Te3结的组合物,具体是Ge2Sb2Te5(GST),以及Au,In-掺杂的碲化Sb(AIST)。这些材料可以在激光加热或电流脉冲条件下,在高电导率、高反射率晶相和低电导率、低反射率无定型相之间以约10纳秒的时间标度循环。
虽然一些传统材料如GST和AIST对于非挥发性记忆应用具有良好的性质,但是具有更快的相转变和/或更长的写入/再写入能力的相变记忆材料将是有利的。
发明概述
本发明的实施方式是基于GeAsTe的用于相变记忆应用但是不属于正规GeSbTe系统的组合物。而且,由于某些GeAsTe组合物可以制成大块玻璃,所以GeAsTe无定型相的稳定性似乎大于不可能形成大块玻璃的GeSbTe类似物的稳定性。这种特性会导致写入/再写入循环的数量增加,同时不会使电导率/反射率反差(contrast)变差,并且导致更长的数据保持。
本发明的一种实施方式是包括晶化薄膜的制品,该晶化薄膜包含具有至少一种六方相的组合物,或者包含在晶化形式下能具有至少一种六方相的可晶化组合物。
本发明的另一种实施方式是一种方法,该方法包括提供包含相变记忆无定型材料的薄膜,以及将该相变记忆无定型材料转化成六方晶相。
本发明的另一种实施方式是一种方法,该方法包括提供包含具有六方晶相的相变记忆材料的薄膜,以及将该六方晶相转化成无定型相。
以下详细说明中将提出本发明的其他特性和优点,这些特性和优点的一部分对于本领域普通技术人员而言通过说明书而容易理解,或者通过按照说明书及其权利要求书以及附图所述实施本发明而了解。
应该理解,以上一般说明和以下详细说明都仅仅是对本发明的例证,意图为理解本发明要求权利的性质和特性提供概况或框架。
包括附图以提供对本发明的进一步理解,附图结合在说明书中并构成说明书的一部分。附图说明本发明的一种或多种实施方式,与说明书一起用于解释本发明的原理和操作。
附图简要描述
单独地通过以下详细说明,或者结合附图,能理解本发明。
图1是GeAsTe材料的组成图。
图2是根据一种实施方式的材料的反射率数据的图。
图3是根据一种实施方式的材料的反射率数据的图。
图4和图5是常规相变记忆材料的X射线衍射数据的图。
图6和图7是根据本发明的相变记忆材料的X射线衍射数据的图。
发明详述
详细参考本发明的各种实施方式。只要有可能,在所有附图中使用相同的附图标记表示相同或类似的特征。
本发明的一种实施方式是包括晶化薄膜的制品,该晶化薄膜包含具有至少一种六方相的组合物,或者包含在晶化形式下能具有至少一种六方相的可晶化组合物。
根据一些实施方式,所述组合物按原子百分比包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
根据一些实施方式,所述组合物按原子百分比包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
所述组合物可进一步包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。在一些实施方式中,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。在一些实施方式中,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于15%。
根据一种实施方式,所述薄膜设置于基材之上。根据一种实施方式,可以将薄膜沉积在基材上。在一些实施方式中,所述基材包括玻璃、玻璃陶瓷、陶瓷、聚合物、金属或其组合。
GeAsTe玻璃及其晶体类似物有可能作为相变材料,因为其特征性的玻璃态比传统相变材料如GST和AIST的玻璃态更稳定。根据本发明的宽范围的GeAsTe玻璃在加热时能转化成反射率大于上述传统材料的晶相。对于基于Te-GeAs2结的玻璃,已经在包含45-65的原子百分比的Te的组合物中证明了这种现象。许多这些材料在晶化时由至少两种相组成:要么是两种晶相,要么是一种晶相加上一种残留的玻璃相。
但是,可以使具有基于As2Te3-GeTe结的组成的玻璃晶化成单相,这样能在玻璃态和晶态之间表现出最大的电导率/反射率反差。这种玻璃可以掺杂有与晶相相容的成分,例如Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合,而不会在加热态中形成第二相。
GeAs2Te4 | Ge2As2Te5 | Ge3As2Te6 | GeAs1.2Sb0.8Te4 | GeAs1.4Sb0.6Te4 | GeAs1.9Bi0.1Te4 | |
Ge | 14.3 | 22.2 | 27.3 | 14.3 | 14.3 | 14.3 |
As | 28.6 | 22.2 | 18.2 | 17.1 | 20 | 27.1 |
Sb | - | - | - | 11.4 | 8.57 | - |
Bi | - | - | - | - | - | 1.43 |
Te | 57.1 | 55.6 | 54.5 | 57.1 | 57.1 | 57.1 |
Si | ||||||
Ga | ||||||
In | ||||||
P |
表1.
Ge0.9Si0.1As2Te4 | Ge0.9Ga0.05P0.05As2Te4 | Ge0.9In0.05P0.05As2Te4 |
Ge | 13.21 | 13.21 | 13.21 |
As | 28.57 | 28.57 | 28.57 |
Sb | - | - | - |
Bi | - | - | - |
Te | 57.14 | 57.14 | 57.14 |
Si | 1.07 | - | - |
Ga | - | 0.54 | - |
In | - | - | 0.54 |
P | - | 0.54 | 0.54 |
表2.
根据本发明的示例性组合物如表1和表2中所示。
本发明的另一种实施方式是一种方法,该方法包括提供包含相变记忆无定型材料的薄膜,以及将该相变记忆无定型材料转化成六方晶相。
从无定型相到六方晶相的相变可以通过将无定型材料加热到能促进成核、晶体形成、和然后晶化的温度来实现。
将相变记忆无定型材料转化成六方晶相的步骤可包括加热。可利用等温加热,例如利用电阻和/或感应加热的电加热;激光加热等加热薄膜以引发相变。
根据一种实施方式,相变记忆无定型材料按原子百分比包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
根据一些实施方式的相变记忆无定型材料按原子百分比包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
相变记忆无定型材料可进一步包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。在一些实施方式中,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于15%。
本发明的另一种实施方式是一种方法,该方法包括提供包含具有六方晶相的相变记忆材料的薄膜,以及将该六方晶相转化成无定型相。
这种到无定型相的相变可以通过在将晶像加热到高于相变记忆材料的熔化温度来实现。
在一些实施方式中,将具有六方晶相的相变记忆材料转化成无定型相的步骤包括加热。可以利用等温加热,例如利用电阻和/或感应加热的电加热;激光加热等,加热薄膜以引发相变。
根据一些实施方式,相变记忆材料按原子百分比包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
根据一些实施方式,相变记忆材料按原子百分比包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
相变记忆材料可进一步包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。在一些实施方式中,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比可等于或小于15%。
可以对大块的GeAsTe玻璃(如图1中实心圆10所示)进行热结晶,形成高反射率的相或相集合。在玻璃具有基于As2Te3-GeTe结12的组成的情况中,这种相是混合的层配混物的一种同质(homologous)系列,该配混物可以用下式表示:As2Te3(GeTe)n,其中n是整数。例如,对于圆14表示的材料,Ge∶As的比例为1∶2时,这种相是GeAs2Te4,即n=1。这些大块玻璃可以用安瓿熔融的硫属化物玻璃加工技术制备。
对于固态记忆中的应用,这些材料以薄膜形式使用。可以通过各种技术制造薄膜,例如磁控管溅射、热蒸发和脉冲激光沉积。可以将这些薄膜沉积在基材上,可以用于相变记忆装置中。
根据一种实施方式,薄膜厚度等于或小于2微米,例如等于或小于1微米,例如等于或小于0.5微米。在一些实施方式中,薄膜的厚度为20纳米至1微米,例如40纳米至1微米,例如50纳米至1微米。虽然指出了具体的范围,但是在其他实施方式中,厚度可以是这些范围内包括小数点的任意数值。
实施例
利用脉冲激光沉积,使用248纳米准分子激光源和高真空(10-6托)沉积室,将GeAs2Te4薄膜(在这个实施例中是Ge14.3As28.6Te57.1)沉积在Eagle XGTM玻璃基材上。从靶烧蚀,用9000-36000个脉冲制备基本连续的薄膜。随后将薄膜制品的一部分以250℃在空气中加热10-180分钟(对加热处理温度进行选择,与通过差示扫描量热法在10℃/分钟加热速率下测得的大块玻璃的峰值结晶温度一致)。
目视检查加热的制品显示反射率增大。通过定量数据证实这种观察结果,如图2中所示,随着加热时间增加到10分钟以上,500-700纳米的反射率从约40%增大到约60%。通过观察刚沉积的制品(线16),加热了10分钟的制品(线18),加热了30分钟的制品(线20),加热了60分钟的制品(线24)和加热了180分钟的制品(线22),反射率的增大很明显。
入射余角X射线衍射证明,对于加热了等于或大于30分钟的样品,由于GeAs2Te4的晶化,反射率增大。
对于常规相变记忆材料(GeSb2Te4和GeAsSbTe4)的X射线衍射数据分别如图4和图5中所示。这些膜的晶化形式中的相是立方相。这是根据只存在d-间距值在3.5、3.1、2.1和1.7埃附近的四个峰而推断的;这是所谓“岩盐”或NaCl结构的特征。
根据本发明的材料(GeAs2Te4和GeAs1.9Bi0.1Te4)的X射线衍射数据分别如图6和图7中所示。与立方材料相比峰的数量增加证明根据本发明的材料包含六方晶相。发现表1中所示另外的组合物的X射线衍射数据的峰与六方晶相一致。
将另外的Ge14.3As28.6Te57.1薄膜制品以350℃在空气中加热1-10分钟。这些制品的反射率数据如图3中所示。通过观察刚沉积的制品(线26),加热了1分钟的制品(线28),加热了5分钟的制品(线30)和加热了10分钟的制品(线32),反射率的增大很明显。
对于表1中所述的组合物重复这种方法,得到类似的结果。对于从具有基于As2Te3-GeTe结的组成的样品得到的其他薄膜,以及对于从具有以下大致组成(按原子百分比:5-45%的Ge,5-40%的As,和45-65%的Te)的其他GeAsTe玻璃,以及对于还包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的组合物,预期也获得类似的结果。这些额外的修改不应使这些材料的相变特性变差。
对于本领域普通技术人员显而易见的是,可以在不偏离本发明精神或范围的条件下对本发明进行各种修改和变化。因此,本发明涵盖这些修改和变化,前提是这些修改和变化落在所附权利要求及其等同项的范围之内。
Claims (23)
1.一种制品,其包含:
a.包含具有至少一种六方相的组合物的晶化薄膜;或
b.在晶化形式下能具有至少一种六方相的可晶化组合物。
2.如权利要求1所述的制品,其特征在于,按原子百分比计,所述组合物包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
3.如权利要求2所述的制品,其特征在于,按原子百分比计,所述组合物包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
4.如权利要求2所述的制品,其特征在于,所述组合物进一步包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。
5.如权利要求2所述的制品,其特征在于,所述Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。
6.如权利要求5所述的制品,其特征在于,所述Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于15%。
7.如权利要求1所述的制品,其特征在于,所述薄膜设置在基材上。
8.如权利要求7所述的制品,其特征在于,所述基材包括玻璃、玻璃陶瓷、陶瓷、聚合物、金属或其组合。
9.一种方法,其包括:
提供包含相变记忆无定型材料的薄膜;和
将该相变记忆无定型材料转化成六方晶相。
10.如权利要求9所述的方法,其特征在于,所述将相变记忆无定型材料转化成六方晶相的步骤包括加热。
11.如权利要求10所述的方法,其特征在于,按原子百分比计,所述相变记忆无定型材料包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
12.如权利要求11所述的方法,其特征在于,按原子百分比计,所述组合物包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
13.如权利要求11所述的方法,其特征在于,所述相变记忆无定型材料还包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。
14.如权利要求13所述的方法,其特征在于,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。
15.如权利要求13所述的方法,其特征在于,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于15%。
16.一种方法,其包括:
提供包含具有六方晶相的相变记忆材料的薄膜;和
将该六方晶相转化成无定型相。
17.如权利要求16所述的方法,其特征在于,所述将具有六方晶相的相变记忆材料转化成无定型相的步骤包括加热。
18.如权利要求16所述的方法,其特征在于,按原子百分比计,所述相变记忆材料包含以下组分:
5-45的Ge;
5-40的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
45-65的Te。
19.如权利要求18所述的方法,其特征在于,按原子百分比计,所述相变记忆材料包含以下组分:
10-30的Ge;
15-30的As,或者As和Sb的组合,其中As的原子百分比大于Sb的原子百分比;和
50-60的Te。
20.如权利要求18所述的方法,其特征在于,所述相变记忆材料还包含Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合。
21.如权利要求20所述的方法,其特征在于,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于20%。
22.如权利要求21所述的方法,其特征在于,Al、Si、Ga、Se、In、Sn、Tl、Pb、Bi、P、S或其组合的原子百分比等于或小于15%。
23.一种相变记忆装置,其包括如权利要求1所述的制品。
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