5-1 結論
根據前四章的論敘、模擬與討論的內容,將結論歸納如下:
1. 本文利用有限元素模擬BTO
[ ]
100 /CFO[ ]
100 纖維複合材料之結果,與Li 和 Dunn[1]使用 Mori-Tanaka 模式模擬BTO[ ]
001/CFO[ ]
001之結果相吻合,並 利用相同的有限元素設定模式,模擬不同極化方向組合之壓電壓磁複合材 料。2. 從
BTO
/CFO
模擬結果得知,BTO[ ]
001/CFO[ ]
001 最佳之磁電電壓係數α E * , 11
為-0.0306V/cmOe 與α * E , 33
為1.1494V/cmOe。經由改變材料之極化方向,得到最佳之
α E * , 11
為-1.3441V/cmOe 與α * E , 33
為-5.8250V/cmOe,因此α E * , 11
提升約 44 倍而α * E , 33
提升約5 倍。3. 從
CFO
/BTO
模擬結果得知,CFO[ ]
001/BTO[ ]
001 最佳之磁電電壓係數α E * , 11
為-0.0244V/cmOe 與α * E , 33
為1.2288V/cmOe。經由改變材料之極化方向,得到最佳之
α E * , 11
為-2.4823V/cmOe 與α * E , 33
為-6.2357V/cmOe,因此α E * , 11
提升約101 倍而
α * E , 33
提升約5 倍。4. 由正方形單位晶胞與正六邊形單位晶胞模擬結果得知,正六邊形單位晶胞較 接近Mori-Tanaka 模式之結果。其原因是因為 Mori-Tanaka 模式可以模擬從 f = 0 到 f = 1 的複合材料,因此可以模擬內含物較緊密之複合材料。正六邊 形單位晶胞亦屬於緊密堆積,它的體積比最大約f = 0.906,而正方形單位晶 胞的體積比僅能達至f = 0.785。所以正六邊形單位晶胞的結果會比正方形單 位晶胞來的接近Mori-Tanaka 模式。
95
5-2 未來展望
1. 本文是選擇圓形纖維結構為模擬對象,然而,實驗中多數採用方形纖維結構 置入母材內,觀察磊晶薄膜生長情形並量測其磁電耦合效應。因此未來為了 符合實驗的幾何形狀,可利用本文之有限元素模型模擬方形纖維複合材料之 行為。
2. 複合材料除了纖維結構的複合形式外,還有層板結構與顆粒結構複合形式。
因此未來可探討材料極化方向對壓電壓磁顆粒複合材料磁電耦合行為之影 響,並求其最佳之磁電耦合效應。
96
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101
附錄 A BiFeO 3 置入 CoFe 2 O 4 之壓電壓磁複合材料
近年來由於技術提升,Oh[51]提出鐵電磁材料 BiFeO 3
可於其他方向之晶面 得到比以往更佳之磁電耦合效應。為了得知BFO/CFO 複合材料之耦合效應,本 文將探討BFO[ ]
001 和BFO[ ]
100 材料性質所產生的磁電耦合效應。利用BFO[ ]
001 材料性質(表 A-1),可得到BFO[ ]
001和BFO[ ]
100 的矩陣形式。另外本文模擬 BFO/CFO 複合材料於任意極化方向之最佳磁電耦合效應,其結果可作為實驗參 考基準。表A-1 BFO 材料性質[39, 52]
BiFeO
3
) GPa
11 (
C 298 e
33( C m
2) − 979 . 88
) GPa
12 (
C 162 κ
11( C
2Nm
2) 8 . 85 × 10
−10)
GPa
44 (
C 62 q
15( m A ) 0
) m C
(
2e
154 . 058 q
31( m A ) 0
) m C
(
2e
16− 40 . 582 q
33( m A ) 0
) m C
(
2e
2240 . 582 μ
11( Ns
2C
2) 5 × 10
−6)
m C
(
2e
31− 503 . 04
102
BFO[001]
=L
BFO[100]
=L
103
會參與耦合的行為,而CFO 材料中的
q 15
項因應BFO 的耦合行為產生平均磁場H 1
,因此可以得到複合材料的λ ;如果提供平均電場* 11 E 2
,則BFO 中的e 項24
會參與耦合的行為,而CFO 材料中的q 項會因應 BFO 的耦合行為產生平均磁24
場H 2
,因此會得到複合材料的λ ;如果提供平均電場* 22 E 3
,則BFO 中的e 31
、e 32
、e 33
項會參與耦合的行為,而CFO 材料中的q 31
、q 32
、q 33
項會因應BFO 的 耦合行為產生平均磁場H 3
,因此會得到複合材料的λ * 33
。本節模擬得到BFO
[ ]
001 /CFO[ ]
001 之等效磁電電壓係數α * E
,其模擬結果(圖 A-2)發現當內含物的體積比 f = 0.67 時,在 MT 的曲線中得到α* E , 11
的極值-0.0116V/cmOe。由於材料晶格對稱性的關係,
λ * 22
會等於λ * 11
且κ * 22
會等於κ 11 *
, 所以α* E , 22
也會與αE * , 11
同在f = 0.67 時,極值達至-0.0116V/cmOe。當 f = 0.25,αE * , 33
達至極值1.4394V/cmOe,且α* E , 33
幾乎為一個定值;當f = 0.91,λ* 11
、λ* 22
極值約 -7.3969×10-12
Ns/VC。當 f = 0.50,λ* 33
極值約2.5327×10-7
Ns/VC。104
Volume Fraction of Inclusion C
*(P a)
Volume Fraction of Inclusion e
*(C /m
2)
Volume Fraction of Inclusion κ
*(C
2/Nm
2)
Volume Fraction of Inclusion q
*(N /A m )
Volume Fraction of Inclusion μ
*(N s
2/C
2)
Volume Fraction of Inclusion
λ
*(N s/ V C )
105
(a)
α * E , 11
,α E * , 22
(b)
α * E , 33
圖A-2
α * E
與 f 之關係0 0.2 0.4 0.6 0.8 1
-0.014 -0.012 -0.01 -0.008 -0.006 -0.004 -0.002 0
Volume Fraction of Inclusion
α E, * 11 , α * E, 2 2 ( V /c m O e)
MT SQU HEX
BFO[001]/CFO[001]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0 0.5 1 1.5
Volume Fraction of Inclusion
α * E, 3 3 ( V /c m O e)
MT SQU HEX
BFO[001]/CFO[001]
106
A-2
BFO[ ]
100 置入CFO[ ]
001根據(A.2)式得知BFO
[ ]
100 之性質,對BFO[ ]
100 /CFO[ ]
001 提供平均電場 E 時,則 BFO 中的1
e 、11
e 、12 e 13
項會參與耦合的行為,從(3.3)式知道 CFO 材 料性質中的q 31
、q 32
、q 33
項因應BFO 的耦合行為產生平均磁場 H ,因此得到3
複合材料的λ * 13
;如果提供平均電場 E ,則 BFO 中的2
e 、22 e 23
項會參與耦合的 行為,而CFO 材料中的q 32
、q 33
項會因應BFO 的耦合行為產生平均磁場 H ,3
因此得到複合材料的λ * 23
;如果提供平均電場 E ,則 BFO 中的3 e 34
、e 35
項會參 與耦合的行為,相對的CFO 中的q 、24 q 15
項因應BFO 的耦合行為產生平均磁場H 與
2
H ,因此得到複合材料的1 λ * 32
與λ * 31
。由等效性質(圖 A-3)得知BFO
[ ]
100 /CFO[ ]
001的λ * 13
於f = 0.44,極值達到 -6.2007×10-11
Ns/VC。λ * 23
於f = 0.44,極值達到 6.2007×10-10
Ns/VC。λ * 31
於f = 0.99,極值達到5.6523×10
-9
Ns/VC。λ * 32
於f = 0.86,極值達到-1.9761×10-9
Ns/VC。另外因為λ 、
* 11
λ 、* 22 λ * 33
數值小於λ * 13
、λ * 23
、λ * 31
與λ * 32
的10 − 3
倍,因此視作為 0。由於沒有λ * 11
、λ * 22
和λ* 33
等主對角線上的值,所以主對角線上的α * E
為0;而因 為κ13 *
、κ23 *
、κ31 *
和κ32 *
為0 所以α* E , 13
、α* E , 23
、αE * , 31
和αE * , 32
不存在。107
Volume Fraction of Inclusion C
*(P a)
Volume Fraction of Inclusion e
*(C /m
2)
Volume Fraction of Inclusion κ
*(C
2/Nm
2)
Volume Fraction of Inclusion q
*(N /A m )
Volume Fraction of Inclusion μ
*(N s
2/C
2)
Volume Fraction of Inclusion
λ
*(N s/ V C )
108
A-3
BFO[ ]
001置入CFO[ ]
100由(4.3)式知道CFO
[ ]
100 的材料性質,因此當對BFO[ ]
001/CFO[ ]
100 提供平 均電場 E ,則 BFO 中的1 e 15
、e 16
項會參與耦合的行為,而CFO 材料性質中的q 35
、q 26
項因應BFO 的耦合行為產生平均磁場 H 、3
H ,因此得到複合材料的2 λ * 13
與λ ;如果提供平均電場* 12
E ,則 BFO 中的2
e 、21
e 項會參與耦合的行為,而22
CFO 材料中的q 、11
q 項會因應 BFO 的耦合行為產生平均磁場12
H ,因此會得1
到複合材料的λ ;如果提供平均電場* 21
E ,則 BFO 中的3 e 31
、e 32
、e 33
項會參與 耦合的行為,相對的CFO 中的q 、11
q 、12 q 13
項因應BFO 的耦合行為產生平均磁 場 H ,因此得到複合材料的1 λ * 31
。由等效性質(圖 A-4)得知BFO
[ ]
001/CFO[ ]
100 的λ 於 f = 0.92,極值達到* 12
3.3276×10-10
Ns/VC。λ * 13
於f = 0.46,極值達到 1.9103×10-7
Ns/VC。λ 於 f = 0.91,* 21
極值達到2.6019×10-9
Ns/VC。λ * 31
於f = 0.92,極值達到-8.0830×10-10
Ns/VC。另外因為λ 、
* 11
λ 、* 22 λ * 33
數值小於λ 、* 12 λ 13 *
、λ 與* 21 λ * 31
的10 − 3
倍,所以視作為 0。由於沒有λ 11 *
、λ * 22
和λ* 33
等主對角線上的值,所以主對角線上的α * E
也為0; κ 、12 *
*
κ
13
、κ 和21 *
κ31 *
為0,所以αE * , 12
、α* E , 13
、αE * , 21
和αE * , 31
不存在。109
Volume Fraction of Inclusion C
*(P a)
Volume Fraction of Inclusion e
*(C /m
2)
Volume Fraction of Inclusion κ
*(C
2/Nm
2)
Volume Fraction of Inclusion q
*(N /A m )
Volume Fraction of Inclusion μ
*(N s
2/C
2)
Volume Fraction of Inclusion
λ
*(N s/ V C )
110
A-4
BFO[ ]
100 置入CFO[ ]
100由於在A-2 與 A-3 節得BFO
[ ]
100 /CFO[ ]
001與BFO[ ]
001/CFO[ ]
100 等複合材 料之耦合結果,因此當BFO 與 CFO 之極化方向皆是[ ]
100 的時候,則BFO[ ]
100 /[ ]
100CFO 將得到新的耦合性質。
當對BFO
[ ]
100 /CFO[ ]
100 提供平均電場 E ,則 BFO 中的1
e 、11
e 、12 e 13
項會 參與耦合的行為,而CFO 材料性質中的q 、11
q 、12 q 13
項因應BFO 的耦合行為產 生平均磁場 H ,因此得到複合材料的1
λ ;如果提供平均電場* 11
E ,則 BFO 中2
的e 26
項會參與耦合的行為,而CFO 材料中的q 26
項因應BFO 的耦合行為產生平 均磁場 H ,因此會得到複合材料的2
λ ;如果提供平均電場* 22
E ,則 BFO 中的3 e 35
項會參與耦合的行為,相對的CFO 中的q 35
項因應BFO 的耦合行為產生平均 磁場 H ,因此得到複合材料的3 λ * 33
。由等效性質(圖 A-5)得知BFO
[ ]
100 /CFO[ ]
100 的λ 、* 11
λ 、* 22 λ * 33
與κ 、11 *
κ 、22 * κ 33 *
, 也就可以得知αE * , 11
、α* E , 22
與αE * , 33
(圖 A-6)。當 f = 0.96 時,αE * , 11
極值達至0.4574V/cmOe。當 f = 0.51 時,α
E * , 22
極值達至-1.0089V/cmOe。當 f = 0.39 時,αE * , 33
極值達至-1.5657V/cmOe;當 f = 0.99 時,λ 11 *
極值達至3.9077×10-9
Ns/VC。當 f = 0.86 時,λ 極值達至-3.7611×10* 22 -10
Ns/VC。當 f = 0.53 時,λ * 33
極值達至-5.2171×10
-9
Ns/VC。111
Volume Fraction of Inclusion C
*(P a)
Volume Fraction of Inclusion e
*(C /m
2)
Volume Fraction of Inclusion κ
*(C
2/Nm
2)
Volume Fraction of Inclusion q
*(N /A m )
Volume Fraction of Inclusion μ
*(N s
2/C
2)
Volume Fraction of Inclusion
λ
*(N s/ V C )
112
(a)
α * E , 11
(b)
α E * , 22
(c)
α E * , 33
圖A-6
α * E
與 f 之關係0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-0.1 0 0.1 0.2 0.3 0.4 0.5 0.6
Volume Fraction of Inclusion α
* E,11( V /cm O e)
MT SQU HEX
BFO[100]/CFO[100]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0
Volume Fraction of Inclusion α
* E,22( V /cm O e)
MT SQU HEX
BFO[100]/CFO[100]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2
Volume Fraction of Inclusion α
* E,33( V /cm O e)
MT SQU HEX
BFO[100]/CFO[100]
113
A-5 結論
綜合 A-1 到 A-4 節之模擬,可將得到的磁電電壓係數、磁電耦合係數分別 整理成表A-2 與表 A-3。其中可發現當極化方向為
[ ]
100 時,等效磁電電壓係數* 11 ,
α
E
從-0.0116V/cmOe 提升到 0.4574V/cmOe,提升了 39 倍;等效磁電電壓係 數α* E , 22
從-0.0116V/cmOe 提升到-1.0089V/cmOe,提升了 86 倍;等效磁電電壓 係數αE * , 33
從1.4394V/cmOe 提升到-1.5657V/cmOe,提升了 1.08 倍。表A-2 極化方向與磁電電壓係數之關係
Ⅰ Ⅱ Ⅲ Ⅳ
(V/cmOe)
* 11 ,
α E
-0.0116( f = 0.67)
- - 0.4574
( f = 0.96)
(V/cmOe)
* 22 ,
α E
-0.0116( f = 0.67)
- -
-1.0089 ( f = 0.51)
(V/cmOe)
* 33 ,
α E
1.4394( f = 0.25)
- - -1.5657
( f = 0.39)
Ⅰ:BFO
[ ]
001 /CFO[ ]
001 Ⅱ:BFO[ ]
100 /CFO[ ]
001Ⅲ:BFO
[ ]
001 /CFO[ ]
100 Ⅳ:BFO[ ]
100 /CFO[ ]
100114
表A-3 極化方向與磁電耦合係數之關係
Ⅰ Ⅱ Ⅲ Ⅳ
) Ns/VC
* (
λ
11
-7.3969×10-12
( f = 0.91)- - 3.9077×10
-9
( f = 0.99)) Ns/VC
* (
2
λ
1
- - 3.3276×10-10
( f = 0.92)
-
) Ns/VC
*
(3
λ 1
- -6.2007×10-11
( f = 0.44)1.9103×10
-7
( f = 0.46)-
) Ns/VC
* (
λ
21
- - 2.6019×10-9
( f = 0.91)
-
) Ns/VC
* (
λ
22
-7.3969×10-12
( f = 0.91)- - -3.7611×10
-10
( f = 0.86)) Ns/VC
*
(λ 23
- 6.2007×10-10
( f = 0.44)- -
) Ns/VC
*
(λ 31
- 5.6523×10-9
( f = 0.99)-8.0830×10
-10
( f = 0.92)-
) Ns/VC
*
(λ 32
- -1.9761×10-9
( f = 0.86)- -
) Ns/VC
*
(λ 33
2.5327×10-7
( f = 0.50)
- - -5.2171×10
-9
( f = 0.53)Ⅰ:BFO
[ ]
001 /CFO[ ]
001 Ⅱ:BFO[ ]
100 /CFO[ ]
001Ⅲ:BFO
[ ]
001 /CFO[ ]
100 Ⅳ:BFO[ ]
100 /CFO[ ]
100115
附錄 B CoFe 2 O 4 置入 BiFeO 3 之壓電壓磁複合材料
B-1
CFO[ ]
001 置入BFO[ ]
001利用Mori-Tanaka 微觀力學模型與有限元素法探討CFO
[ ]
001置入BFO[ ]
001 形式之複合材料(圖 B-1),並將其表示為CFO[ ]
001 /BFO[ ]
001。當f = 0 時代表材 料還未置入任何的CFO,材料呈現完整的 BFO 性質,而當 f = 1 時,則代表 CFO 完全取代BFO,呈現完整的 CFO 性質。在這過程中,複合材料的等效性質會由 原本BFO 的性質慢慢轉變為 CFO 的性質。[ ]
001CFO /BFO
[ ]
001之耦合情形,當對其提供平均磁場H 1
時,CFO 中的q 15
項會參與耦合的行為,而BFO 材料中的e 15
項因應CFO 的耦合行為產生電場 E ,1
因此可以得到複合材料的λ ;如果提供平均磁場* 11
H ,則 CFO 中的2
q 項會參24
與耦合的行為,而BFO 中的e 會因應 CFO 的耦合行為產生均電場24
E ,因此2
會得到複合材料的λ ;如果提供平均磁場* 22
H ,則 CFO 中的3 q 31
、q 32
、q 33
項 會參與耦合的行為,BFO 材料中的e 31
、e 32
、e 33
會因應CFO 的耦合行為產生平CFO /BFO