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Research article

Analytical subthreshold swing model of junctionless elliptic gate-all-around (GAA) FET

  • An analytical subthreshold swing (SS) model has been presented to determine the SS of an elliptic junctionless gate-all-around field-effect transistor (GAA FET). The analysis of a GAA FET with an elliptic cross-section is essential because it is difficult to manufacture a GAA FET with an accurate circular cross-section during the process. The SS values obtained using the proposed SS model were compared with 2D simulation values and other papers to confirm good agreement. Using this analytical SS model, SS was analyzed according to the eccentricity of the elliptic cross-section structure. As a result, it was found that the carrier control ability within the channel improved as the eccentricity increased due to a decrease in the effective channel radius by a decrease in the minor axis length and a decrease in the minimum potential distribution within the channel, and thus the SS decreased. There was no significant change in SS until the eccentricity increased to 0.75 corresponding to the aspect ratio (AR), the ratio of the minor and major axis lengths, of 1.5. However, SS significantly decreased when the eccentricity increased to 0.87 corresponding to AR = 2. As a result of the SS analysis for changes in the device parameters of the GAA FET, changes in the channel length, radius, and oxide film thickness significantly affected the changing rate of SS with eccentricity.

    Citation: Hakkee Jung. Analytical subthreshold swing model of junctionless elliptic gate-all-around (GAA) FET[J]. AIMS Electronics and Electrical Engineering, 2024, 8(2): 211-226. doi: 10.3934/electreng.2024009

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  • An analytical subthreshold swing (SS) model has been presented to determine the SS of an elliptic junctionless gate-all-around field-effect transistor (GAA FET). The analysis of a GAA FET with an elliptic cross-section is essential because it is difficult to manufacture a GAA FET with an accurate circular cross-section during the process. The SS values obtained using the proposed SS model were compared with 2D simulation values and other papers to confirm good agreement. Using this analytical SS model, SS was analyzed according to the eccentricity of the elliptic cross-section structure. As a result, it was found that the carrier control ability within the channel improved as the eccentricity increased due to a decrease in the effective channel radius by a decrease in the minor axis length and a decrease in the minimum potential distribution within the channel, and thus the SS decreased. There was no significant change in SS until the eccentricity increased to 0.75 corresponding to the aspect ratio (AR), the ratio of the minor and major axis lengths, of 1.5. However, SS significantly decreased when the eccentricity increased to 0.87 corresponding to AR = 2. As a result of the SS analysis for changes in the device parameters of the GAA FET, changes in the channel length, radius, and oxide film thickness significantly affected the changing rate of SS with eccentricity.



    Riemann-Liouville fractional integral given by

    Iαa+ξ()=1Γ(α)χa(χ)α1ξ()dt.

    Many different concepts of fractional derivative maybe found in [9,10,11]. In [12] studied a conformable derivative:

    αf()=limϵ0f(+ϵ1α)f()ϵ.

    The time scale conformable derivatives was introduced by Benkhettou et al. [17].

    Further, in recent years, numerous mathematicians claimed that non-integer order derivatives and integrals are well suited to describing the properties of many actual materials, such as polymers. Fractional derivatives are a wonderful tool for describing memory and learning. a variety of materials and procedures inherited properties is one of the most significant benefits of fractional ownership. For more concepts and definition on time scales see [13,14,15,16,17,18,19,33,34,35].

    Continuous version of Steffensen's inequality [7] is written as: For 0g()1 on [a,b]. Then

    bbλf()dtbaf()g()dta+λaf()dt, (1.1)

    where λ=bag()dt.

    Supposing f is nondecreasing gets the reverse of (1.1).

    Also, the discrete inequality of Steffensen [6] is: For λ2n=1g()λ1. Then

    n=nλ2+1f()n=1f()g()λ1=1f(). (1.2)

    Recently, a large number of dynamic inequalities on time scales have been studied by a small number of writers who were inspired by a few applications (see [1,2,3,4,8,28,29,30,31,32,36,37,40,41,42,44,48,49,50,51,52,53]).

    In [5] Jakšetić et al. proved that, if ˆμ([c,d])=[a,b]g()dˆμ(), where [c,d][a,b]. Then

    [a,b]f()g()dˆμ()[c,d]f()g()dˆμ()+[a,c](f()f(d))g()dˆμ(),

    and

    [c,d]f()dˆμ()[d,b](f(c)f())g()dˆμ()[a,b]f()g()dˆμ().

    Anderson, in [3], studied the inequality:

    bbλϕ()baϕ()ψ()a+λaϕ(), (1.3)

    In [47] the authors have proved, for

    m+λ1mζ()d=kmζ()g()d,

    and

    nnλ2ζ()d=nkζ()g()d.

    If there exists a constant A such that r()/ζ()At is monotonic on the intervals [m,k], [k,n], and

    nmtq()g()d=m+λ1mtq()d+nnλ2tq()d,

    then

    nmr()g()dm+λ1mr()d+nnλ2r()d.

    In particularly, Anderson [3] proved

    nnλr()nmr()g()m+λmr().

    where m,nTκ with m<n, r, g:[m,n]TR are -integrable functions such that r is of one sign and nonincreasing and 0g()1 on [m,n]T and λ=nmg(), nλ,m+λT.

    We prove the next two needed results:

    Theorem 1.1. Assume q>0 with 0g()ζ() [m,n]T and λ is given from nmg()Δα=m+λmζ()Δα, then

    nmr()g()Δαm+λmr()ζ()Δα. (1.4)

    Also, provided with 0g()ζ() and nnλζ()Δα=nmg()Δα, we have

    nnλr()ζ()Δαnmr()g()Δα. (1.5)

    We get the reverse inequalities of (1.4) and (1.5) when assuming r/ζ is nondecreasing.

    Theorem 1.2. Assume ψ is integrable on time scales interval [m,n], with ζ()ψ()g()ψ()0[m,n]T and m+λmζ()Δα=nmg()Δα=nnλζ()Δα and g, r and ζ are Δα-integrable functions, ζ()g()0, we have

    nnλr()ζ()Δα+nm|(r()r(nλ))ψ()|Δαnmr()g()Δαm+λmr()ζ()Δαnm|(r()r(m+λ))ψ()|Δα, (1.6)

    and

    nnλr()ζ()Δαnnλ[r()ζ()(r()r(nλ))][ζ()g()]Δαnmr()g()Δαm+λm[r()ζ()(r()r(m+λ))][ζ()g()]Δαm+λmr()ζ()Δα. (1.7)

    Proof. The proof techniques of Theorems 1.6 and 1.7 are like to that in [4] and is removed.

    Several authors proved conformable Hardy's inequality [20,21], conformable Hermite-Hadamard's inequality [22,23,24], conformable inequality of Opial's [26,27] and conformable inequality of Steffensen's [25]. In [45] Anderson proved the followong results:

    Theorem 1.3. [45] Suppose α(0,1] and r1, r2R such that 0r1r2. Suppose :[r1,r2][0,) and Γ:[r1,r2][0,1] are α-fractional integrable functions on [r1,r2] with Π is decreasing, we get

    r2r2Π(ζ)dαζr2r1Π(ζ)Γ(ζ)dαζr1+r1Π(ζ)dαζ,

    where =α(r2r1)rα2rα1r2r1Γ(ζ)dαζ[0,r2r1].

    In [46] the authors gave an extension for Theorem 1.8:

    Theorem 1.4. Assume α(0,1] and r1, r2R such that 0r1r2. Suppose ,Γ,Σ:[r1,r2][0,) are integrable on [r1,r2] with the decreasing function Π and 0ΓΣ, we get

    r2r2Σ(ζ)Π(ζ)dαζr2r1Π(ζ)Γ(ζ)dαζr1+r1Σ(ζ)Π(ζ)dαζ,

    where =(r2r1)r2r1Σ(ζ)dαζr2r1Γ(ζ)dαζ[0,r2r1].

    In this paper, we prove and explore several novel speculations of the Steffensen inequality obtained in [47] through the conformable integral containing time scale concept. We furthermore recover certain known results as special cases of our results.

    Lemma 2.1. Assume ζ>0 is rd-continuous function on [m,n]T, g, r be rd-continuous on [m,n]T such that r/ζ nonincreasing function and 0g()1 [m,n]T. Then

    (Λ1)

    nmr()g()Δαm+λmr()Δα, (2.1)

    where λ is given by

    nmζ()g()Δα=m+λmζ()Δα.

    (Λ2)

    nnλr()Δαnmr()g()Δα, (2.2)

    such that

    nnλζ()Δα=nmζ()g()Δα.

    (2.1) and (2.2) are reversed when r/ζ is nondecreasing.

    Proof. Putting g()ζ()g() and r()r()/ζ() in (1.4), (1.5) to get (Λ1) and (Λ2) simultaneously.

    Lemma 2.2. Under the same hypotheses of Lemma 2.1. with ψ be integrable functions on [m,n]T and 0ψ()g()1ψ() for all [m,n]T. Then

    nnλr()Δα+nm|(r()ζ()r(nλ)ζ(nλ))ζ()ψ()|Δαnmr()g()Δαm+λmr()Δαnm|(r()ζ()r(m+λ)ζ(m+λ))ζ()ψ()|Δα,

    where λ is obtained from

    m+λmh()Δα=nmζ()g()Δα=nnλζ()Δα.

    Proof. Putting g()ζ()g(), r()r()/h() and ψ()ζ()ψ() in (1.6).

    Lemma 2.3. Under the same conditions of Lemma 2.1. Then

    nnλr()Δαnnλ(r()[r()ζ()r(nλ)ζ(nλ)]ζ()[1g()])Δαnmr()g()Δαm+λm(r()[r()ζ()r(a+λ)ζ(m+λ)]ζ()[1g()])Δαm+λmr()Δα,

    where λ is obtained from

    m+λmζ()Δα=nmg()Δα=nnλζ()Δα.

    Proof. Taking g()ζ()g() and r()r()/ζ() in (1.7).

    Theorem 2.1. Under the same conditions of Lemma 2.3 such that k(m,n) and λ1, λ2 are given from

    (Λ3)

    m+λ1mζ()Δα=kmζ()g()Δα,
    nnλ2ζ()Δα=nkζ()g()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=m+λ1mϕ()ζ()Δα+nnλ2ϕ()ζ()Δα, (2.3)

    then

    nmrσ()g()Δαm+λ1mrσ()Δα+nnλ2rσ()Δα. (2.4)

    (2.4) is reversed if rσ/ζAHk2[m,n] and (2.3).

    (Λ4)

    kkλ1ζ()Δα=kmζ()g()Δα,
    k+λ2kζ()Δα=nkζ()g()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=k+λ2kλ1ϕ()ζ()Δα, (2.5)

    then

    nmrσ()g()Δαk+λ2kλ1rσ()Δα. (2.6)

    If rσ/ζAHk2[m,n] and (2.5) satisfied, then we reverse (2.6).

    (Λ5) If λ1, λ2 be the same as in (Λ3) and rσ/ζAHk1[m,n] so that

    nmϕ()ζ()g()Δα=m+λ1m(ϕ()ζ()[ϕ()mλ1]ζ()[1g()])Δα+nnλ2(ϕ()ζ()[ϕ()n+λ2]ζ()[1g()])Δα, (2.7)

    then

    nmrσ()g()Δαm+λ1m(rσ()|rσ()ζ()rσ(m+λ1)ζ(m+λ1)|ζ()[1g()])Δα+nnλ2(rσ()|rσ()ζ()rσ(nλ2)ζ(nλ2)|ζ()[1g()])Δα. (2.8)

    If rσ/ζAHk2[m,n] and (2.7) satisfied, the inequality in (2.8) is reversed.

    (Λ6) If λ1, λ2 be defined as in (Λ4) and rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=kkλ1(ϕ()ζ()[ϕ()k+λ1]ζ()[1g()])Δα=m+λ1m(ϕ()ζ()[ϕ()k+λ2]ζ()[1g()])Δα, (2.9)

    then

    nmrσ()g()Δαkkλ1(rσ()[rσ()ζ()rσ(kλ1)ζ(kλ1)]ζ()[1g()])Δα+k+λ2k(rσ()[rσ()ζ()rσ(k+λ2)ζ(k+λ2)]ζ()[1g()])Δα. (2.10)

    If rσ/ζAHk2[m,n] and (2.9) satisfied, we reverse (2.10).

    Proof. (Λ3) Consider rσ/ζAHk1[m,n], and R1()=rσ()Aϕ()ζ(), since A is given in Definition 2.1. Since R1/ζ:[m,k]TR, using Lemma 2.1(Λ1), we deduce

    0m+λ1mR1()ΔαkmR1()g()Δα=m+λ1mrσ()Δαkmrσ()g()ΔαA(m+λ1mϕ()ζ()Δαkmϕ()ζ()g()Δα). (2.11)

    As R1/ζ:[k,n]TR is nondecreasing, using Lemma 2.1(Λ2), we obtain

    0nkR1()g()Δαnnλ2R1()Δα=nkrσ()g()Δαnnλ2rσ()ΔαA(nkϕ()ζ()g()Δαnnλ2ϕ()ζ()Δα). (2.12)

    (2.11) and (2.12) imply that

    m+λ1mrσ()Δα+nnλ2rσ()Δαnmrσ()g()ΔαA(m+λ1mϕ()ζ()Δα+nnλ2ϕ()ζ()Δαnmϕ()ζ()g()Δα)

    Hence, if (2.3) is hold, then (2.4) holds. For rσ/ζAHk2[m,n], we get the some steps.

    (Λ4) Let rσ/ζAHk1[m,n], also R1(x)=rσ(x)Aϕ(x)ζ(x), where A as in Definition 2.1. R1/ζ:[m,k]TR is nonincreasing, so from Lemma 2.1(Λ1) we obtain

    0kmrσ()g()Δαkkλ1rσ()ΔαA(kmϕ()h()g()Δαkcλ1ϕ()ζ()Δα). (2.13)

    Using Lemma 2.1(Λ1) we have

    0k+λ2krσ()Δαnkrσ()g()ΔαA(k+λ2kϕ()ζ()Δαnkϕ()ζ()g()Δα). (2.14)

    Thus, from (2.13), (2.14), we get

    nmrσ()g()Δαk+λ2kλ1rσ()ΔαA(nmϕ()ζ()g()Δαk+λ2kλ1ϕ()ζ()Δα)

    Therefore, if nmϕ()ζ()g()Δα=k+λ2kλ1ϕ()ζ()Δα is satisfied, then (2.8) holds. Follow the same steps for rσ/ζAHk2[m,n].

    Using Lemma 2.3 and repeat the steps of Theorem 2.1(Λ3) and Theorem 2.1(Λ4) in the proof of (Λ5) and (Λ6) respectively.

    Corollary 2.1. The inequalities (2.4), (2.6), (2.8) and (2.10) of Theorem 2.1 letting T=R takes

    (i)nmfσ()g()dαm+λ1mrσ()dα+nnλ2rσ()dα. (2.15)
    (ii)nmrσ()g()dαk+λ2kλ1rσ()dα. (2.16)
    (iii)nmrσ()g()dαm+λ1m(rσ()[rσ()ζ()rσ(m+λ1)ζ(m+λ1)]ζ()[1g()])dα+nnλ2(rσ()[rσ()ζ()rσ(nλ2)ζ(nλ2)]ζ()[1g()])dα. (2.17)
    (iv)nmrσ()g()dαkkλ1(rσ()[rσ()ζ()rσ(kλ1)ζ(kλ1)]ζ()[1g()])dα+k+λ2k(rσ()[rσ()ζ()rσ(k+λ2)ζ(k+λ2)]ζ()[1g()])dα. (2.18)

    Corollary 2.2. We get [47,Theorems 8,10,21 and 22], if we put α=1 and ϕ()= in Corollary 2.1 [(i),(ii),(iii),(iv)] simultaneously.

    Corollary 2.3. In Corollary 2.1 taking T=Z, the results (2.15)–(2.18) will be equivalent to

    (i)n1=mr(+1)g()α1m+λ11=mr(+1)+n1=nλ2r(+1)α1.
    (ii)n1=mr(+1)g()α1k+λ21=kλ1r(+1)α1.
    (iii)n1=mr(+1)g()α1m+λ11=m(r(+1)[r(+1)ζ()r(a+λ1+1)ζ(m+λ1)]ζ()[1g()])α1+n1=nλ2(r(+1)[r(+1)ζ()r(nλ2+1)ζ(nλ2)]ζ()[1g()])α1.
    (iv)n1=mr(+1)g())α1k1=kλ1(r(+1)[r(+1)ζ()r(kλ1+1)ζ(kλ1)]ζ()[1g()]))α1+k+λ21=k(r(+1)[r(+1)ζ()r(k+λ2+1)ζ(k+λ2)]ζ()[1g()]))α1.

    Theorem 2.2. Under the assumptions in Lemma 2.1 with 0g()ζ() and λ1, λ2 be defined as

    (Λ7)

    m+λ1mζ()Δα=kmg()Δα,
    nnλ2ζ()Δα=nkg()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()g()Δα=m+λ1mϕ()ζ()Δα+nnλ2ϕ()ζ()Δα, (2.19)

    then

    nmrσ()g()Δαm+λ1mrσ()ζ()Δα+nnλ2rσ()ζ()Δα. (2.20)

    (Λ8)

    kkλ1ζ()Δα=kmg()Δα,
    k+λ2kζ()Δα=nkg()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()g()Δα=k+λ2kλ1ϕ()ζ()Δα, (2.21)

    then

    nmrσ()g()Δαk+λ2kλ1rσ()ζ()Δα. (2.22)

    If rσ/ζAHk2[m,n] and (2.19), (2.21) satisfied, we get the reverse of (2.20) and (2.22).

    Proof. By using Theorem 2.1 [(Λ3),(Λ4)] and by putting gg/h and ffh, we get the proof of (Λ7) and (Λ8).

    Corollary 2.4. In Theorem 2.2 [(Λ7),(Λ8)], assuming T=R, the following results obtains:

    (i)nmrσ()g()dαm+λ1mrσ()ζ()dα+nnλ2rσ()ζ()dα. (2.23)
    (ii)nmrσ()g()dαk+λ2kλ1rσ()ζ()dα. (2.24)

    Corollary 2.5. In Corollary 2.4 [(i),(ii)], when we put α=1 and ϕ()= then [47,Theorems 16 and 17] gotten.

    Corollary 2.6. In (2.23) and (2.24) letting T=Z, gets

    (i)n1=mr(+1)g()α1m+λ11=mr(+1)h()+n1=nλ2r(+1)h()α1.
    (ii)n1=mr(+1)g()α1k+λ21=kλ1r(+1)ζ()α1.

    Theorem 2.3. Using the same conditions in Lemma 2.3. Letting w:[m,n]TR be integrable with 0g()w() [m,n]T and

    (Λ9)m+λ1mw()ζ()Δα=kmζ()g()Δα,
    nnλ2w()ζ()Δα=nkζ()g()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=m+λ1mϕ()w()ζ()Δα+nnλ2ϕ()w()ζ()Δα, (2.25)

    then

    nmrσ()g()Δαm+λ1mrσ()w()Δα+nnλ2rσ()w()Δα. (2.26)
    (Λ10)kkλ1w()ζ()Δα=kmζ()g()Δα,
    k+λ2kw()ζ()Δα=nkζ()g()Δα.

    If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=k+λ2kλ1ϕ()w()ζ()Δα, (2.27)
    nmrσ()g()Δαk+λ2kλ1rσ()w()Δα. (2.28)

    The inequalities in (2.26) and (2.28) are reversible if rσ/ζAHc2[a,b] and (2.25), (2.27) hold.

    Proof. In Theorem 2.1 [(Λ3),(Λ4)], ζ changes wq, g changes g/w and r changes rw.

    Corollary 2.7. In (2.26) and (2.28). Letting T=R, we have

    (i)nmrσ()g()dαm+λ1mrσ()w()dα+nnλ2rσ()w()dα. (2.29)
    (ii)nmrσ()g()dαk+λ2kλ1rσ()w()dα. (2.30)

    Corollary 2.8. In Corollary 2.7 [(i),(ii)], letting α=1 and ϕ()= we get [47,Theorems 18 and 19].

    Corollary 2.9. In (2.29) and (2.30), crossing T=Z, gets

    (i)n1=mr(+1)g()α1m+λ11=mr(+1)w()+n1=nλ2r(+1)w()α1.
    (ii)n1=mr(+1)g()α1k+λ21=kλ1r(+1)w()α1.

    Theorem 2.4. Using the same conditions in Lemma 2.1, and Theorem 2.1 [(Λ3),(Λ4)] with ψ:[m,n]TR be a integrable: 0ψ()g()1ψ().

    (Λ11) If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=m+λ1mϕ()ζ()Δαkm|ϕ()mλ1|ζ()ψ()Δα+nnλ2ϕ()ζ()Δα+nk|ϕ()n+λ2|ζ()ψ()Δα, (2.31)

    then

    nmrσ()g()Δαm+λ1mrσ()Δαkm|rσ()ζ()rσ(m+λ1)ζ(m+λ1)|ζ()ψ()Δα+nnλ2rσ()Δα+nk|rσ()ζ()rσ(nλ2)ζ(nλ2)|ζ()ψ()Δα. (2.32)

    (Λ12) If rσ/ζAHk1[m,n] and

    nmϕ()ζ()g()Δα=kkλ1ϕ()ζ()Δαkm|ϕ()k+λ1|ζ()ψ()Δα+nk|ϕ()kλ1|ζ()ψ()Δα, (2.33)

    then

    nmrσ()g()Δαk+λ2kλ1rσ()Δα+km|rσ()ζ()rσ(kλ1)ζ(kλ1)|ζ()ψ()Δαnk|rσ()ζ()rσ(k+λ2)ζ(k+λ2)|ζ()ψ()Δα. (2.34)

    If rσ/ζAHk2[m,n] and (2.31) and (2.33) satisfied, we get the reverse of (2.32) and (2.34).

    Proof. The same steps of Theorem 2.1 [(Λ3),(Λ4)] with Lemma 2.1, R1/ζ:[m,k]TR nonincreasing, R1/ζ:[k,n]TR nondecreasing.

    Corollary 2.10. In Theorem 2.4 [(Λ11),(Λ12)], letting T=R we get:

    (i)nmrσ()g()dαm+λ1mrσ()dαkm|rσ()ζ()rσ(m+λ1)ζ(m+λ1)|ζ()ψ()dα+nnλ2rσ()dα+nk|rσ()ζ()rσ(nλ2)ζ(nλ2)|ζ()ψ()dα. (2.35)
    (ii)nmrσ()g()dαk+λ2kλ1rσ()dα+km|rσ()ζ()rσ(kλ1)ζ(kλ1)|ζ()ψ()dαnk|rσ()ζ()rσ(k+λ2)ζ(k+λ2)|ζ()ψ()dα. (2.36)

    Corollary 2.11. In (2.35) and (2.36), we put α=1, with ϕ()= we get [47,Theorems 23 and 24].

    Corollary 2.12. Our results (2.35) and (2.36), by using T=Z gets

    (i)n1=mr(+1)g()α1m+λ11=mr(+1)α1k1=m|r(+1)ζ()r(m+λ1+1)ζ(m+λ1)|ζ()ψ()ˆ+n1=nλ2r(+1)α1+n1=k|r(+1)ζ()r(nλ2+1)ζ(nλ2)|ζ()ψ()α1.
    (ii)n1=mr(+1)g()α1k+λ21=kλ1r(+1)α1+k1=m|r(+1)ζ()r(kλ1+1)ζ(kλ1)|ζ()ψ()α1n1=k|r(+1)ζ()r(k+λ2+1)ζ(k+λ2)|h()ψ()α1.

    In this work, we explore new generalizations of the integral Steffensen inequality given in [38,39,43] by the utilization of the α-conformable derivatives and integrals, A few of these results are generalised to time scales. We also obtained the discrete and continuous case of our main results, in order to gain some fresh inequalities as specific cases.

    The authors extend their appreciation to the Research Supporting Project number (RSP-2022/167), King Saud University, Riyadh, Saudi Arabia.

    The authors declare no conflict of interest.



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