Data for the Publication "Two-carrier model-fitting of Hall effect in semiconductors with dual-band occupation: A case study in GaN two-dimensional hole gas"

Derivation of Two-Carrier Hall effect Equations Joseph E. Dill, Chuan F. C. Chang, Debdeep Jena, Huili Grace Xing Cornell University Introduction This Mathematica notebook presents a derivation of Eqs. 3-13 in the manuscript “Two-carrier modelfitting of Hall effect in semiconductors with dual-band occupation: A case study in GaN two-dimensional hole gas” (doi: 10.1063/5.0248998 ) The final results that are printed in the manuscript are marked with Blue text. Derivation In the presence of m Drude-like parallel-conducting channels, the XX and XY conductivities are given by [Eq. (3a), (3b)] σxx(B) = ∑i mqiniμi 1+(μiB)2 σxy(B) = ∑i mσxx,i(μiB) Here, we consider the case of two conducting channels In[1]:= σxx1 =q1 n1 μ1 1+ (μ1 B)2; σxx2 =q2 n2 μ2 1+ (μ2 B)2; σxx =σxx1 +σxx2; σxy1 = (μ1 B)σxx1; σxy2 = (μ2 B)σxx2; σxy =σxy1 +σxy2; Here, and in the text, we adopt the convention that q and μ always carry the same sign, while n is always positively signed. This means that σxx is always positive, while σxy,i will take the same sign as qi (by way of μ i ). Printed by Wolfram Mathematica Student Edition From the conductivity tensor equation, [Eq. (4)] J=σxx σxy -σxy σyy E⟺E=ρxx ρxy -ρxy ρyy J, The XX ans XY resistivities are given by Eq. (5a) In[7]:= ρxx =σxx σxx2+σxy2// FullSimplify Out[7]= n2 q2 1+B2μ12μ2+n1 q1 μ11+B2μ22 n12q12μ12+2 n1 n2 q1 q2 μ1μ2+n22q22+B2(n1 q1 +n2 q2)2μ12μ22 Eq. (5b) In[8]:= ρxy =σxy σxx2+σxy2// FullSimplify Out[8]= B n1 q1 μ12+Bn2 q2 +B2(n1 q1 +n2 q2)μ12μ22 n12q12μ12+2 n1 n2 q1 q2 μ1μ2+n22q22+B2(n1 q1 +n2 q2)2μ12μ22 To express ρxx and ρxy in terms of σ1, σ2, μ1, and μ2. We will define a casting set that replaces ni with σi qiμi . In[9]:= Castσ=n1 σ1 q1 μ1 , n2 σ2 q2 μ2; Eq. (6a) In[10]:= ρxx /. Castσ// FullSimplify Out[10]= σ1+B2μ22σ1+σ2+B2μ12σ2 1+B2μ22σ12+1+B2μ12σ22+2σ1σ2+B2μ1μ2σ2 Eq. (6b) In[11]:= ρxy /. Castσ// FullSimplify Out[11]= Bμ1σ1+B2μ22σ1+μ2σ2+B2μ12μ2σ2 1+B2μ22σ12+1+B2μ12σ22+2σ1σ2+B2μ1μ2σ2 It is also convenient to express ρxx(B) and ρxy(B) in their series-expanded forms, as seen below: Eq. (7a) 2 2carrier-equations.nb Printed by Wolfram Mathematica Student Edition In[12]:= Series[ρxx, {B, 0, 5}] // FullSimplify Out[12]= 1 n1 q1 μ1+n2 q2 μ2 +n1 n2 q1 q2 μ1(μ1-μ2)2μ2 B2 (n1 q1 μ1+n2 q2 μ2)3 - n1 n2 q1 q2 (n1 q1 +n2 q2)2μ13(μ1-μ2)2μ23B4 (n1 q1 μ1+n2 q2 μ2)5 +O[B]6 Eq. (7b) In[13]:= Series[ρxy, {B, 0, 6}] // FullSimplify Out[13]= n1 q1 μ12+n2 q2 μ22B (n1 q1 μ1+n2 q2 μ2)2 -n1 n2 q1 q2 (n1 q1 +n2 q2)μ12(μ1-μ2)2μ22B3 (n1 q1 μ1+n2 q2 μ2)4 + n1 n2 q1 q2 (n1 q1 +n2 q2)3μ14(μ1-μ2)2μ24B5 (n1 q1 μ1+n2 q2 μ2)6 +O[B]7 ρxx has strictly even B-dependence, while ρxy is strictly odd, both with alternating sign. We denote the j - th series coefficient by cj as follows: ρxx(B)≈c0+c2B2-c4B4+c6B6- … ρxy(B)≈c1B-c3B3-c5B5+ … We can calculate the value of each series coefficient in terms of q1, q 2 , n1, n2, μ1, μ 2 as follows: In[14]:= C0 =Rsh =SeriesCoefficient[ρxx, {B, 0, 0}] // FullSimplify Out[14]= 1 n1 q1 μ1+n2 q2 μ2 In[15]:= C1 =RH =SeriesCoefficient[ρxy, {B, 0, 1}] // FullSimplify Out[15]= n1 q1 μ12+n2 q2 μ22 (n1 q1 μ1+n2 q2 μ2)2 In[16]:= C2 =SeriesCoefficient[ρxx, {B, 0, 2}] // FullSimplify Out[16]= n1 n2 q1 q2 μ1(μ1-μ2)2μ2 (n1 q1 μ1+n2 q2 μ2)3 In[17]:= C3 = -SeriesCoefficient[ρxy, {B, 0, 3}] // FullSimplify Out[17]= n1 n2 q1 q2 (n1 q1 +n2 q2)μ12(μ1-μ2)2μ22 (n1 q1 μ1+n2 q2 μ2)4 In[18]:= C4 = -SeriesCoefficient[ρxx, {B, 0, 4}] // FullSimplify Out[18]= n1 n2 q1 q2 (n1 q1 +n2 q2)2μ13(μ1-μ2)2μ23 (n1 q1 μ1+n2 q2 μ2)5 2carrier-equations.nb 3 Printed by Wolfram Mathematica Student Edition In[19]:= C5 =SeriesCoefficient[ρxy, {B, 0, 5}] // FullSimplify Out[19]= n1 n2 q1 q2 (n1 q1 +n2 q2)3μ14(μ1-μ2)2μ24 (n1 q1 μ1+n2 q2 μ2)6 We can define a casting set to express ρxx and ρxy in terms of c0, c1, c2, c3. In[20]:= CastCj =Solve[{c0 C0, c1 C1, c2 C2, c3 C3},{n1, n2, μ1, μ2}]〚1〛// FullSimplify Out[20]= n1 2 c23-c1 c2 c3 +c3 c0 c3 -c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32 2c22-c1 c3c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32q1 , n2  -2 c23+c1 c2 c3 -c3 c0 c3 +c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32 2c22-c1 c3c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32q2, μ1c1 c2 +c0 c3 +c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32 2 c0 c2 , μ2- -c1 c2 -c0 c3 + (c1 c2 +c0 c3)2+4 c0 c2 c22-c1 c3 2 c0 c2  Doing so, we obtain the following simplification for ρxx and ρxy, which is very convenient for fitting measured data Eq. (9a) In[21]:= ρxx /. CastCj // FullSimplify Out[21]= c0 +B2c23 c22+B2c32 Eq. (9b) In[22]:= ρxy /. CastCj // FullSimplify Out[22]= B c1 -B3c22c3 c22+B2c32 The series coefficients follow the recursive condition Eq. (8) cj≥2=c3j-2 c2j-3 exemplified below with c4 and c 5 In[23]:= C4 C34-2 C24-3 Out[23]= True In[24]:= C5 C35-2 C25-3 Out[24]= True 4 2carrier-equations.nb Printed by Wolfram Mathematica Student Edition So, we can define a generic cj variable as follows: In[25]:= Cj =C3j-2 C2j-3 Out[25]= n1 n2 q1 q2 (n1 q1 +n2 q2)μ12(μ1-μ2)2μ22 (n1 q1 μ1+n2 q2 μ2)4 -2+jn1 n2 q1 q2 μ1(μ1-μ2)2μ2 (n1 q1 μ1+n2 q2 μ2)3 3-j We’ll confirm that it works for various j≥2. In[26]:= (Cj /. j 2)C2 Out[26]= True In[27]:= (Cj /. j 3)C3 Out[27]= True In[28]:= (Cj /. j 4)C4 Out[28]= True In[29]:= (Cj /. j 5)C5 Out[29]= True In the manuscript, we give an equation for cj≥2 in terms of β=n2 n1+n2 and γ=μ2 μ1 , derived below. In[30]:= Castβγ =Solve rsh Rsh, rh RH, β  n2 n1 +n2 , γ  μ2 μ1 ,{n1, n2, μ1, μ2}〚1〛// FullSimplify Out[30]= n1 (-1+β)q1 (-1+β) - q2 βγ2 rh (q1 -q1 β+q2 βγ)2, n2 βq1 -q1 β+q2 βγ2 rh (q1 -q1 β+q2 βγ)2, μ1rh (q1 -q1 β+q2 βγ) rsh q1 -q1 β+q2 βγ2,μ2rh γ(q1 -q1 β+q2 βγ) rsh q1 -q1 β+q2 βγ2 Case: q1 = q 2 In[31]:= Cj /. Castβγ /.q1q2 // Expand // FullSimplify Out[31]= - rsh (-1+β)β(-1+γ)2-rh3(-1+β)β(-1+γ)2γ2 rsh21+β-1+γ23j-rh2(-1+β)β(-1+γ)2γ rsh 1+β-1+γ22-j γ 2carrier-equations.nb 5 Printed by Wolfram Mathematica Student Edition In[32]:= - rsh (-1+β)β(-1+γ)2 -rh3(-1+β)β(-1+γ)2γ2 rsh21+β-1+γ23 -rh2(-1+β)β(-1+γ)2γ rsh 1+β-1+γ22 j γ Out[32]= - rsh (-1+β)β(-1+γ)2rh γ rsh 1+β-1+γ2j γ Eq. (8) [q1=q2case] In[33]:= rsh (1-β)β(-1+γ)2rhjγj γrshj1+β-1+γ2j Out[33]= rhjrsh1-j(1-β)β(-1+γ)2γ-1+j1+β-1+γ2-j Case: q1 = -q2 In[34]:= Cj /. Castβγ /.q1-q2 // Expand // FullSimplify Out[34]= rsh (-1+β)β(-1+γ)2rh3(-1+β)β(-1+2β) (-1+γ)2γ2 rsh2-1+β+βγ23jrh2(-1+β)β(-1+γ)2γ rsh -1+β+βγ22-j (1-2β)2γ In[35]:= rsh (-1+β)β(-1+γ)2 rh3(-1+β)β(-1+2β) (-1+γ)2γ2 rsh2-1+β+βγ23 rh2(-1+β)β(-1+γ)2γ rsh -1+β+βγ22 j (1-2β)2γ Out[35]= rsh (-1+β)β(-1+γ)2rh (-1+2β)γ rsh -1+β+βγ2j (1-2β)2γ In[36]:= rsh (-1+β)β(-1+γ)2rhjγj(2β-1)j (1-2β)2γrshj-1+β+βγ2j Out[36]= rhjrsh1-j(-1+β)β(-1+2β)j(-1+γ)2γ-1+j-1+β+βγ2-j (1-2β)2 Eq. (8) [q1= -q2case] In[37]:= rhj rshj-1(-1)j-1(1-β)β(-1+γ)2γj-1(1-2β)j-2 βγ2+1-1j Out[37]= (-1)-1+jrhjrsh1-j(1-2β)-2+j(1-β)β(-1+γ)2γ-1+j-1+β1+γ2-j Note that in the case of q1= -q2, the sign of all cj∈odd terms is set by whether carrier 1 or carrier 2 has higher conductivity. 6 2carrier-equations.nb Printed by Wolfram Mathematica Student Edition In[38]:= nAP =1 q RH // FullSimplify Out[38]= (n1 q1 μ1+n2 q2 μ2)2 qn1 q1 μ12+n2 q2 μ22 In[39]:= μAP =RH Rsh // FullSimplify Out[39]= n1 q1 μ12+n2 q2 μ22 n1 q1 μ1+n2 q2 μ2 In[40]:= nAP /.{q2 -q1, q q1} /.{μ2 γμ1} /.{n2  βn1} // FullSimplify Out[40]= -n1 (-1+βγ)2 -1+βγ2 Eq. (11) We define the following substitution for use in Eq. 10: c*=c22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32 In[41]:= cs =Sqrtc22c12+4 c0 c2-2 c0 c1 c2 c3 +c02c32; Eq. (10) In[42]:= CastnμToCj =n1 q1 q2 2 c23-c1 c2 c3 +c3 (c0 c3 +Sign[q2]cs) 2 q c1 c3 -c22cs , n2 -2 c23-c1 c2 c3 +c3 (c0 c3 -Sign[q2]cs) 2 q c1 c3 -c22cs , μ1c1 c2 +c0 c3 -Sign[q2]cs 2 c0 c2 , μ2c1 c2 +c0 c3 +Sign[q2]cs 2 c0 c2 ; We will validate Eqs. 10a-c by substituting in values for n1, n2, μ1, and μ 2 and confirming that we get the same output In[43]:= ({n1, n2, μ1, μ2} /. CastnμToCj) /.{c0 C0, c1 C1, c2 C2, c3 C3} /. {n1 4, n2 1, μ1200, μ2800, q1 +1, q2 +1} /. q 1 Out[43]= {4, 1, 200, 800} In[44]:= ({n1, n2, μ1, μ2} /. CastnμToCj) /.{c0 C0, c1 C1, c2 C2, c3 C3} /. {n1 4, n2 1, μ1-200, μ2800, q1 -1, q2 +1} /. q 1 Out[44]= {4, 1, -200, 800} 2carrier-equations.nb 7 Printed by Wolfram Mathematica Student Edition In[45]:= ({n1, n2, μ1, μ2} /. CastnμToCj) /.{c0 C0, c1 C1, c2 C2, c3 C3} /. {n1 4, n2 1, μ1+200, μ2-800, q1 +q, q2 -q} /. q 1 Out[45]= {4, 1, 200, -800} In[46]:= ({n1, n2, μ1, μ2} /. CastnμToCj) /.{c0 C0, c1 C1, c2 C2, c3 C3} /. {n1 4, n2 1, μ1-200, μ2-800, q1 -q, q2 -q} /. q 1 Out[46]= {4, 1, -200, -800} For equations 12 and 13, we will define a new casting set with β, γ, n, and μ as the free parameters. These equations are explicitly only in the case of q1=q2. In[47]:= Castβγnμ=Solve μ  n1 μ1+n2 μ2 n1 +n2 , nn1 +n2, β  n2 n1 +n2 , γ  μ2 μ1 ,{n1, n2, μ1, μ2}〚1〛// FullSimplify Out[47]= n1 n-nβ, n2 nβ,μ1μ 1+β(-1+γ),μ2γμ 1+β(-1+γ) Eq. (12) In[48]:= nApparent =1 q2 C1 /. Castβγnμ/.{q1 q2} // FullSimplify %n(1+β(γ-1))2 1+βγ2-1// FullSimplify Out[48]= n(1+β(-1+γ))2 1+β-1+γ2 Out[49]= True Eq. (13) In[50]:= μApparent =RH Rsh /. Castβγnμ/.{q1 q2} // FullSimplify % μ 1+βγ2-1 (1+β(γ-1))2// FullSimplify Out[50]= μ+β-1+γ2μ (1+β(-1+γ))2 Out[51]= True 8 2carrier-equations.nb Printed by Wolfram Mathematica Student Edition