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3. Quantitative Solution

3. Quantitative Solution

1. Electrostatics: Assumptions & Definitions

  • 가정
    • 1차원, 평면 접합 부분만을 두고 계산
    • Ohmic Contact - 반도체 사이는 매우 작은 저항값을 갖는 linear한 I-V 특성을 갖고 있음
    • metallurgical junction을 x=0인 좌표로 설정
    • VAV_A : p - n 사이에 인가된 전압 - built-in potential은 VAV_A와 반대방향

2. Electrostatics of a Step Junction in Equilibrium

  • VA=0V_A=0일 때 Step Junction Electrostatics
    • Electric Field

      • ρ(x){qNA:xpx<0qND:0xxn\rho(x)\simeq\begin{cases} -qN_A : -x_p\leq x<0\\ qN_D : 0\leq x\leq x_n \end{cases}, qNAxp=qNDxnqN_Ax_p=qN_Dx_n
      • dE(x)dx=ρ(x)ϵ0ϵs{qNA/ϵ0ϵs:xpx<0qND/ϵ0ϵs:0xxn\frac{dE(x)}{dx}=\frac{\rho(x)}{\epsilon_0\epsilon_s}\simeq\begin{cases} -qN_A/\epsilon_0\epsilon_s : -x_p\leq x<0\\ qN_D/\epsilon_0\epsilon_s : 0\leq x\leq x_n \end{cases}

        E(x){qNA[xp+x]/ϵ0ϵs:xpx<0qND[xnx]/ϵ0ϵs:0xxn\rarr E(x)\simeq\begin{cases} -qN_A[x_p+x]/\epsilon_0\epsilon_s : -x_p\leq x<0\\ qN_D[x_n-x]/\epsilon_0\epsilon_s : 0\leq x\leq x_n \end{cases}
      • Boundary Condition : E(xp)=E(xn)=0E(-x_p)=E(x_n)=0
    • Step Junction Electric Potential

      • dV(x)dx=E(x){qNA[xp+x]/ϵ0ϵs:xpx<0qND[xnx]/ϵ0ϵs:0x<xn-\frac{dV(x)}{dx}=E(x)\simeq\begin{cases} -qN_A[x_p+x]/\epsilon_0\epsilon_s : -x_p\leq x<0\\ qN_D[x_n-x]/\epsilon_0\epsilon_s : 0\leq x < x_n \end{cases}

        V(x){qNA[xp+x]2/2ϵ0ϵs:xpx<0VBIqND[xnx]2/2ϵ0ϵs:0x<xnVBI:xxn\rarr V(x)\simeq\begin{cases} -qN_A[x_p+x]^2/2\epsilon_0\epsilon_s : -x_p\leq x<0\\ V_{BI}-qN_D[x_n-x]^2/2\epsilon_0\epsilon_s : 0\leq x< x_n\\ V_{BI} : x\geq x_n \end{cases}
      • Boundary Condition : V(xp)=0, V(xn)=VBIV(-x_p)=0,\ V(x_n)=V_{BI}
    • Depletion Region

      • VBI=qNA2ϵ0ϵsxp2+qND2ϵ0ϵsxn2V_{BI}=\frac{qN_A}{2\epsilon_0\epsilon_s}x_p^2+\frac{qN_D}{2\epsilon_0\epsilon_s}x_n^2
      • 연속 조건에 의해 NAxp=NDxnN_Ax_p=N_Dx_n이므로
        VBI=q2ϵ0ϵs[NA+NA2ND]xp2=qNA2ϵ0ϵs[ND+NAND]xp2V_{BI}=\frac{q}{2\epsilon_0\epsilon_s}[N_A+\frac{N_A^2}{N_D}]x_p^2=\frac{qN_A}{2\epsilon_0\epsilon_s}[\frac{N_D+N_A}{N_D}]x_p^2
      • xp=2ϵ0ϵsqNDNA[NA+ND]VBIx_p=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_D}{N_A[N_A+N_D]}V_{BI}}
      • xn=NANDxp=2ϵ0ϵsqNAND[NA+ND]VBIx_n=\frac{N_A}{N_D}x_p=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_A}{N_D[N_A+N_D]}V_{BI}}
      • W=xp+xn=2ϵ0ϵsqNA+NDNANDVBIW=x_p+x_n=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_A+N_D}{N_AN_D}V_{BI}}

3. Electrostatics of a Step Junction with VA0V_A\not ={0}

  • 다이오드에 Bias VAV_A가 인가된 경우
    • p, n, 양 단자에서는 전압 강하가 거의 없음
    • depletion region에서 대부분 전압 강하 VAV_A가 일어남
  • VBIVA=qNA2ϵ0ϵsxp2+qND2ϵ0ϵsxn2V_{BI}-V_A=\frac{qN_A}{2\epsilon_0\epsilon_s}x_p^2+\frac{qN_D}{2\epsilon_0\epsilon_s}x_n^2
    • VBIkTqln(NANDni2)V_{BI}\simeq\frac{kT}{q}ln(\frac{N_AN_D}{n_i^2})
  • xp<x0-x_p<x\leq0에서
    • E(x)=qNAϵ0ϵs(xp+x)E(x)= -\frac{qN_A}{\epsilon_0\epsilon_s}(x_p+x)
    • V(x)=qNA2ϵ0ϵs(xp+x)2V(x)=\frac{qN_A}{2\epsilon_0\epsilon_s}(x_p+x)^2
    • xp=2ϵ0ϵsqNDNA[NA+ND][VBIVA]x_p=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_D}{N_A[N_A+N_D]}[V_{BI-V_A]}}
  • 0<xxn0<x\leq x_n에서
    • E(x)=qNDϵ0ϵs(xnx)E(x)= -\frac{qN_D}{\epsilon_0\epsilon_s}(x_n-x)
    • V(x)=[VBIVA]qND2ϵ0ϵs(xnx)2V(x)=[V_{BI}-V_A]-\frac{qN_D}{2\epsilon_0\epsilon_s}(x_n-x)^2
    • xp=2ϵ0ϵsqNAND[NA+ND][VBIVA]x_p=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_A}{N_D[N_A+N_D]}[V_{BI-V_A]}}
  • W=xp+xn=2ϵ0ϵsqNA+NDNAND[VBIVA]W=x_p+x_n=\sqrt{\frac{2\epsilon_0\epsilon_s}{q}\frac{N_A+N_D}{N_AN_D}[V_{BI}-V_A]}

  • Voltage Dependence
    • Depletion Width, Electric Field : VBIVA\sqrt{V_{BI}-V_A}에 비례
    • Electrostatic Potential : VBIVAV_{BI}-V_A에 비례
    • Energy Band Diagram
      • VA=0V_A=0 (Equilibrium) : 모든 영역에서 페르미 준위 동일 (np=ni2np=n_i^2)
      • VA>0V_A>0 (Forward Bias) : FN>FPF_N>F_P, FNFP=qVAF_N-F_P=qV_A, 접점 근처에서 np>ni2np>n_i^2
      • VA<0V_A<0 (Reverse Bias) : FN<FPF_N<F_P, FNFP=qVAF_N-F_P=qV_A, 접점 근처에서 np<ni2np<n_i^2

4. Electrostatics of a Linearly Graded Junction

  • Metallurgical Junction 근처에서 농도 NANDN_A-N_D가 선형으로 증가
  • 가정
    • depletion region 밖의 전계는 0, 전위는 일정
    • VBI=kTqln[p(xp)n(xn)ni2]V_{BI}=\frac{kT}{q}ln[\frac{p(-x_p)n(x_n)}{n_i^2}]
  • Electrostatics
    • ρ(x)=qax=q[ND(x)NA(x)] (W/2xW/2)\rho(x)=qax=q[N_D(x)-N_A(x)]\ (-W/2\leq x\leq W/2)
    • E(x)=qa2ϵ0ϵs[x2(W2)2] (W/2xW/2)E(x)=\frac{qa}{2\epsilon_0\epsilon_s}[x^2-(\frac{W}{2})^2]\ (-W/2\leq x\leq W/2)
      • E(0)qaW28ϵ0ϵsE(0)\simeq-\frac{qaW^2}{8\epsilon_0\epsilon_s}
    • V(x)={qa6ϵ0ϵs[2(W2)3+3(W2)2xx3] (W/2xW/2)qa12ϵ0ϵsW3 (x>W)V(x)=\begin{cases} \frac{qa}{6\epsilon_0\epsilon_s}[2(\frac{W}{2})^3+3(\frac{W}{2})^2x-x^3]\ (-W/2\leq x\leq W/2)\\ \frac{qa}{12\epsilon_0\epsilon_s}W^3\ (x>W) \end{cases}
    • Bias가 없을 때 VBI=qa12ϵ0ϵsW03V_{BI}=\frac{qa}{12\epsilon_0\epsilon_s}W_0^3
      • VBI=kTqln[p(W0/2)n(W0/2)ni2]kTqln[a2W024ni2]V_{BI}=\frac{kT}{q}ln[\frac{p(-W_0/2)n(W_0/2)}{n_i^2}]\simeq\frac{kT}{q}ln[\frac{a^2W_0^2}{4n_i^2}]
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