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|
{
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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter5:JUNCTIONS IN SEMICONDUCTORS:P-N DIODES"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex5.1:pg-169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Nd = 1.0*10**16 #initialising value of donor atoms in centimeter\n",
"Na= 1.0*10**18 #initialising value of accepter atoms in centimeter\n",
"Nc= 2.8*10**19 #initialising value of conduction band effective density\n",
"Nv= 1.0*10**19 #initialising value of valence band effective density\n",
"kbT = 0.026 #initializing value of kbT at 300K\n",
"Eg = 1.1 #initializing value of forbidden energy gap\n",
"#NOTE: nn=Nd and pp=Na\n",
"eVbi = Eg+(kbT*log(Na/Nv))+((kbT*log(Nd/Nc)))\n",
"print\"built in voltage is ,eVbi = \",\"{:.2e}\".format(eVbi),\" eV\"\n",
"apsilen = 11.9*8.85*10**-12 #initializing value of relative permitivity\n",
"e = 1.6*10**-19 #initializing value of charge of electron\n",
"Vbi=eVbi/e\n",
"Nd = 1.0*10**22 #initialising value of donor atoms in metrers\n",
"Na= 1.0*10**24 #initialising value of accepter atomsin meters\n",
"Wp_Vbi = sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))\n",
"print\"depletion width at p side is ,Wp_Vbi =\",\"{:.2e}\".format(Wp_Vbi),\" m\"\n",
"Wn_Vo = 100*sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))\n",
"print\"depletion width at n side is ,Wn_Vo = \",\"{:.2e}\".format(Wn_Vo),\" m\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"built in voltage is ,eVbi = 8.34e-01 eV\n",
"depletion width at p side is ,Wp_Vbi = 3.30e-09 m\n",
"depletion width at n side is ,Wn_Vo = 3.30e-07 m\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.2:pg-172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Nd = 1.0*10**16 #initialising value of donor atoms in centimeter\n",
"Na= 1.0*10**18 #initialising value of accepter atoms in centimeter\n",
"ni = 1.5*10**10 #initializing value of intrinsic carrier concentration\n",
"#NOTE: nn=Nd and pp=Na\n",
"R= 10*10**-6 #initializing value of radius of pn diode\n",
"A = math.pi*(R**2)\n",
"pn = ni**2/Nd\n",
"print\"concentration of electron in p type is ,pn = \",\"{:.2e}\".format(pn),\" cm**-3\"\n",
"kbT = 0.026 #initializing value of kbT at 300K\n",
"eVbi = (kbT*log(Na/pn))\n",
"print\"built in voltage is ,eVbi =\",\"{:.2e}\".format(eVbi),\" V\"\n",
"apsilen = 11.9*8.84*10**-12 #initializing value of relative permitivity\n",
"e = 1.6*10**-19 #initializing value of charge of electron\n",
"#NOTE: for reverse bias Vr = 0 V,\n",
"Wp_4 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*1.0*10**6)/((Na*1.0*10**6)*((Nd*1.0*10**6)+(Na*1.0*10**6)))))\n",
"print\"depletion width at p side is ,Wp_4 =\",\"{:.2e}\".format(Wp_4),\"m\"\n",
"Wn_4 = Wp_4*100\n",
"print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
"\n",
"#for calculation purpose and for differentiating part (I), equating\n",
"Vbi_4 = eVbi\n",
"Vbi_4=round(Vbi_4,3)\n",
"print(\"\")# for spacing \n",
"Vbi_2 = Vbi_4 + 2\n",
"Vbi_2=round(Vbi_2,3)\n",
"#NOTE: for reverse bias Vr = 2 V,\n",
"Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_2 = \",\"{:.2e}\".format(Wp_2),\"m\"\n",
"Wn_2 = Wp_2*100\n",
"print\"depletion width at n side is ,Wn_2 = \",\"{:.2e}\".format(Wn_2),\"m\"\n",
"\n",
"print(\"\")# for spacing\n",
"Vbi_3 = Vbi_4 + 5\n",
"Vbi_3=round(Vbi_3,2)\n",
"#NOTE: for reverse bias Vr = 5 V,\n",
"Wp_3 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_3 = \"\"{:.2e}\".format(Wp_3),\"m\"\n",
"Wn_3 = Wp_3*100\n",
"print\"depletion width at n side is ,Wn_3 = \",\"{:.2e}\".format(Wn_3),\"m\"\n",
"print(\"\")# for spacing\n",
"Vbi_4 = Vbi_4 + 10\n",
"Vbi_4=round(Vbi_4,2)\n",
"#NOTE: for reverse bias Vr = 10 V,\n",
"\n",
"Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
"Wn_4 = Wp_4*100\n",
"print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
"\n",
"print(\"\")# for spacing\n",
"Vbi_5 = Vbi_4 - 0.5\n",
"Vbi_5=round(Vbi_5,2)\n",
"#NOTE: for forward bias Vf = 0.5 V,\n",
"\n",
"Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_5 =\",\"{:.2e}\".format(Wp_5),\"m\"\n",
"Wn_5 = Wp_5*100\n",
"print\"depletion width at n side is ,Wn_5 = \",\"{:.2e}\".format(Wn_5),\"m\"\n",
"\n",
"#CALCULATION FOR PEAK FIELD :\n",
"F = - e*Nd*(10**6)*Wn_4/apsilen\n",
"print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
"F = - e*Nd*(10**6)*Wn_2/apsilen\n",
"print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
"F = - e*Nd*(10**6)*Wn_3/apsilen\n",
"print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
"F = - e*Nd*(10**6)*Wn_4/apsilen\n",
"print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
"F = - e*Nd*(10**6)*Wn_5/apsilen\n",
"print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
"#calculation for \n",
"Q = e*(Nd*10**6)*Wn_4*A#charge in depletion region for Vr = 0V\n",
"print\"Q = \",\"{:.2e}\".format(Q),\"C\"\n",
"#due to approximation taken by author in the textbook .... the values of Vbi_2, Vbi_3, Vbi_4 and \n",
"#the values of depletion width(Wp_4, Wp_2,Wp_3, Wp_4, Wn_4, Wn_2, Wn_3, Wn_4) differ from the above solution\n",
"# The answer of the textbook can be different due to different precision \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"concentration of electron in p type is ,pn = 2.25e+04 cm**-3\n",
"built in voltage is ,eVbi = 8.17e-01 V\n",
"depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
"depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
"\n",
"depletion width at p side is ,Wp_2 = 6.06e-09 m\n",
"depletion width at n side is ,Wn_2 = 6.06e-07 m\n",
"\n",
"depletion width at p side is ,Wp_3 = 8.71e-09 m\n",
"depletion width at n side is ,Wn_3 = 8.71e-07 m\n",
"\n",
"depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
"depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
"\n",
"depletion width at p side is ,Wp_5 = 3.19e-09 m\n",
"depletion width at n side is ,Wn_5 = 3.19e-07 m\n",
"F= -4.96e+06 v/m\n",
"F= -9.21e+06 v/m\n",
"F= -1.32e+07 v/m\n",
"F= -4.96e+06 v/m\n",
"F= -4.84e+06 v/m\n",
"Q = 1.64e-13 C\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.3:pg-173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import cmath\n",
"Nd = 1.0*10**16\n",
"Na= 1.0*10**18\n",
"ni = 1.5*10**10\n",
"#NOTE: nn=Nd and pp=Na\n",
"R= 10*10**-6\n",
"A = math.pi*(R**2)\n",
"pn = ni**2/Nd\n",
"print\"concentration of electron in p type is ,pn = \",\"{:.2e}\".format(pn),\" cm**-3\"\n",
"kbT = 0.026\n",
"eVbi = (kbT*log(Na/pn))\n",
"print\"built in voltage is ,eVbi= \",\"{:.2e}\".format(eVbi),\" V\"\n",
"apsilen = 11.9*8.84*10**-12\n",
"e = 1.6*10**-19\n",
"#NOTE: for reverse bias Vr = 0 V,\n",
"Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))\n",
"print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
"Wn_4 = Wp_4*100\n",
"print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
"#for calculation purpose and for differentiating part (I), equating\n",
"Vbi_4 = eVbi\n",
"print(\"\")# for spacing \n",
"Vbi_2 = Vbi_4 + 2\n",
"#NOTE: for reverse bias Vr = 2 V,\n",
"Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_2 = \",\"{:.2e}\".format(Wp_2),\"m\"\n",
"Wn_2 = Wp_2*100\n",
"print\"depletion width at n side is ,Wn_2 = \",\"{:.2e}\".format(Wn_2),\"m\"\n",
"print(\"\")# for spacing\n",
"Vbi_3 = Vbi_4 + 5\n",
"#NOTE: for reverse bias Vr = 5 V,\n",
"Wp_3 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_3 = \",\"{:.2e}\".format(Wp_3),\"m\"\n",
"Wn_3 = Wp_3*100\n",
"print\"depletion width at n side is ,Wn_3 = \",\"{:.2e}\".format(Wn_3),\"m\"\n",
"print(\"\")# for spacing\n",
"Vbi_4 = Vbi_4 + 10\n",
"#NOTE: for reverse bias Vr = 10 V,\n",
"Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
"Wn_4 = Wp_4*100\n",
"print\"depletion width at n side is ,Wn_4 =\",\"{:.2e}\".format(Wn_4),\"m\"\n",
"print(\"\")# for spacing\n",
"Vbi_5 = Vbi_4 - 0.5\n",
"#NOTE: for forward bias Vf = 0.5 V,\n",
"Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)\n",
"print\"depletion width at p side is ,Wp_5 =\",\"{:.2e}\".format(Wp_5),\"m\"\n",
"Wn_5 = Wp_5*100\n",
"print\"depletion width at n side is ,Wn_5 =\",\"{:.2e}\".format(Wn_5),\"m\"\n",
"#CALCULATION FOR PEAK FIELD :\n",
"Fm = - e*Nd*(10**6)*Wn_4/apsilen\n",
"print\"peak Field For Vr = 0V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
"Fm = - e*Nd*(10**6)*Wn_2/apsilen\n",
"print\"peak Field for Vr = 2V, Fm =\",\"{:.2e}\".format(Fm),\"V/m\"\n",
"Fm = - e*Nd*(10**6)*Wn_3/apsilen\n",
"print\"peak Field For Vr = 5V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
"Fm = - e*Nd*(10**6)*Wn_4/apsilen\n",
"print\"peak Field For Vr = 10V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
"print\"By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"concentration of electron in p type is ,pn = 2.25e+04 cm**-3\n",
"built in voltage is ,eVbi= 8.17e-01 V\n",
"depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
"depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
"\n",
"depletion width at p side is ,Wp_2 = 6.06e-09 m\n",
"depletion width at n side is ,Wn_2 = 6.06e-07 m\n",
"\n",
"depletion width at p side is ,Wp_3 = 8.70e-09 m\n",
"depletion width at n side is ,Wn_3 = 8.70e-07 m\n",
"\n",
"depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
"depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
"\n",
"depletion width at p side is ,Wp_5 = 3.19e-09 m\n",
"depletion width at n side is ,Wn_5 = 3.19e-07 m\n",
"peak Field For Vr = 0V, Fm = -4.96e+06 V/m\n",
"peak Field for Vr = 2V, Fm = -9.21e+06 V/m\n",
"peak Field For Vr = 5V, Fm = -1.32e+07 V/m\n",
"peak Field For Vr = 10V, Fm = -4.96e+06 V/m\n",
"By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.4:pg-178"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"A= 10**-7\n",
"Na=10**18*10**6\n",
"Nd=10**16*10**6\n",
"Dp1 = 7.8*10**-4\n",
"Dn2 = 7.3*10**-4\n",
"Tn = 10**-6\n",
"Tp = 10**-6\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"ni = 1.5*10**16\n",
"#NOTE: nn=Nd and pp=Na\n",
"Lp = sqrt(Dp1*Tp)\n",
"print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"m\"\n",
"Ln = sqrt(Dn2*Tn)\n",
"print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"m\"\n",
"# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
"# assume that the doants are fully ionised \n",
"Io = A*e*((((Dn2)/(Ln))*(ni**2/Na))+(((Dp1)/(Lp))*(ni**2/Nd)))\n",
"print\"The prefactor current is ,Io =\",\"{:.1e}\".format(Io),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hole diffusion length is ,Lp = 2.79e-05 m\n",
"The electron diffusion length is ,Ln = 2.70e-05 m\n",
"The prefactor current is ,Io = 1.0e-14 A\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.5:pg-181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Na=10**17\n",
"Nd=10**17\n",
"Dp1 = 12.5\n",
"Dn1 = 35\n",
"Dp2 = 10\n",
"Dn2 = 220\n",
"Tn = 10**-8\n",
"Tp = 10**-8\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"pn1 = 2.25*10**3\n",
"np1 = 2.25*10**3\n",
"pn2 = 3.38*10**-5\n",
"np2 = 3.38*10**-5\n",
"#Note : since value of holes and electrons in n- type and p type are not given for silicon and germanium thus we have assume it as above\n",
"#NOTE: nn=Nd and pp=Na\n",
"Lp1 = sqrt(Dp1*Tp)\n",
"Ln1 = sqrt(Dn1*Tn)\n",
"Lp2 = sqrt(Dp2*Tp)\n",
"Ln2 = sqrt(Dn2*Tn)\n",
"# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
"# assume that the doants are fully ionised \n",
"Jo1 = e*((((Dn1)/(Ln1))*np1)+(((Dp1)/(Lp1))*pn1))\n",
"print\"The prefactor current density for silicon is ,Jo1 =\",\"{:.2e}\".format(Jo1),\"A/cm**2\"\n",
"Jo2 = e*((((Dn2)/(Ln2))*np2)+(((Dp2)/(Lp2))*pn2))\n",
"print\"The prefactor current density for GaAs is ,Jo2 =\",\"{:.2e}\".format(Jo2),\"A/cm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The prefactor current density for silicon is ,Jo1 = 3.40e-11 A/cm**2\n",
"The prefactor current density for GaAs is ,Jo2 = 9.73e-19 A/cm**2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex5.6:pg-182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Na=5*10**16\n",
"Nd=5*10**17\n",
"Dp = 15\n",
"Dn = 30\n",
"Tn = 10**-8\n",
"Tp = 10**-7\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"ni = 1.84*10**6\n",
"np=ni**2/Na\n",
"print\"The electron conc in p type is ,np=\",\"{:.2e}\".format(np),\"cm**-3\"\n",
"pn=ni**2/Nd\n",
"print\"The holes conc in n type is ,pn=\",\"{:.2e}\".format(pn),\"cm**-3\"\n",
"Lp = sqrt(Dp*Tp)\n",
"print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
"Ln = sqrt(Dn*Tn)\n",
"print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"cm\"\n",
"Gamma_inj = ((e*Dn*np)/(Ln))/(((e*Dn*np)/(Ln))+((e*Dp*pn)/(Lp)))\n",
"print\"The efficiency of diode is ,Gamma_inj =\",round(Gamma_inj,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron conc in p type is ,np= 6.77e-05 cm**-3\n",
"The holes conc in n type is ,pn= 6.77e-06 cm**-3\n",
"The hole diffusion length is ,Lp = 1.22e-03 cm\n",
"The electron diffusion length is ,Ln = 5.48e-04 cm\n",
"The efficiency of diode is ,Gamma_inj = 0.98\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.7:pg-182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"A= 0.1*10**-2\n",
"Vf= 1.0\n",
"E= 1.43\n",
"Na=5.0*10**16\n",
"Nd=5.0*10**17\n",
"Dp = 15.0\n",
"Dn = 30.0\n",
"Tn =1*10**-8\n",
"Tp = 1.0*10**-7\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"kni = 1.84*10**6\n",
"np=ni**2/Na\n",
"pn=ni**2/Nd\n",
"Ln = sqrt(Dn*Tn)\n",
"In = ((e*A*Dn*np)/Ln)*(exp(Vf/kbT)-1)\n",
"print\"The electron current is ,In =\",\"{:.1e}\".format(In),\"A\"\n",
"In_by_e = In/e\n",
"print\"The electron generation rate is ,In_by_e = \",\"{:.1e}\".format(In_by_e),\"s**-1\"\n",
"power = In*E\n",
"print\"The optical power of photon is ,power =\",\"{:.1e}\".format(power),\"W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron current is ,In = 3.0e-05 A\n",
"The electron generation rate is ,In_by_e = 1.9e+14 s**-1\n",
"The optical power of photon is ,power = 4.3e-05 W\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex5.8:pg-183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"apsilen = 11.9*8.85*10**-14\n",
"GL= 1.0*10**22\n",
"A= 1.0*10**-4\n",
"Vr= 15.0\n",
"Na=2.0*10**16\n",
"Nd=1.0*10**16\n",
"Dp = 12.0\n",
"Dn = 20.0\n",
"Tn = 1.0*10**-8\n",
"Tp = 1.0*10**-8\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"ni = 1.5*10**10\n",
"Ln = sqrt(Dn*Tn)\n",
"print\"The electron diffusion length is ,Ln =\",\"{:.1e}\".format(Ln),\"cm\"\n",
"Lp = sqrt(Dp*Tp)\n",
"print\"The hole diffusion length is ,Lp = \",\"{:.2e}\".format(Lp),\"cm\"\n",
"Vbi = kbT*log((Na*Nd)/ni**2)\n",
"print\"The built in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
"W = sqrt(((2*apsilen)/e)*((Na+Nd)/(Na*Nd))*(Vbi+Vr))\n",
"print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"cm\"\n",
"IL = e*A*GL*(W+Ln+Lp)\n",
"print\"The photo current is IL =\",\"{:.2e}\".format(IL),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron diffusion length is ,Ln = 4.5e-04 cm\n",
"The hole diffusion length is ,Lp = 3.46e-04 cm\n",
"The built in voltage is ,Vbi = 7.15e-01 V\n",
"The depletion width is ,W = 1.76e-04 cm\n",
"The photo current is IL = 1.55e-04 A\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.9:pg-185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"e = 1.6*10**-19\n",
"A=1.0*10**-7\n",
"Na=10**18*10**6\n",
"Nd=10**16*10**6\n",
"Dp = 7.8*10**-4\n",
"Dn = 7.3*10**-4\n",
"ni = 1.5*10**16\n",
"Wln = 5.0*10**-6\n",
"Wlp = Wln\n",
"#NOTE: nn=Nd and pp=Na\n",
"pn = ni**2/Nd\n",
"np = ni**2/Na\n",
"# from example 5.4 and 5.2 we get the value of diffusion length and zero bias depletion widths\n",
"Lp = 27.9*10**-6\n",
"print\"The electron diffusion length is ,Lp= \",\"{:.2e}\".format(Lp),\"m\"\n",
"Ln = 27*10**-6\n",
"print\"The electron diffusion length is ,Ln= \",\"{:.2e}\".format(Ln),\"m\"\n",
"Wp = 3.2*10**-10\n",
"print\"The zero bias depletion widths is ,Wp= \",\"{:.2e}\".format(Wp),\"m\"\n",
"Wn = 0.0000003\n",
"print\"The zero bias depletion widths is ,Wn \",\"{:.2e}\".format(Wn),\"m\"\n",
"# for short diode the prefactor current is given as\n",
"Io = e*A*(((Dp*pn)/(Wln-Wn))+((Dn*np)/abs(Wlp-Wp)))\n",
"print\"The prefactor current is ,Io = \",\"{:.2e}\".format(Io),\"A\"\n",
"# The prefactor current of short diode is approximately increase by a factor of 5.6 from that of long diode\n",
"# Note : due to different precisions taken by me and the author ... my answer differ "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron diffusion length is ,Lp= 2.79e-05 m\n",
"The electron diffusion length is ,Ln= 2.70e-05 m\n",
"The zero bias depletion widths is ,Wp= 3.20e-10 m\n",
"The zero bias depletion widths is ,Wn 3.00e-07 m\n",
"The prefactor current is ,Io = 6.03e-14 A\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.10:pg-188"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"e = 1.6*10**-19\n",
"A= 10**-7\n",
"ni = 1.5*10**16\n",
"T = 1.0*10**-6\n",
"# from example 5.2 we get the value of zero bias depletion widths\n",
"W = 0.32*10**-6\n",
"Io_GR = (e*A*W*ni)/(2*T)\n",
"print\"The prefactor of the is ,generation recombination currentIo_GR = \",\"{:.2e}\".format(Io_GR),\"A\"\n",
"#let V = .2 V \n",
"V = 0.2\n",
"I_GR = Io_GR*(exp(V/(2*kbT))-1)\n",
"print\"The diode current is ,I_GR =\",\"{:.2e}\".format(I_GR),\"A\"\n",
"#let V = 0.6 V \n",
"V = 0.6\n",
"I_GR = Io_GR*(exp(V/(2*kbT))-1)\n",
"print\"The diode current is ,I_GR = \",\"{:.2e}\".format(I_GR),\"A\"\n",
"# The generation-recombination prefactor is much larger than prefactor due to diffusion term\n",
"#In forward bias the diffusion current is initially much smaller than the generation recombination term but at high forward bias diffusion current will start to dominate\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The prefactor of the is ,generation recombination currentIo_GR = 3.84e-11 A\n",
"The diode current is ,I_GR = 1.76e-09 A\n",
"The diode current is ,I_GR = 3.94e-06 A\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex5.11:pg-189"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"A= 1.0*10**-8\n",
"Na=1.0*10**23\n",
"Nd=1.0*10**23\n",
"Dp = 10.0*10**-4\n",
"Dn = 30.0*10**-4\n",
"Tn = 1.0*10**-7\n",
"Tp = 1.0*10**-7\n",
"tau = 1.0*10**-8\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"ni = 1.5*10**16\n",
"apsilen = 11.9*8.85*10**-12\n",
"#NOTE: nn=Nd and pp=Na\n",
"Lp = sqrt(Dp*Tp)\n",
"print\"The hole diffusion length is ,Lp = \",\"{:.2e}\".format(Lp),\"m\"\n",
"Ln = sqrt(Dn*Tn)\n",
"print\"The electron diffusion length is ,Ln = \",\"{:.2e}\".format(Ln),\"m\"\n",
"# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
"np=ni**2/Na\n",
"print\"The electron conc in p type is ,np=\",\"{:.2e}\".format(np),\"m**-3\"\n",
"pn=ni**2/Nd\n",
"print\"The holes conc in n type is ,pn=\",\"{:.2e}\".format(pn),\"m**-3\"\n",
"Vbi = kbT*log((Na*Nd)/ni**2)\n",
"print\"The built in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
"Io = e*((((Dn)*np)/(Ln))+((Dp*pn)/(Lp)))\n",
"print\"The prefactor in the ideal diode is ,Io =\",\"{:.2e}\".format(Io),\"A\"\n",
"#let Vf = 0.5 V \n",
"Vf = 0.5\n",
"W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))\n",
"print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"m\"\n",
"Io_GR = e*A*W*ni/(2*tau)\n",
"print\"prefactor for recombination generation current, Io_GR = \",\"{:.2e}\".format(Io_GR),\"A\"\n",
"I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))\n",
"I_V1 = I\n",
"print\"Current, I =\",\"{:.2e}\".format(I_V1),\"A\"\n",
"#let V = 0.6 V \n",
"Vf = 0.6\n",
"W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))\n",
"print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"m\"\n",
"Io_GR = e*A*W*ni/2/tau\n",
"print\"prefactor for recombination generation current, Io_GR = \",\"{:.1e}\".format(Io_GR),\"A\"\n",
"I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))\n",
"I_V2 = I\n",
"print\"Current, I =\",\"{:.2e}\".format(I_V2),\"A\"\n",
"V1 = 0.5\n",
"V2 = 0.6\n",
"n = e*(V2-V1)/kbT/log(I_V2/I_V1)\n",
"print\"Ideallity factor ,n =\",\"{:.2e}\".format(n)\n",
"#note: in the text book the value of \n",
"#-+\"prefactor of ideal diode equation, Io\" \n",
"#calculated by author is wrong thus it efect the overall calculation of the solution ,so the answer in the are wrong of some of the values\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hole diffusion length is ,Lp = 1.00e-05 m\n",
"The electron diffusion length is ,Ln = 1.73e-05 m\n",
"The electron conc in p type is ,np= 2.25e+09 m**-3\n",
"The holes conc in n type is ,pn= 2.25e+09 m**-3\n",
"The built in voltage is ,Vbi = 8.17e-01 V\n",
"The prefactor in the ideal diode is ,Io = 9.84e-08 A\n",
"The depletion width is ,W = 9.14e-08 m\n",
"prefactor for recombination generation current, Io_GR = 1.10e-10 A\n",
"Current, I = 2.21e+01 A\n",
"The depletion width is ,W = 7.56e-08 m\n",
"prefactor for recombination generation current, Io_GR = 9.1e-11 A\n",
"Current, I = 1.04e+03 A\n",
"Ideallity factor ,n = 1.60e-19\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.12:pg-195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"apsilen = 11.9*8.85*10**-14\n",
"Na=1.0*10**19\n",
"Nd=1.0*10**16\n",
"e = 1.6*10**-19\n",
"Fcrit1 = 4.0*10**5\n",
"Fcrit2 = 1.0*10**7\n",
"VBD_Si = (apsilen*Fcrit1**2)/(2*e*Nd)\n",
"print\"The breakdown field for silicon is ,VBD_Si =\",\"{:.2e}\".format(VBD_Si),\" V\"\n",
"VBD_C = (apsilen*Fcrit2**2)/(2*e*Nd)\n",
"print\"The breakdown field for diomond is ,VBD_C =\",\"{:.2e}\".format(VBD_C),\" V\"\n",
"# Note : In the textbook answer of breakdown voltage of silicon is wrong due to which breakdown voltage of diomand also differ\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The breakdown field for silicon is ,VBD_Si = 5.27e+01 V\n",
"The breakdown field for diomond is ,VBD_C = 3.29e+04 V\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.13:pg-199"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import numpy as np\n",
"print\"Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage\"\n",
"Icv = 0.68\n",
"Vbi = Icv\n",
"print\"Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv\"\n",
"dIcv=2.1*10**23\n",
"C = 7.0*10**-13\n",
"#NOTE: The above mentioned values are taken from the figure given in the question in textbook\n",
"apsilen = 11.9*8.85*10**-12\n",
"e = 1.6*10**-19\n",
"kbT = 0.026\n",
"A = 10**-7\n",
"ni = sqrt(2.25*10**20)\n",
"Neff = 2/(A**2*e*apsilen*dIcv)\n",
"print\"The thickness of n region is ,Neff =\",\"{:.2e}\".format(Neff),\" m**-3\"\n",
"Neff = Neff/10**6\n",
"print\"The thickness of n region is ,Neff =\",\"{:.2e}\".format(Neff),\" cm**-3\"\n",
"NaNd = exp(Vbi/kbT)*ni**2\n",
"print\"NaNd =\",\"{:.2e}\".format(NaNd),\" cm**-6\"\n",
"# solving for Na and Nd by creating a quadratic equation using the equations mentioned in the book\n",
"p1 = [Neff*NaNd, -NaNd, Neff]\n",
"#Neff*NaNd - NaNd*X + Neff*X**2 \n",
"p1=p1[0]\n",
"#p2=p1[1]\n",
"print(p1)\n",
"R= roots(p1)\n",
"#s=roots(p2)\n",
"Na= R\n",
"#Nd= s\n",
"#format ('e',10)\n",
"#print\"Na = \"\"{:.2e}\".format(Na),\"cm**-3\"\n",
"#print\"Nd = \"\"{:.2e}\".format(Nd),\"cm**-3\"\n",
"W = (apsilen*A)/C\n",
"print\"The thickness of n region is ,W =\",\"{:.1e}\".format(W),\" m\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage\n",
"Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv\n",
"The thickness of n region is ,Neff = 5.65e+19 m**-3\n",
"The thickness of n region is ,Neff = 5.65e+25 cm**-3\n",
"NaNd = 5.14e+31 cm**-6\n",
"The thickness of n region is ,W = 1.5e-05 m\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.14:pg-201"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import cmath\n",
"e = 1.6*10**-19\n",
"I= 1*10**-3\n",
"kbT = 0.026\n",
"Tp = 10**-6\n",
"Gs = (I)/(kbT)\n",
"print\"The diode conductance is Gs = \",\"{:.2e}\".format(Gs),\"A/V\"\n",
"Cdiff = (I*Tp)/(2*kbT)\n",
"print\"The diffusion capacitance is Cdiff = \",\"{:.2e}\".format(Cdiff),\"F\"\n",
"# The diffusion capacitance is much larger than junction capacitance hence neglecting junction capacitance\n",
"Y = Gs+(1j*2*math.pi*10**6*Cdiff)\n",
"print\"The admittance of the diode is Y =\",Y,\"A/V\"\n",
"# Note : due to different precisions taken by me and the author ... my answer differ \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diode conductance is Gs = 3.85e-02 A/V\n",
"The diffusion capacitance is Cdiff = 1.92e-08 F\n",
"The admittance of the diode is Y = (0.0384615384615+0.120830486677j) A/V\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.15:pg-207"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Vr= 10.0\n",
"R= 10.0*10**3\n",
"#The junction capacitance is 20pF at zero bias and 10 pF at full reverse bias so\n",
"Cavg= ((20+10)/2)\n",
"Tp = 10**-7\n",
"Ir = (Vr)/(R)\n",
"Tsd = Tp*log(2)\n",
"print\"The storage delay time is Tsd = \",\"{:.2e}\".format(Tsd),\" s\"\n",
"Tt = 2.3*R*Cavg*10**-12\n",
"print\"The time Tt = \",\"{:.2e}\".format(Tt),\" s\"\n",
"T = Tsd+Tt\n",
"print\"The total diode recovery time is T = \",\"{:.2e}\".format(T),\" s\"\n",
"# Note : due to different precisions taken by me and the author ... my answer differ \n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The storage delay time is Tsd = 6.93e-08 s\n",
"The time Tt = 3.45e-07 s\n",
"The total diode recovery time is T = 4.14e-07 s\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}
|
