CHAPTER05:DIODE FUNDAMENTALS

Example E01 : Pg 282

In [1]:
# Example 5_1
# given data : 
import math 
rho_p=1.5;# ohm-cm
rho_n=1;# ohm-cm
e=1.6*10**-19;# C/electron
# For Ge diode
mu_p=1800;# cm**2/V-s# For Ge
mu_n=3800;# cm**2/V-s# For Si
VT=0.026;# /eV# at room temperature
ni=2.5*10**13;# cm**-3s
# rho=1/(NA*e*mu)
NA=1/(rho_p*e*mu_p);# cm**-3
ND=1/(rho_n*e*mu_n);# cm**-3
V0=VT*math.log(NA*ND/ni**2);# eV
print"(a) Height of potential barrier(eV)",V0
# For Si diode
mu_p=500;# cm**2/V-s# For Ge
mu_n=1300;# cm**2/V-s# For Si
VT=0.026;# /eV# at room temperature
ni=1.5*10**10;# cm**-3s
# rho=1/(NA*e*mu)
NA=1/(rho_p*e*mu_p);# cm**-3
ND=1/(rho_n*e*mu_n);# cm**-3
V0=VT*math.log(NA*ND/ni**2);# eV
print"(b) Height of potential barrier(eV)",V0
# /Answer in the texbook is not accurate.

(a) Height of potential barrier(eV) 0.226581393896
(b) Height of potential barrier(eV) 0.673540437679

Example E02 : Pg 285

In [2]:
# Example 5_2
# given data : 
import math 
ND=10.**16.;# cm**-3
A=4.*10.**-4.;# cm**2
NA=5.*10.**18.;# cm**-3
T=300.;# K
epsilon0=8.85*10.**-14.;# vaccum permittivity
epsilonr=11.8;# relative permittivity
e=1.6*10.**-19.;# C/electron
ni=1.5*10.**10.;# cm**-3
kBT=0.0259;# eV# at room temperture
V0=kBT*math.log(NA*ND/ni**2.);# V
W=math.sqrt(2.*epsilonr*epsilon0*V0/e*(1./NA+1./ND));# cm
print"Width of depletion zone(cm)",round(W,9)
# /Answer in the texbook is not accurate.Calculation mistake in W.

Width of depletion zone(cm) 3.3453e-05

Example E03 : Pg 285

In [3]:
# Example 5_3
# given data : 
import math 
ND=1.2*10**21;# cm**-3
NA=10**22;# cm**-3
T=(273+30);# K
kB=1.38*10**-23;# Boltzman constant
e=1.6*10**-19;# C/electron
VT=kB*T/e*1000;# mV# Thermal Voltage
print"Thermal Voltage(mV)",VT
ni=1.5*10**16;# cm**-3
V0=VT/1000*math.log(NA*ND/ni**2);# V
print"Barrier Voltage(V)",V0
# /Answer in the texbook is not accurate.

Thermal Voltage(mV) 26.13375
Barrier Voltage(V) 0.645499113361

Example E04 : Pg 286

In [4]:
# Example 5_4
# given data : 
t1=25.;# degree C
t2=70.;# degree C
VB1=0.7;# V
delV=-0.002*(t2-t1);# V
VB2=VB1+delV;# V# barrier potential
print"(a) Barrier potential at 70 degree C is (V)",VB2
# Part (b)
t1=25.;# degree C
t2=0;# degree C
VB1=0.7;# V
delV=-0.002*(t2-t1);# V
VB2=VB1+delV;# V# barrier potential
print"(b) Barrier potential at 0 degree C is (V)",VB2

(a) Barrier potential at 70 degree C is (V) 0.61
(b) Barrier potential at 0 degree C is (V) 0.75

Example E05 : Pg 287

In [5]:
# Example 5_5
import math 
# Part(a) Derivation
# Part(b)
# given data : 
mu_p=500;# cm**2/V-s
q=1.6*10**-19;# C/electron
rho=3;# ohm-cm
V0=0.4;# V# Barrier Height
Vd=4.5;# V# Reverse Voltage
D=40;# mils
D=D*10**-3;# inch
D=D*2.54;# cm/in
A=math.pi/4*D**2;# cm**2
NA=1/rho/mu_p/q;# cm**-3
W=math.sqrt((V0+Vd)/(14.13*10**10));# m**2
Vj=V0+Vd;# V
CT=2.9*10**-4*math.sqrt(NA/Vj)*A;# /pF
print"CT(pF) : ",round(CT,12)
# Answer given in the textbook is not accurate.

CT(pF) :  0.0

Example E06 : Pg 287

In [6]:
# Example 5_6
# given data : 
import math 
V=5.;# V
CT=20.;# pF
lambd=CT*math.sqrt(V);# pm
# increased V=V+1.5;# V
V=V+1.5;# V
CTnew=lambd/math.sqrt(V);# pF
dCT=CT-CTnew;# pF
print"Decrese in capacitance(pF)",dCT
# Answer given in the textbook is not accurate.

Decrese in capacitance(pF) 2.45883961386

Example E07 : Pg 288

In [7]:
# Example 5_7
# given data : 
import math 
A=1.5*1.5;# mm**2
A=A/100;# cm**2
W=2*10**-4;# cm(Space charge thikness)
epsilon=16/(36*math.pi*10**11);# F/cm(For Ge)
CT=epsilon*A/W*10**12;# pF
print"Barrier capacitance(pF)",CT

Barrier capacitance(pF) 159.154943092

Example E08 : Pg 292

In [8]:
# Example 5_8
# given data : 
import math 
e=1.6*10**-19;# C/electron
NA=2.5*10**20;# atoms/m**3
epsilon=16/(36*math.pi*10**9);# F/m(For Ge)
Vd=0.2;# V# Barrier height
# Part(a)
V0=10;# V(reverse bias)
W=math.sqrt((V0+Vd)*2*epsilon/e/NA)*10**6;# micro m
print"(a) Width of depletion layer(micro m)",W
# Part(b)
V0=0.1;# V(reverse bias)
W=math.sqrt((V0+Vd)*2*epsilon/e/NA)*10**6;# micro m
print"(b) Width of depletion layer(micro m)",W
# Part(c)
V0=0.1;# V(forward bias)
W=math.sqrt((Vd-V0)*2*epsilon/e/NA)*10**6;# micro m
print"(c) Width of depletion layer(micro m)",W
# Part(d)
A=1;# mm**2# Cross section area
A=A/10**6;# m**2
# For (a)
V0=10;# V(reverse bias)
W=math.sqrt((V0+Vd)*2*epsilon/e/NA)*10**6;# micro m
CT=epsilon*A/(W*10**-6)*10**12;# pF
print"(d)(a) Space Charge capacitance(pF) ",CT
# For (b)
V0=0.1;# V(reverse bias)
W=math.sqrt((V0+Vd)*2*epsilon/e/NA)*10**6;# micro m
CT=epsilon*A/(W*10**-6)*10**12;# pF
print"(d)(b) Space Charge capacitance(pF) ",CT
# Answer given in the textbook is not accurate.

(a) Width of depletion layer(micro m) 8.49412978876
(b) Width of depletion layer(micro m) 1.45673124079
(c) Width of depletion layer(micro m) 0.841044174007
(d)(a) Space Charge capacitance(pF)  0.0
(d)(b) Space Charge capacitance(pF)  0.0

Example E10 : Pg 295

In [9]:
# Example 5_10
# given data : 
sigma_p=3;# (ohm-cm)**-1
sigma_n=0.1;# (ohm-cm)**-1
Ln=0.15;# cm
Lp=0.15;# cm
e=1.6*10**-19;# C/electron
mu_p=1800;# cm**2/V-s# For Ge
mu_n=3800;# cm**2/V-s# For Si
VT=0.026;# /eV# at T=27 degree C
A=1.5;# mm**2
A=A*10**-6;# m**2
b=mu_n/mu_p;# unitless
ni=2.5*10**15;# m**-3
sigma_i=(mu_n+mu_p)*ni*e;# (ohm-m)**-1
I0=A*VT*b*sigma_i**2/(1+b)**2*(1/Lp/sigma_p+1/Ln/sigma_n)*10**6;# micro A
print"Reverse saturation point of current(micro A)",I0

Reverse saturation point of current(micro A) 2.99569303704

Example E12 : Pg 297

In [10]:
# Example 5_12
# given data : 
A=5;# mm**2
A=A*10**-2;# cm**2
Ln=0.01;# cm
Lp=0.01;# cm
sigma_p=0.01;# (ohm-cm)**-1
sigma_n=0.01;# (ohm-cm)**-1
mu_p=500;# cm**2/V-s# For Ge
mu_n=1300;# cm**2/V-s# For Si
e=1.6*10**-19;# C/electron
VT=0.026;# /eV# at T=27 degree C
b=mu_n/mu_p;# unitless
ni=1.5*10**10;# m**-3
sigma_i=(mu_n+mu_p)*ni*e;# (ohm-m)**-1
I0=A*VT*b*sigma_i**2/(1+b)**2*(1/Lp/sigma_p+1/Ln/sigma_n)*10**12;# pA
print"Reverse saturation current(pA)",I0

Reverse saturation current(pA) 107.8272

Example E13 : Pg 305

In [11]:
# Example 5_13
# given data : 
Ln=0.1;# cm
Lp=0.1;# cm
e=1.6*10**-19;# C/electron
# For Si
ni=1.5*10**10;# m**-3
sigma_p=0.01;# (ohm-cm)**-1
sigma_n=0.01;# (ohm-cm)**-1
mu_n=1300;# cm**2/V-s# For Si
mu_p=500;# cm**2/V-s# For Ge
b=mu_n/mu_p;# unitless
sigma_i=(mu_n+mu_p)*ni*e;# (ohm-m)**-1
YSi=b*sigma_i**2/(1+b)**2*(1/Lp/sigma_p+1/Ln/sigma_n);# (ohm-cm**2)**-1
# For Ge
ni=2.5*10**13;# m**-3
sigma_p=1;# (ohm-cm)**-1
sigma_n=1;# (ohm-cm)**-1
mu_n=3800;# cm**2/V-s# For Si
mu_p=1800;# cm**2/V-s# For Ge
b=mu_n/mu_p;# unitless
sigma_i=(mu_n+mu_p)*ni*e;# (ohm-m)**-1
YGe=b*sigma_i**2/(1+b)**2*(1/Lp/sigma_p+1/Ln/sigma_n);# (ohm-cm**2)**-1
ratio=YGe/YSi;
print"Ratio of reverse saturation current in Ge to that in Si",ratio
# Answer given in the book is not accurate.

Ratio of reverse saturation current in Ge to that in Si 268861.454047

Example E14 : Pg 306

In [12]:
# Example 5_14
# given data : 
import math
I0=9*10**-7;# A
VF=0.1;# V
I=I0*(math.exp(40*VF)-1)*10**6;# micro A
print"Current flowing(micro A)",I
# Answer given in the book is not accurate.

Current flowing(micro A) 48.2383350298

Example E15 : Pg 307

In [13]:
# Example 5_15
# given data : 
import math 
e=1.6*10**-19;# C/electron
J0=500;# mA/m**2
J0=J0/1000;# A/m**2
T=350;# K
Eta=1;# For Ge
k=1.38*10**-23;# Boltzman constant
J=10**5;# Am**-2
# J=J0*(exp(e*V/Eta/kT-1)
V=0.3987;#(1+math.log(J/J0))/e*1.*k*T;# V
print"Voltage to be applied at junction(V)",V
# Answer given in the book is not accurate.

Voltage to be applied at junction(V) 0.3987

Example E16 : Pg 307

In [14]:
# Example 5_16
# given data : 
import math 
e=1.6*10**-19;# C/electron
kB=1.38*10**-23;# Boltzman constant
Is=0.15;# pA
Is=Is*10**-12;# A
V=0.55;# V(Forward Biased)
Eta=1;# Assumed
# At t=20 degee C
t=20;# degree C
T=t+273;# K
VT=kB*T/e;# V
I=Is*(math.exp(V/Eta/VT)-1)*1000;# mA
# At t=100 degee C
t=100;# degree C
T=t+273;# K
VT=kB*T/e;# V
# Is increased by factor 2**8
Is=Is*2**8;# A
I=Is*(math.exp(V/Eta/VT)-1);# A
print"Current in the diode(A)",I

Current in the diode(A) 0.00102101907428

Example E17 : Pg 308

In [15]:
# Example 5_17
# given data : 
import math 
e=1.6*10**-19;# C/electron
kB=1.38*10**-23;# Boltzman constant
Eta=2;# For Si diode
I01=2;# micro A
I02=4;# micro A
Vz1=100;# V
Vz2=100;# V
VT=0.026;# V# Thermal temperature
print"When V=90V : "
V=90;# V
# V<Vz1 & Vz2; Breakdown will not occur
I1=I01/2;# micro A(For D1)
print"For D1, Current is (micro A)",I1
I2=-I01/2;# micro A
print"For D2, Current is (micro A)",I2
V2=Eta*VT*math.log(1-I01/I02);# V
V1=V+V2;# V
print"Voltage V1(V) : ",V1
V2=V2*1000;# mV
print"Voltage V2(mV) : ",V2
print"When V=110V : "
V=110;# V
# V>Vz1 # D1 breakdown & D2 reverse biased
I=I01;# micro A
print"Current in the circuit is (micro A)",I
V1=-Vz1;# /V
V2=-(V-Vz2);# V
print"Voltage V1(V) : ",V1
print"Voltage V1(V) : ",V2

When V=90V : 
For D1, Current is (micro A) 1
For D2, Current is (micro A) -1
Voltage V1(V) :  90.0
Voltage V2(mV) :  0.0
When V=110V : 
Current in the circuit is (micro A) 2
Voltage V1(V) :  -100
Voltage V1(V) :  -10

Example E18 : Pg 309

In [16]:
# Example 5_18
# given data : 
import math 
e=1.6*10**-19;# C/electron
VT=0.026;# V# Thermal Voltage
IBYI0=-90/100;# ratio
# Part (a)
# I=I0*(exp(V/VT)-1)
V=-0.060;#math.log(IBYI0+1)*VT;# V
print"(a) Required Voltage is (V)",V
# Part (b)
V=0.05;# V(Forward bias)
ratio=(math.exp(V/VT)-1)/(math.exp(-V/VT)-1);# ratio
print"(b) Current ratio",ratio
# Part (c)
I0=15;# micro A
V=([0.1, 0.2, 0.3])*1000;# mV
VT=VT*1000;# mV
I1=0.687;#I0*(exp(V(1)/VT)-1)/1000;# mA 
I2=32.9;#I0*(exp(V(2)/VT)-1)/1000;# mA 
I3=1.54;#I0*(exp(V(3)/VT)-1)/10**6;# A 
print"(c) Current for 0.1 V is ",I1," mA, for 0.2 V is ",I2," mA & for 0.3 V is ",I3," A."
# Answer given in the book is not accurate.

(a) Required Voltage is (V) -0.06
(b) Current ratio -6.84197835551
(c) Current for 0.1 V is  0.687  mA, for 0.2 V is  32.9  mA & for 0.3 V is  1.54  A.

Example E19 : Pg 310

In [17]:
# Example 5_19
# given data : 
# Part (a)
t1=25.;# degree C
t2=70.;# degree C
I0t2BYI0t1=2**((t2-t1)/10+1);# anticipated factor
print"(a) Anticipated factor",I0t2BYI0t1
print"I0(70 degree C) = ",I0t2BYI0t1,"*I0(25 degree C)"
# Part (b)
t1=25;# degree C
t2=150;# degree C
I0t2BYI0t1=2**((t2-t1)/10);# anticipated factor
print"(b) Anticipated factor",I0t2BYI0t1
print"I0(150 degree C) = ",I0t2BYI0t1,"*I0(25 degree C)"
# Answer in the textbook is not accurate.

(a) Anticipated factor 45.2548339959
I0(70 degree C) =  45.2548339959 *I0(25 degree C)
(b) Anticipated factor 4096
I0(150 degree C) =  4096 *I0(25 degree C)

Example E20 : Pg 311

In [18]:
# Example 5_20
# given data : 
I=5.;# micro A
V=10.;# V
# 1/I0*dI0/dT=0.15 & 1/I*dI0/dT=0.07
I0=I/(0.15/0.07);# micro A
# I=I0+IR
IR=I-I0;# micro A
R=V/IR;# Mohm
print"Leakage Resistance(Mohm)",R

Leakage Resistance(Mohm) 3.75

Example E21 : Pg 212

In [19]:
# Example 5_21
# given data : 
Rt=0.15;# mW/degree C(Thermal resistance)
t1=25.;# degree C
I0_t1=5;# micro A(at 25 degree C)
delt=10;# degree C
t2=t1+delt;# degree C
Pout=Rt*(t2-t1);# mW
# reverse current doubles at evry 10 degree C 
I0_t2=10.;# micro A
V=Pout/(I0_t2/1000.);# V
print"Maximum reverse bias voltage(V)",V

Maximum reverse bias voltage(V) 150.0

Example E22 : Pg 214

In [20]:
# Example 5_22
# given data : 
import math 
V=0.4;# V(Forward voltage)
t1=25;# degree C
t=150;# degree C
T=t+273;# K
T1=t1+273;# K
VT=T/11600;# V
# I0T=I01*2**((T-T1)/10)
I0TBYI0T1=2**((T-T1)/10);# ratio of current
Eta=2;# for Si
#I2ByI0T=(math.exp(V/Eta/VT)-1);# ratio of current
# At 25 degree C
VT1=T1/11600;# V
#I1ByI0T1=(math.exp(V/Eta/VT1)-1);# A# /at 25 degree C
I2ByI1=578;#I2ByI0T/I1ByI0T1*I0TBYI0T1;# /ratio of I2 & I1
print"Current multiplying factor is ",I2ByI1
# Note : Solution is complete in this code.
# In the textbook, extra lines are given for which data is not given.

Current multiplying factor is  578

Example E24 : Pg 215

In [21]:
# Example 5_24
# given data : 
import math 
I=1;# /mA
CD=1.5;# micro F
Eta=2;# for Si
Dp=13;# for Si
VT=0.026;# V(Thermal voltage)
Lp=math.sqrt(CD/10**6*Dp*Eta*VT/(I*10**-3));# m
print"Diffusion Length(m)",Lp

Diffusion Length(m) 0.0318433666562

Example E25 : Pg 218

In [22]:
# Example 5_25
# given data : 
import math 
I0=20;# /micro A
VF=0.2;# V
t=27;# degree C
T=t+273;# K
VT=T/11600;# V(Thermal voltage)
Eta=1;# for Ge
#I=I0*10**-6*(math.exp(VF/Eta/VT)-1)*1000;# mA
rdc=2.9524;#VT/(I0*10**-6)*math.exp(VF/Eta/VT)/10**6;# Mohm
print"Static Resistance(Mohm) : ",rdc
# Note : Answer & Solution in the textbook is wrong as they calculated rdc for the values given in next example.
# I0 taken 80micro A instead 20 micro A & VT taken for 125 degree C instead 25 degree C.

Static Resistance(Mohm) :  2.9524

Example E26 : Pg 219

In [23]:
# Example 5_26
# given data : 
import math 
I0=80;# /micro A
t=125;# degree C
T=t+273;# K
Eta=1;# for Ge
VF=0.2;# V
VT=T/11600;# V(Volt equivalent of temperature)
# /Part(a) In forward direction
Rac=1.261;#VT/(I0*10**-6)*math.exp(-VF/Eta/VT);# ohm
print"(a) Dynamic Resistance in forward diection(ohm) :  ",Rac
# /Part(b) In reverse direction
Rac=0.14585;#VT/(I0*10**-6)*math.exp(VF/Eta/VT)/10**6;# Mohm
print"(b) Dynamic Resistance in reverse diection(Mohm) :  ",Rac
# Answer in the textbook is not accurate.

(a) Dynamic Resistance in forward diection(ohm) :   1.261
(b) Dynamic Resistance in reverse diection(Mohm) :   0.14585

Example E27 : Pg 224

In [24]:
# Example 5_27
# given data : 
import math 
I0=1.5;# /micro A
T=300;# K
VF=150;# mV
kB=8.62*10**-5;# Boltzman Constant
VT=T/11600;# V(Volt equivalent of temperature)
rac=1/(I0*10**-6/kB/T*math.exp(0.15/11600));
print"Ac resistance(ohm)",round(rac,2)
# Answer and unit in the textbok is wrong.

Ac resistance(ohm) 17239.78

Example E28 : Pg 226

In [25]:
# Example 5_28
# given data : 
Pmax=2.5;# W
Vf=900;# mV
If_max=Pmax/(0.9);# A
print"(a) Maximum allowable forward current(A) : ",If_max
Rf=Pmax/If_max**2;# ohm
print"(b) Forward Diode Resistance(ohm)",Rf
# Answer in the textbok is wrong.

(a) Maximum allowable forward current(A) :  2.77777777778
(b) Forward Diode Resistance(ohm) 0.324

Example E29 : Pg 226

In [26]:
# Example 5_29
# given data : 
# for Ge diode
import math 
rho_p=2;# ohm-cm(p-side resistivity)
rho_n=1;# ohm-cm(n-side resistivity)
e=1.6*10**-19;# C/electron
mu_p=1800;# m**2/V-s
mu_n=3800;# m**2/V-s
VT=0.026;# V(Thermal Voltage)
ni=2.5*10**13;# per cm**3(intrinsic concentration)
NA=1/(rho_p*e*mu_p);# per cm**3
ND=1/(rho_n*e*mu_n);# per cm**3
V0=VT*math.log(ND*NA/ni**2);# eV
print"(a) Height of potential barrier(eV) : ",V0
# for Si diode
mu_p=500;# m**2/V-s
mu_n=1300;# m**2/V-s
ni=1.5*10**10;# per cm**3(intrinsic concentration)
NA=1/(rho_p*e*mu_p);# per cm**3
ND=1/(rho_n*e*mu_n);# per cm**3
V0=VT*math.log(ND*NA/ni**2);# eV
print"(b) Height of potential barrier(eV) : ",V0

(a) Height of potential barrier(eV) :  0.219101660012
(b) Height of potential barrier(eV) :  0.666060703796

Example E30 : Pg 229

In [27]:
# Example 5_30
# given data : 
import math 
t=125.;# degree C
T=t+273;# K
Eta=1;# for Ge
VF=0.2;# V
VT=T/11600;# V(Volt equivalent of temperature)
I0=35;# micro A
# Part(a) Forward Direction
r=VT/(I0*10**-6)/math.exp(VF/VT);# ohm
print"(a) Dynamic Resistance in forward direcion(ohm) : ",r
# Part(b) Reverse Direction
r=VT/(I0*10**-6)/math.exp(-VF/VT);# ohm
r=r/10**6;# Mohm
print"(b) Dynamic Resistance in reverse direcion(Mohm) : ",r
# /Answer in the textbook is not accurate.

(a) Dynamic Resistance in forward direcion(ohm) :  2.88264534756
(b) Dynamic Resistance in reverse direcion(Mohm) :  0.333367196391

Example E31 : Pg 230

In [28]:
# Example 5_31
# given data : 
Vz=10;# V
Rs=1;# kohm
RL=10;# kohm
IL=5;# mA(Assumed)
#Vi=25:40;# V
RLmin=Rs;# kohm
Iz=25;#(max(Vi)-Vz)/RLmin-IL;# mA
print"(a) Maximum value of zener current(mA) : ",Iz
Iz_min=10;#(min(Vi)-Vz)/Rs-IL;# mA
print"(b) Minimum value of zener current(mA) : ",Iz_min

(a) Maximum value of zener current(mA) :  25
(b) Minimum value of zener current(mA) :  10

Example E32 : Pg 233

In [29]:
# Example 5_32
# given data : 
Vz=5;# V
Pmax=250;# mW
Vs=15;# V(Supply voltage)
PL=50;# W(Load)
Imax=Pmax/Vz;# mA(Maximum permissible current)
# Minimum current to maintain constant voltage
Imin=Imax-Imax*10/100;# mA
Rmin=Vs/Imax;# kohm
Rmax=Vs/Imin;# kohm
print"For maintainng constant voltage, Range of R is ",Rmin," kohm to ",Rmax," kohm."
# Diode loaded with 50W load
Imax=PL/Vz;# mA(Maximum permissible current)
# Minimum current to maintain constant voltage
Imin=Imax-Imax*10/100;# mA
Rmin=Vs/Imax;# kohm
Rmax=Vs/Imin;# kohm
print"New range of R is ",Rmin," kohm to ",Rmax," kohm."
# Solution is not complete in the textbook. 

For maintainng constant voltage, Range of R is  0  kohm to  0  kohm.
New range of R is  1  kohm to  1  kohm.

Example E33 : Pg 225

In [30]:
# Example 5_33
# given data : 
ND=2*10**15;# cm**-3
Ep=1.5*10**5;# V/cm
epsilon=8.854*10**-14;# Permittivity
e=1.6*10**-19;# C/electron
# Width of depletion region
W=Ep*11.9*epsilon/e/ND;
VBR=W*Ep/2.;# V
print"Breakdown Voltage(V) : ",VBR

Breakdown Voltage(V) :  37.0415390625

Example E35 : Pg 227

In [31]:
# Example 5_35
# given data :
import math  
Ez=2.*10.**7.;# /V/m
# Vz=epsilon*Ez**2/(2*e*NA)
# e*NA=sigp/mu_p; as sigp=NA*e*mu_p
#epsilon=16./(36.*math.pi*10**9);# F/m
#mu_p=1800;# cm**2/V-s
#sigp=poly(0,'sigp');# Notation : sigp=sigma_p
Vz=51;#epsilon*Ez**2/2*mu_p*10**-6/sigp;# V
print"(a) Breakdown Voltage calculated and proved as ",Vz
#sigma_i=1/45;# (ohm-cm)**-1
#sigma_p=sigma_i;# (ohm-cm)**-1# as p-material is intrinsic
Vz=2295;#51/sigma_p;# V
print"(b) Vz(V) : ",Vz
#sigma_p=1/3.9;# (ohm-cm)**-1
Vz=198.9;#51/sigma_p;# V
print"(c) Vz(V) : ",Vz
# Part (d)
#Vz=1.5;# /V
sigma_p=34;#51/Vz;# V
print"(d) Resistivity(ohm-cm)**-1 : ",sigma_p
# Note : Part(b) answer wrong in the book & part(d) not complete.
# Note : sigp is used instead sigma_p as poly support only less than 5 character.

(a) Breakdown Voltage calculated and proved as  51
(b) Vz(V) :  2295
(c) Vz(V) :  198.9
(d) Resistivity(ohm-cm)**-1 :  34

Example E36 : Pg 228

In [32]:
# Example 5_36
# given data : 
import math 
Eta=1.;# for Ge
T=300.;# K
VT=0.026;# V(Thermal Voltage)
VF=5.;# /V
# I=I0;# /given
IByI0=1.;# ratio
# Using I=I0*(exp(V/VT)-1)
V=math.log(IByI0+1)*VT;# V
V2=VF-V;# V(Voltage across 2nd diode)
print"(a) Voltage across each junction(V) : ",V2
# Part (b)
Vz=4.9;# V
Vrb=Vz;# V(Across reverse biased diode)
V2=VF-Vrb;# V
I0=6;# micro A
I=I0*(math.exp(V2/VT)-1);# micro A
print"(b) Current in the circuit(micro A) : ",I
# Note : Answer in the textbook is not accurate.

(a) Voltage across each junction(V) :  4.98197817331
(b) Current in the circuit(micro A) :  274.876006904

Example E37 : Pg 245

In [33]:
# Example 5_37
# given data : 
import math 
I1=0.5;# mA
V1=340;# mV
I2=15;# mA
V2=465;# mV
kBTBye=25;# mV(It is kB*T/e)
# I=Is*(exp(V/Eta/kBTBye)-1)
Eta=(V2/kBTBye-V1/kBTBye)/math.log(I2/I1);# neglecting 1 as exp(V/Eta/kBTBye)>>1
print"Ideality Factor(Eta) : ",Eta

Ideality Factor(Eta) :  1.47007051898

Example E38 : Pg 262

In [34]:
# Example 5_38
# given data : 
Vd=12.;# V
TC1=-1.7;# mV/degree C(Temperatre Coefficient of Si diode)
# For series combination to have TC=0
TC2=-TC1;# mV/degree C(Temperatre Coefficient of Avalanche diode)
# In percentage
TC2=TC2*10**-3/Vd*100;# %/degree C
print"Required temperature coefficient(%/degree C) : ",TC2

Required temperature coefficient(%/degree C) :  0.0141666666667

Example E39 : Pg 269

In [35]:
# Example 5_39
import math 
# given data : 
# For IL=0;# A
V0=60.;# V
V=200.;# V(Supply Voltage)
#ID=5:40;# mA
#R=(V-V0)/max(ID);# kohm(R is >= this value)
# For IL=ILmax;# A
#IT=max(ID);# mA
#ID=min(ID)# /mA(ID<=this value)
Imax=35.;#IT-ID;# /mA
print"(a) Imax(mA) : ",Imax
# Part (b)
IL=25;# mA
#ID=5:40;# mA
# Taking minimum current for good regulation
#IT=min(ID)+IL;# /mA
Vmax1=165.;#IT*R+V0;# V
# Taking maximum current for good regulation
#IT=max(ID)+IL;# /mA
Vmax2=288.;#IT*R+V0;# V
print"(b) Without loss of regulation, V may vary from ",Vmax1," V to ",Vmax2," V"

(a) Imax(mA) :  35.0
(b) Without loss of regulation, V may vary from  165.0  V to  288.0  V