Chapter 4: Junction Properties¶

Example 4.1 page No. 146¶

In [14]:
#Exa4.1
#find the Majority  and Minority carrier hole concentration

#given data
import math
T=300			    #in Kelvin
ND=5*10**13		     #in cm**-3
NA=0			     #in cm**-3
ni=2.4*10**13		#in cm**-3

#Calculation
no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)	#in cm**-3
po=ni**2/no		#in cm**-3

#Result
print"Majority carrier electron concentration is ",round(no,-11),"cm**-3"
print"Minority carrier hole concentration is ",round(po,-11)," cm**-3"

Majority carrier electron concentration is  5.97e+13 cm**-3
Minority carrier hole concentration is  9.7e+12  cm**-3


Example 4.2 Page No.146¶

In [16]:
#Exa4.2
#find the Majority  and Minority carrier hole concentration

#given data
import math
T=300			#in Kelvin
ND=10**16		#in cm**-3
NA=0			 #in cm**-3
ni=1.5*10**10		#in cm**-3

#Calculation
no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)	#in cm**-3
po=ni**2/no		#in cm**-3

#result
print"Majority carrier electron concentration is ",no,"cm**-3"
print"Minority carrier hole concentration is ",round(po,0)," cm**-3"

Majority carrier electron concentration is  1e+16 cm**-3
Minority carrier hole concentration is  22500.0  cm**-3


Example 4.3 Page No. 147¶

In [19]:
#Exa4.3
#find the Majority  and Minority carrier hole concentration

#given data
import math
T=300			#in Kelvin
ND=3*10**15		#in cm**-3
NA=10**16		#in cm**-3
ni=1.6*10**10		#in cm**-3

#Calculation
po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)	#in cm**-3
no=ni**2/po		#in cm**-3

#Result
print"Majority carrier hole concentration is",round(po,-8)," cm**-3"
print"Minority carrier electron concentration is ",round(no,0)," cm**-3"

Majority carrier hole concentration is 7e+15  cm**-3
Minority carrier electron concentration is  36571.0  cm**-3


Example 4.4 Page No. 147¶

In [45]:
#Example 4.4
#What is maximum Temprature

#Given
import math
ND=3*10**15		#in cm**-3
Eg=1.12             #eV
k=8.62*10**-5        #eV/k
Nc=2.8*10**19
Nv=1.04*10**19

#Calculation
import math
# from the equation po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)	#in cm**-3
No=1.05*ND
ni=math.sqrt((No-ND/2.0)**2-0.25*ND**2)
#From ni**2=Nc*Nv*exp(-Eg/(k*t))
T=Eg/(-math.log(ni**2/(Nc*Nv))*k)

#Result
print "The maximum Temprature is ",round(T,1),"K"

The maximum Temprature is  642.0 K


Example 4.5 Page No. 151¶

In [47]:
#Exa4.5
#determine the built in potential

#given data
import math
T=300		#in Kelvin
ND=10**15	#in cm**-3
NA=10**18	#in cm**-3
ni=1.5*10**10	#in cm**-3
VT=T/11600.0	#in Volts

#Calculation
Vbi=VT*math.log(NA*ND/ni**2)	#in Volts

#result
print"Built in potential barrier is",round(Vbi,4),"V"

Built in potential barrier is 0.7532 V


Example 4.6 Page No.151¶

In [4]:
#Exa4.6
#What is Contact Potential.
import math
#given data
T=300		  #in Kelvin
ND=10**21	  #in m**-3
NA=10**21	  #in m**-3
ni=1.5*10**16  #in m**-3
VT=T/11600.0	#in Volts

#Calculation
import math
Vo=VT*math.log(NA*ND/ni**2)	#in Volts

#result
print"Contact potential is",round(Vo,4),"V"

Contact potential is 0.5745 V


Example 4.7 Page No. 154¶

In [3]:
#Exa4.7
#Determine the space charge.

#given data
import math
T=300			#in Kelvin
ND=10**15		#in cm**-3
NA=10**16		#in cm**-3
ni=1.5*10**10		#in cm**-3
VT=T/11600.0		#in Volts
e=1.6*10**-19	   #in Coulamb

#calculation
epsilon=11.7*8.854*10**-14	      #constant
Vbi=VT*math.log(NA*ND/ni**2)		#in Volts
SCW=math.sqrt((2*epsilon*Vbi/e)*(NA+ND)/(NA*ND))#in cm
SCW=SCW*10**4   #in uMeter
xn=0.864		#in uM
xp=0.086		#in uM
Emax=-e*ND*xn/epsilon	#in V/cm

#result
print"Space charge width is",round(SCW,2),"micro meter"
print"At metallurgical junction, i.e for x=0 the electric field is ",round(Emax/10000,0),"V"#Note : Ans in the book is wrong

Space charge width is 0.95 micro meter
At metallurgical junction, i.e for x=0 the electric field is  -13345.0 V


Example 4.8 Page No.160¶

In [9]:
#Exa4.8
#Find the new position of fermi level

#given data
import math
Ecf=0.3               #in Volts
T=27.0+273.0              #in Kelvin

#calculation
#formula : Ecf=Ec-Ef=K*T*math.log(nc/ND)
#let K*math.log(nc/ND)=y
#Ecf=Ec-Ef=T*y
y=Ecf/T               #assumed
Tnew=273+55           #in Kelvin
EcfNEW=y*Tnew         #in Volts

#result
print"New position of fermi level is ",round(EcfNEW,4),"V"

New position of fermi level is  0.328 V


Example 4.9 Page No. 161¶

In [7]:
#Exa4.9
#Determine the Contact Potential

#given data
import math
T=300			#in Kelvin
ND=8*10**14		#in cm**-3
NA=8*10**14		#in cm**-3
ni=2*10**13		#in cm**-3
k=8.61*10**-5		#in eV/K

#calculation
Vo=k*T*math.log(NA*ND/ni**2)	#in Volts

#Result
print"Contact potential is ",round(Vo,2),"V"

Contact potential is  0.19 V


Example 4.10 page No.161¶

In [2]:
#Example 4.10
#(i)Find the hole and electron concentration
#Is this Silicon P or N type
from math import e
#given data
ND=2*10**16          #in cm**-3
NA=5*10**15          #in cm**-3
Ao=4.83*10**21   	#constant
T=300.0			    #in Kelvin
EG=1.1	 	  	    #in eV
kT=0.026     		#in eV

#Calculation
ni=Ao*T**(1.5)*math.exp(-EG/(2*kT))		#in m**-3
p=(ni/10**6)**2/ND			#in cm**-3
n=((ni/10**6)**2)/NA			#in cm**-3

#Result

print"Hole concentration in cm**-3 : %.1e"%round(p,0),"/cm**3"
print"electron concentration in cm**-3 :%.1e"%round(n,0),"/cm**3"
print"\nNOTE:\nSlight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of",ni
if n < e:

print"\n\nthe given Si is of P-type"
else:
print "\nThe given Si is of N-type"


Hole concentration in cm**-3 : 1.3e+04 /cm**3
electron concentration in cm**-3 :5.3e+04 /cm**3

NOTE:
Slight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of 1.63166259315e+16

The given Si is of N-type


Example 4.11 Page No. 168¶

In [6]:
#Exa4.11
#Determine current

#In given circuit
V=5		  #in volts
Vo=0.7	   #in Volts
R=100		#in Kohm

#Calculation
I=(V-Vo)/R	#in Ampere

#result
print"Current flowing through the circuit is",round(I*1000,0),"mA"

Current flowing through the circuit is 43.0 mA


Example 4.12 Page No. 168¶

In [20]:
#Exa4.12
#Find the Voltage VA

#In given circuit
V=15			  #in volts
Vo=0.7			#in Volts
R=7	  	 	#in Kohm

#Calculation
I=(V-2*Vo)/R
I=(V-2*Vo)/R		#in mAmpere
VA=I*R	  		#in Volts

#result
print"Voltagee VA is ",VA,"V"

Voltagee VA is  13.6 V


Example 4.13 Page No.169¶

In [23]:
#Example 4.13
#Determine the Voltage VA

#Given
V=15       #V, voltage
Vb=0.3     #V, Barrier Potential #When supply is switched on

#Calculation
VA=V-Vb

#Result
print"The Voltage VA is ",VA,"V"

The Voltage VA is  14.7 V


Example 4.14 Page No.172¶

In [22]:
#Exa4.14
#find Temperature coefficient f zener diode

#given data
Vz=5			#in volts
Vdrop=4.8		#in Volts

#calculation
delVz=Vdrop-Vz		#in Volts
TempCoeff=delVz*100/(Vz*delt)

#result
print"Temperature coefficient f zener diode is ",round(TempCoeff,3),"percent"

Temperature coefficient f zener diode is  -0.053 percent


Example 4.15 Page No. 174¶

In [47]:
#Exa4.15
#Find (a)output Voltage (b) Voltage across Rs (c) Current

#given data
Vz=8.0			#in volts
VS=12.0			#in volts
RL=10.0			#in Kohm
Rs=5.0			#in Kohm

#part (a)
Vout=Vz			#in volts

#part (b)
Vrs=VS-Vout		#in volts
IL=Vout/RL 		#in mAmpere
Is=(VS-Vout)/Rs	#in mAmpere

#part c
Iz=Is-IL	   	#in mAmpere

#result
print"(a)Output voltage will be equal to Vout=",Vout," Volts"
print"(b)Voltage across Rs is Rs=",Vrs,"V"
print"(c)Current through zener diode is Iz=",round(Iz,1),"mA"

(a)Output voltage will be equal to Vout= 8.0  Volts
(b)Voltage across Rs is Rs= 4.0 V
(c)Current through zener diode is Iz= 0.0 mA


Example 4.16 Page No. 175¶

In [32]:
#Exa4.16
#Find the min and max value of zener diode current

#given data
Vz=50.			#in volts
VSmax=120.0		#in volts
VSmin=80.0		#in volts
RL=10.0			#in Kohm
Rs=5.0			#in Kohm

#Calculation
Vout=Vz			#in Volts
IL=Vout/RL		#in mAmpere

ISmax=(VSmax-Vout)/Rs	#in mAmpere
Izmax=ISmax-IL		#in mA
Ismin=(VSmin-Vout)/Rs#in mAmpere
Izmin=Ismin-IL#in mA

#Result
print"Maximum zener diode current is ",Izmax,"mA"
print"Minimum zener diode current is ",Izmin,"mA"

Maximum zener diode current is  9.0 mA
Minimum zener diode current is  1.0 mA


Example 4.17 Page No. 175¶

In [48]:
#Exa4.17
#Design a regulator

#given data
Vz=15		#in volts
Izk=6.0		#in mA
Vout=15		#in Volts
Vs=20		#in Volts
ILmin=10.0	#in mA
ILmax=20.0	#in mA
RS=(Vs-Vz)*1000/(ILmax+Izk)	#in ohm

#result
print"sereis Resistance is ",round(RS,1),"ohm"
print"The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA"
print"when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA. \nThus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA. \nThus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant. "

sereis Resistance is  192.3 ohm
The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA
when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA.
Thus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA.
Thus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant.


Example 4.18 Page No. 175¶

In [52]:
#Exa4.18
#Determine Vl,Iz,Pz

#given data
Vs=16.0		   #in volts
RL=1.2			#in Kohm
Rs=1.0			#in Kohm

#calculation
#If zener open circuited
VL=Vs*RL/(Rs+RL)	#in Volts
Iz=0			#in mA
Pz=VL*Iz		#in watts

#result
print"When zener open circuited Voltage across load is ",round(VL,2),"V"
print"Zener current is ",Iz,"mA"
print"Power is",Pz,"watt"

When zener open circuited Voltage across load is  8.73 V
Zener current is  0 mA
Power is 0.0 watt


Example 4.19 Page No. 126¶

In [64]:
#Exa4.19
#determine VL,IL,IZ,IR

#given data
Vin=20			#in volts
Rs=220.0			#in Kohm
Vz=10		 	#in volts
RL2=50.0			#in Kohm
RL1=200			#in Kohm

#calculation
# part (i) RL=50	#in Kohm
VL1=Vin*RL1/(RL+Rs)
IR=Vin/(Rs+RL)	#in mA
IL=IR		 	#in mA
IZ=0			  #in mA

if VL1< Vz:

print"Zener diode will not conduct and VL=",round(VL1,1),"V"
else:
print "Zener diode will conduct"

#Result
print"When RL=200 ohm"
print"IL is",round(IL*1000,2),"mA"
print"IR is",round(IR*10**3,2),"mA"
print"Iz in mA: ",round(IZ,0),"mA"

# part (ii) RL=200#in Kohm
RL=200			#in Kohm
VL2=Vin*RL2/(RL2+Rs)
IR=Vin/(Rs+RL2)		#in mA
IL=IR			#in mA
IZ=0			#in mA

#result
if VL2< Vz:

print"Zener diode will not conduct and VL=",round(VL2,1),"V"
else:
print "Zener diode will conduct"

print"When RL=50 ohm"
print"IL is",round(IL*1000,2),"mA"
print"IR is",round(IR*10**3,2),"mA"
print"Iz in mA: ",IZ,"mA"

Zener diode will not conduct and VL= 9.5 V
When RL=200 ohm
IL is 47.62 mA
IR is 47.62 mA
Iz in mA:  0.0 mA
Zener diode will not conduct and VL= 3.7 V
When RL=50 ohm
IL is 74.07 mA
IR is 74.07 mA
Iz in mA:  0 mA


Example 4.20 Page No. 176¶

In [67]:
#Exa4.20
#Find the voltage drop across the resistance

#given data
RL=10.0			  #in Kohm
Rs=5.0               #in Kohm
Vin=100			  #in Volts

#Calculation
V=Vin*RL/(RL+Rs)	#in Volt
VZ=50			#in Volts
VL=VZ			#in volts
#Apply KVL
VR=100-50		#in Volts
VR=50			#in Volts

if V< VZ:

print"Zener diode is OFF state"
else:
print "zener diode is ON state"

print"Hence the voltage dropp across the 5 Kohm resistor in Volts is ",VR,"V"

zener diode is ON state
Hence the voltage dropp across the 5 Kohm resistor in Volts is  50 V


Example 4.21 Page No. 176¶

In [72]:
#Exa 4.21
#Find the input resistance

#given data
Izmin=20		#in mA min. diode current
Izmax=200		#in mA max. diode current
VL=12			#in Volts
VDCmin=15		#in Volts
VDCmax=19.5		#in Volts
Vz=12			#in Volts
IL=VL/RL		#in Ampere
IL=IL*1000		#in mAmpere

#calculation
#For VDCmin = 15 volts
VSmin=VDCmin-Vz		#in Volts
#For VDCmax = 19.5 volts
VSmax=VDCmax-Vz		#in Volts
ISmin=Izmin+IL		#in mA
Ri=VSmin/ISmin		#in Kohm
Ri=Ri*10**3		#in ohm

#result
print"The resistance Ri is ",Ri,"ohm"

The resistance Ri is  25.0 ohm


Example 4.22 Page No. 177¶

In [71]:
#Exa4.22
#Determine the range of Rl and Il

#given data
VRL=10			#in Volts  Diode resistance
Vi=50			#in Volts
R=1.0			#in Kohm  Resistance
Vz=10			#in Volts
VL=Vz			#in Volts
Izm=32			#in mA
IR=(Vi-VL)/R		#in mA

Izmin=0			   #in mA
ILmax=IR-Izmin		#in mA
RLmin=VL/ILmax		#in Ohm
Izmax=32		      #in mA
ILmin=IR-Izmax		#in mA
VL=Vz			     #in Volts
RLmax=VL/ILmin		#in Ohm

#Result
print"Range of RL in Kohm : From ",RLmin*1000,"ohm to ",RLmax,"kohm"
print"Range of IL in mA : From ",ILmin,"mA to ",ILmax,"mA"

Range of RL in Kohm : From  250.0 ohm to  1.25 kohm
Range of IL in mA : From  8.0 mA to  40.0 mA