# Chapter 4: Junction Properties¶

### Example 4.1 page No. 146¶

In [14]:

#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]:

#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]:

#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]:

#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]:


#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 [52]:

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 [2]:

#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]:

#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]:


#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 [42]:

#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]:

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]:

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]:

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]:

#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]:

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]:

#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]:

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]:

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]:

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]:

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]:

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]:

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

In [ ]: