## Example 2-5, Page 49¶

In :
Tj=100               #junction temperature 1(C)
Vb=0.7               #Barrier potential(V)
Tamb=25              #Ambient temperature(C)
T1=0                 #junction temperature 2(C)

Vd=-0.002*(Tj-Tamb)  #Change in barrier potential(V)
Vbn=Vb+Vd            #Barrier potential(V)
Vd1=-0.002*(T1-Tamb) #Change in barrier potential(V)
Vb1n=Vb+Vd1          #Barrier potential(V)

print 'Change in barrier potential = Vd =',Vd,'V'
print 'Brrier potential = Vbn =',Vbn,'V'
print 'For junction temperature 0 C'
print 'Change in barrier potential = Vd1 =',Vd1,'V'
print 'New barrier potential = Vb1n =',Vb1n,'V'

Change in barrier potential = Vd = -0.15 V
Brrier potential = Vbn = 0.55 V
For junction temperature 0 C
Change in barrier potential = Vd1 = 0.05 V
New barrier potential = Vb1n = 0.75 V


## Example 2-6, Page 51¶

In :
T2=100    #Temperature(C)
T1=25     #Temperature(C)
Isat1=5   #Current(nA)

Td=T2-T1          #Temperature change(C)
Is1=(2**7)*Isat1  #Current(nA)
Is2=(1.07**5)*Is1 #Current(nA)

print 'Change in temperature = Td = T2-T1 =',Td,'C'
print 'For first 70 C change seven doublings are there.'
print 'Is1 = (2^7)*Isat1 =',Is1,'nA'
print 'For additional 5 C, 7% per C'
print 'Is2 = ',round(Is2,2),'nA'

Change in temperature = Td = T2-T1 = 75 C
For first 70 C change seven doublings are there.
Is1 = (2^7)*Isat1 = 640 nA
For additional 5 C, 7% per C
Is2 =  897.63 nA


## Example 2-7, Page 52¶

In :
Isl=2*(10**-9)     #surface leakage current(nA)
Vr=25              #Reverse voltage(V)
Vr1=35             #Reverse voltage(V)

Rsl=Vr/Isl             #surface leakage reistance(Ohm)
Isl1=(Vr1/Rsl)*(10**9) #surface leakage current(nA)

print 'surface-leakage reistance Rsl = Vr/Isl =',Rsl*10**-6,'MOhm'
print 'for Vr1 = 35 V,'
print 'surface-leakage reistance Isl1 = Vr1/Rsl =',Isl1,'nA'

surface-leakage reistance Rsl = Vr/Isl = 12500.0 MOhm
for Vr1 = 35 V,
surface-leakage reistance Isl1 = Vr1/Rsl = 2.8 nA

In [ ]: