Chapter 4 Electrical Breakdown of Gases

Example 4.10.1 pgno:139

In [1]:
#Claculate alpha and No. of electrons emmited
from math import log,e

#Claculate 
#(a)alpha
d2 = 0.01
d1 = 0.005
I2 = 2.7*10**-7
I1 = 2.7*10**-8
alpha = 1/(d2-d1)*log(I2/I1) 
#(b)number of electrons emmited from cathode per second
I0 = I1*e**(-alpha*d1)
n0 = I0/(1.6*10**-19)
print"Part (a)\n alpha = %.1f m^-1"%alpha
print"Part (b)\n I0 = %.1e "%I0
print"No of electrons emitted = %.3e electrons/s"%n0
#Answer may vary due to round off error

Part (a)
 alpha = 460.5 m^-1
Part (b)
 I0 = 2.7e-09 
No of electrons emitted = 1.688e+10 electrons/s

Example 4.10.2 pgno:140

In [2]:
#Chapter 4, Exmaple 2, page 140
#Claculate electrode space
from math import log
#based on the values of example 1
d2 = 0.01
d1 = 0.005
I2 = 2.7*10**-7
I1 = 2.7*10**-8
a = 1/(d2-d1)*log(I2/I1) # alpha
#10**9 = %e**a(a*d) 
#multiplying log on bith sides log(10**9) = a*d
ad = log(10**9)
print"a*d =  ",ad
d = ad/a
print"electrode space = %.3f m"%d

a*d =   20.7232658369
electrode space = 0.045 m

Example 4.10.3 pgno:140

In [3]:
#Chapter 4, Exmaple 3, page 140
#Claculate size of developed avalanche
from scipy import integrate

a = 4*10**4
b = 15*10**5
#Rewriting equation 4.2
x0=0;x1=0.0005;
def fun1(x):
    y=a-b*(x)**0.5
    return y
X=integrate.quad(fun1,x0,x1);
expX=6765.964568 
As = expX # Avelanche size
print"Avalanche size = %.0f m"%As

#Answers may vary due to round of error

Avalanche size = 6766 m

Example 4.10.4 pgno:141

In [4]:
#Claculate distance to produce avalanche
#Rewrite equation 4.2 , using the values of a and b from previous example
from numpy import array,roots

p=array([7.5*10**5, -4*10**4, 59.97])
r=roots(p)
#obtaining the roots
print"\n %f m or %f m away from the cathode"%(round(r[0],4),round(r[1],5))
#Answer may vary due to round of error.

 0.051800 m or 0.001540 m away from the cathode

Example 4.10.5 pgno:141

In [5]:
#Claculate minimum distance to produce avalanche of size 10^19
#Rewriting equation 4.2 and converting it into quadratic equation
#x=poly(0,"x");
from numpy import array,roots

p=array([7.5*10**5, 4*10**4, 43.75])
r=roots(p)
#obtaining the roots
print"\n Minimum distance =  m",round(-r[1],4)# other root is disregarded
#Answer may vary due to round of error.

 Minimum distance =  m 0.0011

Example 4.10.7 pgno:142

In [6]:
#Claculate secondary coefficient
from math import exp
#Using equation 3.15
E = 9*10**3/0.002
T = 11253.7 # m^-7*kPa^-1
B = 273840 # V/mkPa
p = 101.3 # kPa or 1 atm
d = 0.002 # m
alpha = p*T*exp(-B*p/E)
Y = 1/(exp(alpha*d)-1)
print"\n E = %.1e V/m"%E
print"Alpha = %.2f m^-1"%alpha
print"Total secondary coefficient of ionization = %f "%round(Y,5)
#Answer may vary due to round off error

 E = 4.5e+06 V/m
Alpha = 2397.29 m^-1
Total secondary coefficient of ionization = 0.008340

Example 4.10.8 pgno:143

In [7]:
#Claculate first and secondary ionization coefficient
from math import log
from math import exp

#(a)first ionization coefficient
#Using equation 4.7a
d1 = 0.005
a1d1 = log(1.22)
a1 = a1d1/d1
d2 = 0.01504
a2d2 = log(1.82)
a2 = a2d2/d2
d3 = 0.019 # wrong value used in the text
a3d3 = log(2.22)
a3 = a3d3/d3
print"Alpha 1 = %.1f m^-1"%round(a1,1)
print"Alpha 2 = %.1f m^-1"%round(a2,1)
print"Alpha 3 = %.2f m^-1"%round(a3,2)
print"From the above results we can understand that ionization mechanism must be acting at d3 "
#secondary ionization coefficient
I = 2.22
e = exp(a1*d3)
Y = (I-e)/(I*(e-1.))
print"secondary ionization coefficient = ",round(Y,4)
#Answer may vary due to round off error.

Alpha 1 = 39.8 m^-1
Alpha 2 = 39.8 m^-1
Alpha 3 = 41.97 m^-1
From the above results we can understand that ionization mechanism must be acting at d3 
secondary ionization coefficient =  0.0363

Example 4.10.9 pgno:144

In [8]:
#Claculate distance and voltage
from math import log

a = 39.8 # alpha
Y = 0.0354 # corfficient
p = 0.133 # kPa
Ep = 12000 # E/P , unit : V/m*kPa

d = (1/a)*(log(1/Y + 1)) # distance
E = Ep*p
V = E*d

print"Distance = %.3f m"%round(d,3)
print"E = %d V/m"%E
print"Volatge  = %.2f V"%round(V,2)

#Answers may vary due to round off error

Distance = 0.085 m
E = 1596 V/m
Volatge  = 135.37 V

Example 4.10.10 pgno:144

In [9]:
#Claculate (a)Raether's criterion (b)Meek and Lobe's criterion
from math import log
#(a)Raether's criterion
# as assumed by Raether and based equation 3.3, 3.50, 4.22 and 4.23
d = 0.001 # m
alpha = 10792.2 # m**-1
p = 101.3 #kPa**-1
ap = 106.54 # alpha/p Unit: m**-1*kPa**-1
T = 11253.7 # m**-1*kPa**-1
B = 273840 # V/m*kPa
Ep = 58764.81 # E/p Unit:V/m*kPa

ad = 17.7 + log(d)
E = Ep*p
Vs = E*d*10**-3 #Voltage breakdown
print"E = %.4e V/m"%E
print"Voltage breakdown = %.2f kV"%round(Vs,2)

#(b)Meek and Loeb's criterion
#Using equation 4.11 and based on 4.24 & 4,25 
#+ we get Er = 468*10**4 V/m
Er = 468*10**4 # V/m
Vs2 = Er*0.001*10**-3
print"\nVoltage breakdown = %.2f kV"%Vs2

# Answers may vary due to round of error

E = 5.9529e+06 V/m
Voltage breakdown = 5.95 kV

Voltage breakdown = 4.68 kV

Example 4.10.11 pgno:146

In [10]:
#Claculate the first Townsend's ionization coefficient

t = 0.2*10**-6 # transit time of electrons in seconds
d = 0.05 # m
ve = d/t
TC = 35*10**-9 # Time constant
a = 1/(ve*TC)
print"Electron drift velocity =%.2e m/s"%ve
print"alpha = %.1f m**-1"%round(a,1)

Electron drift velocity =2.50e+05 m/s
alpha = 114.3 m**-1

Example 4.10.12 pgno:146

In [11]:
#Travel time and maximum frequency
from math import pi
from math import asin

#(a)Determine the travel time
Ea = 200*(2)**0.5*10**3/0.1
x = 1.4*10**-4*2828.4*10**3/(2*pi*50)
d = 0.1
print"\nEa = %.4e V/m"%Ea
print"x = %.3f*sin(3.14*t)"%x
#obtaining t from x
t = asin(d/x)/3.14
print"t = %.3f ms"%t# answer mentioned in the text is wrong
#(b)Determine the maximum frequency
k = 1.4*10**-4
fmax = k*Ea/(2*pi*d)
print"fmax = %.1f Hz"%round(fmax,1)

Ea = 2.8284e+06 V/m
x = 1.260*sin(3.14*t)
t = 0.025 ms
fmax = 630.2 Hz