# Chapter 16 High Voltage Generation¶

## Example 16.7.1 pgno:556¶

In :
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor
from math import sqrt

I1 = 5*10**-3 # A
C2 = 0.05*10**-6 # F
C1 = 0.01*10**-6 # F
Vs = 100 # kV
f = 50 # Hz

# (a) Ripple voltage
print"Part (a)"
delV = I1/(C2*f)
print"Ripple Voltage = %d V"%delV
# (b) Voltage drop
print"\nPart (b)"
Vd = I1/f*((1/C1)+(1/(2*C2)))
print"Voltage drop = %.1f V"%Vd
# (c) Average output voltage
print"\nPart (c)"
Vav = 2*Vs*sqrt(2)-Vd*10**-3
print"Average output voltage = %.1f kV"%round(Vav,1)
# (d) Ripple factor
print"\nPart (d)"
RF = Vd*10**-3/(2*Vs*sqrt(2))
print"Ripple Factor in percentage = %.2f "%round(RF*100,2)

Part (a)
Ripple Voltage = 2000 V

Part (b)
Voltage drop = 11000.0 V

Part (c)
Average output voltage = 271.8 kV

Part (d)
Ripple Factor in percentage = 3.89


## Example 16.7.2 pgno:556¶

In :
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor
from math import sqrt

I1 = 5*10**-3 # A
C3 = 0.10*10**-6 # F
C2 = 0.05*10**-6 # F
C1 = 0.01*10**-6 # F
Vs = 100 # kV
f = 50 # Hz

# (a) Ripple voltage
print"\nPart (a)"
delV = I1/f*((2/C1)+(1/C3))
print"Ripple Voltage = %d kV"%(delV*10**-3)
# (b) Voltage drop
print"\nPart (b)"
Vd = I1/f*((1/C2)+(1/C1)+(1/(2*C3)))
print"Voltage drop = %.1f kV"%round(Vd*10**-3,1)
# (c) Average output voltage
print"\nPart (c)"
Vav = 3*Vs*sqrt(2)-Vd*10**-3
print"Average output voltage = %.2f kV"%round(Vav,2)
# (d) Ripple factor
print"\nPart (d)"
RF = Vd*10**-3/(3*Vs*sqrt(2))
print"Ripple Factor in percentage = %.2f "%round(RF*100,2)

# Answers may vary due to round off error

Part (a)
Ripple Voltage = 21 kV

Part (b)
Voltage drop = 12.5 kV

Part (c)
Average output voltage = 411.76 kV

Part (d)
Ripple Factor in percentage = 2.95


## Example 16.7.3 pgno:557¶

In :
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor (e)optimum number of stages
from math import sqrt

I1 = 5*10**-3 # A
C = 0.15*10**-6 # F
Vs = 200. # kV
f = 50 # Hz
n = 12

# (a) Ripple voltage
print"\nPart (a)"
delV = I1*n*(n+1)/(f*C*2)
print"Ripple Voltage = %d kV"%(delV*10**-3)
# (b) Voltage drop
print"\nPart (b)"
a = I1/(f*C)
Vd = a*((2./3*n**3)+(n**2/2)-(n/6)+(n*(n+1)/4))
print"Voltage drop = %.1f kV"%(Vd*10**-3)
# (c) Average output voltage
print"\nPart (c)"
Vav = 2*n*Vs*sqrt(2)-Vd*10**-3
print"Average output voltage = %.1f kV"%Vav
# (d) Ripple factor
print"\nPart (d)"
RF = Vd*10**-3/(2*n*Vs*sqrt(2))
print"Ripple Factor in percentage = %.1f"%(RF*100)
# (e) Optimum number of stages
print"\nPart (e)"
nopt = sqrt(Vs*sqrt(2)*10**3.*f*C/I1)
print"Optimum number of stages = %d stages"%nopt

# Answers may vary due to round off error

Part (a)
Ripple Voltage = 52 kV

Part (b)
Voltage drop = 840.7 kV

Part (c)
Average output voltage = 5947.6 kV

Part (d)
Ripple Factor in percentage = 12.4

Part (e)
Optimum number of stages = 20 stages


## Example 16.7.4 pgno:558¶

In :
#Determine the input voltage and power
from math import pi

Vc = 500*10**3        # V
A = 4                 # A
Xl = 8./100.          # in percentage
kV = 250.
Xc = Vc/A             # Reactance of the cable
XL = Xl*(kV**2/100)*10**3 # Leakage reactance of the transformer
Radd = Xc-XL          # Additional series reactance
Ind = Radd/(2*pi*XL)  # Inductance of required series inductor
R = 3.5/100.*(kV**2/100)*10**3  # Total circuit resistance
Imax = 100./250.      # maximum current that can be supplied by the transformer
Vex = Imax*R          # Exciting voltage of transformer secondary
Vin = Vex*220/kV      # Input voltage of transformer primary
P = Vin*100./220.     # Input power of the transformer
print"Reactance of the cable = %d k ohm"%(Xc*10**-3)
print"Leakage reactance of the transformer = %d k ohm"%(XL*10**-3)
print"Additional series reactance = %d k ohm"%(Radd*10**-3)
print"Inductance of required series inductor = %.1f H"%(Ind*10**3)
print"Total circuit resistance = %.2f k ohm"%(R*10**-3)
print"maximum current that can be supplied by the transformer = %.1f A"%(Imax)
print"Exciting voltage of transformer secondary = %.2f kV"%(Vex*10**-3)
print"Input voltage of transformer primary = %.1f V"%(Vin*10**-3)
print"Input power of the transformer = %.1f kW"%(P*10**-3)

# Answers may vary due to round off error

Reactance of the cable = 125 k ohm
Leakage reactance of the transformer = 50 k ohm
Additional series reactance = 75 k ohm
Inductance of required series inductor = 238.7 H
Total circuit resistance = 21.88 k ohm
maximum current that can be supplied by the transformer = 0.4 A
Exciting voltage of transformer secondary = 8.75 kV
Input voltage of transformer primary = 7.7 V
Input power of the transformer = 3.5 kW


## Example 16.7.5 pgno:559¶

In :
#Determine the charging current and potential difference

ps = 0.5*10**-6 # C/m**2
u = 10 # m/s
w = 0.1 # m
I = ps*u*w
Rl = 10**14 # ohm
V = I*Rl*10**-6
print"Charging current= %.1f micro A"%(I*10**6)
print"Potential difference = %d MV"%V

# Answers may vary due to round off error

Charging current= 0.5 micro A
Potential difference = 50 MV


## Example 16.7.6 pgno:560¶

In :
#Determine the wave generated
from math import sqrt

# With refrence to table 16.1
C1 = 0.125*10**-6 # F
C2 = 1*10**-9 # F
R1 = 360. # ohm
R2 = 544. # ohm
V0 = 100. # kV

theta = sqrt(C1*C2*R1*R2)
neta = 1./(1+(1+R1/R2)*C2/C1)
alpha = R2*C1/(2*theta*neta)
print"Theta = %.2f micro S"%(theta*10**6)
print"Neta = %.2f"%neta
print"Alpha = %.2f "%alpha
# Coresponding to alpha the following can be deduced from Fig 16.12
T2 = 10.1*theta*10**6
T1 = T2/45
imp = T1/T2 # generated lighting impulse
# From equations 16.41 and 16.42
a1 = (alpha-sqrt(alpha**2-1))*10**-6/(theta)
a2 = (alpha+sqrt(alpha**2-1))*10**-6/theta
print"\nT1 = %.2f microS"%T1
print"T2 = %.2f microS"%T2
print"Generated lighting impulse = %f wave"%imp
print"alpha1 = %.3f microS"%a1
print"alpha2 = %.2f microS"%a2
# According to equation 16.40
et = neta*(alpha*V0)/sqrt(alpha**2-1)
print"e(t) = %.3f * (e**%.3ft - f**%.3ft)"%(et,round(-a1,3),round(-a2,2)) # Equation of the wave form generated by the impulese

#Answers may vary due to round off error

Theta = 4.95 micro S
Neta = 0.99
Alpha = 6.96

T1 = 1.11 microS
T2 = 49.97 microS
Generated lighting impulse = 0.022222 wave
alpha1 = 0.015 microS
alpha2 = 2.80 microS
e(t) = 99.722 * (e**-0.015t - f**-2.800t)


## Example 16.7.7 pgno:561¶

In :
#Determine the wave generated
from math import sqrt

C1 = 0.125*10**-6 # F
C2 = 1*10**-9 # F
R1 = 360. # ohm
R2 = 544. # ohm
V0 = 100. # kV

theta = sqrt(C1*C2*R1*R2)
neta = 1/(1+R1/R2+C2/C1)
alpha = R2*C1/(2*theta*neta)
print"Theta = %.2f micro S"%(theta*10**6)
print"Neta = %.2f"%neta
print"Alpha =  %.2f"%alpha
# Coresponding to alpha the following can be deduced from Fig 16.12
T2 = 16.25*theta*10**6
T1 = T2/120.
# From equations 16.41 and 16.42
a1 = (alpha-sqrt(alpha**2-1))*10**-6/(theta)
a2 = (alpha+sqrt(alpha**2-1))*10**-6/theta
print"T1 = %.2f microS"%T1 # Answer given in the text is wrong
print"T2 = %.2f microS"%T2
print"alpha1 = %.4f microS"%a1
print"alpha2 = %.2f microS"%a2
# According to equation 16.40
et = neta*(alpha*V0)/sqrt(alpha**2-1)
print"e(t) = %.3f * (e**%.3ft - f**%.3ft)"%(et,round(-a1,4),round(-a2,2)) # Equation of the wave form generated by the impulese

#Answers may vary due to round off error

Theta = 4.95 micro S
Neta = 0.60
Alpha =  11.47
T1 = 0.67 microS
T2 = 80.40 microS
alpha1 = 0.0088 microS
alpha2 = 4.63 microS
e(t) = 60.117 * (e**-0.009t - f**-4.630t)


## Example 16.7.8 pgno:562¶

In :
#Determine the circuit efficiency
from math import sqrt

C1 = 0.125*10**-6 # F
C2 = 1*10**-9 # F
T2 = 2500.
T1 = 250.

# Bsaed on Figure 16.12
T2T1 = T2/T1
a = 4. # alpha
theta = T2/6.
# From table 16.1
X = (1/a**2)*(1+C2/C1)
R1 = (a*theta*10**-6/C2)*(1-sqrt(1-X))
R2 = (a*theta*10**-6/(C1+C2))*(1+sqrt(1-X))
neta = 1/(1+(1+R1/R2)*C2/C1)
print"Theta = %.2f micro S"%theta
print"X = %.4f "%X
print"R1 = %.1f k Ohm"%(R1*10**-3)
print"R2 = %.1f k Ohm"%(R2*10**-3)
print"neta = %.3f "%round(neta,3)

# Answers may vary due to round off error

Theta = 416.67 micro S
X = 0.0630
R1 = 53.4 k Ohm
R2 = 26.0 k Ohm
neta = 0.976


## Example 16.7.9 pgno:563¶

In :
#Determine the maximum output voltage and energy rating
from math import sqrt

n = 8.
C1 = 0.16/n # micro F
C2 = 0.001 # micro F
T2 = 50.
T1 = 1.2

# beased on figure 16.12
a = 6.4 # alpha
theta = T2/9.5
X = (1./a**2)*(1.+C2/C1)
R1 = (a*theta*10**-6/C2)*(1.-sqrt(1-X))
R2 = (a*theta*10**-6/(C1+C2))*(1+sqrt(1-X))
R1n = R1/n
R2n = R2/n
V0 = n*120
neta = 1/(1+(1+R1/R2)*C2/C1)
V = neta*V0
E = 1./2.*C1*V0**2
print"Theta = %.2f micro S"%theta
print"X = %.3f "%X
print"V0 = %.2f "%V0
print"R1 = %.2f Ohm"%(R1*10**6)
print"R2 = %.2f Ohm"%(R2*10**6)
print"R1/n = %.2f Ohm"%(R1n*10**6)
print"R2/n = %.2f Ohm"%(R2n*10**6)
print"neta = %.2f "%neta
print"Maximum output voltage = %.2f kV"%V
print"Energy rating = %.2f kJ"%(E/1000)

# Answers greatly vary due to round off error

Theta = 5.26 micro S
X = 0.026
V0 = 960.00
R1 = 434.55 Ohm
R2 = 3187.33 Ohm
R1/n = 54.32 Ohm
R2/n = 398.42 Ohm
neta = 0.95
Maximum output voltage = 908.39 kV
Energy rating = 9.22 kJ


## Example 16.7.10 pgno:564¶

In :
#Determine the from and tail times
from math import sqrt

n = 12.
C1 = 0.125*10**-6/n # micro F
C2 = 0.001*10**-6 # micro F
R1 = 70.*n # ohm
R2 = 400.*n # ohm

# beased on figure 16.15
theta = sqrt(C1*C2*R1*R2)
neta = 1./(1+R1/R2+C2/C1)
a = R2*C1/(2*theta*neta) # alpha
T2 = 7.*theta*10**6
T1 = T2/25
print"R1 = %.2f Ohm"%R1
print"R2 = %.2f Ohm"%R2
print"Theta = %.2f microS"%(theta*10**6)
print"Neta = %.2f"%neta
print"Alpha = %.2f "%a
print"T1 = %.2f microS"%round(T1,2)
print"T2 = %.2f microS"%round(T2,2)

# Answers greatly vary due to round off error

R1 = 840.00 Ohm
R2 = 4800.00 Ohm
Theta = 6.48 microS
Neta = 0.79
Alpha = 4.90
T1 = 1.81 microS
T2 = 45.37 microS


## Example 16.7.11 pgno:564¶

In :
#Determine the equation generated by impulse
from math import sqrt

w = 0.02*10**6 # s**-1 obtained by solving eq 16.47 iteratively
R = sqrt(4-(sqrt(8*8*4)*0.02)**2) # solved the simplified equation
L = 8*10**-6
V = 25*10**3
# In equation 16.46
y = R/(2.*L)
# Deriving the equation
a = V/(w*L)
print"R = %.2f ohm"%R
print"y = %.2e s**-1"%y
print"I(t) =  %.2f* exp(%.2et) * sin(%.2ft) A"%(a/10000,round(-y,1),w)

# Answers may vary due to round off error

R = 1.97 ohm
y = 1.23e+05 s**-1
I(t) =  15.62* exp(-1.23e+05t) * sin(20000.00t) A