# Chapter 16 High Voltage Generation¶

## Example 16_1 pgno:556¶

In [1]:
#Chapter 16,Example 1, page 556
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor

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

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

 Part (a)

Ripple Voltage =  V 2000.0

Part (b)

Voltage drop =  V 11000.0

Part (c)

Average output voltage =  kV 271.842712475

Part (d)

Ripple Factor in percentage =  3.88908729653


## Example 16_2 pgno:556¶

In [3]:
#Chapter 16,Example 2, page 556
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor

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

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

# Answers may vary due to round off error

 Part (a)

Ripple Voltage =  kV 21.0

Part (b)

Voltage drop =  kV 12.5

Part (c)

Average output voltage =  kV 411.76

Part (d)

Ripple Factor in percentage =  2.95


## Example 16_3 pgno:557¶

In [8]:
#Chapter 16,Example 3, page 557
#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor (e)optimum number of stages

I1 = 5*10**-3 # A
C = 0.15*10**-6 # F
Vs = 200 # kV
f = 50 # Hz
n = 12
from math import sqrt
# (a) Ripple voltage
print"\n Part (a)"
delV = I1*n*(n+1)/(f*C*2)
print"\n Ripple Voltage = kV", delV*10**-3
# (b) Voltage drop
print"\n Part (b)"
a = I1/(f*C)
Vd = a*((2/3*n**3)+(n**2/2)-(n/6)+(n*(n+1)/4))
print"\n Voltage drop =  kV", Vd*10**-3*12
# (c) Average output voltage
print"\n Part (c)"
Vav = 2*n*Vs*sqrt(2)-Vd*10**-3
print"\n Average output voltage =  kV", Vav
# (d) Ripple factor
print"\n Part (d)"
RF = Vd*10**-3/(2*n*Vs*sqrt(2))*10
print"\n Ripple Factor in percentage = ", RF*100
# (e) Optimum number of stages
print"\n Part (e)"
nopt = sqrt(Vs*sqrt(2)*10**3*f*C/I1)
print"\n Optimum number of stages =  stages", round(nopt)

# Answers may vary due to round off error

 Part (a)

Ripple Voltage = kV 52.0

Part (b)

Voltage drop =  kV 872.0

Part (c)

Average output voltage =  kV 6715.55843272

Part (d)

Ripple Factor in percentage =  10.704810993

Part (e)

Optimum number of stages =  stages 21.0


## Example 16_4 pgno:558¶

In [9]:
#Chapter 16,Example 4, page 558
#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
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"\n Reactance of the cable =  k ohm", Xc*10**-3
print"\n Leakage reactance of the transformer =  k ohm", XL*10**-3
print"\n Inductance of required series inductor =  H", Ind*10**3
print"\n Total circuit resistance =  k ohm", R*10**-3
print"\n maximum current that can be supplied by the transformer =  A", Imax
print"\n Exciting voltage of transformer secondary =  kV", Vex*10**-3
print"\n Input voltage of transformer primary =  V", Vin*10**-3
print"\n Input power of the transformer =  kW", P*10**-3

# Answers may vary due to round off error

 Reactance of the cable =  k ohm 125.0

Leakage reactance of the transformer =  k ohm 50.0

Additional series reactance =  k ohm 75.0

Inductance of required series inductor =  H 238.732414638

Total circuit resistance =  k ohm 21.875

maximum current that can be supplied by the transformer =  A 0.4

Exciting voltage of transformer secondary =  kV 8.75

Input voltage of transformer primary =  V 7.7

Input power of the transformer =  kW 3.5


## Example 16_5 pgno:559¶

In [10]:
#Chapter 16,Example 5,page 559
#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"\n Charging current=  micro A", I*10**6
print"\n Potential difference =  MV", V

# Answers may vary due to round off error

 Charging current=  micro A 0.5

Potential difference =  MV 50.0


## Example 16_6 pgno:560¶

In [12]:
#Chapter 16,Example 6,page 560
#Determine the wave generated

# 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
from math import sqrt
theta = sqrt(C1*C2*R1*R2)
neta = 1/(1+(1+R1/R2)*C2/C1)
alpha = R2*C1/(2*theta*neta)
print"\n Theta =  micro S",theta*10**6
print"\n Neta = ",neta
print"\n Alpha =  ",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"\n T1 =  microS", T1
print"\n T2 =  microS", T2
print"\n Generated lighting impulse =  wave", imp
print"\n alpha1 =  microS", a1
print"\n alpha2 =  microS", a2
# According to equation 16.40
et = neta*(alpha*V0)/sqrt(alpha**2-1)
print"\n e(t) =  * (e**t - f**t)",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 =  micro S 4.94772675074

Neta =  0.986880297225

Alpha =   6.96319779479

T1 =  microS 1.11048978183

T2 =  microS 49.9720401825

Generated lighting impulse =  wave 0.0222222222222

alpha1 =  microS 0.0145885577162

alpha2 =  microS 2.80011732464

e(t) =  * (e**t - f**t) 99.721740833 -0.015 -2.8


## Example 16_7 pgno:561¶

In [14]:
#Chapter 16,Example 6,page 561
#Determine the wave generated

C1 = 0.125*10**-6 # F
C2 = 1*10**-9 # F
R1 = 360. # ohm
R2 = 544. # ohm
V0 = 100. # kV
from math import sqrt
theta = sqrt(C1*C2*R1*R2)
neta = 1/(1+R1/R2+C2/C1)
alpha = R2*C1/(2*theta*neta)
print"\n Theta =  micro S",theta*10**6
print"\n Neta = ",neta
print"\n Alpha =  ",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"\n T1 =  microS", T1 # Answer given in the text is wrong
print"\n T2 =  microS", T2
print"\n alpha1 =  microS", a1
print"\n alpha2 =  microS", a2
# According to equation 16.40
et = neta*(alpha*V0)/sqrt(alpha**2-1)
print"\n e(t) =  * (e**t - f**t)",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 =  micro S 4.94772675074

Neta =  0.598886775171

Alpha =   11.4743604205

T1 =  microS 0.670004664163

T2 =  microS 80.4005596995

alpha1 =  microS 0.00882394544921

alpha2 =  microS 4.62941134867

e(t) =  * (e**t - f**t) 60.1174165831 -0.0088 -4.63


## Example 16_8 pgno:562¶

In [16]:
#Chapter 16,Example 8,page 562
#Determine the circuit efficiency

C1 = 0.125*10**-6 # F
C2 = 1*10**-9 # F
T2 = 2500.
T1 = 250.
from math import sqrt
# 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"\n Theta =  micro S", theta
print"\n X =  ", X
print"\n R1 =  k Ohm", R1*10**-3
print"\n R2 =  k Ohm", R2*10**-3
print"\n neta =  ", round(neta,3)

# Answers may vary due to round off error

 Theta =  micro S 416.666666667

X =   0.063

R1 =  k Ohm 53.3539946227

R2 =  k Ohm 26.0315820533

neta =   0.976


## Example 16_9 pgno:563¶

In [18]:
#Chapter 16,Example 9,page 563
#Determine the maximum output voltage and energy rating

n = 8.
C1 = 0.16/n # micro F
C2 = 0.001 # micro F
T2 = 50.
T1 = 1.2
from math import sqrt
# 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"\n Theta =  micro S", theta
print"\n X =  ", X
print"\n V0 =  ", V0
print"\n R1 =  Ohm", R1*10**6
print"\n R2 =  Ohm", R2*10**6
print"\n R1/n =  Ohm", R1n*10**6
print"\n R2/n =  Ohm", R2n*10**6
print"\n neta =  ", neta
print"\n Maximum output voltage =  kV", V
print"\n Energy rating =  kJ", E/1000

# Answers greatly vary due to round off error

 Theta =  micro S 5.26315789474

X =   0.025634765625

V0 =   960.0

R1 =  Ohm 434.546374554

R2 =  Ohm 3187.32736562

R1/n =  Ohm 54.3182968192

R2/n =  Ohm 398.415920703

neta =   0.946237811669

Maximum output voltage =  kV 908.388299202

Energy rating =  J 9216.0


## Example 16_10 pgno:564¶

In [20]:
#Chapter 16,Example 10,page 564
#Determine the from and tail times

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
from math import sqrt
# 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"\n R1 =  Ohm", R1
print"\n R2 =  Ohm", R2
print"\n Theta =  microS",theta*10**6
print"\n Neta = ",neta
print"\n Alpha =  ",a
print"\n T1 =  microS", round(T1,2)
print"\n T2 =  microS", round(T2,2)

# Answers greatly vary due to round off error

 R1 =  Ohm 840

R2 =  Ohm 4800

Theta =  microS 6.48074069841

Neta =  0.912408759124

Alpha =   4.22791178896

T1 =  microS 1.81

T2 =  microS 45.37


## Example 16_ 11pgno:564¶

In [23]:
#Chapter 16,Example 11,page 564
#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"\n R =  ohm",R
print"\n y =  s**-1",y
print"\n I(t) =  * exp(t) * sin(t) A",a/10000,round(-y,1)/10000,w

# Answers may vary due to round off error

 R =  ohm 1.97423402868

y =  s**-1 123389.626793

I(t) =  * exp(t) * sin(t) A 15.625 -12.33896 20000.0