# chapter08:Microwave Tubes and Circuits¶

## Example 8.1, Page number 336¶

In [1]:
#Calculate dc electron velocity,dc Phase Constant,Plasma Frequency,Reduced Plasma Frequency,dc beam current density,instantaneous beam current density
import math

#Variable declaration
Vo = 14.5*10**3      #beam voltage(V)
i = 1.4              #beam current(A)
f = 10*10**9         #frequency(Hz)
rho_o = 10**-6       #dc electron charge density(c/m^3)
rho = 10**-8         #RF charge density(c/m^3)
V = 10**5            #velocity perturbations(m/s)
eo = 8.854*10**-12
R = 0.4

#Calculations
#Part a
vo = 0.593*10**6*math.sqrt(Vo)        #dc electron velocity

#Part b
w = 2.*math.pi*f
ip = w/vo                             #dc phase current

#Part c
wp = math.sqrt((1.759*10**11*rho_o)/eo)

#Part d
wq = R*wp

#Part e
Jo = rho_o * vo

#Part f
J = rho*vo+rho_o*V

#Results
print "dc electron velocity =",round((vo/1E+8),3),"*10**8 m/sec"
print "dc phase curent =",round(ip,2),"rad/sec (Calculation mistake in the textbook)"
print "Reduced plasma frequency =",round((wq/1E+8),3),"*10**8 rad/sec"
print "dc beam current density =",round(Jo,1), "A/m^2"
print "instantaeneous beam current density =",round(J,3),"A/m^2"

dc electron velocity = 0.714 *10**8 m/sec
dc phase curent = 879.92 rad/sec (Calculation mistake in the textbook)
plasma frequency = 1.41 *10**8 rad/sec
Reduced plasma frequency = 0.564 *10**8 rad/sec
dc beam current density = 71.4 A/m^2
instantaeneous beam current density = 0.814 A/m^2


## Example 8.2, Page number 337¶

In [2]:
#Calculate input rms voltage,output rms voltage,output power

import math

#Variable declaration
Av = 15.              #voltage gain(dB)
Pin = 5*10**-3       #input power(W)
Rsh_in = 30*10**3    #Rsh of input cavity(Ohms)
Rsh_out = 20.*10**3   #Rsh of output cavity(Ohms)

#Calculations
#Part a
V1 = math.sqrt(Pin*Rsh_in)       #input rms voltage

#Part b
#Av = 20log(V2/V1) db
V2 = V1*10**(Av/20)             #deriving V2 from above equation

#Part c
Pout = (V2**2)/Rsh_out           #output power

#Results
print "input rms voltage =",round(V1,2),"V"
print "output rms voltage =",round(V2,2),"V"
print "output power =",round((Pout/1E-3),1),"mW"

input rms voltage = 12.25 V
output rms voltage = 68.87 V
output power = 237.2 mW


## Example 8.3, Page number 338¶

In [3]:
#Calculate input power output power,efficiency

import math

#Variable declaration
n = 2               #no. of modes
Vo = 300            #beam voltage(V)
Io = 20*10**-3      #beam current(A)
J1X = 1.25

#Calculations
#Part a
Pdc = Vo*Io         #input power

#Part b
Pac = (2*Pdc*J1X)/(2*math.pi*n-(math.pi/2))

#Part c
N = (Pac/Pdc)*100.  #efficiency

#Results
print "Input power =",round(Pdc,2),"W"
print "Output power =",round(Pac,2),"W"
print "Efficiency =",round(N,1),"%"

Input power = 6.0 W
Output power = 1.36 W
Efficiency = 22.7 %


## Example 8.4, Page number 338¶

In [4]:
#Calculate Electron velocity,dc transit time of electrons,Maximum input voltage,Volatge gain
import math

#Varaible declaration
Vo = 900           #beam voltage(V)
Io = 30*10**-3     #beam current(A)
f = 8*10**9        #frequency(Hz)
d = 1*10**-3       #gap spacing in either cavity(m)
L = 4*10**-2       #spacing between centers of cavities(m)
Rsh = 40*10**3     #effective shunt impedance(Ohms)
J1X = 0.582
X = 1.841

#Calculations
#Part a
vo = 0.593*10**6*math.sqrt(Vo)

#Part b
To = L/vo

#Part c
w = 2*math.pi*f
theta_o = w*To
theta_g = (w*d)/vo
Bo = math.sin(theta_g/2)/(theta_g/2)
V1_max = (Vo*3.68)/(Bo*theta_o)

#Part d
Ro = Vo/Io
Av = ((Bo**2)*theta_o*J1X*Rsh)/(Ro*X)

#Results
print "Electron velocity =",round((vo/1E+6),2),"*10**6 m/sec"
print "dc transit time of electrons =",round((To/1E-8),3),"*10**-8 sec"
print "Maximum input voltage =",round(V1_max,3),"V"
print "Volatge gain =",round(Av,3),"V"

Electron velocity = 17.79 *10**6 m/sec
dc transit time of electrons = 0.225 *10**-8 sec
Maximum input voltage = 41.923 V
Volatge gain = 23.278 V


## Example 8.5, Page number 339¶

In [5]:
#Calculate input microwave voltage V1 in order to generate maximum output voltage,

import math

#Variable declaration
Vo = 1200.         #beam voltage(V)
Io = 28*10**-3     #beam current(A)
f = 8*10**9        #frequency(Hz)
d = 1*10**-3       #gap spacing in either cavity(m)
L = 4.*10**-2      #spacing between centers of cavities(m)
Rsh = 40*10**3     #effective shunt impedance(Ohms)
J1X = 0.582
X = 1.841
Go = 23.3*10**-6

#Calculations
#Part a
vo = 0.593*10**6*math.sqrt(Vo)
w = 2*math.pi*f
theta_o = (w*L)/vo
theta_g = (w*d)/vo
Bo = math.sin(theta_g/2)/(theta_g/2)
V1_max = (Vo*3.68)/(Bo*theta_o)

#Part b
Ro = Vo/Io
Av = ((Bo**2)*theta_o*J1X*Rsh)/(Ro*X)

#Part c
V2 = 2*Io*J1X*Bo*Rsh
N = ((0.58*V2)/Vo)*100

#Part d
Gb = (Go*((Bo**2)-(Bo*math.cos(theta_g))))/2
Rb = 1/Gb

#Results
print "The input microwave voltage V1 in order to generate maximum output voltage is",round(V1_max,2),"V"
print "The voltage gain (reflecting beam loading in the output cavity) is",round(Av,3)
print "The value of",round((Rb/1E+3)),"K Ohms is very much comparable to Rsh and cannot be neglected because theta_g is quite high"

The input microwave voltage V1 in order to generate maximum output voltage is 58.71 V
The voltage gain (reflecting beam loading in the output cavity) is 17.058
The value of 73.0 K Ohms is very much comparable to Rsh and cannot be neglected because theta_g is quite high


## Example 8.6, Page number 341¶

In [6]:
#Calculate value of repeller voltage,dc necesaary to give the microwave gap of voltage of 200V,elctron efficiency

import math

#Variable declaration
Vo = 500.          #beam voltage(V)
Rsh = 20*10**3     #effective shunt impedance(Ohms)
f = 8*10**9        #frequency(Hz)
L = 1.*10**-3      #spacing between centers of cavities(m)
n = 2
e_m = 1.759*10**11
V1 = 200
J1X = 0.582

#Calculations
#Part a
w = 2*math.pi*f
x = (e_m*((2*math.pi*n)-(math.pi/2))**2)/(8*(w**2)*(L**2))
y = math.sqrt(Vo/x)
Vr = y+Vo

#Part b
Bo = 1              #Assumption
Io = V1/(2*J1X*Rsh)

#Part c
vo = 0.593*10**6*math.sqrt(Vo)
theta_o = (w*2*L*vo)/(e_m*(Vr+Vo))
Bi = 1                             #Assumption
X_dash = (V1*theta_o)/(2*Vo)
X = 1.51                           #from graph
J1X = 0.84
N = ((2*J1X)/((2*math.pi*n)-(math.pi/2)))*100

#Results
print "The value of repeller voltage is",round(Vr,2),"V (Calculation mistake in the textbook)"
print "The dc necesaary to give the microwave gap of voltage of 200V is",round((Io/1E-3),2),"mA"
print "The elctron efficiency is", round(N,2),"%"

The value of repeller voltage is 1189.36 V (Calculation mistake in the textbook)
The dc necesaary to give the microwave gap of voltage of 200V is 8.59 mA
The elctron efficiency is 15.28 %


## Example 8.7, Page number 342¶

In [7]:
#Calculate efficiency of the reflex klystron,total power output,elctron efficiency

import math

#Variable declaration
n = 1               #no. of modes
Pdc = 40*10**-3     #input power(W)
V1_Vo = 0.278          #ratio

#Calculations
#Part a
N = (V1_Vo*3*math.pi)/4

#Part b
Pout = (8.91*Pdc)/100

#Part c
Pl = (Pout*80)/100

#Results
print "The efficiency of the reflex klystron is",round(N,3)
print "The total power output is",round((Pout/1E-3),3),"W"
print "The power delivered to the load is",round((Pl/1E-3),2),"W"

The efficiency of the reflex klystron is 0.655
The total power output is 3.564 W
The power delivered to the load is 2.85 W


## Example 8.8, Page number 343¶

In [8]:
#calculate Hull cut-off voltage,Cut-off magnetic flux density,Cyclotron frequency
#chapter-8 page 342 example 8.8
#For a circular magnetron
import math
B=1.2*10**(-3)##magnetic flux density in Wb/sqm
x=1.759*10**11##Value of e/m in C/kg
V=6000.##beam voltage in V

#CALCULATION
V0=((x/8.)*(B**2.)*(b**2.)*(1.-(a/b)**2.)**2.)/1000.##Hull cut-off voltage in kV
Bc=((math.sqrt(8.*(V/x)))/(b*(1.-(a/b)**2.)))*1000.##Cut-off magnetic flux density in mWb/sqm
fc=((x*B)/(2.*(math.pi)))/10.**9.##Cyclotron frequency in GHz

#OUTPUT
print '%s %2.3f %s %s %3.3f %s %s %.4f %s' %('\nHull cut-off voltage is V0=',V0,'kV','\nCut-off magnetic flux density is Bc=',Bc,'mWb/sqm','\nCyclotron frequency is fc=',fc,'GHz')#

#Hull cut-off voltage is V0=5.066 kV
#Cut-off magnetic flux density is Bc=1.305953 mWb/sqm
#Cyclotron frequency is fc=0.0336 GHz

Hull cut-off voltage is V0= 5.066 kV
Cut-off magnetic flux density is Bc= 1.306 mWb/sqm
Cyclotron frequency is fc= 0.0336 GHz


## Example 8.9, Page number 343¶

In [9]:
#Calculate Axial phase velocity, anode voltage
import math

#Variable declaration
d = 2*10**-3         #diameter of helical TWT(m)
n = 50.              #no. of turns per cm
v = 3*10**8          #velocity of light(m/s)
m = 9.1*10**-31      #mass of electron
e = 1.6*10**-19      #charge on electron

#Calculations
p = 1/n*10**-2       #pitch(m)
c = math.pi*d        #circumference(m)
Vp = (v*p)/c

Vo = (m*(Vp**2))/(2*e)

#Results
print "Axial phase velociity =",round(Vp,2),"m/sec"
print "Anode voltage =",round(Vo,2),"V(Calculation mistake in the textbook)"

Axial phase velociity = 9549296.59 m/sec
Anode voltage = 259.32 V(Calculation mistake in the textbook)


## Example 8.10, Page number 344¶

In [10]:
#Calculate dc electron velocity,Transit time,Input voltage,Voltage gain

import math

#Variable declaration
Vo = 900              #beam voltage(V)
Io = 30.*10**-3       #beam current(A)
f = 8.*10**9          #frequency(Hz)
d = 1.*10**-3         #gap spacing in either cavity(m)
L = 4.*10**-2         #spacing between centres of cavity(m)
Rsh = 40.*10**3       #effective shunt impedance(Ohms)

#Calculations
#Part a
vo = 0.593*10**6*math.sqrt(Vo)

#Part b
Tt = d/vo

#Part c
w = 2*math.pi*f
theta_g = (w*d)/vo
Bo = math.sin(theta_g/2)/(theta_g/2)   #Beam coupling coefficient
theta_o = (w*L)/vo                     #dc transit angle
#For maximum o/p volltage,
J1X = 0.582
X = 1.841
V1max = (2*Vo*X)/(Bo*theta_o)

#Part d
Av = (Bo**2*theta_o*J1X*Rsh)/(Io*X)

#Results
print "dc electron velocity =",round((vo/1E+7),1),"*10**7 m/sec"
print "Transit time =",round((Tt/1E-10),2),"*10^-10 s"
print "Input voltage for maximum output voltage =",round(V1max,2),"V"
print "Voltage gain =",round((Av/1E+6),2),"dB"

dc electron velocity = 1.8 *10**7 m/sec
Transit time = 0.56 *10^-10 s
Input voltage for maximum output voltage = 41.95 V
Voltage gain = 23.28 dB


## Example 8.11, Page number 345¶

In [11]:
#Calculate dc electron velocity,dc phase constant,plasma frequency ,Reduced plasma frequency ,dc beam current density,instantaeneous beam current density

import math

#Variable declaration
Vo = 20*10**3        #beam voltage(V)
Io = 2               #beam current(A)
f = 9*10**9          #frequency(Hz)
rho_o = 10**-6       #dc electron charge density(c/m^3)
rho = 10**-8         #RF charge density(c/m^3)
V = 10**5            #velocity perturbations(m/s)
eo = 8.854*10**-12
R = 0.5

#Calculations
#Part a
vo = 0.59*10**6*math.sqrt(Vo)

#Part b
w = 2.*math.pi*f
ip = w/vo                             #dc phase current

#Part c
wp = math.sqrt((1.759*10**11*rho_o)/eo)

#Part d
wq = R*wp

#Part e
Jo = rho_o * vo

#Part f
J = rho*vo-rho_o*V

#Results
print "dc electron velocity =",round((vo/1E+7),3),"*10**7 m/sec"
print "dc phase constant =",round(ip,2),"rad/sec (Calculation mistake in the textbook)"
print "Reduced plasma frequency =",round((wq/1E+8),3),"*10**8 rad/sec"
print "dc beam current density =",round(Jo,2), "A/m^2"
print "instantaeneous beam current density =",round(J,4),"A/m^2"

dc electron velocity = 8.344 *10**7 m/sec
dc phase constant = 677.73 rad/sec (Calculation mistake in the textbook)
plasma frequency = 1.41 *10**8 rad/sec
Reduced plasma frequency = 0.705 *10**8 rad/sec
dc beam current density = 83.44 A/m^2
instantaeneous beam current density = 0.7344 A/m^2


## Example 8.12, Page number 345¶

In [12]:
#Calculate Transit angle
import math

#Variable declaration
f = 5*10**9         #frequency(Hz)
Vo = 1000           #operating voltage(V)
n = 1.75            #no. of turns
Vr = -500           #repeller voltage(V)
d = 2*10**-3        #cavity gap(m)

#Calculations
w = 2*math.pi*f
uo = 5.93*10**5*math.sqrt(Vo)
theta_g = (w*d)/uo

#Results
print "\nThe length of drift region cannot be computed as the value of F is not given"

Transit angle = 3.351 radians

The length of drift region cannot be computed as the value of F is not given


## Example 8.13, Page number 346¶

In [13]:
#Calculate Input RF voltage ,Voltage gain ,efficiency
#Variable declaration
import math
f = 10*10**9         #frequency(Hz)
Vo = 1200            #beam voltage(V)
Io = 30*10**-3       #beam current(A)
d = 1*10**-3         #diameter(m)
Rsh = 40*10**3       #shunt resistance(Ohms)
L = 4*10**-2         #length(m)
X = 1.84

#Calculations
#Part a
vo = 0.59*10**6*math.sqrt(Vo)
w = 2*math.pi*f
theta_o = (w*L)/vo
V1 = (2*X*Vo)/theta_o
theta_g = (theta_o*d)/L
Bi = (math.sin(theta_g/2))/(theta_g/2)
V1max = V1/Bi

#Part b
J1X = 0.58      #from table
I2 = 2*Io*J1X
V2 = Bi*I2*Rsh
A = V2/V1
Av = 20*math.log10(A)

#Part c
N = ((0.58*V2)/Vo)*100

#Results
print "Input RF voltage is",round(V1max,3),"V"
print "Voltage gain is",round(Av,2),"dB"
print "efficiency is",round(N,2),"%"

Input RF voltage is 55.231 V
Voltage gain is 28.03 dB
efficiency is 43.75 %


## Example 8.14, Page number 347¶

In [14]:
#Calculate Cyclotron angular frequency,Hull cut-off voltage,Cut-off magnetic flux density

import math

#Variable declaration
Vo = 30*10**3        #beam voltage(V)
Io = 80              #beam current(A)
Bo = 0.01            #Wb/m**2
a = 4*10**-2         #length of magnetron(m)
b = 8*10**-2         #breadth of magnetron(m)
e = 1.6*10**-19      #charge on electron(C)
m = 9.1*10**-31      #mass of electron

#Calculations
#Part a
w = (e*Bo)/m

#Part b
Vhc = (e*(Bo**2)*(b**2)*((1-((a/b)**2))**2))/(8*m)

#PArt c
Bc = ((8*Vo*(m/e))**0.5)/(b*(1-((a/b)**2)))

#Results
print "Cyclotron angular frequency =",round((w/1E+9),3),"*10**9 rad/s"
print "Hull cut-off voltage =",round((Vhc/1E+3),4),"kV"
print "Cut-off magnetic flux density =",round((Bc/1E-3),3),"mWb/m**2"

Cyclotron angular frequency = 1.758 *10**9 rad/s
Hull cut-off voltage = 7.9121 kV
Cut-off magnetic flux density = 19.472 mWb/m**2


## Example 8.15, Page number 348¶

In [15]:
#Calculate Input power,Output power,Efficiency

import math

#Variable declaration
n = 2            #mode
Vo = 280         #beam volatge(V)
Io = 22*10**-3   #beam current(A)
V1 = 30          #signal voltage(V)

#Calculations
#Part a
Pdc = Vo*Io

#Part b
J1X = 1.25       #from table
Pac = (2*Pdc*J1X)/((2*n*math.pi)-(math.pi/2))

#Part c
N = (Pac/Pdc)*100

#Results
print "Input power =",round(Pdc,2),"W"
print "Output power =",round(Pac,2),"W"
print "Efficiency =",round(N,2),"%"

Input power = 6.16 W
Output power = 1.4 W
Efficiency = 22.74 %


## Example 8.16, Page number 348¶

In [16]:
#Calculate

import math

#Variable declaration
f = 8*10**9        #frequency(Hz)
Vo = 300           #beam voltage(V)
Rsh = 20*10**3     #shunt resistance(Ohms)
L = 1*10**-3       #length(m)
V1 = 200           #gap voltage(V)
e_m = 1.759*10**11
n = 2              #mode

#Calculations
#Part a
w = 2*math.pi*f
x = (e_m*((2*math.pi*n)-(math.pi/2))**2)/(8*(w**2)*(L**2))
y = math.sqrt(Vo/x)
Vr = y+Vo

#Part b
Bo = 1             #assumption
J1X = 0.582        #from table
Io = V1/(2*J1X*Rsh)

#Results
print "Repeller voltage =",round(Vr,3),"V"
print "Beam current =",round((Io/1E-3),2),"mA"

Repeller voltage = 833.98 V
Beam current = 8.59 mA