Ch-5 : Microwave tubes Klystrons

Pages Number: 288 Example 5.1

In [2]:
from __future__ import division 
from math import sqrt, pi
#Given
f=10e9  #Hz
v=9e3  #V
i=40e-3  #A
l=3  #cm
l1=l/100  #m
G=2e-6  #mho
bet=0.92 
j1x=0.582 
x=1.841 
ebym=1.7e11  #J

#Maximum voltage
w=2*pi*f 
v0x=sqrt(2*ebym) 
thet=(w*l1)/(v0x*sqrt(v)) 

av=(bet**2*thet*i*j1x)/(x*v*G) 
print 'Maximum voltage:%0.3f'%av,'V'

#Power Gain
ic=2*i*j1x 
v2=(bet*ic)/G 
pout=bet*ic*v2 
pin=2*i*v 

#Efficiency
eet=pout/pin 
print 'Power gain: %0.3f'%(eet*100), '%'

#Answer for effciency comes out to be wrong, it is calculted wrongly in book

Maximum voltage:20.262 V
Power gain: 127.420 %

Page Number: 288 Example 5.2

In [20]:
 
#Given
l=2  #cm
l1=l/100 #m
f=5e9  #Hz
i=25e-3  #A
n=21/4  
e=1.6e-19 
m=9.1e-31 
thetag=0 
bet=1 
j1x=0.582 
x=1.841 

#(i) Beam Voltage
v0=(m*l1*l1*f*f)/(2*e*n*n) 
print 'Beam voltage: %0.3f'%v0, 'V'

#(ii) Input voltage
v1=x*v0/(pi*bet*n) 
print 'Input voltage: %0.3f'%v1,'V'

#(iii) Output voltage
v2=0.25*v0 
print 'Output voltage %0.3f'%v2,'V'

#(iv) Power output
pmax=i*v0*j1x 
print 'Maximum power output: %0.3f'%pmax, 'W'

#(v) Efficiency
eet=j1x*bet*v2/v0 
print 'Efficiency:' ,eet*100,'%'

Beam voltage: 1031.746 V
Input voltage: 115.164 V
Output voltage 257.937 V
Maximum power output: 15.012 W
Efficiency: 14.55 %

Page Number: 289 Example 5.3

In [22]:
from math import sin
#Given
r0=45e3  #W
j0=25e-3  #A
V=1500  #V
f=5e9  #hz
d=1  #mm
d1=d/1000  #m
l=3.5  #cm
l1=l/100  #m
rsh=32e3  #ohms
j1x=0.582 
x=1.841 

#(i) Input gap voltage
w=2*pi*f 
v0=(5.93e5*sqrt(V)) 
thetag=(w*d1)/v0 
bet=sin(thetag/2)/(thetag/2) 
theta0=(w*l1)/v0 
v1=(2*V*x)/(bet*theta0) 
print 'Input gap voltage: %0.3f'%v1

#(ii) Voltage gain
av=(bet**2*theta0*j1x*rsh)/(r0*x) 
print 'Voltage gain %0.3f'% av

Input gap voltage: 124.871
Voltage gain 9.186

Page Number: 290 Example 5.4

In [27]:
from math import cos
#Given
V=1000   #V
r0=40e3  #ohm
i0=25e-3  #A
f=3e9  #Hz
d=1  #mm
d1=d/1000  #m
l=4  #cm 
l1=4/100  #m
j1x=0.582 
x=1.841 
rsh=30e3  #ohm

#(i) Input gap voltage
w=2*pi*f 
v0=(5.93e5*sqrt(V)) 
thetag=(w*d1)/v0 
bet=sin(thetag/2)/(thetag/2) 
theta0=(w*l1)/v0 
vmax=(2*V*x)/(bet*theta0) 
print 'Input gap voltage: %0.3f'%vmax, 'V'

#(ii) Voltage gain
av=(bet*bet*theta0*j1x*rsh)/(r0*x) 
print 'Voltage gain: %0.3f'%av

#(iii) Efficiency
v2=bet*2*i0*j1x*rsh 
eet=(bet*2*i0*j1x*v2)/(2*i0*V) 
print 'Efficiency: %0.3f'%(eet*100),'%'

#(iv) Beam loading conductance
gbl=(i0/(2*V))*((bet*bet)-(bet*cos(thetag/2))) 
print 'Beam loading conductance:%0.3e'%gbl

#Answer for beam loading conductance is calculated wrong in book

Input gap voltage: 95.547 V
Voltage gain: 8.757
Efficiency: 46.672 %
Beam loading conductance:9.835e-07

Page Number: 291 Example 5.5

In [28]:
 
#Given
f=3e9  #hz
v=900  #V
i=30e-3  #A
d=4  #cm
d1=d/100  #m
gap=1  #mm
gap1=1/1000  #m
rsh=40e3  #ohm
x=1.841 
j1x=0.582 
r=40e3  #ohm
ebym=1.758e11  #J

#(i) Electron velocity
v0=sqrt(2*ebym*v) 
print 'Electron velocity: %0.3f'%v0,'m/s'

#(ii) Electron transit time
t=d1/v0 
print 'Electron transit time: %0.3f'%t,'s'

#(iii) Input voltage gap
w=2*pi*f 
theta0=(w*d1)/v0 
thetag=(w*gap1)/v0 
bet=sin(thetag/2)/(thetag/2) 
v2=(2*v*x)/(bet*theta0) 
print 'Input voltage gap: %0.3f'%v2, 'V'

#(iv) Voltage gain
av=(bet**2*theta0*j1x*rsh)/(x*r) 
print 'Voltage gain: %0.3f'%av

#Values of v and f are changed in question and answer, hence vaules used in answer are taken.
#Also second part has not been done in book

Electron velocity: 17788760.496 m/s
Electron transit time: 0.000 s
Input voltage gap: 81.964 V
Voltage gain: 12.192

Page Number: 292 Example 5.6

In [29]:
from math import sqrt, log10 
#Given
f=8e9  #hz
i=2.5  #A
v=20e3  #V
bet=1 
amp=10*sqrt(2)  #V
rsh=10e3  #ohm
rsho=30e3  #ohm
dc=1e-6  #c/m**3
rf=0.5 
e=1.6e-19 
ee=8.854e-12 
m=9.1e-31  #kg

#(i) Induced current
w=2*pi*f 
wq=rf*sqrt((e*dc)/(m*ee)) 

#Amplitude of induced current
ic=(i*w*(bet**2)*amp)/(2*v*wq) 
print 'Induced current: %0.3f'%ic,'A'

#Induced voltage
icrms=ic/sqrt(2) 
v2rms=icrms*rsho 
print 'Induced voltage: %0.3f'%v2rms,'V'

#(ii) Power gain
pg=(((i*w)**2)*(bet**4)*rsh*rsho)/(4*((v*wq)**2)) 
pgdb=10*log10(pg) 
print 'Power gain: %0.3f'%pgdb,'dB'

#(iii) Electronic efficiency
eeta=((icrms**2)*rsho)/(i*v) 
print 'Electronic efficiency: %0.3f'%(eeta*100),'%'

Induced current: 0.631 A
Induced voltage: 13376.164 V
Power gain: 57.755 dB
Electronic efficiency: 11.928 %

Page Number: 294 Example 5.7

In [30]:
 
#Given
f=3e9  #hz
l=4  #cm
l1=4/100  #m
d=0.1  #cm
d1=d/100  #m
V=900  #V
i0=30e-3  #A
rsh=25e3 #ohm
x=1.841 
j1x=0.582 

#(i) Input voltage for maximum output
v0=0.593e6*sqrt(V) 
w=2*pi*f 
theta0=w*l1/v0  #rad
thetag=w*d1/v0  #rad
bet=sin(thetag/2)/(thetag/2) 
v1max=2*V*x/(bet*theta0)  #v
print 'Input voltage for maximum output: %0.3f'%v1max,'V'

#(ii) Voltage gain
r0=V/i0 #ohm
av=((bet**2)*theta0*j1x*rsh)/(x*r0) #V
print 'Voltage gain: %0.3f'%av, 'V'

#(iii) Efficiency
ic=2*i0*j1x  #A
v2=bet*ic*rsh  #V
eet=bet*ic*v2/(2*i0*V) 
print 'Efficiency: %0.3f'%(eet*100), '%'

#(iv) Beam loading conductance
gb=(i0/(V*2))*(bet**2-(bet*cos(thetag/2))) #ohm
print 'Beam loading conductance: %0.3e'%gb,'ohm'

Input voltage for maximum output: 81.969 V
Voltage gain: 10.159 V
Efficiency: 51.366 %
Beam loading conductance: 1.446e-06 ohm

Page Number: 295 Example 5.8

In [33]:
 
#Given
f=5e9  #hz
v0=10e3  #V
d=1  #mm
d1=d/1000  #m
v1=100  #V

#(i) Gap transit time
vv0=0.593e6*sqrt(v0) #m/sec
tau=d1/vv0 #sec
print 'Gap transit time: %0.3e'%tau,'sec'

#Gap transit angle
w=2*pi*f 
thetag=w*tau #rad
print 'Gap transit angle: %0.3f'%thetag, 'rad'

#(ii) Beam coupling coefficient
betin=sin(thetag/2)/(thetag/2) 
print 'Beam coupling coefficient: %0.3f'%betin

#(iii) Velocity of electron leaving buncher gap
vig=vv0*(1+((betin*v1)/(2*v0))) #m/sec
print 'Velocity of electron leaving buncher gap: %0.3f'%vig, 'm/sec'

#(iv) Depth of modulation
m=betin*v1/v0 
print 'Depth of modulation: %0.3f'%m

Gap transit time: 1.686e-11 sec
Gap transit angle: 0.530 rad
Beam coupling coefficient: 0.988
Velocity of electron leaving buncher gap: 59593044.746 m/sec
Depth of modulation: 0.010

Page Number: 296 Example 5.9

In [37]:
 
#Given
f=10e9  #hz
v0=15e3  #V
i0=2.5e-3  #A
d=1  #cm
d1=d/100  #m
vrms=10  #V
bet=1 
p=1e-8  #C/m**3
rf=0.6 
e=1.6e-19 
m=9.1e-31 
ee=8.854e-12 

#(i) DC electron beam phase cobstant
vv0=(0.593e6*sqrt(v0)) 
w=2*pi*f 
bete=w/vv0  #rad/m
print 'DC electron beam phase constant: %0.3f'%bete,'rad/m'

#(ii) Reduced plasma frequency and reduced plasma phase constant
wq=rf*sqrt(e*p/(m*ee)) #rad/m
print 'Reduced plasma frequency: %0.3f'%wq,'rad/m'
betq=wq/vv0 #rad/sec
print 'Reduced plasma phase constant: %0.3f'%betq,'rad/sec'

#(iii) Gap transit time
tau=d1/vv0 #sec
vtg=vv0*(1+(bet*vrms*sin(w*tau)/(2*v0))) #m/sec
print 'Gap transit time: %0.3f'%vtg, 'm/sec'

DC electron beam phase constant: 865.126 rad/m
Reduced plasma frequency: 8455139.683 rad/m
Reduced plasma phase constant: 0.116 rad/sec
Gap transit time: 72644284.673 m/sec

Page Number: 296 Example 5.10

In [40]:
 
#Given
f=4e9 #hz
v0=10e3 #V
i0=0.75 #A
v1=2  #V
bet=1 
rsh=10e3 #ohm
p=5e-5 #C/m**3
r=0.6 
rsht=4e3 #ohm
e=1.6e-19 
m=9.1e-31 
ee=8.854e-12 

#(i) Induced current and voltage in output cavity
w1=sqrt(e*p/(m*ee)) #rad/sec
w=2*pi*f 
wq=0.5*w1 #rad/sec
rr=w/wq 

i4=((i0**3)*(rr**3)*(bet**6)*v1*(rsh**2))/(8*(v0**3))  #A
print 'Induced current: %0.3f'%i4,'A'
v4=i4*rsht #V
print 'Induced voltage: %0.3f'%(v4/1000), 'kV'

#(ii) Power output
pout=(i4**4)*rsht #W
print 'Power output: %0.3f'%pout, 'W'

#Answer for Pout should be 13.43 kW but it is given as 10.89kW as value of I4 is calculated as 1.289 but it comes out to be 1.35

Induced current: 1.354 A
Induced voltage: 5.415 kV
Power output: 13438.135 W

Page Number: 297 Example 5.9

In [42]:
 
#Given
f=8e9 #hz
v0=500 #V
l=1.2 #mm
l1=l/1000 #m
rsh=18e3 #ohm
ebym=1.759e11 
ee=8.854e-12 

#(i) Repeller voltage
n=1+(3/4) 
v11=(ebym*n*n)/(8*(l1**2)*(f**2)) 
vr=sqrt(v0/v11)-v0 
print 'Repeller voltage: %0.3f'%vr,'V'

#(ii) Required dc current
v2=200 #V
j1x=0.582 
i=v2/(2*rsh*j1x) #A
print 'Required dc current: %0.3f'%(i*1000), 'mA'

#Answer for repeller voltage is calculated wrong in book

Repeller voltage: 327.238 V
Required dc current: 9.546 mA

Page Number: 298 Example 5.12

In [44]:
 
#Given
f=9e9 #hz
v0=361 #V
i0=30e-3 #A
l=0.1 #cm
l1=l/100 #m
x=2.408 
j1x=0.582 
ebym=1.759e11 

#Maximum power output
n=1 
pout=2*i0*v0*x*j1x/(2*pi*(n+(3/4))) #W
print 'Maximum power output: %0.3f'%pout,'W'

#Operating repeller voltage
vr=((6.744e-6*sqrt(v0)*l1*f)/(n+(3/4)))-v0 #v
print 'Operating repeller voltage: %0.3f'%vr,'V'

Maximum power output: 2.761 W
Operating repeller voltage: 297.985 V

Page Number: 298 Example 5.13

In [45]:
 
#Given
f=9e9 #hz
v0=250 #V
l=0.5 #cm
l1=l/100 #m

#Bandwidth
n=3 
df=(n+(3/4))/(6.774e-6*l1*sqrt(v0)) #hz
print 'Bandwidth: %0.3f'%(df/10**6),'Mhz'

Bandwidth: 7.002 Mhz

Page Number: 299 Example 5.14

In [46]:
 
#Given
f=10e9 #hz
v0=600 #V
vr=250 #V
ebym=1.759e11 

#Repeller space
n=1 
l=sqrt((ebym*(n+(3/4))**2*(vr+v0)**2)/(8*f**2*v0)) #m
print 'Repeller space: %0.3f'%(l*1000),'mm'

Repeller space: 0.900 mm

Page Number: 299 Example 5.15

In [47]:
 
#Given
v0=300 #V
i0=20e-3 #A
v1=40 #V
n=2 
x=2.408 
j1x=0.52 

#(i) Input power
pin=i0*v0 #W
print 'Input power:' ,pin,'W'

#(ii) Output power
pout=(2*v0*i0*x*j1x)/((2*pi*n)-(pi/2)) #W
print 'Output power: %0.3f'%pout, 'W'

#Efficiency
eet=pout/pin 
print 'Efficiency: %0.3f'%(eet*100), '%'

#Answer for output power in book is 0.7 which is wrong, it should be 1.3W
#Hence answer of efficiency also changes

Input power: 6.0 W
Output power: 1.367 W
Efficiency: 22.776 %

Page Number: 300 Example 5.16

In [48]:
 
#Given
f=10e9 #hz
v0=600 #V
l=0.1 #cm
l1=l/100 #m
bet=0.9 
ebym=1.759e11 
n=2 
j1x=0.575 #from standard table


#(i) Repeller voltage
vr=((6.744e-6*sqrt(v0)*l1*f)/(n-(1/4)))-v0 #V
print 'Repeller voltage:',round(vr), 'V'

#(ii) Bunching parameter
v1=200 #V
x=bet*v1*2*pi*(n-(1/4))/(2*v0) 
print 'Bunching parameter: %0.3f'%x

#(iii) Required DC current
rsh=20e3 #ohm
i=v1/(2*rsh*j1x) #A
print 'Required DC current: %0.3f'%(i*1000),'mA'

#(iv) Electronic efficiency
eet=2*x*j1x/(2*pi*(n-(1/4))) 
print 'Electronic efficiency:',eet*100, '%'

Repeller voltage: 344.0 V
Bunching parameter: 1.649
Required DC current: 8.696 mA
Electronic efficiency: 17.25 %

Page Number: 301 Example 5.17

In [49]:
 
#Given
f=10e9 #hz
v0=300 #V
j0=0.3 #A/cm
i0=45e-3 #A

rb=sqrt(i0/(pi*j0)) #mm
print 'Electron beam radius: %0.3f'%(rb*10),'mm'
r=rb*(120/100) #mm
print 'Radius of cathode disc:%0.3f'%(r*10),'mm'
d=sqrt(2.335e-6*(300)**(3/2)/j0) #mm
print 'Cathode anode spacing: %0.3f'%(d*10),'mm'
#Anode hole has to be 15% larger than cathode disc
ra=r*1.15 #mm
print 'Anode hole: %0.3f'%(ra*10),'mm'

Electron beam radius: 2.185 mm
Radius of cathode disc:2.622 mm
Cathode anode spacing: 2.011 mm
Anode hole: 3.015 mm

Page Number: Example 5.18

In [50]:
from math import tan, exp
#Given
f=9e9 #hz
v0=300 #V
vr=125 #V
bet=0.9 
c=3e8  #m/s
w=2*pi*f 
br=2.18 #mm
e0=8.854e-12 
ebym=1.7e11 

#From sin(theta)/theta table, thetag is found out to be
thetag=0.25*pi  
d=(2*thetag*0.593e6*sqrt(v0))/w 
print 'Distance: %0.3f'%(d*1000),'mm'

#Axial cavity length
l=c/(10*f) #m
print 'Axial cavity length: %0.3f'%(l*1000),'mm'

#Ratio of outer to inner conductor
a=1.5*br 
a1=a/1000 
x=d/(w*e0*a1*a1*60*tan((w*l)/c)) 
bbya=exp(x) 
print 'Ratio of outer to inner conductor: %0.3f'%bbya

#radii of outer and inner conductor
print 'Radius of outer conductor:',a, 'mm'

b=1.52*a #mm
print 'Radius of inner conductor:' ,b,'mm'

#Repeller spacing
lopt=sqrt(ebym*(19/4)**2*(v0+vr)**2/(8*f**2*v0)) #m
print 'Repeller spacing: %0.3f'%(lopt*1000),'mm'

#Answer for radii of outer and inner conductor have wrong calculations in book
#Also ratio of outer to inner conductor is also calculated wrong

Distance: 0.285 mm
Axial cavity length: 3.333 mm
Ratio of outer to inner conductor: 3.396
Radius of outer conductor: 3.27 mm
Radius of inner conductor: 4.9704 mm
Repeller spacing: 1.888 mm