4.3.7.

Solved Problems

1. A two cavity klystron amplifier has the following paraimeiers:

Beam voltage V. =1000v
DC beam resistance =4O K
Beam current I,25 mnA
Operating frequency f=3GHz
Gap spacing in either cavity d = 1 mm

Spacing between the two cavities L= 4 cm

Effective shunt impedance, excluding beam loading R 30K
Compute
(a) DC electron velocity
(b) Angular frequency
(c) Gap transit angle
(d) Beam coupling coefficient
(e) DC transit angle
()Inputgap voltage to give maximum votage V2
(gVoltage gain by neglecting the beam loading in the output cavity
(h) Efficiency of the amplifier by neglecting beam loading
() DC beam conductance
)Beam loading conductance
(k) Beam loading resistance
Solution:
(a) DC electron velocity

2ev
e = 0.593x10°,V =0.593x10° V1000 = 1.88x10 m/s
m

v=1.88x10'm/s
(b) Angular frequency
-2xTx3x10° =18.8x10 rad/s
2Tf

(c) Gap transit angle

bd (18.8x10" )x(1x10l rad
rad 1.88x10'

The gap transit angle e in degrees

180 1x180 57.325"
e, =0,( in rads)x
O=57.325°

Beam coupling coefficient

(d)
sinf28.662)
sin(0./2)sin(57.325/2 =0.9592
B- o 012 1/2 1/2

(e) DC transit angle

rad
Vo
@=2xTx3x10
=
18.8x10'rad/s
L_18.8x10 )x(4x10)
= 40 rad
1.88x10
040rads
Input gap voltage to give
maximum voltage V2
(
=0.582
V2 is maximum when X=1.841, i.e J(X)

Vma
Tmax
2,X 2x1000x1.8495.965 V
BO 0.9592x 40

(a) The votage gain by neglecting the beam loading in the output cavity

B B0,J(X),. -.9592x40x 0.582 x30x10

30x10 = 8.7258
AvR X 40x10
= 8.7258
1.841

(h) Efficiency of an amplifier by neglecting beam loading

Efficiency 7= Pll2
21.
 I, =21,J,(X)=2x(25x10 )x0.582 =0.0291= 29.1mA
V, =
B.1R =0.9592x(29.1x10 )x(30x10) 837.39 V =

B x100 =-(0.9592)x (29.1x10 )x (837.39)

21 x100 46.748%

DC beam conductance
2x(25x10)x(10°)
G Ra = = 25
40x10
Beam loading conductance

G, -B,cos
(25x10) 0.9592-0.9592cos57.325'-0.98x10 =0.98
2510
2 2

(k) Beam loading resistance

1 = 0.10204x10' = 1.0204MQ
Ra G 0.98x10

2. A two cavity klystron amplifier has the following parameters:

Beam voltage V 2 0 KV

Beam current 2 A
f=8 GHz
Operating frequency
Beam coupling coefficient B, P =1

DC electron beam current density P=10 C/m3

Signal voltage K10 (rms)

Shunt resistance of the cavity R10KQ
Total shunt resistance including load R 30K2

Compute
(a) Plasma frequency
Reduced plasma frequency for R =0.5
(b)
(c) Angular frequency
(d)
(d) Induced current in the output cavity
(e) Induced voltage in the output cavity
() Output power delivered to the load
(g) Power delivered from the input cavity
(h) Power gain
i) Electronic efficiency
 Soiution

(a) Plasma frequency

1
1.602 x 10
m
9.109x10-./59x 10" C/Kg; E8.854x10-2

ePo
m E
1.759x10)x{101.4lx10 rad/s
8.854x102

=1.41x10 rad/s

(b) Reduced plasma frequency for R =0.5

o,= Ro, =0.5x(1.41x10") 0.705x10 =

rad/s

(c) Angular frequency o

2 f =2xTx(8x10°)=0.5024x10 rad/s

d Induced current in the output cavity

The ratio between angular frequency (o) and reduced plasma frequency (o,
0.5024x10 =713
0.705x100

713x1 x1003565 A
2V. 2 20x10
e) Induced voltage in the output cavity

V=R =0.3565x(30x10") =
10.71Kv
( Output power delivered to the load
Paa- R =0.3565 x(30x10") = 3.813 KW
(g) Power gain

Power Gain=ou BRRh

2
-2
4 20x10*/15 'x(10x10")«(30>x10")=3.8313x10"
Power Gain(dB) = 10logo(3.8313x10') = 55.8dB
Power Gain (dB) = 65.8 dB
 Electronic efficiency

ou x100 =- 3.813x10x100=
9.532%
P

P Vo 2x(20x10')
43.8. Tutorial

A two cavity Kiystron amplifier has the following parameters:

Beam voltage V. =1200ov

Beam current 28 mA
Operating frequency S=8 GHz
Gap spacing in either cavity d =1mmn
Spacing between the two cavities L=4cmn
Effective shunt impedance, excluding beam loading R30 KQ
Compuste

a)DC electron velocity =0.205 x10° m/s

D) Angular frequency @=0.502 x10 rad/s
c) Gap transit angle 2.4488 rads; 0, =140.38°
dBeam coupling coefficient B=B, =0.7686
(e) DC transit angle 94.95 rads
Input gap voltage to give maximum voltage V2 Vimas 58.690 V; R, = 42.857 K
(g Voltage gain by neglecting the beam loading in the output cavity A, = 17.073; 24.65dB
(h)Efficiency of the amplifier by neglecting beam loading
32.6mA; V =1.00022KV; n-37.35%

0 DC beam conductance G 33.33/

0 Beam loading conductance G, 3.85S
) Beam loading resistance R 0.25974MQ
2. A two cavity klystron amplifier has the following parameters

Voltage gain Ay =15 dB

Input power Pn-5mW
Total Shunt impedance of input cavity R 30 KN
Total Shunt impedance of output cavity R 40 KQ.
Load impedance at output cavity R,=40KQ
Compute
(a) The input voltage(RMS) V(RMS) -12.25V
(b) The output voltage(RMS)
to the load
V(RMS)=68.89V
Dower delivered
(c) The Pu118.65mW
 4.5.6. Solved Problems
1. A reflex klystron operates under the following conditions
Beam voltage V =600 V
Operating frequency S =
9 GHz
Spacing between the two cavities L=1mm
Equivalent shunt resistance R=15 KQ
The tube is oscillating at f, at the peak of the n=2 mode or 14 mode. Assume that the

transit time through the gap and beam loading is neglected.

(a) Find the value of the repeller voltage V,.
(b) Find the direct current necessary to give a microwave gap voltage of 200V.
(c)What is the electronic efficiency underthis condition?
Solution
(a)Find the value of the repeller voltage V

N=n-2=1.75
4 4

= N27T =1.75x2xT =10.9956

N27 10.9956 =1.9444 x10-10
2xTx9x10

T, 2mLv V+V)2mbv

eV.+V) eT

V, 2mLYo-V
eTo
where v
=
0.593x10'x VVo

2x1x10x0.593x10xV600 1 - 600 where=1.759x10"

m
1.9444x10-10 1.759x10
V, =249.3739 V

Find the direct current necessary to give a microwave gap voltage of

(b)
200V.

V=1,R =21,(x)R
200
2,(X), 2x(0.582)x(15%10")
=11.45 mA
Therefore, the direct current I, =11.45 mA

(c) What is the electronic efficiency under this condition?

2xJX
2n- 2
2x1.841x0.582-x100
(2x2x)
n 22.24 %
 Areflex Klystron operates at the peak mode of n= 2 with

Beam voltage V300 V

Beam current I20 mA
Signal Voltage V-40V
Let assume B, P =1
Determine

(a) Input power in watts

(b) Output power in watts
(c) Efficiency
Solution
(a) The input power in watts
PPe =1,V, =300x 20x10 =6 Watts
(b) The output power in watts

PhdoBX)
=o,=2nm

2 2,

(x40
2x300
2x2x7
X = 0.733,

J(0.733)= 0.345
P=oB(X)
40x20x10x1xJ (0.733)
x0.345 = 0.276
40x20x10
P=0.276Watts
(c) Efficiency
0.276
-x100
n=x100=
in
6
7 4.6%
 4.5.7.Tutorial
. A reflex klystron operates under the following conditions
V 500V

Beam voltage
S 8GHz
Operating frequency L=1 mm
Spacing between the two cavities R20 KQ
Equivalent shunt resistance
he tube is oscillating at , atthe peak of the n=2 mode or 1 mode

(a) Find the value of the repeller voltage ,

to give a microwave gap voltage of 200V.
() Find the direct current necessary
condition?
(c) What is the electronic efficiency under this
(c) 7=28.02%
(a) V, =189.2167 (b)I,=9.09mA

mode. The dc power input is 40

2. A reflex klystron operates at the peak of then=2
by the beam is dissipated in the
mW and -=0.278. If 20 % of the power delivered

cavity walls, find the power delivered to the load.

n, =15.65%;P6.26mW; PL =5mW]
conditions
3. A reflex klystron operates underthe following
Beam voltage V= 700 V
S. =11GHz
Operating frequency
Spacing between the two cavities L=2mmm
Equivalent shunt resistance R20 K2
The tube is oscillating at s, at the peak of the n=4 mode or 1 mode. Find the

value of the repeller voltageV,. V, =346.5483

4. A reflex klystron operates underthe following conditions:

Beam voltage V =650 V
Operating frequency S, =10GHz
Spacing between the two cavities
L=1.5mm
Equivalent shunt resistance
R20 KQ
The tube is oscillating at S, at the peak of the n=2 mode or 1 mode. Find U

value ofthe repeller voltage, V,823.4274

4.7.6.1. Solved Problems
. An X band pulsed cvlindrical maanetron has the following
parametei
Anode voltage V =
26 Kv
Beam current 1 27 A

Magnetic flux density B0.336 Wb/m2

Radius of cathode a=5cm
cylinder
Radius of vane edge to center b 10 cm

Compute
(a) Cyclotron angular frequency
(b) Cyclotron frequency
(c) Hull cutoff voltage for a fixed B,
(d) Hull cutoff magnetic flux density for a fixed V
Solution
(a) Cyclotron angular frequency

a B -=1.759 x10
m

5.91x100 rads
=(1.759x10" )x0.336
=

o
(b) Cyclotron frequency
5.91x100
5.91x109.4111x10°
27
=9.4111 GHz

Hull cutoff voltage for a fixed B,

(c)

1.759x10)*0.336x(10x10|1-(5x10
Voc 8m 10x10 )
Yoc =139.63 KV
 (d) Hull cutoff magnetic flux density fora fixed Vo

8mV
Boc
e 8(1.759x10")«(26x 10') = 14.499 mWb/m

10x102 10)'
(10x102)
. An X band pulsed cylindrical magnetron has the following operating parameters:

Anode voltage Vo-5.5 KV

Beam current 4.5 A
Operating frequency S=9 GHz

Resonator conductance G, -2x10 T

Load conductance G 2.5x 10 U

Vane capacitance C 2.5 pF

Duty cycle DC 0.002
Power loss PLoss18.50 KW
Compute
(a) Angular resonant frequency
(6) Unloaded quality factor
(c) Loaded quality factor
(d) External quality factor
(e) Circuit efficiency
() Electronic efficiency
Solution
(a) Angular resonant frequency
a 2 f =2xrx9xl0" = 56.549x10 rads

(b) Unloaded quality factor

,C(56.549x10" )x(2.5x10) = 706.8583

G, 2x10
(c) Loaded quality factor

C.S6.549x10")x(2.5x 10-12
QG.+G (2x10)+(2.5x10
= 628.3185
 External quality factor

o,C_(56.549x10')x(2.5x10")= 5654.9
2.5x10
le Circuit efficiency
1 1
x100= 1 + 5 6 5 4 . 9 X 0 0 = 1 1 . 1 1 %

I+ 1+
706.8583
Electronic efficiency
P e o o - o x100-55x10°)x(4.5)-(18.50x10") 25.25%
1 Pa
(5.5x10)x(4.5)
4.7.6.2. Tutorial
1. A normal cylindrical magnetron has the following parameters
Inner radius R=0.15 meter
R, 0.45 meter
Outer radius
B, =
1.2 Wb/m
Magnetic flux density
Compute
@ 2.1108 x10" rads
(a) Cyclotron angularfrequency S33.594MHz
(b) Cyclotron frequency
B, fixed Voc=5.699 KV
Hull cut-off voltage for
a
(c) 6000
flux density if the
beam voltage =

(d) Hull cutoff magnetic

Boc1.3 mWblm*|
magnetron has the following parameters:
2. An X band pulsed cylindrical
V 3 2 KV
Anode voltage
844
Beam current
B 0.01 Wb/m
Magnetic flux density
cylinder
a 6 cm
cathode
Radius of b= 12 cm
to center
v a n e edge
Radius of

Compute
(a)Cyclotron angularfrequency =1.7590x 10 rads
(b) Cyclotron frequency S. =0.28010GHz
(b) voltage
for a fixed B, Voc 17.810 KV
Hull cutoff
(c) TIUK Sensity for a fixed . Boc 13.4 mWb/m*
magnetic
(d) Hull cutoff
4.7.7.1. Solved Problems

1. A linear magnetron has the following operation parameters

Anode voltage
V =15 KV
1, =
1.2 A
Cathode current
f =
8 GHz
Operating frequency B, = 0.015Wb/m
Magnetic flux density
h 2.77 cm
Hub thickness
Distance between anode and cathode
d 5 cmn

Computee
(a) Hull-cut off voltage for a fixed B
for fixed V
(b) Hull-cut off magnetic flux density
a

(c) Electron velocity at the hub surface

(d) Phase velocity for synchronism
(e) Hartree anode voltage
Solution
Hull-cut off voltage for a fixed B,
(a)
1
B 4 =<x(1.759x10" )x(0.015) «(5x10*)=49.472 KV
Voc2 m
Voc 2
(b) Hull-cut off magnetic flux density for a fixed V%
=8.259 mWb/m?
Boe V1.759x10Tx(15x10")
Sx10
(c) Electron velocity at the hub surface
10")x(0.015)x(2.77x102)= 73.086 MVolts
B,h
,(h m =(1.759x
(d) Phase velocity for synchronism
ph Bh=(1.759x
m
10")x(0.015)x(2.77x10)=73.086MVols
(e) Hartree anode voltage
ph B,h=1.759x10" x0.015x 2.77x102 = 73.086x 10
m

Voa=B.d. l mo

1
-(73.086x10 )x0.015x(5x10)-x10(73.086x10"
2 1.759x1oT*{73.086x10*

Va 39.631KV
2. A linear magnetron has the following operation parameters:
Anode voltage V20 KV
Cathode current , =17A
Operating frequency S=9 GH:
Magnetic flux density
B, 0.01wb/m
Hub thickness h 2.77 cm
Distance between anode and cathode d 5 cm
Compute
(a) Hull-cut off voltage for a fixed B,
(b) Hull-cut off magnetic flux density for a fixed
V
(c) Electron velocity at the hub surface
(d) Phase velocity for synchronism
(e) Hartree anode voltage
Solution
(a) Hull-cut off voltage for a fixed B,
Ve 2m B'-5*(1.759x10")x(0.01)' «(6x10)-31.662 KV
 Hull-cut off magnetic flux density for a fixedV,
Hul

1 1
6x10 2x 1.759x10T
10TX(20x10") =
0.0079476
=7.9476 mWblm
Electron velocity at the hub surface
b
,( Bh=(1.759x10 x(0.01)x(2.77x10)=48.724 MVolts
Phase velocity for synchronism

Vph Bh=(1.759 x10")x(0.01)x(2.77x1o*)=48.724 MVolts

d Hartree anode voltage

=B,h =1.759x10x0.01x2.77x102 = 48.724x10°

Yph

VaB,d-2
=(48.724 x10*)x0.01x(6x10*)-x51TX48.724x10")
VO22.486KV

4.7.7.2 Tutorial
1. A linear magnetron has the following operation parameters:

Anode voltage V=32K

Cathode current ,60A
Operating frequencyy f-10 GHz

Magnetic flux density B=0.01Wb/m*

Hub thickness
h 3 cm
Distance between
anode and cathodee d=6cm

Compute
for a fixed B, Voc 31.662 KV
(a) Hull-cut off voltage
flux density for a fixed
(b) Hull-cut off magnetic Boc=0.0101 mWb/m2
velocity at the hub surface
Electron V,(h)= 52.77MV
(c)
for synchronism
(d) Phase velocity Vph 52.77MV
anode voltage
(e) Hartree
OA23.747 KV
 4.7.10.1. Solved Problems
1. An inverted coaxial maqnetron has the following operation parameters.
Anode voltage V =10 KV

Cathode current T 2A

Operating frequency S=9 GHz

Magnetic flux density B, =0-01Wb/m
Anode radius a=3cmn
Cathode radius b= 4 cm
Compute
(a) Cyclotron angular frequency
(b) Hull cut-off voltage for a fixed B,
(c) Hull cut-off magnetic flux density for a fixed V

Solution

(a) Cyclotron angular frequency

1.602x10-19

0
m

m
9.109x104T=1.759x10!1

o B=(1.759x10 )x0.01 =1.7590x10' rads

(b) Hull cut-off voltage for a fixed B,

1.759x10
Voc Ba1 8
x0.01'«(3x10*)14x10)
8m
Yoc 1.1971KV
(3x10
 ) HHull-cut off magnetic flux density for a fixed V

8
1.759x10T X10x103
e
Boc - =-0.0289

a 1 3x1031 4x102
3x10
Boc-28.9 mWblm
4.7.10.2. Tutorial
1. An inverted coaxial magnetron has the following operation parameters:

Anode voltage V 30 KV
Cathode current 25 A
Operating frequency S=9 GHz
Magnetic flux density B, 0.01b/m?
Anode radius a 2.5 cm

Cathode radius b 5cm

Compute
(a) Cyclotron angular frequency @ =1.759 x 10 rads
fixed Voc12.368KV
(b) Hull cut-off voltage for a B,
) Hull cut-off magnetic flux density for a fixed V. Boc -15.6 mWb Im?
parameters:
2. An inverted coaxial magnetron has the following operation
Anode voltage V=5KV
Cathode current 24
f=9 GH
Operating frequency
B =0.01Wb/m2
Magnetic flux density
a=3 cm
Anode radius
Cathode radius b=4cm
Compute @1.759x10 rads
) Cyclotron angular frequency
Voc1.1971KV
6) Hull cut-off voltage for a fixed B
c) Hull cut-off magnetic flux density for a fixed , Boc 20.4 mWblm
4.9. Dackward-Wave Crossed-Field Amplifier
a l e d AMPLITRON Pair of pins and the cavity is excited
in opposite Pa
interact in the resOnan
e Strap line. The electron beam and EM waves
circuits as shown in Figure 4.38.
It is operated in S-band
frequencies.
It is used in high data-rate transmission systems.
Magnetic fux Cathode
density B
Anode
cavity

Contact
points

Output
(Strap line) d
Input
(Strap line)

of Amplitron
Figure 4.38 Schematic

Derivation
for the Amplitron is given by
The basic circuit and electronic equation
2
(-)n-[UA-*+R]-Bzs|uA-1)*/
Circuit propagation constant

Where wave propagation

constant
Harmonic

Electron beam phase constant

Vo
 2eVo d.c electron beam velocity
m

= -mB Cyclotron angular frequency

Vo Cyclotron phase constant

BeB

mvo
B Crossed-magnetic-flux density

C- 4
The gain parameter

H =2(1+a)p°c3
Aexp(-irv)+Bexp(irv)
a=-
Aexp(-jry)-Bexp(jYy)
P(°IIY)+Bexp( t r j L 4

y-jp(1+P)

p-ti H

HaC?,
M-type device has lower gain parameter than O-type devices.
410.
Backward-Wave Crossed-Field Oscillator
# uses backward waves for oscillation. Another name is carcinotron.

Ypes
(a) Linear M-Carcinotron
(b) Circular M-Carcinotron

Linear M-Carcinotron

In this device, the interaction between the electrons and the SWS takes plae

a space of crossed field.

fiela s
The SWS is in parallel with an electrode known as sole. A dc electric
sole as
maintained between the grounded slow-wave structure and the negative
shown in Figure 4.39.

.The electrons emitted from the cathode are bent through

90° angle by the
harmonic
magnetic field. The electron interacts with a backward-wave space
direction of the
of the circuit and the energy in the circuit flows opposite to the
electron motion.

The SNS is terminated in the collector end, and the RF signal output is
removed at the electron-gunn end as shown in Figure 4.40.

Characteristics
Crossed field device

Very high efficiency

RF
RF signal
output
termination
Accelerator Slow-wave structure

Electron beam

B EV

Sole Collector
Cathode

Figure 4.39 Schematic of Linear M-Carcinotron

 Circuit

TITT

D
G
Solc

Figure 4.40 Beam electrons and electric field of carcinotron Schema

matic
(b) Circular M-Carcinotron

It is constructed in circular reentrant cavity. The slow-wave structure and soie .

are
circular in shape as shown in Figure 4.41.

RF attenuator Collector Grid

Accelerating anode
Cathode
RF output

wwwi

Slow-wave
structure
Sole
Electron beam
RF wave

Figure 4.41. Schematic of Circular M-Carcinotron

In this circuit, the delay line is terminated at the collector end by sprayingu
aftenuating material on the surface of the conductors.
he output is taken from the gun end of the delay line, which is an interdign
line.

ne electron drift velocity has to be in synchronism with a backward-wave

vave space

harmonic.