iv/iv b.tech regular degree examination, november 2017 ...1.h) write hull cut off voltage...
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IV/IV B.Tech Regular Degree Examination, November 2017
Scheme of Evaluation
Subject code:- 14EC703
Subject Name:- Microwave Theory and Techniques
Faculty: Head of Department
D. Gopi chand
Assistant Professor
9494357618
Answer all questions:
1.a)Mention the advantage of microwaves?
a) High Bandwidth (1% BW at 600MHz is 6MHz at 60GHz is 600MHz).
b) High Antenna gain and Directivity (π ππ’π§ βπ
ππ β
π
ππππππ ).
c) Line of sight Transmission and reception.
d) High penetration capability helps in large distance
communications.(Ionosperic propagation and satellite communication)
f) Less power requirements of transmitter at microwave frequencies.
Note: Any 2 points from above points.
1.b)What is circulator?
A circulator is a multiport device in which the wave incident at port 1 is
coupled to port 2 only, a wave incident in port 2 is coupled to port 3 only, and
so on.
Note: Even circulator diagram with power flow indications can be
awarded marks.
1.c)What is Joint in microwave components?
As a waveguide system cannot built in a single piece always, sometimes it is
necessary to join different waveguides. This joining must be carefully done
to prevent problems such as Reflection effects, creation of standing waves,
and increasing the attenuation etc., The waveguide joints besides avoiding
irregularities, should also take care of E and H field patterns by not affecting
them. There are many types of waveguide joints such as bolted flange,
flange joint, choke joint etc.
1.d) Why slotted sections are used in waveguide systems?
The purpose of using Slotted sections in waveguide systems is to find the
position of voltage minima and maxima. The measurement of impedance,
VSWR, guide wavelength can be done with the help of these slotted
sections.
1.e) Give the applications of H plane Tee?
It can be used as power divider (3 dB splitter) and power combiner.
1.f) Define velocity modulation in Klystrons?
The variation of electron velocity in drift space is known as velocity
modulation. In klystrons drift space is the space between the two cavity
resonators.
1.g) Compare the efficiencies of two cavity klystron and Reflex
Klystron?
The maximum electronic efficiency of two cavity klystron is 58% .The
maximum electronic efficiency of reflexive klystron is 22.7%.
In general efficiency of two cavity klystron is about 40%. In general
efficiency of reflex klystron is about 30%.
Note: general efficiency is also can be considered and awarded marks.
1.h) Write Hull cut off voltage expression?
The expression for Hull cut off voltage is given by
π½ππ = π
π
π
ππ©ππππ π β
ππ
ππ π
1.i) Give the significance of operating a magnetron in a Ο mode?
For sustaining the oscillation in a magnetron, the phase difference between
two adjacent anode plates should equal (2Οn/N) where N is the number of
cavities and n is any integer, this gives rise to Ο mode operation when n =
N/2. Also magnetron operating in the Ο mode has greater power output and
hence most commonly used.
1.j) Expand TRAPATT and IMPATT?
TRAPATT: Trapped Plasma Avalanche Triggered Transit
IMPATT: Impact ionization Avalanche Transit Time
1.k) What is Schottky diode?
The Schottky diode is a semi conductor diode formed by junction of semi
conductor with a metal. It has low forward voltage drop and a very fast
switching action.
1.l) Define attenuation with respect to microwave bench?
Attenuation is defined as the ratio of input power to the output power in dB.
π¨ππππππππππ ππ π π© = πππ₯π¨π (π·π
π·π)
Where P1 = power detected by the load without attenuator in the line.
P2 = power detected by the load with the attenuator in the line.
Unit-1
2. a) Explain with suitable examples, about the advantages of Microwaves?
Microwave Properties and Advantages:-
High Bandwidth: At high frequencies more bandwidth (information carrying
capacity) can be realized. A 1% percent bandwidth at 600MHz is 6MHz(the
bandwidth of one TV channel), and at a 60GHz a 1% bandwidth is 600MHz(
bandwidth of 100 TV channels).
Antenna Gain and Directivity: The gain of antenna is directly proportional to
electrical size. The beamwidth of an antenna is inversely proportional to the
electrical size of its maximum dimension. Thus shorter wavelengths at microwave
frequencies allow for smaller antennas.
π ππ’π§ βπ
ππ β
π
ππππππ
Effect of Ionosphere: When low frequency waves are directed upward into the
atmosphere, they experience significant reflection due to ionosphere. The high
frequency waves which pass through ionosphere with little effect and are therefore
utilized in satellite communications and space transmissions.
Line of sight Transmission/Reception: the microwave receiving antenna must be
with the line of sight of the transmission antenna. Long distance communication on
earth requires that microwave relay stations be used.
Electromagnetic Noise Characteristics: The EM noise level in nature over 1-
10GHz frequency is small. This allows for the detection of low signal levels using
sensitive receivers.
Target reflection of Electromagnetic waves (Radar cross section): In general,
electrically large conducting radar targets reflect more energy. Thus the higher
frequencies of microwaves are preferred for radar systems.
Power Requirements: At microwave frequencies, the power requirements of the
transmitter become very small as compared to that at MF/HF, due to high gain of
the antennas at microwaves.
Note: Any four advantages with explanation carries 6M first two points are
mandatory.
2.b) Derive the expression for cut off frequency of circular cavity resonators?
Circular-cavity resonator:
A circular-cavity resonator is a circular waveguide with two ends closed by
a metal wall. The wave function in the circular resonator should satisfy Maxwell's
equations, subject to the same boundary conditions described for a rectangular-
cavity resonator. It is merely necessary to choose the harmonic functions in z to
satisfy the boundary conditions at the remaining two end walls. These can be
achieved if
where n = 0, 1, 2, 3, ... is the number of the periodicity in the Ο direction
p = 1, 2, 3, 4, ... is the number of zeros of the field in the radial direction
q = 1, 2, 3, 4, . . . is the number of half-waves in the axial direction
Jn = Bessell's function of the first kind
Hoz = amplitude of the magnetic field
Where n = 0, l, 2, 3, .. .
p = 1, 2, 3, 4, .. .
q = 0, l, 2, 3, .. .
Eoz = amplitude of the electric field
The separation equations for TE and TM modes are given by
Substitute k2=Ο
2Ρμ
TM110 mode is dominant where 2a>d and the TE111 is dominant when d>=2a.
Note: Circular cavity resonator equations 4M and diagram and explanation
2M.
3.a) Explain construction and working of directional coupler?
A directional coupler is a four-port waveguide junction. It consists of a primary
waveguide 1-2 and a secondary waveguide 3-4.
When all ports are terminated in their characteristic impedances, there is free
transmission of power, without reflection, between port 1 and port 2, and there is
no transmission of power between port 1 and port 3 or between port 2 and port 4
because no coupling exists between these two pairs of ports. The degree of
coupling between port 1 and port 4 and between port 2 and port 3 depends on the
structure of the coupler.
The characteristics of a directional coupler can be expressed in terms of its
coupling factor and its directivity. Assuming that the wave is propagating from
port l to port 2 in the primary line, the
Coupling factor and the directivity are defined as
Coupling factor is defined as the ratio of the incident power pi to the forward
coupled power Pf measured in dB. Coupling factor is a measure of how much of
the incident power is being sampled.
πΆππ’πππππ ππππ‘ππ ππ΅ = 10 log ππ
ππ = 10log(
π1
π4)
Directivity is defined as the ratio of forward coupled power Pf to the back power Pb
expressed in dB.
π·πππππ‘ππ£ππ‘π¦ ππ΅ = 10 log ππ
ππ = 10 log(
π4
π3)
Isolation: it is defined as the ratio of the incident power Pi to the power Pb.
πΌπ ππππ‘πππ ππ΅ = 10 log ππππ = 10log(
π1
π3)
where P1 = power input to port 1,P2 = power input to port2,P3 = power output from
port 3 ,P4 = power output from port 4.
Note: Construction 3M working 3M
3.b) Explain the Faraday rotation principle using an isolator?
Faraday rotation: When a DC or static magnetic field B0 is applied along the z-
direction. A plane TEM wave that is already linearly polarized along x-axis at t=0
is made to propagate through the ferrite placed in the z-direction. The plane of
polarization of this wave will rotate with distance. This phenomenon is called
Faraday rotation.
Any linearly polarized wave can be regarded as the vector sum of two
oppositely rotating circularly polarized waves (E0/2 vectors in above figure).
The ferrite material offers different characteristics to these waves with the result
that the phase change in one wave is larger than the other wave resulting in
rotation βΞΈβ of linearly polarized wave at z=l.
The angle of rotation βΞΈβ is given by
ΞΈ=π
2(π½+ β π½β)
l= length of the ferrite rod, Ξ²+= phase shift of the right polarized wave, Ξ²-=
phase shift of left polarized wave.
The Faraday rotation principle is shown by the ferrite materials those exhibits
the property if magnetic anisotropy, non reciprocal property, insulator nature.
These ferrite devices are used in designing the non reciprocal devices like
Gyrator, Isolator, and circulator.
Isolator:
An isolator is a two-port device which provides negligible attenuation for
transmission of signal from port-1 to port-2 but provides high attenuation for
transmission from port-2 to port-1.
Construction:
The Cut away view of Faraday rotation isolator is shown above. It consists of
circular waveguide carrying dominant mode i.e. TE11 mode, with transitions to
a standard rectangular waveguide with dominant mode TE10 at both ends but the
output end transition is twisted by 450. A thin circular ferrite is placed inside the
circular guide, supposed by poly foam, and the waveguide is surrounded by a
permanent magnet which generates DC magnetic field in the axial direction of
ferrite rod. The input resistive attenuator/ card placed along larger dimension of
rectangular guide and the output resistive attenuator/ card are displaced by 450.
The function of resistive attenuator is to absorb any wave whose plane of
polarization is parallel to the plane of resistive attenuator.
Principle of Operation:
An input TE10 dominant mode is incident on port 1 of the isolator. As the wave
is perpendicular to the resistive card, it passes through the ferrite rod without
any attenuation and the wave is rotated 450 clockwise due to Faraday rotation.
As a result of rotation, the wave arrives at output port i.e. port 2 without
attenuation.
If a wave tries to propagate from port 2 to port 1, it passes through resistive card
placed at port 2 as electric field is perpendicular to the plane of card and the
wave is rotated 450 clockwise direction(looking from port 1) due to Faraday
rotation by the ferrite rod. Since the wave coming towards port 1 is parallel to
the resistive card, the wave is absorbed by it.
Thus any reflected wave from output port will not reach input port.
The typical performance of these isolators is about 1-dB insertion loss in
forward transmission and about 20-30 dB isolation in reverse direction.
Note: Faraday rotation principle 2M isolator diagram 2M working 2M
Unit-II
4.a) Derive the S-matrix of a magic Tee?
The simplest type of power divider is a T-junction. A magic tee is a
combination of E-plane Tee and H-plane Tee. The magic tee has following
characteristics.
1. If two waves of equal magnitude and the same phase are fed into port 1 and
port 2, the output will be zero at port 3 and additive at port 4.
2. If a wave is fed into port 4 (H arm), it will be divided equally between port 1
and port 2 of the collinear arms and will not appear at port 3 (E arm).
3. If a wave is fed into port 3 (E arm), it will produce an output of equal
magnitude and opposite phase at port 1 and port 2. Output at port 4 is zero
i.e S43 = S34 = 0.
4. If a wave is fed into one of the collinear arms at port 1 or port 2, it will not
appear in the other collinear arm at port 2 or port 1 because the E arm causes
a phase delay while the H arm causes the phase advance. i.e S12 = S21 = 0.
For the purpose of double the power at one port and by taking the input power
from the two transmitters situated at two other ports, the property 1 is used.
The S-matrix for magic Tee is given by
[S] =
π11 π12
π21 π22
π13 π14
π23 π24
π31 π32
π41 π42
π33 π34
π43 π44
Due to E-plane Tee:
S13 = -S23
Due to H-plane Tee:
S14 = S24
Due to geometry, port 3 and port 4 are isolated
S34 = S43 = 0
From the Magic Property:
S12 = S21 = 0
If the port 3 and port 4 are matched to the junction
S33 = S44 = 0
From the Symmetry Property, Sij = Sji
S13 = S31; S14 = S41;
S23 = S32; S24 = S42;
The S-matrix by substituting above results are:
[S] =
π11 00 π22
βπ23 π14
π23 π14
βπ23 π32
π14 π14
0 00 0
From the unitary Property, [S] [S]* = [U]
π11 00 π22
βπ23 π14
π23 π14
βπ23 π23
π14 π14
0 00 0
*
π11β 00 π22
ββπ23
β π14β
π23β π14
β
βπ23β π23
β
π14β π14
β0 00 0
=
1 00 1
0 00 0
0 00 0
1 00 1
R1C1: |S11|2 + |S23|
2 + |S14|
2 = 1 eqn (1)
R2C2: |S22|2 + |S23|
2 + |S14|
2 = 1 eqn(2)
R3C3: |S23|2 + |S23|
2 = 1 eqn (3)
R4C4: |S14|2 + |S14|
2 = 1 eqn (4)
From the above equations, we get
S23 = 1
2 and S14 =
1
2
S13 = - S23 = β1
2
Then S14 = S24 = 1
2
Substituting S23, S14 in above equations 1&2 gives S11 = S22 = 0
The final S-matirx will be shown below by substiting above values
[S] =
0 00 0
π13 π14
π23 π24
π31 π32
π41 π42
0 00 0
[S]=
0 00 0
β1
2
1
21
2
1
2β1
2
1
21
2
β1
2
0 00 0
Note: Diagram 2M,initial conditions 2M, correct S-matrix 2M. initial
conditions are specified using properties of magic Tee
4.b)What are the evaluation parameters of a S matrix?
The behavior of the network in terms of reflected and injected power waves can be
described by a set of linear equations. For the 2-port case, the outputs can be
replaced to the inputs by
π1 = π11π1 + π12π2
π2 = π21π1 + π22π2
π11 = π1
π1 π2=0
=πππππππ‘ππ πππ€ππ π€ππ£π ππ‘ ππππ‘ 1
ππππππππ‘ πππ€ππ ππ‘ ππππ‘1
π21 = π2
π1 π2=0
=π‘ππππ πππ‘π‘ππ πππ€ππ π€ππ£π ππ‘ ππππ‘ 2
ππππππππ‘ πππ€ππ ππ‘ ππππ‘1
π22 = π2
π2 π1=0
=πππππππ‘ππ πππ€ππ π€ππ£π ππ‘ ππππ‘ 2
ππππππππ‘ πππ€ππ ππ‘ ππππ‘2
π12 = π1
π2 π1=0
=π‘ππππ πππ‘π‘ππ πππ€ππ π€ππ£π ππ‘ ππππ‘ 1
ππππππππ‘ πππ€ππ ππ‘ ππππ‘2
S parameters can only determined under conditions of perfect matching at the input
or output port.
Determination of S11 and S21 parameters: S11, S21 can be computed if port 2 is
terminated with matched load(Z0).
Load impedance (ZL) = characteristic impedance(Z0)
By defining Zin as input impedance given ZL = Z0
π11 = πππ ππΏ=π0
β π0
πππ ππΏ=π0+ π0
From the above equation, it is known that reflection coefficient(S11 is the method
of specifying the input impedancezin).
S11 = b1/a1 is the input reflection coefficient (when a2 = 0)
S21 = b2/a1 is the forward transmission gain (when a2 = 0)
Determination of S22 and S12 parameters: S22, S12 can be computed if port 1 is
terminated with matched load(Z0).
input impedance (Zin) = characteristic impedance(Z0)
π22 = πππ’π‘ πππ =π0
β π0
πππ’π‘ πππ =π0+ π0
From the above equation, it is known that reflection coefficient(S22 is the method
of specifying the output impedance zL).
S22 = b2/a2 is the output reflection coefficient (when a1 = 0)
S12 = b1/a2 is the reverse transmission gain (when a1 = 0)
Note: formulas of S-parameters 4M, diagram explanation 2M
5.a) Derive the S-matrix of a H plane Tee?
H-plane tee (shunt tee): An H-plane tee is a waveguide tee in which the axis of its
side arm is "shunting" the E field or parallel to the H field of the main guide as
shown below
Characteristics of H-plane Tee:
If the H plane junction is completely symmetrical and waves enter through side
arm, the waves that leave through collinear ports are equal in magnitude and phase
Therefore π13 = π23
S matrix is of order 3X3
π =
π11 π12 π13
π21 π22 π23
π31 π32 π33
Because of plane of symmetry of the junction π13 = π23
If port 3 is perfectly matched, S33 = 0
From symmetry property, Sij = Sji
S12 = S21 ,S13 = S31 , S23 = S32 = S13
With all above properties [S] becomes,
π =
π11 π12 π13
π12 π22 π13
π13 π13 0
From unitary property π [π ]β = [π]
π11 π12 π13
π12 π22 π13
π13 π13 0β
π11β π12
β π13β
π12β π22
β π13β
π13β π13
β 0=
1 0 00 1 00 0 1
R1C1: |S11|2 + |S12|
2 + |S13|
2 = 1
R2C2: |S12|2 + |S22|
2 + |S13|
2 = 1
R3C3: |S13|2 + |S13|
2 = 1
R3C1: π13π11β + π13π12
β = 0
S11 = S22; π13 = 1
2 ;
π13 π11β + π12
β = 0 ; π13 β 0 ;
π11β = βπ12
β
π12 = βπ11
π11 2 + π11
2 + 1
2= 1;π11 =
1
2 π12 =
β1
2 π22 =
1
2
π =
1
2
β1
2
1
2β1
2
1
2
1
21
2
1
20
Note: Diagram 2M, initial conditions 2M, correct S-matrix 2M. initial
conditions are specified using properties of H plane Tee
5.b) Compare H plane Tee with E plane Tee?
H-plane Tee E-plane Tee
In an H-plane Tee, the axis of its side
arm is parallel to the magnetic field or
shunting the electric field of the main
waveguide.
In an E-plane Tee, the axis of its side
arm is parallel to the electric field of the
main waveguide.
An H plane Tee is also called a parallel
or shunt Tee.
The E-plane Tee is also called series
Tee.
When the power is fed at port-3, that is
at the side arm, the resulting power is
equally divided between port 1 and port
2 with in phase.
When the power is fed at port3, that is at
the side arm, the resulting power is
equally divided between port1 and 2,
but phase shift of 1800 is introduced
between the two outputs.
When the equal input power is fed to
both ports 1 and 2, the maximum power
is obtained at port 3.
When the equal input power is fed to
both ports 1 and 2, the maximum power
is obtained at port3.
When the input is applied at any one of
the collinear ports i.e., port1 and port2,
the resulting power is obtained at port3.
when the input is applied at any one of
the collinear ports i.e., port1 or port2,
the resulting power is obtained at port3.
Note: diagrams can also be drawn as a difference. Any 4 valid points with neat
sketches can be awarded 6M.
Unit III
6.a) Derive the equation of motion of an electron in the repeller region and
hence find the round trip transit time and transit angle of the centre of the
bunch of electrons in the reflex klystron?
Let us assume that space charge effects of electron motion are negligible. The
electron velocity π£0 before entering cavity gap at z=0 and time t0 is given by
π£0 = 2ππ0
π= 0.593π 106 π0ππ‘π /π ππ
V0 = DC beam voltage
The same electron leaves the cavity gap at z=d at the time t1 with velocity
π£ π‘1 = π£0[1 + π½1π1
2π0sin(ππ‘1 β
ππ2
)
The same electron is forced back to cavity by the repeller at z=d and time t2 by the
retarding electric field E, given by
πΈ =πππ‘ππ π£πππ‘πππ
πΏππππ‘π=ππ + π0 + π1 sin(ππ‘)
πΏ
E exists in the z direction and is constant in z direction.
The force equation for the electron in the repeller region is
ππ2π
ππ‘ 2= βππΈ = βπ
ππ+π0
πΏ where |π1sin(ππ‘) βͺ (ππ + π0) is assumed.
Integrating above equation twice
ππ§
ππ‘=
βπππ + π0
ππΏ ππ‘π‘
π‘1
= βπ ππ + π0
ππΏ π‘ β π‘1 + πΆ1
At π‘ = π‘1 ,ππ§
ππ‘= π£ π‘1 = πΆ1; π‘πππ
π§ =βπ(ππ + π0)
ππΏ π‘ β π‘1 ππ‘π‘
π‘1
+ π£(π‘1) ππ‘π‘
π‘1
π§ =βπ(ππ + π0)
2ππΏ(π‘ β π‘1)2 + π£ π‘1 π‘ β π‘1 + πΆ2
At π‘ = π‘1 , π§ = π = πΆ2; then
π§ =βπ(ππ + π0)
2ππΏ(π‘ β π‘1)2 + π£ π‘1 π‘ β π‘1 + π
Now, on the assumption that the electron leaves the cavity gap at z=d and time t1,
with a velocity of v(t1) and returns to the gap at z=d and time t2, then at t= t2, z=d
π =βπ(ππ + π0)
2ππΏ(π‘2 β π‘1)2 + π£ π‘1 π‘2 β π‘1 + π
π£ π‘1 =π ππ + π0
2ππΏ(π‘2 β π‘1)
The round trip transit time in the repeller space is given by
πβ² = π‘2 β π‘1 =2ππΏ
π ππ + π0 π£ π‘1 = π0
β² [1 + π½1π1
2π0sin(ππ‘1 β
ππ2
)
Where π0β² =
2ππΏ
π ππ+π0 is the round trip DC transit time
Multiplication of above equation by radian frequency results in
ππ1 = π π‘2 β π‘1 = π0β² + πβ² sin(ππ‘1 β
ππ2
)
π0β² = ππ0
β²
is the round trip DC transit angle and
πβ² =π½1π1
2π0π0β²
is the Bunching Parameter of the reflex klystron oscillator.
Note: Valid expressions and step by step derivation with boundary conditions
included can be awarded 6M. Boundary conditions substitution 2M bunching
parameter 2M, round trip transit time 2M.
6.b) A Reflex Klystron operates at the peak of the n=1 or ΒΎ mode. The dc
power input is 40mW, and the ratio of V1 to V0 is 0.278
(i) Determine the efficiency of the Reflex klystron
(ii)Find the total output in mW
(iii) if 20% of the power delivered by the electron beam is displayed in the
cavity walls, find the power delivered to the load?
Since, Pdc = 40mW n=1 mode π1
π0= 0.278 =
2πβ²
π½π(2ππ β π2
)
Assume Ξ²i = 1, and substitute n=1 we get Xβ =0.6555
J1(0.6555) = 0.31045.
Since 20% of dc power is dissipated in wall net DC power available for conversion
is Pdc = 40mW β 20% of 40mW = 32mW.
The output AC power is πππ = πππ2π β² π½1(π β² )
2ππβπ
2
= 2.740ππ~2.75ππ
Note: correct formula 2M, correct bunching parameter and first order Bessel
function value 2M, correct answers 2M.
7.a) write short notes on the following:
(i) Beam coupling coefficient
(ii) Beam loading
(iii)Different types of electric currents
(iv) Assumptions made to explain the principles of operation of klystrons.
Beam Coupling Coefficient:
It is defined as beam coupling coefficient at the input cavity gap Ξ²i.
By increasing the transit angle ΞΈg decreases the coupling between the
electron beam and the buncher cavity i.e., the velocity modulation of given
microwave signal decreased.
Beam Loading:
The maximum bunching should occur approximately midway between the
catcher grids.
The phase of the catcher gap voltage must be maintained in such a way that
the bunched electrons, as they pass through the grids, encounter a retarding
phase.
When the bunched electron beam passes through the retarding phase, its
kinetic energy is transferred to the field of the catcher cavity.
When the electrons emerge from the grids, they have reduced velocity and
are finally collected by the collector.
Figure shows the output equivalent circuit
Rsho - the wall resistance of the catcher cavity.
RB - the beam loading resistance
RL - the external load resistance
Rsh - the effective shunt resistance
Different types of current:
The current induced by the electron beam in the walls of the catcher cavity is
directly proportional to the amplitude of the microwave input voltage V1 ,
the fundamental component of the induced microwave current in the catcher
is given by
Where Ξ²0 is the beam coupling coefficient of the catcher gap. If the buncher
and catcher cavities are identical, then Ξ²i = Ξ²0 . The fundamental
component of the current induced in the catcher cavity then has the
magnitude.
Assumptions made to explain the principles of operation of klystrons:
1. The electron beam is assumed to have a uniform density in the cross section of
the beam.
2. Space charge effects are negligible.
3. The magnitude of the microwave signal input is assumed to be much smaller
than the dc accelerating voltage.
Note: each bit carries 2M
7.b)Derive the Hull cut off voltage expression of a magnetron?
The Hull cut off condition is obtained under the condition that there is no RF field,
which is in turn defines anode voltage is a function of magnetic field. The
magnetic field tends to prevent the flow of electrons to the anode. On the other
hand, under right circumstances, the electrons leave the hub after getting interacted
with RF wave that is rotating about the cathode , and flow to the anode. It happens
electrons speed is reduced to RF rotation rate. In this process the electrons amplify
the wave and losses the energy.
Force acting on the electron πΉ = π΅ππ£
In the direction of Ο, the force component is given by πΉπ = ππ΅π£π
Where π£π = π£ππππππ‘π¦ ππ π‘ππ ππππππ‘πππ ππ π‘ππ ππππππ πππ π‘ππππ π, from the
center of the cathode cylinder.
Torque in the direction of Ο can be given as
ππ = ππΉπ = πππ£ππ΅
Angular momentum = angular velocity X moment of inertia = ππ
ππ‘= ππ2
Time rate of angular momentum = π
ππ‘(ππ
ππ‘π₯ππ2)
π
ππ‘ ππ
ππ‘π₯ππ2 = πππ£ππ΅
2ππππ
ππ‘+ ππ2
π2π
ππ‘2= ππ΅π
ππ
ππ‘
Integrate above equation with respect to βtβ
2πππ + ππ2ππ
ππ‘= ππ΅
π2
2
For a particular direction, mΟΟ can be considered a constant
ππ2ππ
ππ‘+ πΆ = ππ΅
π2
2
The value of C can be determined by applying boundary conditions(i.e., at surface
of the cathode Ο=a and ππ
ππ‘= 0
πΆ = ππ΅π2
2
Substitute above C value we get
ππ2ππ
ππ‘=ππ΅
2(π2 β π2)
ππ
ππ‘=ππ΅
2π(1 β
π2
π2)
When Ο=a(i.e. at cathode), ππ
ππ‘ ππππππππππ π‘π 0
When π β« π,ππ
ππ‘ ππππππππππ (π)πππ₯
(ππ
ππ‘)πππ₯ = (π)πππ₯ =
ππ΅
2π=ππ΅π2π
Where B= Bc is the cut-off magnetic flux density
We know potential energy of electron = kinetic energy of electron
ππ0 =1
2ππ£2 =
1
2π(π£π
2 + π£π2 )
π£π =ππ
ππ‘ πππ π£π =
πππ
ππ‘
ππ0 =1
2π[
ππ
ππ‘
2+ π2(
ππ
ππ‘)2]
ππ
ππ‘= (π)πππ₯ (1 β
π2
π2)
ππ0 =1
2π[
ππ
ππ‘
2
+ π2 π πππ₯2 1 β
π2
π2
2
]
At anode Ο=b, ππ
ππ‘= 0, substituting these boundary conditions in above equations,
π
2 π2 π πππ₯
2 1 β
π2
π2
2
= ππ0
Substituting above equations we get
π΅πΆ = 8π0π/π 1/2
π(1 βπ2
π2)
The Hull cut-off voltage is given by πππ = 1
8
π
ππ΅0
2π2 1 βπ2
π2 2
Note: Correct derivation and expressions with boundary conditions carries
4M
Unit IV
8.a) Explain the basic modes of operation of a Gunn diode?
Gunn modes of operation:
Depending on the device characteristics and external circuitry, the Gunn diode can
be made to oscillate in any one of these four modes.
Gunn oscillation mode
Limited Space Charge accumulation (LSA mode)
Stable Amplification mode
Bias circuit oscillation mode
The modes of operation are classified based on the condition
π0π >ππ π£ππ|ππ |
Gunn Oscillation modes
This mode has following features:
The electric field is greater than the threshold.
The product of frequency and length is about 107 cm/s
The product of doping and length is more than 1012
/cm2.
Due to the cyclic arrangement of either accumulation layer or the high
voltage domain the Gunn diode is unstable.
There are three types of Gunn oscillation modes. They are
Transit time domain mode
Delayed domain mode
Quenched domain mode
Transit time domain mode: Transit time domain oscillation is the basic mode and
is not depend on the external circuit. When a domain is quenched at the anode the
current peaks are obtained. Then, another is nucleated near the cathode. At any
time, the entire electric field across the device is above the threshold electric field.
The frequency is given by
f=1/Ο=π£π/π π£π ππ π‘ππ πππππ ππ πππππ‘ π£ππππππ‘π¦ l is the effective length and fl is the product of
frequency and length.
The oscillation time Ο0 is equal to the transit time(Ο). It is a low power(<2W), low
efficiency mode, and the operating frequency is between 1 GHz and 18GHz.
Delayed domain mode: This is also called inhibited mode. In this mode, the
domain is collected when the dc bias is less than the threshold electric field(Eth).
The next domain can only be formed when the field again reaches threshold. The
oscillation period is greater than transit time of the critical electric field(Eth), that is
Ο0>Ο. The frequency of oscillation is determined by the resonant circuit. The
efficiency of this mode is limited to 20%. The drift velocity (π£π) or f l lies between
106cm/s and 10
7cm/s.
Quenched domain mode: In this mode, the domain collapses before it reaches the
anode; that is it quenched before it is collected, hence the name quenched mode.
The bias electric field reduces below the sustaining electric field(Es) in the negative
half cycle. When the bias electric field swings back more than the threshold
electric field, another domain is nucleated and the procedure repeats. The operating
frequencies are higher than the transit time frequency. The maximum efficiency of
this mode is 13%.
Limited Space Charge Accumulation mode:
This mode operates by using high-Q resonant cavity with current pulses from the
Gunn diode. The Gunn diode is placed in a resonant cavity, which is tuned to a
frequency of the LSA mode(f0), so that the circuit operates like a negative
resistance device and the domains do not have enough time to form. In this mode,
the device can be biased to several times higher than Eth. When the input field is
increase more than the threshold electric field, the device remains in the negative
resistance region. Oscillation time is obtained by the external circuit, which is
given by
π0 = 2π πΏπΆ + πΏ
π (ππππ‘π
)
Stable Amplification mode: This mode is defined in the region where there
product of frequency times length is about 107cm/s and product of doping times
length is in between 1011
and 1012
/cm2.
Bias circuit oscillation mode: This mode occurs only when there is either Gunn or
LSA oscillation, and it is usually at the region where the product of frequency
times length is too small to appear. The drop in current at threshold can lead to
oscillations in the bias circuit typically 1KHz to 100MHz.
Note: 4 modes and explanation 4M and diagrams 2M
8.b) Using a typical microwave bench, explain the measurement of
attenuation?
Attenuators are used to adjust the power level of the microwave signal. If the
network is perfectly matched, the reflected power is zero and the insertion
loss is similar to the attenuation provided by the microwave device or
component.
Attenuation is defined as the ratio of input power to the output power in dB.
π¨ππππππππππ ππ π π© = πππ₯π¨π (π·π
π·π)
Where P1 = power detected by the load without attenuator in the line.
P2 = power detected by the load with the attenuator in the line. Power Ratio method:- Power ratio is a process of measuring the input and output power with the
device(setup 1) as shown in figure below and without device(or attenuator)
in the circuit shown below.
The P1 and P2 are the powers measured the setup 1 and 2.The attenuation is
the ratio of power (P1 / P2 ) which is expressed in decibels.
Disadvantages of the power ratio method:- The attenuation calculated will not be accurate. Because the powers
measured (P1 & P2 ) is non linear.
It is also true for the networks with low input power.
With this method, we can measure the attenuation up to 20dB
RF substitution method:- The attenuation through the device under test is compared with a standard
microwave attenuator at the same frequency in this method.
Thus the output power βPβ is measured by this method.
The drawback of power ratio method can be overcome by this method as the
attenuation is measured at single power position.
In setup2 the attenuator is adjusted to get the same output power βPβ as in
setup1. the attenuation value is measured in the precision attenuator.
Note: 2methods each method carries 3M
9.a) Explain the method of measuring VSWR using microwave bench setup?
VSWR stands for voltage standing wave ratio. In a perfectly matched
system, there is no variation in the field strength along the waveguide.
A mismatch leads to reflected waves, there by leading to standing waves
along the length of the guide,
Standing guide are the indication of the quality of the transmission. VSWR
= 1 for perfectly matched system. The ratio of maximum to the minimum
voltage gives the VSWR.
ππππ =ππππ₯ππππ
= 1 + π€
1 β π€
Where Ξ is the reflection coefficient.
LOW VSWR MEASUREMENT:
VSWR values below 10 are very easily measurement by this method.
The VSWR meter direct displays these values. The attenuator is adjusted to
give a maximum reading on the meter. Then, the minimum reading on the
meter is obtained by adjusting the probe on the slotted line. Thus VSWR is
defined as the ratio of first reading to the second reading.
The meter can be calibrated in terms of VSWR. Here attenuator is adjusted
so that travelling probe gives maximum deflection on the VSWR meter.
The VSWR of 1 corresponds to full scale deflection(i.e. 10mV in the meter).
By adjusting the travelling probe the minimum reading can be obtained on
the meter. The variation in deflections and the corresponding VSWR values
are tabulated.
Deflection
in the
meter
VSWR
5mV 2
3.33mV 3
2.5mV 4
2mV 5
1.69mV 6
HIGH VSWR MEASUREMENT:
VSWR greater than 10 can be measured by double minimum method. In this
method, the probe is inserted to a depth where the minimum value can be read
easily.
Then the probe should be moved to a point where the power is twice the
minimum ( Pmin = 2V2
min /RL i.e. Pmin= 2P).
Let d1 be the position. Then again, the probe is moved to twice the power
point on the other side of the minimum(say d2) as shown in fig below
For the dominant mode TE10 mode rectangular waveguide, Ξ»0, Ξ»g , and Ξ»c are
related as below. 1
π02 =
1
ππ2
+1
ππ2
Ξ»0 is free space wavelength
Ξ»g is guide wavelength
Ξ»c is cutt off wavelength
For the TE10 mode Ξ»c = 2a where βaβ is the broad dimension of the
waveguide
π0 = πΆπ
ππππ = ππ
π(π2 β π1)
Note: 2 methods and each method carries 3M
9.b) Explain Varactor and Crystal diode?
Crystal detectors
Point-contact diodes, commonly called CRYSTALS, are the
oldest microwave semiconductor devices.
Unlike the PN-junction diode, the point-contact diode depends on the pressure of
contact between a point and a semiconductor crystal for its operation.. One section
of the diode consists of a small rectangular crystal of n-type silicon. A fine
berylium-copper, bronze-phosphor, or tungsten wire called the CATWHISKER
presses against the crystal and forms the other part of the diode. During the
manufacture of the point contact diode, a relatively large current is passed from the
catwhisker to the silicon crystal. The result of this large current is the formation of
a small region of p-type material around the crystal in the vicinity of the point
contact. Thus, a PN-junction is formed which behaves in the same way as a normal
PN-junction.
Varactor diode:
A varactor diode is a variable capacitor junction diode. These two terminal devices
also called as varicaps. This is a special type of PN junction that is designed to
operate in a microwave range. It works on the principle of voltage variable nature
of the depletion capacitance.
Working: In a PN junction, due to density gradient, holes diffuse N side and
Electrons diffuse P side. This causes a few ions on either side of the junction to be
depleted of mobile charges. This region is known as depletion region or space
charge region.
If forward bias is applied, the carriers move toward the junction which reduces the
depletion width.
If reverse bias is applied, the carriers move away from the junction: as a result the
depletion width increases.
The variation of width with voltage may be considered a capacitive effect with the
depletion region as the dielectric, and P and N region as parallel plates. This
capacitance is known as transition capacitance or junction capacitance.
πΆπ =ππ π΄
π
Note: both diagrams and symbol 3M working explanation 3M
Prepared by:
Gopi chand. Dasari
Asst. Prof
Dept of ECE
BEC, Bapatla.