МИНОБРНАУКИ РОССИИ

САНКТ-ПЕТЕРБУРГСКИЙ ГОСУДАРСТВЕННЫЙ

ЭЛЕКТРОТЕХНИЧЕСКИЙ УНИВЕРСИТЕТ

«ЛЭТИ» ИМ. В.И. УЛЬЯНОВА (ЛЕНИНА)

Кафедра РТЭ

Отчет по индивидуальному заданию №1

По дисциплине «Микроволновая электроника»

Студент гр. 9201		Рауан М.
Преподаватель		Иванов В.А.

Санкт-Петербург

2022

Task 1.

1. Calculate the quantum energy of microwave radiation with a frequency of: GSM +Groupe [GHz] standard f0, 5G+ Ngroup [GHz] standard f0

1.1 To what temperature should an absolutely black body be heated

so that the maximum radiation is at the received frequencies?

1.2 Compare this energy with the energy of thermal vibrations of molecules at T = 300K.

1.3 Compare this energy with the binding energy of clusters in water. To what temperature should the water be heated to destroy its cluster structure? to ionize?

1.4 Can microwave radiation change the chemical structure of a living cell? At the expense of what?

Solution:

1. For GSM, the operating range is 900-1800 MHz, and for 5G, 24-100 GHz.

Sources: GSM Radio Frequency Planning | GSM frequency band-ARFCN (rfwireless-world.com)

5G frequency bands in INDIA,USA,Europe,CHINA,JAPAN,Korea | 5G Bands (rfwireless-world.com)

The quantum energy of microwave radiation for given frequencies [source: Lecture No. 1 MBE]:

Given:

For f0=1.5+1=2.5 GHz – GSM Standard

For f0=27+1=28 GHz – 5G Standard

;

;

1. Wien's displacement law:

T is the absolute temperature in kelvins. b is the displacement constant of the Wine, equal to 2.89 m⋅K

We will find the corresponding temperatures:

For f=2.5 ГГц, ;

For f=28 ГГц, ;

2. The energy of thermal vibrations of molecules is determined by the formula:

E=

Let's compare the energy of thermal vibrations of molecules at T = 300K with the previously obtained energies:

1.3 The binding energy of clusters in water is 0.24 eV.

Sources: 35_11_pr.pdf (lebedev.ru)

Let's compare the binding energy of clusters in water with the previously obtained energies:

The condition of destruction of the cluster structure of water:

; – binding energy

Calculate the temperature to which we need to heat the water to destroy its cluster structure:

The ionization energy of water is 12.6 eV(Sources: 35_11_pr.pdf (lebedev.ru))

Calculate the temperature to which we need to heat the water for its ionization:

3. Microwave radiation range 300 MHz - 300 GHz

Living cells mainly consist of water (about 80%). Thus, the quantum energy is not enough to destroy the structure of water, but at the same time the living cell will heat up (

But at the same time, if there is a long and massive impact, the destruction of the structure of a living cell will begin.

Answers:

1. 

1.1

1.3

Task 2.

2. What is the power density of microwave radiation considered acceptable in everyday life and at work according to the standards of the Russian Federation? According to international standards?

2.1 Estimate how your body temperature will increase at home and at work at a power density level equal to (50 + Nstudent) % of the maximum allowed. In calculations, heat losses for external cooling of the body should be neglected.

2.2 Estimate how many quants of microwave radiation will be required for such heating.

Solution:

2. According to the standards of the Russian Federation, the permissible power of microwave radiation is (Source):

- In everyday life: 10 at a distance of 50 cm

- In production: no more than 8 hours per shift at 25 , at a distance of 50 cm.

According to US and European standards: 200-1000

According to Chinese standards: 2000

1. Conditions:

Body surface area (using Du Bois formula)

The heat changing of the human body:

Heat of radiation:

Then equals:

Specific heat capacity of the human’s body

Now we can determine how much the temperature will rise in an hour in everyday life and on manufacture

1. The number of quanta of microwave radiation:

,

Answers:

Task 3.

3. Compare numerically 2 typical devices: vacuum and semiconductor according to the following parameters:

3.1 The maximum velocity of charged particles.

3.2 The length of the interaction region for the angle of flight -radian.

3.3 Volumetric charge density

3.4 Calculate the microperviance, the "plasma" frequency for the vacuum device.

3.5 For semiconductor: Debye length, plasma frequency.

Compare the values in clauses 3.4. and 3.5. Explain the difference in physical processes in both variants.

Parameters of the vacuum device: current (Nstudent*15)mA, accelerating voltage (Nstudent+Ngroup/2)kV, flow diameter Ngroup*0.5 mm.

Semiconductor: Nstudent doping level* , voltage 25V, current channel thickness 1mkm.

The operating frequency of the devices is Ngroup GHz.

Operating temperature (300+ Nstudent)K.

Operating temperature (300+ Nstudent)K.

3. Parameters of the vacuum device:

Current I0 = 140 mA,

accelerating voltage U0 =6 kV,

the diameter of the flow = 1 mm.

Parameters of the semiconductor surf:

the level of doping = 14 * ,

voltage U = 20V,

current channel thickness = 1mkm.

The operating frequency of the devices is 15 GHz.

Operating temperature 314 K.

3.1 The maximum velocity of charged particles:

Vacuum:

Because all the energy of the accelerating voltage is converted into kinetic energy electron movements:

459.335 Gm/s

Semiconductor:

In transistors, unlike vacuum devices, the maximum

the carrier transfer rate is

3.2 Let's find the length of the interaction area for the span angle -radians. Let 's use the formula of the angle of flight:

L= 15.3 mm

Semiconductor:

L= 3.33 m

3. Let's find the volumetric charge density.

Vacuum device:

I₀=j₀·S

Consequently:

;

we get:

;

where:

where j₀ - current density (А/m), S – cross-sectional area

,

where – charge density (Ql/m³), – the speed of motion of charged particles (m/s)

Semiconductor device:

3. Calculate the microperviance and the "plasma" frequency for the vacuum device. In order to calculate the microperviance, we use the formula:

In order to calculate the "plasma" frequency, we use the formula:

(3.5)

s^-1

5. Calculate the Debye length and plasma frequency for a semiconductor device.

m

Both vacuum devices and semiconductor devices use the concept of plasma frequency to determine the spatial charge forces. But there is a rather important difference between the two concepts from the point of view of the physics of processes. In the vacuum case, the Coulomb action of the charges of the same name leads to longitudinal oscillations. And in semiconductors, opposite charges lead to fluctuations (the initial displacement is provided by thermal motion). Since the semiconductor is quasi-neutral, a sufficiently high electron density can be obtained in such a plasma.

Answer:

3.1 v_v>>v_sem

3.2

3.3

3.4 ; s^-1

3.5 = m;

The plasma frequency of a semiconductor device is much higher than the plasma frequency of a vacuum device. This can be explained by the fact that due to the quasi-neutrality of the semiconductor, a higher concentration of free charge carriers can be achieved in it. And in vacuum devices, it is necessary to compensate for the repulsive force of the same charges due to external influences.

Task 4.

Explain graphically and using formulas how the law

of total current is fulfilled when grouped charges pass through the

drift tube.

Solution:

Ampere 's Law: