- •Task №1
- •1. Present your arguments (logical and numerical) on the assessment
- •3. Comparison of the quantum energy with the binding energy of clusters in water. Calculation of the water heating temperature necessary for the destruction of its cluster structure:
- •4. Comparison of quantum energy with the energy of chemical bonding of atoms in a water molecule:
- •5. Comparison of the quantum energy from a microwave oven with the thermal energy characteristic of protein denaturation:
- •2. Calculation of the number of quanta for heating:
- •Task №3
- •3. Compare numerically 2 typical devices: vacuum and semiconductor according to the following parameters:
- •1. Numerical comparison of the maximum velocities of charged particles:
- •2. Numerical comparison of the length of the interaction region for the span angle of π - radians:
- •3. Numerical comparison of the bulk charge density:
- •4. Calculation of the microperviance and «plasma» frequency for a vacuum device:
- •5. Calculation of the Debye length and plasma frequency for a semiconductor device:
- •2 Балл task №4
- •4. Is it possible to provide high-speed modulation and grouping of charged particles in semiconductor devices using the initial part of the field-velocity characteristic?
- •2 Балл task №5
- •Task №6
- •6. Determine the noise factor of the amplifier in dB if its effective noise temperature is 115 k.
- •2. Calculation of the effective noise temperature of two such devices connected in a cascade:
- •3. Analyze the result:
- •0.75 Балл
- •1. Балл
3. Numerical comparison of the bulk charge density:
3.1 Vacuum device:
From the formula for the current strength [14]:
IV
= nv
vv
q
S
= nv
vv
q
= nv
vv
q
(11)
where nv - bulk charge density; S - the cross-sectional area of the electron beam.
We obtain the expression of the bulk charge density:
=
=
=
=
1,139
1014
m-3
3.2 Semiconductor device:
In a semiconductor device, the bulk charge density is determined by the doping level, which means [15]:
= n =
6
1016
cm-3
= 6
1022
m-3
Thus, the bulk charge density in a semiconductor device in 6 1022/ 1,139 1014 = 5,27 108 times more than in a vacuum device.
4. Calculation of the microperviance and «plasma» frequency for a vacuum device:
Calculation of the microperveance by the formula [16]:
P
=
(12)
Then we get:
P
=
=
=
0,32
Calculation of the «plasma» frequency by the formula [17]:
= 2
(13)
In the end, we get:
= 2
=
2
=
3,781
109
rad/s
5. Calculation of the Debye length and plasma frequency for a semiconductor device:
The formula for the Debye length [17]:
LS
=
(14)
Calculation of the Debye length:
LS
=
=
=
8,544
10-10
m
Calculation of the plasma frequency by the formula (13):
= 2
=
2
=
8,677
1013
rad/s
We conclude that the plasma frequency of the semiconductor device is 8,677 1013/3,781 109 = 22949 times greater than the «plasma frequency» of the vacuum device.
The difference in plasma frequencies of vacuum and semiconductor devices is that in a vacuum device there is an interaction between electrons, that is, between particles of the same charge, and in a semiconductor device there is an interaction between charges of different names (electrons and holes). In the case of a vacuum device, it is not entirely correct to talk about a «plasma frequency», and in the case of a semiconductor device talk about a solid-state plasma [18].
Answer:
Table 1
Parameters |
Maximum velocities, m/s |
Length of the interaction region, m |
Bulk charge density, m-3 |
Plasma frequency, rad/s |
Microperviance,
|
Debye length, m |
Vacuum device |
4,276 107 |
0,00164 |
1,139 1014 |
3,781 109 |
0,32 |
|
Semiconductor device |
105 |
3,846 10-6 |
6 1022 |
8,677 1013 |
|
8,544 10-10 |
2 Балл task №4
4. Is it possible to provide high-speed modulation and grouping of charged particles in semiconductor devices using the initial part of the field-velocity characteristic?
Estimate at what speed the electron will fly through the interaction space with a length of 0.1 microns at an applied voltage pulse of 0,7 [V]? The material is gallium arsenide. Pulse duration 7 10-10 s и 7 10-14 s? When answering, use the concepts of relaxation times in terms of momentum and energy.
Solving:
First of all, it is necessary to determine such concepts as relaxation time by momentum and relaxation time by energy.
The pulse relaxation time is the average free path time of an electron or hole between two consecutive collisions with crystal lattice defects. Thermal vibrations of the crystal lattice (acoustic and optical phonons) can act as defects of the crystal lattice [19].
The energy relaxation time is the time during which the electron gives up all the excess energy and comes into thermal equilibrium with the lattice [20].
At the same time, comparing these characteristics (relaxation time of energy and momentum), it should be noted that these parameters have different values. In simple words, we can explain it this way: the relaxation time of energy is much longer than the relaxation time of the pulse, since a sufficiently large number of collisions is necessary for a charged particle to expend energy. At the same time, considering the relaxation time of the pulse, we can say that even with one collision, the velocity of the particle radically changes. That is, first the speed is lost, and only then the energy.
Since we have begun to consider the relaxation time of energy and momentum, it is worth talking about the free path length in the case of momentum and the length of heating (cooling) in the case of energy. If we write it down formulically, then [21]:
(15)
where
– free path length;
– heating (cooling) length;
– pulse relaxation time;
– energy relaxation time; vT
– RMS thermal velocity.
At the same time, it is worth clarifying that we are talking about the length of heating if we apply an external field, if on the contrary, respectively, we are talking about the length of cooling.
Hence, given that = 10-(12…15) s and = 10-(11…13) [21], and also vT = 105 m/s [12], then we get the free run length and the heating (cooling) length:
= 10-12 105 = 10-8 = 0,1 10-6 m = 0,1 m
= 10-11 105 = 10-8 = 1 10-6 m = 1 m
Thus, we conclude that the modulation of the electron velocity is possible only within the time (length) of the relaxation of the pulse, or, in simple words, the modulation of the velocity is possible at an interval of about 0,1 microns, otherwise the charged particles lose speed further and cannot react to external influences in any way.
Calculation of the electron flight velocity:
First of all, let's pay attention to the pulse duration. In the first case, it is equal to 7 10-10 s. This time value is greater than the energy relaxation time, equal to 10-11 s. That is, from here we can conclude that the pulse will not change the velocity of the electrons in any way, since the electrons will have time to lose energy during this time, in which case the velocity of the electrons will be equal to 105 m/s.
A completely different way can explain the change in particle velocity at a pulse duration of 7 10-14 s. This time is much less than the energy relaxation time, that is, by this time the electrons will not have time to lose energy and their speed will depend on the applied field. This dependency will look like this [21]:
v =
(16)
where - mobility of charges; E - field strength.
Transform the expression for the mobility of charges:
=
(17)
Then, based on formulas (16) and (17) we get:
v =
=
For a pulse duration of 7 10-14 s, we obtain the following electron velocity
v =
=
=
86154 m/s =
m/s
Answer:
The velocity modulation of charged particles in semiconductor devices can be provided at a distance of the pulse relaxation length, that is, of the order of 0,1 m.
Velocity of the electron's passage through the interaction space during the pulse time of 7 10-10 s:
v = 105 m/s
Velocity of the electron's passage through the interaction space during the pulse time of 7 10-14 s:
v = 8,6 104 m/s
