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electrical_engineering_2:the_stationary_electric_flow [2023/03/16 14:55] – mexleadminelectrical_engineering_2:the_stationary_electric_flow [2024/03/29 19:57] (aktuell) – mexleadmin
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-====== 2. The stationary Current Density and Flux ======+====== 2 The stationary Current Density and Flux ======
  
 <callout>  <callout> 
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 In the discussion of the electrostatic field in principle, no charges in motion were considered. Now the motion of charges shall be considered explicitly. In the discussion of the electrostatic field in principle, no charges in motion were considered. Now the motion of charges shall be considered explicitly.
  
-The current density here describes how charge carriers move together (collectively). The stationary current density describes the charge carrier movement if a **direct voltage**  is the cause of the movement. A constant direct current then flows in the stationary electric flow field. Thus, there is no time dependency of the current:+The current density here describes how charge carriers move together (collectively). The stationary current density describes the charge carrier movement if a **direct voltage**  is the cause of the movement. A constant direct current then flows in the stationary electric flow field. Thus, there is no time dependency on the current:
  
 $\large{{{\rm d}I}\over{{\rm d}t}}=0$ $\large{{{\rm d}I}\over{{\rm d}t}}=0$
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 <WRAP> <imgcaption ImgNr01 | part of a Conductor> </imgcaption> {{drawio>charges_in_conductor.svg}} </WRAP> <WRAP> <imgcaption ImgNr01 | part of a Conductor> </imgcaption> {{drawio>charges_in_conductor.svg}} </WRAP>
  
-For this purpose, a packet of charges ${\rm d}Q$ is considered, which will pass the area $A$ during the period ${\rm d}t$ (see <imgref imageNo02>). These charges are located in a partial volume element ${\rm d}V$, which is given by the area $A$ to be traversed and a partial section ${\rm d}x$: ${\rm d}V = A \cdot {\rm d}x$. The amount of charges per volume is given by the charge carrier density, specifically for metals by the electron density $n_\rm{e}$. The electron density $n_\rm{e}$ gives the number of free electrons per unit volume a. For copper, for example, this is approximately $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~\rm{{1}\over{mm^3}}$.+For this purpose, a packet of charges ${\rm d}Q$ is considered, which will pass the area $A$ during the period ${\rm d}t$ (see <imgref imageNo02>). These charges are located in a partial volume element ${\rm d}V$, which is given by the area $A$ to be traversed and a partial section ${\rm d}x$: ${\rm d}V = A \cdot {\rm d}x$. The amount of charges per volume is given by the charge carrier density, specifically for metals by the electron density $n_{\rm e}$. The electron density $n_{\rm e}$ gives the number of free electrons per unit volume a. For copper, for example, this is approximately $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~{\rm {1}\over{mm^3}}$.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 The flowing charges contained in the partial volume element $dV$ are then (with elementary charge $e_0$): The flowing charges contained in the partial volume element $dV$ are then (with elementary charge $e_0$):
  
-\begin{align*} dQ = n_e \cdot e_0 \cdot A \cdot dx \end{align*}+\begin{align*} {\rm d} Q = n_{\rm e} \cdot e_0 \cdot A \cdot  {\rm d}x \end{align*}
  
 The current is then given by $I={{{\rm d}Q}\over{{\rm d}t}}$: The current is then given by $I={{{\rm d}Q}\over{{\rm d}t}}$:
  
-\begin{align*} I = n_e \cdot e_0 \cdot A \cdot {{{\rm d}x}\over{{\rm d}t}} = n_e \cdot e_0 \cdot A \cdot v_e \end{align*}+\begin{align*} I = n_{\rm e} \cdot e_0 \cdot A \cdot {{{\rm d}x}\over{{\rm d}t}} = n_e \cdot e_0 \cdot A \cdot v_{\rm e} \end{align*}
  
 This leads to an electron velocity $v_e$ of: This leads to an electron velocity $v_e$ of:
  
-\begin{align*} v_\rm{e} = {{{\rm d}x}\over{{\rm d}t}} = {{I}\over{n_e \cdot e_0 \cdot A }} \end{align*}+\begin{align*} v_{\rm e} = {{{\rm d}x}\over{{\rm d}t}} = {{I}\over{n_{\rm e} \cdot e_0 \cdot A }} \end{align*}
  
 In contrast to the considerations in electrostatics, the charges now travel with finite velocities.  In contrast to the considerations in electrostatics, the charges now travel with finite velocities. 
-With regard to the electron velocity $v_\rm{e} \sim {{I}\over{A}}$ it is obvious to determine a current density $S$ (related to the area):+With regard to the electron velocity $v_{\rm e} \sim {{I}\over{A}}$ it is obvious to determine a current density $S$ (related to the area):
  
 \begin{align*} \begin{align*}
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 <WRAP> <imgcaption imageNo03 | Field Lines and equipotential Surfaces of the electric Current Field> </imgcaption> <WRAP> {{drawio>fieldlineelectricflowfield.svg}} \\ </WRAP> <WRAP> <imgcaption imageNo03 | Field Lines and equipotential Surfaces of the electric Current Field> </imgcaption> <WRAP> {{drawio>fieldlineelectricflowfield.svg}} \\ </WRAP>
  
-The current density was determined only for a constant cross-sectional area $A$, through which a homogeneous current - thus also a homogeneous current field - passes perpendicularly. Now however, a general approach for the electric current strength has to be found.+The current density was determined only for a constant cross-sectional area $A$, through which a homogeneous current - thus also a homogeneous current field - passes perpendicularly. Now, however, a general approach for the electric current strength has to be found.
  
 For this purpose, instead of a constant current density $S$ over a vertical, straight cross-sectional area $A$, a varying current density $S(A)$ over many small partial areas ${\rm d}A$ is considered. Thus, if the subareas are sufficiently small, a constant current density over the subarea can again be obtained. It then becomes For this purpose, instead of a constant current density $S$ over a vertical, straight cross-sectional area $A$, a varying current density $S(A)$ over many small partial areas ${\rm d}A$ is considered. Thus, if the subareas are sufficiently small, a constant current density over the subarea can again be obtained. It then becomes
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 In concrete terms, this means for the bottleneck: **The resistance increases at the bottleneck. Thus the voltage drop also increases there. Thus there are also more equipotential surfaces there. ** In concrete terms, this means for the bottleneck: **The resistance increases at the bottleneck. Thus the voltage drop also increases there. Thus there are also more equipotential surfaces there. **
  
-This solves the question of why there is a compression of the equipotential surfaces at the bottleneck (cmp. <imgref imageNo02 >). Interestingly, the thought model can now also for a __homogeneous body__  explain the general material law. To do this, one inserts equation $(2.1.2)$ and $(2.1.3)$ into $(2.1.1)$. Then it follows:+This solves the question of why there is a compression of the equipotential surfaces at the bottleneck (see <imgref imageNo02 >). Interestingly, the thought model can now also for a __homogeneous body__  explain the general material law. To do this, one inserts equation $(2.1.2)$ and $(2.1.3)$ into $(2.1.1)$. Then it follows:
  
 \begin{align*}  \begin{align*} 
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 <panel type="info" title="Task 2.1.2 Electron velocity in copper"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 2.1.2 Electron velocity in copper"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A current of $I = 2\{\rm A}$ flows in a conductor made of copper with cross-sectional area $A$. \\ The electron density in copper is $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~\rm{{1}\over{mm^3}}$ and the magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~\rm{As}$+A current of $I = 2~{\rm A}$ flows in a conductor made of copper with cross-sectional area $A$. \\ The electron density in copper is $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~{\rm {1}\over{mm^3}}$ and the magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~{\rm As}$
  
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.5~\rm{mm}^2$? +  - What is the average electron velocity $v_{\rm e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.5~{\rm mm}^2$? 
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.0~\rm{mm}^2$?+  - What is the average electron velocity $v_{\rm e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.0~{\rm mm}^2$?
  
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 <panel type="info" title="Task 2.1.3 Electron velocity in Semiconductors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 2.1.3 Electron velocity in Semiconductors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Similar to task 2.1.2 a current of $I = 2~\rm{A}$ shall flow through a cross-sectional area $A = 1.5~\rm{mm}^2$ of semiconductors. \\ +Similar to task 2.1.2 a current of $I = 2~{\rm A}$ shall flow through a cross-sectional area $A = 1.5~{\rm mm}^2$ of semiconductors. \\ 
-(magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~\rm{As}$)+(magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~{\rm As}$)
  
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons for undoped silicon (electron density $n_{\rm e}({\rm Si})=9.65 \cdot 10^{9} ~\rm{{1}\over{cm^3}}$ at room temperature)? +  - What is the average electron velocity $v_{\rm e,1}$ of electrons for undoped silicon (electron density $n_{\rm e}({\rm Si})=9.65 \cdot 10^{9} ~{\rm {1}\over{cm^3}}$ at room temperature)? 
-  - The electron density of doped silicon $n_\rm{e}(Si\;doped)$ is (mainly) given by the number of dopant atoms per volume. The doping shall provide 1 donator atom per $10^7$ silicon atoms (here: one donator atom shall release one electron). The molar volume of silicon is $12 \cdot 10^{-6}~\rm{{m^3}\over{mol}}$ with $6.022 \cdot 10^{23}$ silicon atoms per $\rm{mol}$. What is the average electron velocity $v_\rm{e,1}$ of electrons for doped silicon?+  - The electron density of doped silicon $n_{\rm e}(Si\; doped)$ is (mainly) given by the number of dopant atoms per volume. The doping shall provide 1 donator atom per $10^7$ silicon atoms (here: one donator atom shall release one electron). The molar volume of silicon is $12 \cdot 10^{-6}~{\rm {m^3}\over{mol}}$ with $6.022 \cdot 10^{23}$ silicon atoms per ${\rm mol}$. What is the average electron velocity $v_{\rm e,1}$ of electrons for doped silicon?
  
  
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 <callout> <callout>
  
-==== Targets ====+==== Goals ====
  
 After this lesson, you should: After this lesson, you should:
  
   - know which quantities are comparable for the electrostatic field and the flow field.   - know which quantities are comparable for the electrostatic field and the flow field.
-  - be able to explain the displacement current on the basis of enveloping surfaces.+  - be able to explain the displacement current based on enveloping surfaces.
   - understand how current can flow "through" a capacitor.   - understand how current can flow "through" a capacitor.
  
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 In transformer stations sometimes water resistors are used as {{wp>Liquid rheostat}}. In this resistor, the water works as a (poor) conductor which can handle a high power loss.  In transformer stations sometimes water resistors are used as {{wp>Liquid rheostat}}. In this resistor, the water works as a (poor) conductor which can handle a high power loss. 
  
-The water resistor consists of a water basin. In the given basin two quadratic plates with the edge length of $l = 80 ~\rm{cm}$ are inserted with the distance $d$ between them.  +The water resistor consists of a water basin. In the given basin two quadratic plates with the edge length of $l = 80 ~{\rm cm}$ are inserted with the distance $d$ between them.  
-The resistivity of the water is $\rho = 0.25 ~\Omega \rm{m}$. The resistor shall dissipate the energy of $P = 4 ~\rm{kW}$ and shall exhibit a homogeneous current field. +The resistivity of the water is $\rho = 0.25 ~\Omega {\rm m}$. The resistor shall dissipate the energy of $P = 4 ~{\rm kW}$ and shall exhibit a homogeneous current field. 
  
-  - Calculate the required distance of the plates in order to get a current density of $S = 25 ~\rm{mA/cm^2}$+  - Calculate the required distance of the plates to get a current density of $S = 25 ~{\rm mA/cm^2}$
   - What are the values of the current $I$ and the voltage $U$ at the resistor, such as the internal electric field strength $E$ in the setup?   - What are the values of the current $I$ and the voltage $U$ at the resistor, such as the internal electric field strength $E$ in the setup?
  
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