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electrical_engineering_2:the_stationary_electric_flow [2023/03/26 15:35]
ott 		electrical_engineering_2:the_stationary_electric_flow [2023/09/19 23:51]
mexleadmin
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-====== 2. The stationary Current Density and Flux ======+====== 2 The stationary Current Density and Flux ======
  
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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 $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~{\rm {1}\over{mm^3}}$.
  
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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:
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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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 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 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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