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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/03 23:16] – mexleadminelectrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2024/10/31 08:26] (aktuell) – mexleadmin
Zeile 4: Zeile 4:
 Another example is a $1.5~\rm V$ battery: If such a battery is short-circuited by a piece of wire, not so much current flows that the piece of wire glows, but noticeably less. Another example is a $1.5~\rm V$ battery: If such a battery is short-circuited by a piece of wire, not so much current flows that the piece of wire glows, but noticeably less.
  
-So it makes sense here to develop the concept of the ideal voltage source further. In addition, we will see that this also opens up the possibility of converting and simplifying more complicated circuits.+So it makes sense here to develop the ideal voltage source concept further. In addition, we will see that this also opens up the possibility of converting and simplifying more complicated circuits.
  
 <WRAP> <imgcaption imageNo1 | passive two-terminal network> </imgcaption> {{drawio>PassiverZweipol.svg}} </WRAP> <WRAP> <imgcaption imageNo1 | passive two-terminal network> </imgcaption> {{drawio>PassiverZweipol.svg}} </WRAP>
Zeile 51: Zeile 51:
 This realization shall now be described with some technical terms: This realization shall now be described with some technical terms:
  
-  * It is called **open circuit**  when no current is drawn from an active two-terminal network: $I_{\rm LL}=0$. \\ The voltage corresponds to the **open circuit voltage**  $U=U_{\rm OC}$ (German: //Leerlaufspannung//). \\ The open circuit power is $P_{\rm OC}=U_{\rm OC} \cdot I_{\rm OC} = 0$. +  * It is called **open circuit**  when no current is drawn from an active two-terminal network: $I_{\rm SC}=0$. \\ The voltage corresponds to the **open circuit voltage**  $U=U_{\rm OC}$ (German: //Leerlaufspannung// $U_{\rm LL}$). \\ The open circuit power is $P_{\rm OC}=U_{\rm OC} \cdot I_{\rm OC} = 0$. 
-  * The term **short circuit**  is used when the terminals of the two-terminal network are bridged without resistance. The current then flowing is called the **short-circuit current**  $I=I_{\rm SC}$ (German: //Kurzschlussstrom//). \\ The short-circuit voltage is $U_{\rm SC}=0~\rm V$. \\ Also, the short-circuit power is $P_{\rm SC}=U_{\rm SC} \cdot I_{\rm SC} = 0$. +  * The term **short circuit**  is used when the terminals of the two-terminal network are bridged without resistance. The current then flowing is called the **short-circuit current**  $I=I_{\rm SC}$ (German: //Kurzschlussstrom// $I_{\rm KS}$). \\ The short-circuit voltage is $U_{\rm SC}=0~\rm V$. \\ Also, the short-circuit power is $P_{\rm SC}=U_{\rm SC} \cdot I_{\rm SC} = 0$. 
-  * In the region between no-load and short-circuit, the active two-terminal network outputs power to a connected load.+  * The active two-terminal network outputs power to a connected load in the region between no-load and short-circuit.
  
-Important: As will be seen in the following, the short-circuit current can cause considerable power loss inside the two-terminal network and thus a lot of waste heat. +Important: As seen in the following, the short-circuit current can cause considerable power loss inside the two-terminal network and thus a lot of waste heat. 
 Not every real two-terminal network is designed for this. Not every real two-terminal network is designed for this.
  
Zeile 87: Zeile 87:
 \begin{align*} U = U_0 - R_{\rm i} \cdot I \end{align*} \begin{align*} U = U_0 - R_{\rm i} \cdot I \end{align*}
  
-The source voltage $U_0$ of the ideal voltage source is to be measured at the terminals of the two-terminal network if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{\rm LL}$.+The source voltage $U_0$ of the ideal voltage source will be measured at the terminals of the two-terminal network if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{\rm OC}$.
  
 \begin{align*} U = U_{\rm OC} - R_{\rm i} \cdot I \end{align*} \begin{align*} U = U_{\rm OC} - R_{\rm i} \cdot I \end{align*}
Zeile 358: Zeile 358:
 ==== The Characteristics: Efficiency and Utilization Rate ==== ==== The Characteristics: Efficiency and Utilization Rate ====
  
-In order to understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again:+To understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again:
  
-The **efficiency**  $\eta$ describes the delivered power (consumer power) in relation to the supplied power (power of the ideal source): +The **efficiency**  $\eta$ describes the delivered power (consumer power) concerning the supplied power (power of the ideal source): 
 \begin{align*} \begin{align*}
 \eta = {{P_{\rm out}}\over{P_{\rm in}}}  \eta = {{P_{\rm out}}\over{P_{\rm in}}} 
Zeile 367: Zeile 367:
 \end{align*} \end{align*}
  
-The **utilization rate**  $\varepsilon$ describes the delivered power in relation to the maximum possible power of the ideal source. +Once we want to get the **relative maximum power** out of a system (so maximum power related to the input power) the efficiency should go towards $\eta \rightarrow 100\%$. This situation close to (1.) in <imgref imageNo14>.  
 + 
 +Application: 
 +  - In __power engineering__ $\eta \rightarrow 100\%$ is often desired: We want the maximum power output with the lowest losses at the internal resistance of the source. Thus, the internal resistance of the source should be low compared to the load $R_{\rm L} \gg R_{\rm i} $.  
 + 
 +The **utilization rate**  $\varepsilon$ describes the delivered power $P_{\rm out}$ concerning the maximum possible power $P_{\rm in, max}$ of the ideal source. 
 Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case:  Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case: 
 \begin{align*} \begin{align*}
Zeile 378: Zeile 383:
 \end{align*} \end{align*}
  
-In __power engineering__  a situation close to (1.) in <imgref imageNo14 > is desired: maximum power output with the lowest losses at the internal resistance of the source.  +In other applications, the **absolute maximum power** has to be taken from the source, without consideration of the losses via the internal resistance. This corresponds to the situation (2.) in <imgref imageNo14>. For this purpose, the internal resistance of the source and the load are matched. This case is called **{{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}}  ** (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$.
-Thus, the internal resistance of the source should be low compared to the load $R_{\rm L} \gg R_{\rm i} $. The efficiency should go towards $\eta \rightarrow 100\%$.+
  
-In __communications engineering__, one situation is different and corresponds to the situation (2.): The maximum power is to be taken from the source, without consideration of the losses via the internal resistance. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. This case is called **{{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}}  ** (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$+Application: 
 +  - In __communications engineering__ the impedance matching of the source (the antenna) and the load (the signal-acquiring microcontroller) uses resistors, capacitors, and inductors.  There, we want to get the maximum power out of an antenna. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. An example can be seen in this {{electrical_engineering_1:anp084a_en_-_impedance_matching_for_near_field_com.pdf#page=4|application note for near field communication}}. 
 +  - Furthermore, also for __photovoltaic cells__ one wants to get the maximum power out. In this case, the concept is often called **{{https://en.wikipedia.org/wiki/Maximum_power_point_tracking|Maximum Power Point Tracking (MPPT)}}  **
  
 The impedance matching/power matching is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|here]] explained in a German video. The impedance matching/power matching is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|here]] explained in a German video.
Zeile 455: Zeile 461:
 </WRAP></WRAP></WRAP></panel> </WRAP></WRAP></WRAP></panel>
  
-#@TaskTitle_HTML@#3.3.3 Power of two pole components \\ <fs medium>(written test, approx. xxx % of a xxx-minute written test, WS2xxx)</fs>#@TaskText_HTML@#+#@TaskTitle_HTML@#3.3.3 Power of two pole components #@TaskText_HTML@#
  
 Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$). Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$).
Zeile 476: Zeile 482:
 \end{align*} \end{align*}
  
-<callout type="success"> 
 As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.\\ As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.\\
 Therefore, a series configuration of the batteries ($2 R_{\rm i} = 0.2~\Omega$) and a parallel configuration of the load (${{1}\over{2}} R_{\rm L}= 0.25~\Omega$) will have the highest output. Therefore, a series configuration of the batteries ($2 R_{\rm i} = 0.2~\Omega$) and a parallel configuration of the load (${{1}\over{2}} R_{\rm L}= 0.25~\Omega$) will have the highest output.
-</callout> 
- 
-A detailed analysis is shown here 
-{{drawio>electrical_engineering_1:diagram333_2.svg}} 
 #@HiddenEnd_HTML~Solution333_2,Solution ~@# #@HiddenEnd_HTML~Solution333_2,Solution ~@#
 +
 +#@HiddenBegin_HTML~Result333_2,Result~@#
 +The following configuration has the maximum output power.
 +
 +{{drawio>electrical_engineering_1:diagram333_3.svg}}
 +#@HiddenEnd_HTML~Result333_2,Result~@#
 +
  
 3. What is the value of the maximum power $P_{\rm L ~max}$? 3. What is the value of the maximum power $P_{\rm L ~max}$?
 +
 +#@HiddenBegin_HTML~Solution333_3,Solution~@#
 +The maximum utilization rate is:
 +\begin{align*}
 +\varepsilon &= {{{{1}\over{2}} R_{\rm L} \cdot 2 R_{\rm i} }   \over { ({{1}\over{2}} R_{\rm L} + 2 R_{\rm i} )^2}} \\
 +            &= { {0.25 ~\Omega           \cdot 0.2 ~\Omega }   \over { (           0.25 ~\Omega + 0.2 ~\Omega )^2}} \\
 +            &= 24.6~\%
 +\end{align*}
 +
 +Therefore, the maximum power is:
 +\begin{align*}
 +            \varepsilon  &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\
 +\rightarrow P_{\rm out}  &= \varepsilon   \cdot P_{\rm in, max} \\
 +                         &= \varepsilon   \cdot {{U_s^2}\over{R_{\rm i}}} \\
 +                         &= 24.6~\%       \cdot {{(3.3~\rm V)^2}\over{0.1~\Omega}} \\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution333_3,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_3,Result~@#
 +\begin{align*}
 +P_{\rm out}  = 26.8 W
 +\end{align*}
 +#@HiddenEnd_HTML~Result333_3,Result~@#
  
 4. Which circuit has the highest efficiency? 4. Which circuit has the highest efficiency?
 +
 +#@HiddenBegin_HTML~Solution333_4,Solution~@#
 +The highest efficiency $\eta$ is given when the output power compared to the input power is minimal. \\
 +A parallel configuration of the batteries (${{1}\over{2}} R_{\rm i} = 0.05~\Omega$) and a series configuration of the load ($2 R_{\rm L}= 1.0~\Omega$) will have the highest efficiency.
 +#@HiddenEnd_HTML~Solution333_4,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_4,Result~@#
 +{{drawio>electrical_engineering_1:diagram333_4.svg}}
 +#@HiddenEnd_HTML~Result333_4,Result~@#
  
 5. What is the value of the highest efficiency? 5. What is the value of the highest efficiency?
 +
 +#@HiddenBegin_HTML~Solution333_5,Solution~@#
 +The efficiency $\eta$ is given as:
 +\begin{align*}
 +\eta  &= { {2 R_{\rm L} }\over{ 2 R_{\rm L}+ {{1}\over{2}} R_{\rm i} }} \\
 +      &= { { 1.0~\Omega }\over{ 1.0~\Omega + 0.05~\Omega }}
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution333_5,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_5,Result~@#
 +\begin{align*}
 +\eta  = 95.2~\%
 +\end{align*}
 +#@HiddenEnd_HTML~Result333_5,Result~@#
 +\\ \\
 +#@HiddenBegin_HTML~Details333,Detailed Comparison~@#
 +{{drawio>electrical_engineering_1:diagram333_2.svg}}
 +
 +#@HiddenEnd_HTML~Details333,Detailed Comparison~@#
 +
  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#