Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/03/22 19:44] – mexleadmin | electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2024/10/31 08:26] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 3. Linear Sources and two-terminal Networks ====== | + | ====== 3 Linear Sources and two-terminal Networks ====== |
| It is known from everyday life that battery voltages drop under heavy loads. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so great that the low beam or radio briefly cuts out.\\ | It is known from everyday life that battery voltages drop under heavy loads. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so great that the low beam or radio briefly cuts out.\\ | ||
| 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 | + | So it makes sense here to develop the ideal voltage source |
| < | < | ||
| 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** | + | * It is called **open circuit** |
| - | * The term **short circuit** | + | * The term **short circuit** |
| - | * In the region between no-load and short-circuit, | + | * 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 | + | The source voltage $U_0$ of the ideal voltage source |
| \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 125: | Zeile 125: | ||
| - __From linear voltage source to linear current source__: < | - __From linear voltage source to linear current source__: < | ||
| - | Given: Source voltage $U_0$, resp. open circuit voltage | + | Given: Source voltage $U_0$, resp. open circuit voltage |
| - | in question: source current $I_0$, resp. short circuit current $I_{SC}$, internal conductance $G_i$ \\ | + | in question: source current $I_0$, resp. short circuit current $I_{\rm SC}$, internal conductance $G_\rm i$ \\ |
| - | $\boxed{I_{SC} = {{U_{OC}}\over{R_i}}}$ , $\boxed{G_i = {{1}\over{R_i}}}$ | + | $\boxed{I_{\rm SC} = {{U_{\rm OC}}\over{R_\rm i}}}$ , $\boxed{G_\rm i = {{1}\over{R_\rm i}}}$ |
| </ | </ | ||
| - __From linear current source to linear voltage source__: < | - __From linear current source to linear voltage source__: < | ||
| - | Given: Source current $I_0$, resp. short-circuit current $I_{SC}$, internal resistance $G_i$ \\ | + | Given: Source current $I_0$, resp. short-circuit current $I_{\rm SC}$, internal resistance $G_\rm i$ \\ |
| - | in question: source voltage $U_0$, resp. open-circuit voltage $U_{OC}$, internal resistance $R_i$ \\ | + | in question: source voltage $U_0$, resp. open-circuit voltage $U_{\rm OC}$, internal resistance $R_\rm i$ \\ |
| - | $\boxed{U_{OC} = {{I_{SC}}\over{G_i}}}$ , $\boxed{R_i = {{1}\over{G_i}}}$ | + | $\boxed{U_{\rm OC} = {{I_{\rm SC}}\over{G_\rm i}}}$ , $\boxed{R_\rm i = {{1}\over{G_\rm i}}}$ |
| </ | </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 196: | Zeile 196: | ||
| < | < | ||
| - | In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[:elektrotechnik_labor: | + | In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[elektrotechnik_labor: |
| \\ | \\ | ||
| In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice? | In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice? | ||
| Zeile 224: | Zeile 224: | ||
| - In the second step, the linear voltage source $U_4$ formed in 1. with $R_4$ can be connected to the resistor $R_3$. From this again a linear current source can be created. This now has a resistance of $R_5 = R_3+R_4$ and an ideal current source with $I_5 = {{U_4}\over{R_3+R_4}}= {{I_4}\cdot{R_4}\over{R_3+R_4}} $. | - In the second step, the linear voltage source $U_4$ formed in 1. with $R_4$ can be connected to the resistor $R_3$. From this again a linear current source can be created. This now has a resistance of $R_5 = R_3+R_4$ and an ideal current source with $I_5 = {{U_4}\over{R_3+R_4}}= {{I_4}\cdot{R_4}\over{R_3+R_4}} $. | ||
| - The circuit diagram that now emerges is a parallel circuit of ideal current sources and resistors. This can be used to determine the values of the ideal equivalent current source and the equivalent resistance: | - The circuit diagram that now emerges is a parallel circuit of ideal current sources and resistors. This can be used to determine the values of the ideal equivalent current source and the equivalent resistance: | ||
| - | - ideal equivalent current source $I_{\rm eq}$: \begin{align*} I_ = I_1 + I_3 + I_5 = I_1 + I_3 + I_4\cdot{{R_4}\over{R_3+R_4}} \end{align*} | + | - ideal equivalent current source $I_{\rm eq}$: \begin{align*} I_{\rm eq} = I_1 + I_3 + I_5 = I_1 + I_3 + I_4\cdot{{R_4}\over{R_3+R_4}} \end{align*} |
| - Substitute conductance $G_{\rm eq}$: \begin{align*} G_{\rm eq} = \Sigma G_i = {{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_5}}={{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_3+R_4}} \end{align*} | - Substitute conductance $G_{\rm eq}$: \begin{align*} G_{\rm eq} = \Sigma G_i = {{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_5}}={{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_3+R_4}} \end{align*} | ||
| Zeile 313: | Zeile 313: | ||
| First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter. | First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter. | ||
| - | In <imgref imageNo12 > the wattmeter with the circuit symbol can be seen as a round network with crossed measuring inputs. The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_S$ and the input power of the load $P_L$. | + | In <imgref imageNo12 > the wattmeter with the circuit symbol can be seen as a round network with crossed measuring inputs. |
| + | The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_\rm S$ and the input power of the load $P_\rm L$. | ||
| < | < | ||
| Zeile 323: | Zeile 324: | ||
| The simulation in <imgref imageNo13 > shows the following: | The simulation in <imgref imageNo13 > shows the following: | ||
| - | * The circuit with linear voltage source ($U_0$ and $R_i$), and a resistive load $R_L$. | + | * The circuit with linear voltage source ($U_0$ and $R_\rm i$), and a resistive load $R_\rm L$. |
| - | * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_S$ (English: shunt) and a voltage measurement for $U_S$. The power is then: $P_L = {{1}\over{R_S}}\cdot | + | * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_\rm S$ (English: shunt) and a voltage measurement for $U_\rm S$. The power is then: $P_\rm L = {{1}\over{R_\rm S}}\cdot |
| * in the oscilloscope section (below). | * in the oscilloscope section (below). | ||
| - | * On the left is the power $P_L$ plotted against time in a graph. | + | * On the left is the power $P_\rm L$ plotted against time in a graph. |
| * On the right is the already-known current-voltage diagram of the current values. | * On the right is the already-known current-voltage diagram of the current values. | ||
| - | * The slider load resistance $R_L$, with which the value of the load resistance $R_L$ can be changed. | + | * The slider load resistance $R_\rm L$, with which the value of the load resistance $R_\rm L$ can be changed. |
| - | Now try to vary the value of the load resistance $R_L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set? | + | Now try to vary the value of the load resistance $R_\rm L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set? |
| < | < | ||
| Zeile 339: | Zeile 340: | ||
| * Diagram top: current-voltage diagram of a linear voltage source. | * Diagram top: current-voltage diagram of a linear voltage source. | ||
| - | * Diagram in the middle: source power $P_S$ and consumer power $P_L$ versus delivered voltage $U_L$. | + | * Diagram in the middle: source power $P_\rm S$ and consumer power $P_\rm L$ versus delivered voltage $U_\rm L$. |
| - | * Diagram below: Reference quantities over delivered voltage $U_L$. | + | * Diagram below: Reference quantities over delivered voltage $U_\rm L$. |
| The two powers are defined as follows: | The two powers are defined as follows: | ||
| - | * source power: $\, \, \large{ | + | * source power: $\, \, \large{ |
| - | * consumer power: $\large{ | + | * consumer power: $\large{ |
| - | - Both power $P_S$ and $P_L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is when the load resistance $R_L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire. | + | - Both power $P_\rm S$ and $P_\rm L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is when the load resistance $R_\rm L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire. |
| - | - If the load resistance becomes just as large as the internal resistance $R_L=R_i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_L = {{1}\over{2}}\cdot U_{OC}$. On the other hand, the current is also half the short-circuit current $I_L=I_{SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case. | + | - If the load resistance becomes just as large as the internal resistance $R_\rm L=R_\rm i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_\rm L = {{1}\over{2}}\cdot U_{\rm OC}$. On the other hand, the current is also half the short-circuit current $I_\rm L=I_{\rm SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case. |
| - | - If the load resistance becomes high impedance $R_L\rightarrow\infty$, | + | - If the load resistance becomes high impedance $R_{\rm L}\rightarrow\infty$, |
| < | < | ||
| - | The whole context can be investigated in this [[https:// | + | The whole context can be investigated in this [[https:// |
| ==== The Characteristics: | ==== The Characteristics: | ||
| - | 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** | + | The **efficiency** |
| + | \begin{align*} | ||
| + | \eta = {{P_{\rm out}}\over{P_{\rm in}}} | ||
| + | = {{R_{\rm L}\cdot I_{\rm L}^2}\over{(R_{\rm L}+R_{\rm i}) \cdot I_{\rm L}^2}} \quad \rightarrow \quad \boxed{ \eta | ||
| + | = {{R_{\rm L}} \over {R_{\rm L}+R_{\rm i}}} } | ||
| + | \end{align*} | ||
| - | The **utilization rate** | + | Once we want to get the **relative maximum power** out of a system (so maximum |
| - | In __power engineering__ | + | Application: |
| + | - In __power engineering__ | ||
| - | In __communications engineering__, one situation | + | The **utilization rate** |
| + | Here, the currently supplied power is not assumed (as in the case of efficiency), | ||
| + | \begin{align*} | ||
| + | \varepsilon = {{P_{\rm out}}\over{P_{\rm in, max}}} | ||
| + | = {{R_{\rm L} \cdot I_{\rm L}^2}\over{{U_0^2}\over{R_i}}} | ||
| + | = {{R_{\rm L}\cdot R_{\rm i} \cdot I_{\rm L}^2}\over {U_0^2}} | ||
| + | = {{R_{\rm L}\cdot R_{\rm i} \cdot \left({{U_0}\over | ||
| + | = {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} | ||
| + | = {{R_{\rm L}} \over {(R_{\rm L}+R_{\rm i})}}\cdot {{R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})}}} | ||
| + | \end{align*} | ||
| + | |||
| + | In other applications, | ||
| + | |||
| + | Application: | ||
| + | - In __communications engineering__ the impedance matching of the source (the antenna) and the load (the signal-acquiring microcontroller) uses resistors, capacitors, and inductors. | ||
| + | - Furthermore, | ||
| The impedance matching/ | The impedance matching/ | ||
| Zeile 371: | Zeile 393: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | <panel type=" | ||
| - | Simplify the following circuits by the Northon theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT). | + | ==== Exercises ==== |
| - | < | + | <panel type=" |
| + | |||
| + | Simplify the following circuits by the Norton theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT). | ||
| + | |||
| + | < | ||
| <button size=" | <button size=" | ||
| Zeile 381: | Zeile 406: | ||
| Shutting down all sources leads to | Shutting down all sources leads to | ||
| \begin{equation*} | \begin{equation*} | ||
| - | R_{i}= 8~\Omega \end{equation*} | + | R_{\rm i}= 8~\Omega \end{equation*} |
| Next, we figure out the current in the short circuit. | Next, we figure out the current in the short circuit. | ||
| Zeile 393: | Zeile 418: | ||
| To substitute the circuit in $b)$ first we determine the inner resistance. | To substitute the circuit in $b)$ first we determine the inner resistance. | ||
| Shutting down all sources leads to | Shutting down all sources leads to | ||
| - | \begin{equation*} R_{i}= 4 ~\Omega \end{equation*} | + | \begin{equation*} R_{\rm i}= 4 ~\Omega \end{equation*} |
| Next, we figure out the voltage at the open circuit. | Next, we figure out the voltage at the open circuit. | ||
| Thus we know the given current flows through the ideal current source as well as the resistor. | Thus we know the given current flows through the ideal current source as well as the resistor. | ||
| The voltage drop on the resistor is | The voltage drop on the resistor is | ||
| - | \begin{equation*} R_{i}= -4~\Omega \cdot 2~A \end{equation*} | + | \begin{equation*} R_{\rm i}= -4~\Omega \cdot 2~A \end{equation*} |
| The voltage at the open circuit is | The voltage at the open circuit is | ||
| - | \begin{equation*} | + | \begin{equation*} |
| </ | </ | ||
| Zeile 412: | Zeile 437: | ||
| </ | </ | ||
| + | <wrap anchor # | ||
| <panel type=" | <panel type=" | ||
| - | For the company „HHN Mechatronics & Robotics“ you shall analyze a competitor product: a simple drilling machine. This contains a battery pack, some electronics, | + | For the company „HHN Mechatronics & Robotics“ you shall analyze a competitor product: a simple drilling machine. This contains a battery pack, some electronics, |
| - | The drill has two speed modes: | + | The drill has two speed-modes: |
| - max power: here, the motor is directly connected to the battery. | - max power: here, the motor is directly connected to the battery. | ||
| - | - reduced power: in this case, a shunt resistor $R_s = 1 ~\Omega$ is connected in series to the motor. | + | - reduced power: in this case, a shunt resistor $R_{\rm s} = 1 ~\Omega$ is connected in series to the motor. |
| {{drawio> | {{drawio> | ||
| Zeile 427: | Zeile 453: | ||
| - What are the efficiencies for both modes? | - What are the efficiencies for both modes? | ||
| - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly half of the maximum power? | - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly half of the maximum power? | ||
| - | - Your company uses the reduced power mode instead of the shunt resistor $R_s$ multiple diodes in series $D$, which generates a constant voltage drop of $U_D = 2.8 ~V$. \\ What are the input and output power, such as the efficiency in this case? | + | - Your company uses the reduced power mode instead of the shunt resistor $R_{\rm s}$ multiple diodes in series $D$, which generates a constant voltage drop of $U_D = 2.8 ~V$. \\ What are the input and output power, such as the efficiency in this case? |
| You can check your results for the currents, voltages, and powers with the following simulation: | You can check your results for the currents, voltages, and powers with the following simulation: | ||
| Zeile 434: | Zeile 460: | ||
| </ | </ | ||
| + | |||
| + | # | ||
| + | |||
| + | Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$). | ||
| + | |||
| + | 1. What are the possible ways to connect these components? | ||
| + | |||
| + | # | ||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| + | 2. Which circuit can provide the maximum power $P_{\rm L ~max}$ at the loads? | ||
| + | |||
| + | # | ||
| + | |||
| + | At the maximum utilization rate $\varepsilon = 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved. \\ | ||
| + | The utilization rate is given as: | ||
| + | \begin{align*} | ||
| + | \varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}} | ||
| + | &= {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} \\ | ||
| + | \end{align*} | ||
| + | |||
| + | 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. | ||
| + | # | ||
| + | |||
| + | # | ||
| + | The following configuration has the maximum output power. | ||
| + | |||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| + | |||
| + | 3. What is the value of the maximum power $P_{\rm L ~max}$? | ||
| + | |||
| + | # | ||
| + | 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 | ||
| + | &= 24.6~\% | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore, the maximum power is: | ||
| + | \begin{align*} | ||
| + | \varepsilon | ||
| + | \rightarrow P_{\rm out} &= \varepsilon | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | P_{\rm out} = 26.8 W | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 4. Which circuit has the highest efficiency? | ||
| + | |||
| + | # | ||
| + | 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. | ||
| + | # | ||
| + | |||
| + | # | ||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| + | 5. What is the value of the highest efficiency? | ||
| + | |||
| + | # | ||
| + | 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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | \eta = 95.2~\% | ||
| + | \end{align*} | ||
| + | # | ||
| + | \\ \\ | ||
| + | # | ||
| + | {{drawio> | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | # | ||
| - | <panel type=" | + | <panel type=" |
| Further German exercises can be found in ILIAS (see [[https:// | Further German exercises can be found in ILIAS (see [[https:// | ||