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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/03/09 10:15] mexleadmin [Bearbeiten - Panel] |
electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/04 00:22] (aktuell) mexleadmin |
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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 load. 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 | + | Another example |
- | 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 a possibility | + | 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 |
< | < | ||
Zeile 22: | Zeile 22: | ||
By the end of this section, you will be able to: | By the end of this section, you will be able to: | ||
- describe the difference between an ideal and a linear voltage or current source. | - describe the difference between an ideal and a linear voltage or current source. | ||
- | - Know and apply the relationship between output voltage, source voltage $U_q$ and internal resistance $R_i$. | + | - Know and apply the relationship between output voltage, source voltage $U_q$, and internal resistance $R_i$. |
- | - know and apply the relationship between the current supplied, the source current $I_q$ and the internal conductance $G_i$. | + | - know and apply the relationship between the current supplied, the source current $I_q$, and the internal conductance $G_i$. |
- represent the voltage curve of the linear (voltage/ | - represent the voltage curve of the linear (voltage/ | ||
- determine the open-circuit voltage and the short-circuit current using two current/ | - determine the open-circuit voltage and the short-circuit current using two current/ | ||
Zeile 39: | Zeile 39: | ||
==== Practical Example of a realistic Source ==== | ==== Practical Example of a realistic Source ==== | ||
- | For the ideal voltage source it was defined that it always supplies the same voltage independent of the load. In <imgref imageNo2 >, in contrast, an example of a " | + | For the ideal voltage source, it was defined that it always supplies the same voltage independent of the load. In <imgref imageNo2 >, in contrast, an example of a " |
- | - This active two-terminal network generates a voltage of $1.5V$ and a current of $0A$ when the circuit is open. | + | - This active two-terminal network generates a voltage of $1.5~\rm V$ and a current of $0~\rm A$ when the circuit is open. |
- | - If a resistor is added, the voltage decreases and the current increases. For example, a voltage of $1.2V$ is applied to the resistor of $2 \Omega$ and a current of $0.6A$ flows. | + | - If a resistor is added, the voltage decreases, and the current increases. For example, a voltage of $1.2~\rm V$ is applied to the resistor of $2 ~\Omega$, and a current of $0.6~A$ flows. |
- | - The terminals of the active two-terminal network can be directly connected via the outer switch. Then a current of $3A$ flows at a voltage of $0V$. | + | - The terminals of the active two-terminal network can be directly connected via the outer switch. Then a current of $3~\rm A$ flows at a voltage of $0~\rm V$. |
< | < | ||
Zeile 51: | Zeile 51: | ||
This realization shall now be described with some technical terms: | This realization shall now be described with some technical terms: | ||
- | * One speaks of **open circuit** | + | * It is called |
- | * The term **short circuit** | + | * The term **short circuit** |
* In the region between no-load and short-circuit, | * 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. Not every real two-terminal network is designed for this. | + | 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. |
+ | Not every real two-terminal network is designed for this. | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ < | ~~PAGEBREAK~~ ~~CLEARFIX~~ < | ||
Zeile 61: | Zeile 62: | ||
What is interesting now is the current-voltage characteristic of the circuit in <imgref imageNo2> | What is interesting now is the current-voltage characteristic of the circuit in <imgref imageNo2> | ||
- | From a purely mathematical point of view, the course can be represented by the basic equation of linear graphs with the y-axis intercept $I_{SC}$ and a slope of $-{{I_{SC}}\over{U_{OC}}}$: | + | From a purely mathematical point of view, the course can be represented by the basic equation of linear graphs with the y-axis intercept $I_{\rm SC}$ and a slope of $-{{I_{\rm SC}}\over{U_{\rm OC}}}$: |
- | \begin{align*} I = I_{SC} - {{I_{SC}}\over{U_{OC}}}\cdot U \tag{3.1.1} \end{align*} | + | \begin{align*} I = I_{\rm SC} - {{I_{\rm SC}}\over{U_{\rm OC}}}\cdot U \tag{3.1.1} \end{align*} |
On the other hand, the formula can also be resolved to $U$: | On the other hand, the formula can also be resolved to $U$: | ||
- | \begin{align*} U = U_{OC} - {{U_{OC}}\over{I_{SC}}}\cdot I \tag{3.1.2} \end{align*} | + | \begin{align*} U = U_{\rm OC} - {{U_{\rm OC}}\over{I_{\rm SC}}}\cdot I \tag{3.1.2} \end{align*} |
- | ~~PAGEBREAK~~ ~~CLEARFIX~~ <callout icon=" | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ <callout icon=" |
+ | If a two-terminal network results in a linear curve between $U_{\rm OC}$ and $I_{\rm SC}$, it is called a **linear source**. This curve describes in good approximation the behavior of many real sources. Often one finds synonymous to the term ' | ||
+ | </ | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
Zeile 80: | Zeile 83: | ||
==== Linear Voltage Source ==== | ==== Linear Voltage Source ==== | ||
- | The linear voltage source consists of a series connection of an ideal voltage source with the source voltage $U_0$ (English: EMF for ElectroMotive Force) and the internal resistance $R_i$. To determine the voltage outside the active two-terminal network, the system can be considered as a voltage divider. The following applies: | + | The linear voltage source consists of a series connection of an ideal voltage source with the source voltage $U_0$ (English: EMF for ElectroMotive Force) and the internal resistance $R_\rm i$. To determine the voltage outside the active two-terminal network, the system can be considered as a voltage divider. The following applies: |
- | \begin{align*} U = U_0 - R_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 $U_0 = U_{LL}$. | + | 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 $U_0 = U_{\rm LL}$. |
- | \begin{align*} U = U_{OC} - R_i \cdot I \end{align*} | + | \begin{align*} U = U_{\rm OC} - R_{\rm i} \cdot I \end{align*} |
- | When the external voltage $U=0$, it is the short circuit case. In this case, $0 = U_{OC} - R_i \cdot I_{SC}$ and transform $R_i = {{U_{OC}}\over{I_{SC}}}$. Thus, equation $(3.1.2)$ is obtained: \begin{align*} U = U_{OC} - {{U_{OC}}\over{I_{SC}}} \cdot I \end{align*} | + | When the external voltage $U=0$, it is the short circuit case. In this case, $0 = U_{\rm OC} - R_{\rm i} \cdot I_{\rm SC}$ and transform $R_{\rm i} = {{U_{\rm OC}}\over{I_{\rm SC}}}$. |
+ | Thus, equation $(3.1.2)$ is obtained: \begin{align*} U = U_{\rm OC} - {{U_{\rm OC}}\over{I_{\rm SC}}} \cdot I \end{align*} | ||
Is this the structure of the linear source we are looking for? To verify this, we will now look at the second linear source. | Is this the structure of the linear source we are looking for? To verify this, we will now look at the second linear source. | ||
Zeile 94: | Zeile 98: | ||
==== Linear Current Source ==== | ==== Linear Current Source ==== | ||
- | The linear current source now consists of a __parallel circuit__ | + | The linear current source now consists of a __parallel circuit__ |
- | \begin{align*} I = I_0 - G_i \cdot U \end{align*} | + | \begin{align*} I = I_0 - G_{\rm i} \cdot U \end{align*} |
- | Here, the source current can be measured at the terminals in the event of a short circuit. The following therefore applies: $I_{SC}= I_0$ | + | Here, the source current can be measured at the terminals in the event of a short circuit. The following therefore applies: $I_{\rm SC}= I_0$ |
- | \begin{align*} I = I_{SC} - G_i \cdot U \end{align*} | + | \begin{align*} I = I_{SC} - G_{\rm i} \cdot U \end{align*} |
- | When the external current $I=0$, it is the no-load case. In this case, $0 = I_{SC} - G_i \cdot U_{OC}$ and transform $G_i = {{I_{SC}}\over{U_{OC}}}$. \\ Thus, equation $(3.1.1)$ is obtained: \begin{align*} I = I_{SC} - {{I_{SC}}\over{U_{OC}}} \cdot U \end{align*} | + | When the external current $I=0$, it is the no-load case. In this case, $0 = I_{\rm SC} - G_{\rm i} \cdot U_{\rm OC}$ and transform $G_{\rm i} = {{I_{\rm SC}}\over{U_{\rm OC}}}$. \\ |
+ | Thus, equation $(3.1.1)$ is obtained: \begin{align*} I = I_{\rm SC} - {{I_{\rm SC}}\over{U_{\rm OC}}} \cdot U \end{align*} | ||
So it seems that the two linear sources describe the same thing. | So it seems that the two linear sources describe the same thing. | ||
Zeile 108: | Zeile 113: | ||
==== Duality of Linear Sources ==== | ==== Duality of Linear Sources ==== | ||
- | Through the previous calculations, | + | Through the previous calculations, |
The <imgref imageNo7 > compares again the two linear sources and their characteristics: | The <imgref imageNo7 > compares again the two linear sources and their characteristics: | ||
- | - The linear voltage source is given by the source voltage $U_0$, or the open circuit voltage $U_{OC}$ and the internal resistance $R_i$. | + | - The linear voltage source is given by the source voltage $U_0$, or the open circuit voltage $U_{\rm OC}$ and the internal resistance $R_{\rm i}$. |
- | - The linear current source is given by the source current $I_0$, or the short-circuit current $I_{SC}$ and the internal conductance $G_i$. | + | - The linear current source is given by the source current $I_0$, or the short-circuit current $I_{\rm SC}$ and the internal conductance $G_{\rm i}$. |
< | < | ||
Zeile 119: | Zeile 124: | ||
The conversion is now done in such a way that the same characteristic curve is obtained: | The conversion is now done in such a way that the same characteristic curve is obtained: | ||
- | - __From linear voltage source to linear current source__: | + | - __From linear voltage source to linear current source__: |
- | - __From linear current source to linear voltage source__: | + | Given: Source voltage $U_0$, resp. open circuit voltage |
+ | in question: source current $I_0$, resp. short circuit current $I_{\rm SC}$, internal conductance $G_\rm i$ \\ | ||
+ | $\boxed{I_{\rm SC} = {{U_{\rm OC}}\over{R_\rm i}}}$ , $\boxed{G_\rm i = {{1}\over{R_\rm i}}}$ | ||
+ | </ | ||
+ | - __From linear current source to linear voltage source__: | ||
+ | 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 $U_{\rm OC}$, internal resistance $R_\rm i$ \\ | ||
+ | $\boxed{U_{\rm OC} = {{I_{\rm SC}}\over{G_\rm i}}}$ , $\boxed{R_\rm i = {{1}\over{G_\rm i}}}$ | ||
+ | </ | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
==== Operating Point of a real Voltage Source ==== | ==== Operating Point of a real Voltage Source ==== | ||
- | <imgref imageNo5 > shows the characteristics of the linear voltage source (left) and a resistive resistor (right). For this purpose, both are connected to a test system in the simulation: In the case of the source with a variable ohmic resistor, in the case of the load with a variable source. The characteristic curves formed in this way were described in the previous chapter. | + | <imgref imageNo5 > shows the characteristics of the linear voltage source (left) and a resistive resistor (right). For this purpose, both are connected to a test system in the simulation: In the case of the source with a variable ohmic resistor, |
< | < | ||
Zeile 131: | Zeile 144: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | The operating point can be determined from both characteristic curves. This is assumed when both the linear voltage source is connected to the ohmic resistor (without the respective test systems). In <imgref imageNo6 > both characteristic curves are drawn in a current-voltage diagram. The point of intersection is just the operating point that sets in. If the load resistance is varied, the slope changes in inverse proportion and a new operating point is established (light grey in the figure). | + | The operating point can be determined from both characteristic curves. This is assumed when both the linear voltage source is connected to the ohmic resistor (without the respective test systems). In <imgref imageNo6 > both characteristic curves are drawn in a current-voltage diagram. The point of intersection is just the operating point that sets in. If the load resistance is varied, the slope changes in inverse proportion, and a new operating point is established (light grey in the figure). |
The derivation of the working point is also [[https:// | The derivation of the working point is also [[https:// | ||
Zeile 141: | Zeile 154: | ||
The variation of the different source parameters will be briefly discussed. \\ \\ | The variation of the different source parameters will be briefly discussed. \\ \\ | ||
- | For the linear | + | For the linear |
- | | + | |
< | < | ||
Zeile 173: | Zeile 186: | ||
- know that any linear circuit with two connections of ohmic resistors and sources can be understood as a linear current source or linear voltage source. | - know that any linear circuit with two connections of ohmic resistors and sources can be understood as a linear current source or linear voltage source. | ||
- Be able to apply source conversion to more complicated circuits with multiple current sources or voltage sources. | - Be able to apply source conversion to more complicated circuits with multiple current sources or voltage sources. | ||
- | - know how to determine the open circuit voltage $U_{OC}$ and the short circuit current $I_{SC}$). | + | - know how to determine the open circuit voltage $U_{\rm OC}$ and the short circuit current $I_{\rm SC}$). |
- | - be able to calculate the parameters of the equivalent voltage source (internal resistance $R_i$ and source voltage $U_q$) of any linear circuit. | + | - be able to calculate the parameters of the equivalent voltage source (internal resistance $R_{\rm i}$ and source voltage $U_{\rm s}$) of any linear circuit. |
- understand and be able to draw the graphical interpretation of voltage and current at the linear two-terminal network in the form of a characteristic curve. | - understand and be able to draw the graphical interpretation of voltage and current at the linear two-terminal network in the form of a characteristic curve. | ||
</ | </ | ||
- | In < | + | In < |
- | < | + | < |
- | In the simulation a measuring current 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 |
\\ | \\ | ||
- | In order to understad | + | In order to understand |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | The circuit in <imgref imageNo8 > shows this circuit again. The ohmmeter is replaced by a current source and a voltmeter, since only the electrical properties are important in the following. In this setup, it can be seen that the current through $G_i$ is just given by $I_i = I_0 + I_\Omega$ (node theorem). Thus, the two sources in the circuit can be reduced. | + | The circuit in <imgref imageNo8 > shows this circuit again. The ohmmeter is replaced by a current source and a voltmeter since only the electrical properties are important in the following. In this setup, it can be seen that the current through $G_{\rm i}$ is just given by $I_{\rm i} = I_0 + I_\Omega$ (node theorem). Thus, the two sources in the circuit can be reduced. |
- | This should make the situation clear with a measuring current of $1mA$. The voltage at the resistor is now given by $U_\Omega = R \cdot (I_0 + I_\Omega)$. Only when $I_\Omega$ is very large does $I_0$ become negligible. The current of a conventional ohmmeter cannot guarantee this for every measurement. | + | This should make the situation clear with a measuring current of $1~\rm mA$. The voltage at the resistor is now given by $U_\Omega = R \cdot (I_0 + I_\Omega)$. |
+ | Only when $I_\Omega$ is very large does $I_0$ become negligible. The current of a conventional ohmmeter cannot guarantee this for every measurement. | ||
< | < | ||
Zeile 199: | Zeile 213: | ||
==== More complex Example ==== | ==== More complex Example ==== | ||
- | This knowledge can now be used for more complicated circuits. In < | + | This knowledge can now be used for more complicated circuits. In < |
**Important here**: Only two-terminal networks can be converted via source duality. This means that only 2 nodes may act as output terminals for selected sections of the circuit. If there are more nodes the conversion is not possible. | **Important here**: Only two-terminal networks can be converted via source duality. This means that only 2 nodes may act as output terminals for selected sections of the circuit. If there are more nodes the conversion is not possible. | ||
- | < | + | < |
- | - As a first step, sources are to be converted in such a way that resistors can be combined after the conversion. In this example this is done by: | + | - As a first step, sources are to be converted in such a way that resistors can be combined after the conversion. In this example, this is done by: |
- converting the linear voltage source $R_1$ and $U_1$ into a linear current source with $I_1={{U_1}\over{R_1}}$ and $R_1$ (or $G_1={{1}\over{R_1}}$) | - converting the linear voltage source $R_1$ and $U_1$ into a linear current source with $I_1={{U_1}\over{R_1}}$ and $R_1$ (or $G_1={{1}\over{R_1}}$) | ||
- converting the linear current source $R_4$ and $I_4$ into a linear voltage source with $U_4={{I_4}\cdot{R_4}}$ and $R_4$ | - converting the linear current source $R_4$ and $I_4$ into a linear voltage source with $U_4={{I_4}\cdot{R_4}}$ and $R_4$ | ||
- 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}} $. | - 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 $I_g$: \begin{align*} | + | - ideal equivalent current source $I_{\rm eq}$: \begin{align*} |
- | - Substitute conductance $G_g$: \begin{align*} | + | - Substitute conductance $G_{\rm eq}$: \begin{align*} |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
Zeile 222: | Zeile 236: | ||
* as a single, linear current source \\ ({{https:// | * as a single, linear current source \\ ({{https:// | ||
- | In <imgref imageNo63 > it can be seen that the three circuits give the same result (voltage / current) with the same load. This is also true when an (AC) source is used instead of the load. </ | + | In <imgref imageNo63 > it can be seen that the three circuits give the same result (voltage/ |
< | < | ||
Zeile 234: | Zeile 248: | ||
If only the equivalent resistance of a more complex circuit is sought, the following approach can be used: | If only the equivalent resistance of a more complex circuit is sought, the following approach can be used: | ||
- | - Replace all ideal voltage sources | + | - Replace all ideal voltage sources |
- | - Replace all ideal current sources | + | - Replace all ideal current sources |
- Add the remaining resistors to an equivalent resistance using the rules for parallel and series connection. | - Add the remaining resistors to an equivalent resistance using the rules for parallel and series connection. | ||
Zeile 283: | Zeile 297: | ||
By the end of this section, you will be able to: | By the end of this section, you will be able to: | ||
- calculate the source power and consumer power. | - calculate the source power and consumer power. | ||
- | - distinguish between the optimisation | + | - distinguish between the optimization |
- calculate the efficiency and utilization rate. | - calculate the efficiency and utilization rate. | ||
Zeile 299: | 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 309: | 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 324: | Zeile 339: | ||
<imgref imageNo14> | <imgref imageNo14> | ||
- | * Diagram top: current-voltage diagram of 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: | ||
Zeile 345: | Zeile 360: | ||
In order to understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again: | In order 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 | ||
+ | = {{R_{\rm L}} \over {R_{\rm L}+R_{\rm i}}} } | ||
+ | \end{align*} | ||
- | The **utilization rate** | + | 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}}} | ||
+ | | ||
+ | | ||
+ | | ||
+ | | ||
+ | | ||
+ | \end{align*} | ||
- | In __power engineering__ | + | In __power engineering__ |
+ | 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__, | + | In __communications engineering__, |
- | The impedance matching / power matching is also [[https:// | + | The impedance matching/ |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | <panel type=" | ||
- | Simplify the following circuits by 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 367: | Zeile 400: | ||
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. |
- | In case of a short circuit we have $2~V$ in a branch which in turn means there must be $−2~V$ on the resistor. | + | In case of a short circuit, we have $2~V$ in a branch which in turn means there must be $−2~V$ on the resistor. |
The current through that branch is | The current through that branch is | ||
\begin{equation*} I_R=\frac{2~V}{8~\Omega} \end{equation*} | \begin{equation*} I_R=\frac{2~V}{8~\Omega} \end{equation*} | ||
- | The asked current is the sum of both the other branches | + | The current |
\begin{equation*} I_S= I_R + 1~A \end{equation*} | \begin{equation*} I_S= I_R + 1~A \end{equation*} | ||
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 398: | Zeile 431: | ||
</ | </ | ||
+ | <wrap anchor # | ||
<panel type=" | <panel type=" | ||
- | For the company „HHN Mechatronics & Robotics“ you shall analyse | + | For the company „HHN Mechatronics & Robotics“ you shall analyze |
- | 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 414: | Zeile 446: | ||
- Calculate the input and output power for both modes. | - Calculate the input and output power for both modes. | ||
- 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 exaclty | + | - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly |
- | - Your company uses for 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: |
{{url> | {{url> | ||
</ | </ | ||
+ | |||
+ | # | ||
+ | |||
+ | 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:// |