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electrical_engineering_1:network_analysis [2023/11/23 02:58] – [Bearbeiten - Panel] mexleadminelectrical_engineering_1:network_analysis [2025/01/29 00:30] (aktuell) – mexleadmin
Zeile 406: Zeile 406:
 For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ to a supply voltage $U_{\rm s}$. For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ to a supply voltage $U_{\rm s}$.
  
-The following simulation shows roughly the situation (the resistor values are not correct).+The following simulation shows roughly the situation. Be aware that the resistor and voltage values are not correct!
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCDMCmC0DsIBskB08CcGkA4AsYGADEQKxFIYjnUh6TVxhgBQATiLHjiAEy95kg-oKLgSJdkL4DpkIqOoSiUpMNlrwSMWN4SWAczkK5ekDpYAlaWG03ePHXTGlzboqlIsAbp0K8oE1h-PjwnCDC+Jx1PQz8MAPlBYISoMwsAd3iAgTEUnNkVDnyZZOxStzB9SFxpEQr6iwAjTl5SJHA8ZOZEgnMWAA9W0kSkRFgMCEgMRGFBJoBLAAcAewAbAEM2AB0AZwBlFYBXNgBjaD3t7YA7AApoVANUPY293ehr3ZW2AEpBkFShAYVSohA6cxAln+zF0REQYF4DF4VTooXA0IRfHgwLaWPBaN4-wwwlIoOYfCQAQhUKGwR4enGrmRYghkH+sBq4AwDBSXPxshp4Bwoymtigk1RsiO12W6y2bwW1wA1lc7g8nnsjgBhH5-SDwPLlepcemyCAWWlgHA8K3deAQK2zNGC4KULk8sCkYHlCH7FgrAHiQrODD0sRIVCKAKuFRAA noborder}} </WRAP>+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCBMCmC0AcUB08AMBmMA2SAWeAnLgKyRZi4jGpUi7pVxhgBQATiFpZHp5djRrVULAEp9wWGl3CREgukJALUSYiwBuIWGAKQQ6VJR16ouBRHOprUFcrUsA5tt37Dx1wcgqWAdxemeDQm+jyUIhwhULywBFjR4eDWIuhYiDJhCVmCLABG2pDE8RQeYG64ECIAHgWkBlgA7NoEEOgETdyUuQCWAA4A9gA2AIZsADoAzgDK-QCubADG0JNjYwB2ABTQSI5Ik8OTE9BrE-1sAJQsNQQE4AQMzMVxdGYgolfgYE3eTWCGUGAaJ1wB9HlB4BAyrc5JRgZAPgREGRfmBiFAGkDXu8amB4G5yAZsAYWi9eLM1n0hqNDt01gBrVabba7SazADC50u6AxzXimVg+CylWUHx08EQuI8DUh8A6WNF2FuugYOmID2ewKm7AkmRk7mUVGS2oyMUFmRo3msTgk+r13gN1W0-Fl4HVmN42PA3FQEEg3yMpMo6BY-U4kgN5kRtk4SESoRFQA noborder}} </WRAP>
  
 Questions: Questions:
  
-1. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition.+1. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition. \\ 
 +  * Determine suitable values for $R_1$, $R_2$, and $R_3$. 
 +  * What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this?
  
 #@HiddenBegin_HTML~1,Solution~@# #@HiddenBegin_HTML~1,Solution~@#
Zeile 456: Zeile 458:
 \end{align*} \end{align*}
  
-The formula $(1)$ is the general formula to calculate the output voltage $U_{\rm O}$ for a changing input voltage $U_{\rm I}$, where the supply voltage $U_{\rm S}" is constant. \\+The formula $(1)$ is the general formula to calculate the output voltage $U_{\rm O}$ for a changing input voltage $U_{\rm I}$, where the supply voltage $U_{\rm S}$ is constant. \\
 </callout> </callout>
  
Zeile 525: Zeile 527:
 \end{align*} \end{align*}
  
-We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_2$ and $R_3$:+We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_3$ and $R_1$:
 \begin{align*} \begin{align*}
 R_3 &= 45 {~\rm k\Omega}\\ R_3 &= 45 {~\rm k\Omega}\\
Zeile 531: Zeile 533:
 R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega}  R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega} 
 \end{align*} \end{align*}
 +
 +Based on the E24 series, the following values are next to the calculated ones:
 +\begin{align*}
 +R_3^0 &= 43 {~\rm k\Omega}\\
 +R_1^0 &= {{1}\over{3}}   (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\
 +R_2^0 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 12 {~\rm k\Omega} 
 +\end{align*}
 +
 #@HiddenEnd_HTML~Result1,Result~@# #@HiddenEnd_HTML~Result1,Result~@#
  
 +2. Find the relationship between $R_1$, $R_2$, and $R_3$ by investigating Kirchhoff's nodal rule for the node where $R_1$, $R_2$, and $R_3$ are interconnected.
 +
 +#@HiddenBegin_HTML~Solution2,Solution~@#
 +
 +The potential of the node is $U_\rm O$. Therefore the currents are:
 +  - the current $I_2$ over $R_2$ is flowing to ground: $I_2 = - {{U_\rm O}\over{R_2}} $ 
 +  - the current $I_1$ over $R_1$ is coming from the supply voltage $U_{\rm S}$ to the nodal voltage $U_{\rm O}$:  $I_1 = {{U_{\rm S} - U_{\rm O}}\over{R_1}}$
 +  - the current $I_4$ over $R_4$ is coming from the input voltage  $U_{\rm I}$ to the nodal voltage $U_{\rm O}$:  $I_4 = {{U_{\rm I} - U_{\rm O}}\over{R_4}}$
 +
 +This led to the formula based on the Kirchhoff's nodal rule: 
 +
 +\begin{align*}
 +\Sigma I = 0 &= I_1 + I_2 + I_3 \\
 +         0 &= {{U_{\rm S} - U_{\rm O}}\over{R_1}} + {{U_{\rm I} - U_{\rm O}}\over{R_4}} - {{U_\rm O}\over{R_2}} 
 +\end{align*}
 +
 +The formula can be rearranged, with all terms containing $ U_{\rm O}$ on the left side: 
 +\begin{align*}
 +    {{U_{\rm O}}\over{R_1}} + {{U_{\rm O}}\over{R_2}} + {{U_{\rm O}}\over{R_4}}         &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\
 +      U_{\rm O}\cdot \left( {{1}\over{R_1}} + {{1}\over{R_2}} + {{1}\over{R_4}} \right) &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\        
 +\end{align*}
 +
 +Both sides can be multiplied by $\cdot R_1$, $\cdot R_2$, $\cdot R_4$  - in order to get rid of the fractions : 
 +\begin{align*}
 +      U_{\rm O}\cdot \left( {{R_1 R_2 R_4 }\over{R_1}} + {{R_1 R_2 R_4 }\over{R_2}} + {{R_1 R_2 R_4 }\over{R_4}} \right) &=  R_1 R_2 R_4 \cdot {{U_{\rm S}}\over{R_1}}  + R_1 R_2 R_4 \cdot {{U_{\rm I}}\over{R_4}}  \\        
 +      U_{\rm O}\cdot \left( R_2 R_4 + R_1 R_4 + R_1 R_2 \right) &=  R_2 R_4 \cdot U_{\rm S}  + R_1 R_2 \cdot U_{\rm I} \\        
 +      U_{\rm O} &= {{R_2}\over{R_2 R_4 + R_1 R_4 + R_1 R_2 }}  \left( R_4 \cdot U_{\rm S}  + R_1 \cdot U_{\rm I} \right)\\        
 +\end{align*}
 +
 +The last formula was just the result we also got by the superposition but by more thinking. \\
 +So, sometimes there is an easier way... 
 +  * Unluckily, there is no simple way to know before, what way is the easiest.
 +  * Luckily, all ways lead to the correct result.
 +
 +#@HiddenEnd_HTML~Solution2,Solution ~@#
 +
 +3. What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)?
 +
 +#@HiddenBegin_HTML~Solution3,Solution~@#
 +
 +\begin{align*}
 +R_{\rm in}(R_1, R_2, R_3) &= R_3 + R_1 || R_2 \\
 +                          &= R_3 + {{R_1  R_2}\over{R_1 + R_2}}
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution3,Solution~@#
 +
 +4. What is the minimum allowed input resistance ($R_{\rm in, min}(R_1, R_2, R_3)$) for the sensor to still deliver current?
 +
 +#@HiddenBegin_HTML~Solution4,Solution~@#
 +
 +\begin{align*}
 +R_{\rm in, min} &= {{U_{\rm sense}}\over{I_{\rm sense, max}}} \\
 +                &= \rm {{15 V}\over{1 mA}} \\
 +                &= 15 k\Omega \\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution4,Solution~@#
  
-  * Find the relationship between $R_1$, $R_2$, and $R_3$ using the star-delta transformation. 
-  * What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)? 
-  * What is the maximum allowed input resistance ($R_{\rm in}(R_1, R_2, R_3)$) for the sensor to still deliver current? 
-  * Determine suitable values for $R_1$, $R_2$, and $R_3$. 
-  * What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this? 
-  *  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#