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circuit_design:uebung_3.5.3 [2021/11/11 16:32] – slinncircuit_design:uebung_3.5.3 [2023/03/28 10:38] (aktuell) – mexleadmin
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-<WRAP pagebreak></WRAP><panel type="info" title="Aufgabe 3.5.3. R-2R-Leiter"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<WRAP pagebreak></WRAP><panel type="info" title="Exercise 3.5.3. R-2R conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-<WRAP right> {{  :elektronische_schaltungstechnik:dac.jpg?400|dac.jpg}}\\+<WRAP right> {{drawio>circuit_design:dac.svg}}\\
 <fs 70%>Abbildung 1 </fs> </WRAP> <fs 70%>Abbildung 1 </fs> </WRAP>
  
 You work in the company "HHN Mechatronics & Robotics", which is supposed to build a battery model for a customer. This model is intended to replicate a real battery. For this purpose, a voltage is to be output which is specified by a software model of the battery. A digital-to-analog converter (DAC) is therefore required. You work in the company "HHN Mechatronics & Robotics", which is supposed to build a battery model for a customer. This model is intended to replicate a real battery. For this purpose, a voltage is to be output which is specified by a software model of the battery. A digital-to-analog converter (DAC) is therefore required.
  
-You have found the DAC7741 for this. On [[https://www.ti.com/lit/ds/symlink/dac7741.pdf#page=12|page 12]] of the data sheet you can see an image of the internal structure - this is similar to the illustration on the right. For an error analysis you now want to understand this structure in more detail.+You have found the DAC7741 for this. On [[https://www.ti.com/lit/ds/symlink/dac7741.pdf#page=12|page 12]] of the datasheet you can see an image of the internal structure - this is similar to the illustration on the right. For an error analysis, you now want to understand this structure in more detail.
  
-In the drawing on the right, the current switch position stands for 000b, so all switches $SW1 … SW3$ are switched to ground. It is advisable to recreate the circuit in the [[http://www.falstad.com/circuit/|Falstad-Circuit]] for a better understanding. In this case it is advisable to measure the individual node voltages $K1 … K3$ as well.+In the drawing on the right, the current switch position stands for 000b, so all switches $\rm SW1 … SW3$ are switched to ground. It is advisable to recreate the circuit in the [[http://www.falstad.com/circuit/|Falstad-Circuit]] for a better understanding. In this case, it is advisable to measure the individual node voltages $\rm K1 … K3$ as well.
  
 . .
  
-  - $SW_3$ = 1, $SW_2$ = 0 und $SW_1$ = 0 should now apply initially - that is, only switch $SW_3$ is switched to $U_{logic}$.+  - $\rm SW_3$ = 1, $\rm SW_2$ = 0 and $\rm SW_1$ = 0 should now apply initially - that is, only switch $\rm SW_3$ is switched to $U_{\rm logic}$.
       - To do this, draw the equivalent circuit diagram without a switch.       - To do this, draw the equivalent circuit diagram without a switch.
       - Simplify this equivalent circuit using an equivalent resistor.       - Simplify this equivalent circuit using an equivalent resistor.
-      - The result is a resistance that lies between the inverting and non-inverting input. The operational amplifier always tries to feed so much current into the resistor network surrounding it that a small differential voltage $U_D$ results. This is also possible with a (not too small) resistance between the inverting and non-inverting input. \\ So what is the gain? +      - The result is a resistance that lies between the inverting and non-inverting input. The operational amplifier always tries to feed so much current into the resistor network surrounding it that a small differential voltage $U_\rm D$ results. This is also possible with a (not too small) resistance between the inverting and non-inverting input. \\ So what is the gain? 
-  - $SW_3$ = 0, $SW_2$ = 1 und $SW_1$ = 0 should now apply initially - that is, only switch $SW_3$ is switched to $U_{logic}$.+  - $\rm SW_3$ = 0, $\rm SW_2$ = 1 und $\rm SW_1$ = 0 should now apply initially - that is, only switch $\rm SW_3$ is switched to $U_{\rm logic}$.
       - Here, too, draw the equivalent circuit diagram without a switch.       - Here, too, draw the equivalent circuit diagram without a switch.
       - Simplify this equivalent circuit using equivalent resistors.       - Simplify this equivalent circuit using equivalent resistors.
-      - The statement about the above-mentioned resistance between the inverting and non-inverting input also applies here. You should also be aware of the tension of node $K_3$. \\ Now draw an equivalent circuit diagram of the left, taking the voltage at the node $K_3$ of the ideal amplifier. +      - The statement about the above-mentioned resistance between the inverting and non-inverting input also applies here. You should also be aware of the voltage of node $\rm K_3$. \\ Now draw an equivalent circuit diagram of the left, taking the voltage at the node $\rm K_3$ of the ideal amplifier. 
-      - Now determine the voltage at node $K_2$. +      - Now determine the voltage at node $\rm K_2$. 
-      - This voltage at node $K_2$ is the output voltage of an inverting amplifier, which starts from node $K_2$ to the right. Now calculate the gain of the resulting network.+      - This voltage at node $\rm K_2$ is the output voltage of an inverting amplifier, which starts from node $\rm K_2$ to the right. Now calculate the gain of the resulting network.
   - The concept should be understood by now. Now specify which input / which switch specifies the [[https://en.wikipedia.org/wiki/Bit_numbering#Least_significant_bit|LSB]].   - The concept should be understood by now. Now specify which input / which switch specifies the [[https://en.wikipedia.org/wiki/Bit_numbering#Least_significant_bit|LSB]].