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--- electrical_engineering_and_electronics_1:uebung_3.5.2 [2025/09/19 14:13] – angelegt mexleadmin
+++ electrical_engineering_and_electronics_1:uebung_3.5.2 [2025/12/13 23:12] (aktuell) – [Bearbeiten - Panel] mexleadmin
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-<WRAP pagebreak></WRAP> <panel type="info" title="Exercise 3.5.2. Variations of the non-inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<WRAP pagebreak></WRAP>
+<panel type="info" title="Exercise 3.5.2. Variations of the non-inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 Below you will find circuits with an ideal operational amplifier, which are similar to the non-inverting amplifier and whose voltage gain $A_V$ must be determined.
@@ Zeile 24: / Zeile 25: @@
  \\
 **__Important__**: As always in your studies, you should try to generalize the knowledge gained from the task.
-<WRAP onlyprint> **Tipps**
-  * How big is the current flow into the inverting and non-inverting input of an ideal operational amplifier? So what voltage drop would there be across a resistor whose one connection only leads to one input of the operational amplifier ($R_3$)?
-  * The operational amplifier always tries to output as much current at the output that the required minimum voltage $U_\rm D$ results between the inverting and non-inverting input. How high can $U_\rm D$ be assumed? Can this voltage also be built up via a resistor ($R_4$)?
-  * Can different resistors (e.g. because they are between the same nodes) be combined?
-</WRAP>
 <WRAP pagebreak></WRAP><WRAP group><WRAP half column>
-<WRAP right><panel type="default"> Abb. 1 \\ {{drawio>pic3_5_2_Aufgabe1.svg}} </panel></WRAP> \\  \\  \\  \\  \\  \\  \\  \\ <WRAP noprint>
+<WRAP right><panel type="default"> Abb. 1 \\ {{drawio>pic3_5_2_Aufgabe1.svg}} </panel></WRAP>
 ++++ Hints|
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 ++++
-</WRAP>
 </WRAP><WRAP half column>
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 <WRAP right><panel type="default"> Abb. 2 \\ {{drawio>pic3_5_2_Aufgabe2.svg}} </panel></WRAP>
- \\  \\  \\  \\  \\  \\  \\  \\ <WRAP noprint>
 ++++ Hints|
@@ Zeile 60: / Zeile 53: @@
 ++++
-</WRAP></WRAP></WRAP>
+</WRAP>
-<WRAP onlyprint> \\  \\  \\  \\  \\  \\  \\  \\ </WRAP> <WRAP group><WRAP half column>
+<button size="xs" type="link" collapse="sol3521">{{icon>eye}} Solution for (1) + (2)</button><collapse id="sol3521" collapsed="true">
+Fig. 1
+  * Between the inverting and non-inverting input, only $U_D \rightarrow 0$ is present. As long as $R_4 > 0$, this small voltage can exist there. Therefore, $R_4$ can be replaced by an open circuit.
+  * Since no current flows through $R_3$, there is no voltage difference across $R_3$. Thus, $R_3$ can be replaced by a short circuit.
+  * This results in a non-inverting amplifier with $A_V = 3/2$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUhnhgwkhGxiTieSokkZCEEaPoAnFihohsbFEZB4qVHZH2HjphyH5Ob8ezRodvtX7MVkOE9-MDYvDjBNKBYPAHM-SOiI2I5regB3RPAw5wCoTLzc82MrAqzWEEJjSrK7Cow2MpLLdIa2Jxa3QpouXxTfeudXNjk2EYKAJXBicacwLxc3GN7wGOtoBHpsQh9+jgUfNIKgA 500,400 noborder}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+Fig. 2
+  * No current flows through $R_3$, since no current can flow into the op-amp. Therefore, there is no voltage difference across $R_3$, and $R_3$ can be replaced by a short circuit.
+  * $R_4$ and $R_2$ are in parallel and yield $R_g = \frac{2}{3} R_1$
+  * The gain thus becomes $A_V = \frac{R_g + R_1}{R_g} = 2.5$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUhkpxCKGsTx1wyUfQBOLeeDBsU2sISpURcDVpo692mhxPxzNW9yccwuqKsiPn7tq-AMQk80MwBzWl8gyI40Oyh6AHcYqwtU72TWbizCSwyWDH9YmzskgrZsa0tKhOSaLmcA528AJTSasGia404VHpgEemxCDiaOBBRR+PyAzuD9SznatKWF8CMEoA 500,400 noborder}}
+</WRAP>
+</collapse>
+</WRAP>
+<WRAP group><WRAP half column>
 <WRAP right><panel type="default"> Abb. 3 \\ {{drawio>pic3_5_2_Aufgabe3.svg}} </panel></WRAP> </WRAP><WRAP half column>
 <WRAP right><panel type="default"> Abb. 4 \\ {{drawio>pic3_5_2_Aufgabe4.svg}} </panel></WRAP> </WRAP></WRAP>
+<button size="xs" type="link" collapse="sol3522">{{icon>eye}} Solution for (3) + (4)</button><collapse id="sol3522" collapsed="true">
+Fig. 3
+  * $R_3$ and $R_4$ form an unloaded voltage divider.
+  * Thus, the input voltage is halved.
+  * This results in a non-inverting amplifier with $A_V = \frac{R_2 + R_1}{R_2} = 0.75$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUhnhgwkhGxiTieSokkZCEEaPoAnFihohsbFEZB4qVHZH2HjphyH5Ob8ezRodvtX7MVkOE9-MDYvDjBNKBYPAHM-SOiI2I5regB3RPAw5wCoTLzc82MrAqzWEEJjSrK7Cow2MpLLdIa2Jxa3QpouXxTfeudXNjk2EYKAJXBicacwLxc3GN7wGOtoBHpsQh9+jgUfNIKgA 500,400 noborder}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+Fig. 4
+  * No current flows through $R_3$, since no current can flow into the op-amp. Therefore, there is no voltage difference across $R_3$, and $R_3$ can be replaced by a short circuit.
+  * Since the input voltage comes from a low-impedance voltage source, $R_4$ is negligible compared to the internal resistance of the source. Therefore, $R_4$ can be replaced by an open circuit.
+  * The overall gain thus becomes $A_V = 3/2$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUnkroUNUoULhko+gCcWckNjYotYQlSoi46zTW26MbHVBWQzNGh2e1XYMG2PxH7z244wDCUqNFMAcwDwYKi0DiN6AHcojys2VKgklmsQQgtWcH8HZL0LfltSy0zkmi5XJxd4+gAlcxBy9JiOuypa5SNxBHpsQhd6jgQURp6shuilSqCQrMWYxcNMoA 500,400 noborder}}
+</WRAP></collapse>
 <WRAP pagebreak></WRAP> <WRAP group><WRAP half column>
@@ Zeile 74: / Zeile 108: @@
 <WRAP right><panel type="default"> Abb. 6 \\ {{drawio>pic3_5_2_Aufgabe6.svg}} </panel></WRAP> </WRAP></WRAP>
-<WRAP onlyprint> \\  \\  \\  \\  \\  \\  \\  \\ </WRAP> <WRAP group><WRAP half column>
+<button size="xs" type="link" collapse="sol3523">{{icon>eye}} Solution for (5) + (6)</button><collapse id="sol3523" collapsed="true">
+Fig. 5
+  * No current flows through $R_3$, since no current can flow into the op-amp. Therefore, there is no voltage difference across $R_3$, and $R_3$ can be replaced by a short circuit.
+  * $R_4$ and $R_2$ are in parallel and yield $R_g = \frac{2}{3} R_1$
+  * The gain thus becomes $A_V = \frac{R_g + R_1}{R_g} = 2.5$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUhkpxCKGsTx1wyUfQBOLDGzBg2rcBkJRVkDS3khs+7VbZURcczRodXtd7vunnnvR44wIxM0JwBzAMNjFw40DiozAHdIry0dfyS0kEIaLNTMmi53GO4zACULXOsouxMqQpUE8QR6bEI3Yo4EFDd4qHpkkqDjA2H+oA 500,400 noborder}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+Fig. 6
+  * No current flows through $R_4$, since no current can flow into the op-amp. Therefore, there is no voltage difference across $R_4$, and $R_4$ can be replaced by a short circuit.
+  * Since the inverting input is thus at 0 V, the output voltage becomes the maximum possible output voltage of the op-amp.
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRwyUfQBOLDG0I05bMGDZURcGcvBraNDqvWTI2mge7nDejfDMWj+wxkLi0WgOZPwL72g5UpgDuOo6sPq7BnFQWVtymAEo62GwoKEop4jEaWVDQCPTYhBxxFggoJQFQ9CFxYL7h9ZH0QA 500,400 noborder}}
+</WRAP></collapse>
+ <WRAP group><WRAP half column>
 <WRAP right><panel type="default"> Abb. 7 \\ {{drawio>pic3_5_2_Aufgabe7.svg}} </panel></WRAP> </WRAP><WRAP half column>
@@ Zeile 80: / Zeile 130: @@
 <WRAP right><panel type="default"> Abb. 8 \\ {{drawio>pic3_5_2_Aufgabe8.svg}} </panel></WRAP> </WRAP></WRAP>
+<button size="xs" type="link" collapse="sol3524">{{icon>eye}} Solution for (7) + (8)</button><collapse id="sol3524" collapsed="true">
+Fig. 7
+  * The input voltage is applied across $R_4$, so a current $I = U_E / R$ flows through it. This current also flows through $R_3$. Therefore, the voltage across both is $2 \cdot U_E$.
+  * The resistors $R_2$, $R_3$, and $R_4$ form $R_g = (R_3 + R_4)||R_2 = 2R||2R = R$
+  * Thus $R_g = R_1$, and the output voltage $U_A$ is twice the voltage across $R_g$, so the overall gain is $A_V = 4$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUoXhDJ2BLwgjR9AE4sMbMGDatwGQlGRw1GrTto0O2tlWWRTNK9yfWLd+I+c3L1g0bQTAHNffUNXFnwjBwB3MxBCGnifWM4qZwjnBwAleOxdFCT86M5wEpgEemxCDkyOBBRajipUiLB-PXbDByA 500,400 noborder}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+Fig. 8
+  * The inverting and non-inverting inputs are shorted, so the output voltage is 0.
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRohFEhiThoyPQBOLDG0I0VbMGDZURcJZpD91R7LuQHlNGh1u172i-oXXHOhxzAZCUFvHoAc09wHxC0DioFAHdOKnsbO0j6ACUjEw0UFA0Mv3i9PPEEemxCOwSOBBQkvPpYxPAPVkbdOrM2ZtMFIA 500,400 noborder}}
+</WRAP>
+</collapse>
 <WRAP pagebreak></WRAP> <WRAP group><WRAP half column>
 <WRAP right><panel type="default"> Abb. 9 \\ {{drawio>pic3_5_2_Aufgabe9.svg}} </panel></WRAP> </WRAP><WRAP half column> </WRAP></WRAP>
+<button size="xs" type="link" collapse="sol3525">{{icon>eye}} Solution for (9)</button><collapse id="sol3525" collapsed="true">
+Fig. 9
+  * Between the inverting and non-inverting input, only $U_D \rightarrow 0$ is present. As long as $R_3 > 0$, this small voltage can exist there. Therefore, $R_3$ can be replaced by an open circuit.
+  * Since no current flows through $R_4$, there is no voltage difference across $R_4$. Thus, $R_4$ can be replaced by a short circuit.
+  * This results in a non-inverting amplifier with $A_V = \frac{R_2 + R_1}{R_2} = 1.5$
+<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAQxExVr1rnZGPCQFowSMPHhRopUnkqJJGQhEWjI9AE4saVQjQ1UwYNnvhrd4Qqe1RkcEzRod7tR-sPWV6uxxdOv8q2hsAcx9wP08WfCsVAHdOKkdwxxUAJVN+bDYwYjZ013i9KPEEemxCBwSOBBQHDioYkO8UTXADKHpYpq0dVhBLep6MlgxM837hs1MwUfbJ806QXLagA 500,400 noborder}}
+</WRAP>
+</collapse>
 </WRAP></WRAP></panel>