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--- circuit_design:uebung_3.5.1 [2023/03/28 10:14] – mexleadmin
+++ circuit_design:uebung_3.5.1 [2023/03/28 11:00] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
 <WRAP pagebreak></WRAP><panel type="info" title="Exercise 3.5.1 inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-For the inverting amplifier, derive the voltage gain. Use the procedure that was used for the non-inverting amplifier.\\
+. Derive the voltage gain $A_{\rm V}= {{U_{\rm O}}\over{U_{\rm I}}}$ for the inverting amplifier. \\
-Take into account that for the differential gain $A_\rm D$ of the ideal OPV applies: $A_\rm D \rightarrow \infty$.\\
+Use the procedure that was used for the non-inverting amplifier.\\
-This also applies: $1/A_\rm D \rightarrow 0$ , **but** it doesn't always apply ${{C}\over{U_x \cdot A_\rm D}} \rightarrow 0$, for an unknown constant $C$ and a voltage $U_x$!
    * What is looking for?
@@ Zeile 9: / Zeile 8: @@
   * Number of necessary equations?
   * Establishing the known equations
-  * Derivation of the voltage gain <wrap onlyprint> \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\ </wrap>
+  * Derivation of the voltage gain
-Which type of amplifier (inverting or non-inverting amplifier) has the lower input resistance? Why?
+Take into account that for the differential gain $A_\rm D$ of the ideal OPV applies: $A_\rm D \rightarrow \infty$. And the following also applies: $1/A_\rm D \rightarrow 0$ \\
+**But** the following doesn't always apply:  ${{C}\over{U_x \cdot A_\rm D}} \rightarrow 0$, for an unknown constant $C$ and a voltage $U_x$!
+<wrap onlyprint> \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\ </wrap> \\
+. Which type of amplifier circuit (inverting or non-inverting amplifier) has the lower input resistance? Why?
 </WRAP></WRAP></panel>