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electrical_engineering_and_electronics_1:uebung_3.5.1 [2025/12/13 22:35] – mexleadminelectrical_engineering_and_electronics_1:uebung_3.5.1 [2025/12/13 22:46] (aktuell) – [Bearbeiten - Panel] mexleadmin
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-<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%>+<WRAP pagebreak></WRAP> 
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 1. Derive the voltage gain $A_{\rm V}= {{U_{\rm O}}\over{U_{\rm I}}}$ for the inverting amplifier. \\ 1. Derive the voltage gain $A_{\rm V}= {{U_{\rm O}}\over{U_{\rm I}}}$ for the inverting amplifier. \\
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 **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 $U_x$! **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 $U_x$!
    
-#@HiddenBegin_HTML~1531T,Solution for "What is required?"~@#+<button size="xs" type="link" collapse="sol3511">{{icon>eye}} Solution for "What is required?" </button><collapse id="sol3511" collapsed="true">
  
 $A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}$  $A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}$ 
  
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-#@HiddenBegin_HTML~1532T,Solution for "Number of variables?"~@#+<button size="xs" type="link" collapse="sol3512">{{icon>eye}} Solution for "Number of variables?" </button><collapse id="sol3512" collapsed="true">
   * 5 voltages: $U_{\rm I}$, $U_{\rm 1}$, $U_{\rm D}$, $U_2$, $U_{\rm O}$     * 5 voltages: $U_{\rm I}$, $U_{\rm 1}$, $U_{\rm D}$, $U_2$, $U_{\rm O}$  
   * 5 currents: $I_1$, $I_{\rm m}$, $I_{\rm p}$, $I_2$, $I_{\rm o}$   * 5 currents: $I_1$, $I_{\rm m}$, $I_{\rm p}$, $I_2$, $I_{\rm o}$
   * --> 10 variables   * --> 10 variables
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 + 
 +<button size="xs" type="link" collapse="sol3513">{{icon>eye}} Solution for "Number of necessary equations?" </button><collapse id="sol3513" collapsed="true">
  
-#@HiddenBegin_HTML~1533T,Solution for "Number of necessary equations?"~@# 
 9, since one equation is to be determined 9, since one equation is to be determined
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   * Fundamental equation: (1) $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$   * Fundamental equation: (1) $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$
   * Golden rules:   * Golden rules:
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     * (8) $R_1 = \frac{U_1}{I_1}$     * (8) $R_1 = \frac{U_1}{I_1}$
     * (9) $R_2 = \frac{U_2}{I_2}$     * (9) $R_2 = \frac{U_2}{I_2}$
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-#@HiddenBegin_HTML~1535T,Solution for "Derivation of the voltage gain"~@#+<button size="xs" type="link" collapse="sol3515">{{icon>eye}} Solution for "Derivation of the voltage gain" </button><collapse id="sol3515" collapsed="true">
 \begin{align*}  \begin{align*} 
 A_V &= \frac{U_{\rm O}}{U_{\rm I}}                                   \quad | \quad \text{using (5) and (6)} \\ A_V &= \frac{U_{\rm O}}{U_{\rm I}}                                   \quad | \quad \text{using (5) and (6)} \\
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 A_V &= -\frac{R_2 }{R_1 }  A_V &= -\frac{R_2 }{R_1 } 
 \end{align*} \end{align*}
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 2. Which type of amplifier circuit (inverting or non-inverting amplifier) has the lower input resistance? Why? 2. Which type of amplifier circuit (inverting or non-inverting amplifier) has the lower input resistance? Why?
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 +<button size="xs" type="link" collapse="sol3516">{{icon>eye}} Solution" </button><collapse id="sol3516" collapsed="true">
 +The input resistance of the **inverting amplifier** is the resistor $R_1$. \\
 +The input resistance of the **non-inverting amplifier** is **larger than the input resistance of the op-amp**. \\
 +Therefore, the inverting amplifier has the lower input resistance. \\
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