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--- electrical_engineering_and_electronics_1:block22 [2025/12/14 23:26] – mexleadmin
+++ electrical_engineering_and_electronics_1:block22 [2025/12/14 23:36] (aktuell) – mexleadmin
@@ Zeile 490: / Zeile 490: @@
 {{page>uebung_3.5.4&nofooter}}
 {{page>uebung_3.5.5&nofooter}}
+<panel type="info" title="Task 22.1 Voltage follower as impedance converter"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A voltage follower is built with an ideal op-amp. \\
+The input is a voltage source $U_{\rm I}=2.0~\rm V$ with internal resistance $R_{\rm S}=10~\rm k\Omega$.   \\
+The output drives a load resistor $R_{\rm L}$ which is varied between $100~\Omega$ and $100~\rm k\Omega$.
+  - Determine the input current drawn from the source for $R_{\rm L}=100~\Omega$ and for $R_{\rm L}=100~\rm k\Omega$.
+  - Explain briefly why the load does not “pull down” the source voltage in this circuit.
+<button size="xs" type="link" collapse="Loesung_22_1_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_1_Tipps" collapsed="true">
+  * The input voltage source sees (ideally) infinite input resistance.
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_1_Endergebnis" collapsed="true">
+  * Input current from the source: $I_{\rm S}\approx 0$ for both load values (ideal model).
+</collapse>
+</WRAP></WRAP></panel>
+<panel type="info" title="Task 22.2 Non-inverting amplifier design"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A non-inverting amplifier should have a voltage gain of $A_{\rm V}=11$.
+  - Choose resistor values $R_1$ and $R_2$ in the kOhm-range.
+  - If $U_{\rm I}=0.25~\rm V$, compute $U_{\rm O}$ (ideal op-amp, no saturation).
+  - What happens to $U_{\rm O}$ if the op-amp supply rails are $\pm 2.5~\rm V$?
+<button size="xs" type="link" collapse="Loesung_22_2_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_2_Tipps" collapsed="true">
+  * Rearrange $1+\frac{R_1}{R_2}=11$ to a resistor ratio.
+  * Check the computed $U_{\rm O}$ against the supply rails (clipping).
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_2_Endergebnis" collapsed="true">
+  * One possible choice: $R_2=10~\rm k\Omega$, $R_1=100~\rm k\Omega$.
+  * Ideal: $U_{\rm O}=11\cdot 0.25~\rm V=2.75~\rm V$.
+  * With $\pm 2.5~\rm V$ rails: $U_{\rm O}$ clips near $+2.5~\rm V$ (model-dependent headroom).
+</collapse>
+</WRAP></WRAP></panel>
+<panel type="info" title="Task 22.3 Inverting amplifier and virtual ground"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+An inverting amplifier is built with $R_1=2.2~\rm k\Omega$ and $R_2=22~\rm k\Omega$.
+The non-inverting input is connected to ground.
+  - Compute the closed-loop gain $A_{\rm V}$.
+  - For an input $U_{\rm I}(t)=0.30~\rm V~\sin(2\pi\cdot 1\,{\rm kHz}\cdot t)$, determine $U_{\rm O}(t)$.
+  - State the potential at the inverting input node (the summing node) in the ideal negative-feedback case.
+<button size="xs" type="link" collapse="Loesung_22_3_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_3_Tipps" collapsed="true">
+  * Use $A_{\rm V}=-\frac{R_2}{R_1}$ for the inverting amplifier.
+  * Virtual ground means $U_{\rm m}\approx U_{\rm p}=0~\rm V$ (not a physical short).
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_3_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_3_Endergebnis" collapsed="true">
+  * $A_{\rm V}=-\frac{22~\rm k\Omega}{2.2~\rm k\Omega}=-10$.
+  * $U_{\rm O}(t)=-3.0~\rm V~\sin(2\pi\cdot 1\,{\rm kHz}\cdot t)$.
+  * Summing node potential: approximately $0~\rm V$ (virtual ground).
+</collapse>
+</WRAP></WRAP></panel>
+<panel type="info" title="Task 22.4 Inverting summing amplifier via superposition"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+An inverting summing amplifier has $R_0=10~\rm k\Omega$, $R_1=10~\rm k\Omega$, and $R_2=20~\rm k\Omega$.
+Two inputs are applied: $U_{\rm I1}=+1.0~\rm V$ and $U_{\rm I2}=-0.5~\rm V$.
+  - Use superposition to compute $U_{\rm O}$.
+  - Compute the same result by writing the sum directly as a weighted sum.
+  - Explain briefly why the resistor between the op-amp inputs carries (ideally) no current.
+<button size="xs" type="link" collapse="Loesung_22_4_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_4_Tipps" collapsed="true">
+  * For each input alone: treat the circuit as an inverting amplifier with gain $-\frac{R_0}{R_i}$.
+  * Superposition: set the other voltage source to $0$ (replace by a short).
+  * With feedback: $U_{\rm D}\rightarrow 0$ implies negligible current through a resistor between inputs.
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_4_Endergebnis" collapsed="true">
+  * $U_{\rm O(1)}=-\frac{R_0}{R_1}U_{\rm I1}=-\frac{10}{10}\cdot 1.0~\rm V=-1.0~\rm V$.
+  * $U_{\rm O(2)}=-\frac{R_0}{R_2}U_{\rm I2}=-\frac{10}{20}\cdot(-0.5~\rm V)=+0.25~\rm V$.
+  * $U_{\rm O}=U_{\rm O(1)}+U_{\rm O(2)}=-0.75~\rm V$.
+</collapse>
+</WRAP></WRAP></panel>
+<panel type="info" title="Task 22.5 Current-to-voltage converter (transimpedance amplifier)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A photodiode is modeled as an ideal current source delivering $I_{\rm I}$ into a transimpedance amplifier.  \\
+The feedback resistor is $R_1=220~\rm k\Omega$ and the non-inverting input is grounded.
+  - Determine the transfer resistance $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm I}}$.
+  - Compute $U_{\rm O}$ for $I_{\rm I}=+2.0~\rm \mu A$.
+  - What sign does $U_{\rm O}$ have for a positive $I_{\rm I}$ (according to the circuit convention)?
+<button size="xs" type="link" collapse="Loesung_22_5_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_5_Tipps" collapsed="true">
+  * For the current-to-voltage converter in the given convention: $U_{\rm O}=-R_1 I_{\rm I}$.
+  * Keep units consistent: $\rm \mu A$ and $\rm k\Omega$.
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_5_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_5_Endergebnis" collapsed="true">
+  * $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm I}}=-R_1=-220~\rm k\Omega$.
+  * $U_{\rm O}=-(220~\rm k\Omega)\cdot (2.0~\rm \mu A)=-0.44~\rm V$.
+  * For $I_{\rm I}>0$: $U_{\rm O}<0$ (with this sign convention).
+</collapse>
+</WRAP></WRAP></panel>
+<panel type="info" title="Task 22.6 Voltage-to-current converter (transconductance)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A voltage-to-current converter should generate an output current proportional to an input voltage, with transfer conductance
+\[
+S=2.0~\rm mA/V.
+\]
+Assume the circuit uses a single resistor $R$ to set the current (ideal op-amp behavior), such that approximately $I_{\rm O}\approx \frac{U_{\rm I}}{R}$.
+  - Determine $R$ for the desired $S$.
+  - Compute $I_{\rm O}$ for $U_{\rm I}=0.6~\rm V$.
+  - Briefly name one application of such a circuit.
+<button size="xs" type="link" collapse="Loesung_22_6_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_6_Tipps" collapsed="true">
+  * If $I_{\rm O}\approx \frac{U_{\rm I}}{R}$, then $S\approx \frac{1}{R}$.
+  * Convert $2.0~\rm mA/V$ into $\rm A/V$ before inverting.
+</collapse>
+<button size="xs" type="link" collapse="Loesung_22_6_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_6_Endergebnis" collapsed="true">
+  * $S=2.0~\rm mA/V=2.0\times 10^{-3}~\rm A/V \Rightarrow R=\frac{1}{S}=500~\Omega$.
+  * $I_{\rm O}=S\,U_{\rm I}=(2.0~\rm mA/V)\cdot 0.6~\rm V=1.2~\rm mA$.
+  * Example application: voltage-controlled current source (e.g. driving an actuator/LED current or biasing a sensor).
+</collapse>
+</WRAP></WRAP></panel>
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