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electrical_engineering_and_electronics_1:block22 [2025/12/13 22:26] – mexleadminelectrical_engineering_and_electronics_1:block22 [2025/12/14 23:36] (aktuell) – mexleadmin
Zeile 4: Zeile 4:
 <callout> <callout>
 After this 90-minute block, you can After this 90-minute block, you can
-  - ... apply the superposition method to operational amplifier circuits. +  * explain why negative feedback “tames” an op-amp with very large open-loop gain $A_{\rm D}$, and use the ideal assumptions ($A_{\rm D}\rightarrow\infty$, $R_{\rm D}\rightarrow\infty$, $R_{\rm O}\rightarrow 0$) to analyze circuits. 
-  - ... know the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like. +  * apply a systematic equation-based workflow (goal $\rightarrow$ variables $\rightarrow$ equations $\rightarrow$ solving) to derive transfer functions such as $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}$ for basic feedback circuits. 
-  - ... name applications for the summing inverter, voltage-to-current converter, and current-to-voltage converter.+  * recognize and use the concept of **virtual ground** (more generally: $U_{\rm D}\rightarrow 0$) at the inverting input in negative-feedback circuits. 
 +  * apply the **superposition method** to linear op-amp circuits with multiple independent sources (e.g. the inverting summing amplifier) and compute $U_{\rm O}(U_{\rm I1},U_{\rm I2},\dots)$. 
 +  * identify the circuit topologies and transfer functions of 
 +      - the voltage follower ($A_{\rm V}=1$), 
 +      - the non-inverting amplifier ($A_{\rm V}=1+\frac{R_1}{R_2}$), 
 +      - the inverting amplifier ($A_{\rm V}=-\frac{R_2}{R_1}$), 
 +      - the inverting summing amplifier ($U_{\rm O}=-\sum_i \frac{R_0}{R_i}U_{{\rm I}i}$), 
 +      - the current-to-voltage converter (transimpedance), $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm in}}=-R_1$, 
 +      - the voltage-to-current converter (transconductance), $S=\frac{I_{\rm out}}{U_{\rm in}}$. 
 +  * name typical applications of these circuits (buffer/impedance converter, programmable gain amplifier, summing/mixing, differential measurement, current sensing, photodiode readout, voltage-controlled current source).
 </callout> </callout>
  
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 ===== 90-minute plan ===== ===== 90-minute plan =====
-  - Warm-up (x min):  +  - Warm-up (10 min): 
-    - ....  +    - Quick recall: ideal op-amp model and “golden rules” in negative feedback: \\ $I_{\rm p}\approx 0$, $I_{\rm m}\approx 0$, and (with feedback) $U_{\rm D}=U_{\rm p}-U_{\rm m}\rightarrow 0$. 
-  - Core concepts & derivations (x min): +    - One-minute concept check: why a voltage follower is useful although $A_{\rm V}=1$ (buffering / impedance conversion). 
-    - ... +  - Core concepts & derivations (55 min): 
-  - Practice (x min): ... +    - Introductory motivation: current sensing (10 min) 
-  - Wrap-up (x min): Summary box; common pitfalls checklist.+      - Why a shunt resistor alone may be insufficient; need for current-to-voltage conversion plus amplification. 
 +      - Link to the idea of “transimpedance” as gain: $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm in}}$. 
 +    - Voltage follower (10 min) 
 +      - Circuit idea: $U_{\rm O}$ fed back directly to the inverting input. 
 +      - Derive $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}=1$ using $U_{\rm D}\rightarrow 0$. 
 +      - Interpretation: decouples input from load (high input resistance, low output resistance) → impedance converter / buffer. 
 + 
 +    - Non-inverting amplifier (10 min) 
 +      - Feedback factor from divider: $k=\frac{R_2}{R_1+R_2}$. 
 +      - Transfer function: $A_{\rm V}=\frac{1}{k}=1+\frac{R_1}{R_2}$. 
 +      - Design notes: practical resistor range, loading, bias currents (qualitative). 
 + 
 +    - Inverting amplifier + virtual ground (15 min) 
 +      - Virtual ground at node $\rm K1$ (inverting input): in ideal feedback $U_{\rm D}\rightarrow 0$, so node is held near ground without a physical short. 
 +      - Derive $A_{\rm V}=-\frac{R_2}{R_1}$ (via divider approach or equal currents through $R_1$ and $R_2$). 
 +      - Input resistance insight: $R_{\rm I}^0\approx R_1$ (ideal), and output resistance reduced by feedback. 
 + 
 +    - Superposition with the inverting summing amplifier (10 min) 
 +      - Linearity check: resistors + (ideal) op-amp behavior under negative feedback. 
 +      - Apply superposition: analyze each source separately (others set to zero), then sum: $ U_{\rm O}=-\left(\frac{R_0}{R_1}U_{\rm I1}+\frac{R_0}{R_2}U_{\rm I2}+\cdots\right). $ 
 +      - Application link: mixers, weighted sums, DAC concept. 
 + 
 +  - Practice (20 min): 
 +    - Mini-problems (individually → pair check): 
 +      - Choose $R_1,R_2$ for a non-inverting amplifier with target gain $A_{\rm V}$. 
 +      - Inverting amplifier: given $R_1,R_2$, compute $U_{\rm O}(t)$ for a supplied $U_{\rm I}(t)$; mark phase inversion. 
 +      - Summing amplifier: compute $U_{\rm O}$ for two sources; verify via superposition. 
 +      - Transimpedance amplifier: compute $U_{\rm O}$ for a given $I_{\rm in}$ and $R_1$; interpret sign. 
 +      - Transconductance circuit: given $U_{\rm in}$, estimate $I_{\rm out}$ (qualitative if full derivation is later). 
 + 
 +  - Wrap-up (5 min): 
 +    - Summary box: “Feedback enforces $U_{\rm D}\rightarrow 0$; resistors set the gain.” 
 +    - Common pitfalls checklist (see below). 
 +    - Outlook: differential amplifier as subtraction / common-mode rejection; application circuits (PGA, instrumentation concepts). 
  
 ===== Conceptual overview ===== ===== Conceptual overview =====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
-  - ...+  * Negative feedback turns a very large (and imperfect) op-amp gain $A_{\rm D}$ into predictable closed-loop behavior: the circuit “chooses” $U_{\rm O}$ so that the differential input voltage $U_{\rm D}=U_{\rm p}-U_{\rm m}$ becomes (almost) zero. 
 +  * In the ideal feedback limit ($A_{\rm D}\rightarrow\infty$) we can treat 
 +      - $I_{\rm p}=I_{\rm m}=0$ (no input currents), 
 +      - $U_{\rm p}\approx U_{\rm m}$ (virtual short between inputs), 
 +      - and the output as an ideal voltage source ($R_{\rm O}\approx 0$). 
 +  * The voltage follower is the simplest example: $A_{\rm V}=1$ but with a huge practical effect—high input impedance and low output impedance (buffering). 
 +  * Resistors in the feedback path set **ratios**, and these ratios define gains: 
 +      - non-inverting: $A_{\rm V}=1+\frac{R_1}{R_2}$, 
 +      - inverting: $A_{\rm V}=-\frac{R_2}{R_1}$ (with virtual ground at the summing node). 
 +  * Multi-source op-amp circuits remain linear, so superposition works: a summing amplifier is just a controlled way of forming weighted sums. 
 +  * Negative-feedback op-amp circuits are not only “voltage amplifiers”: they also realize signal conversions: 
 +      - current $\rightarrow$ voltage (transimpedance, $\frac{U_{\rm O}}{I_{\rm in}}$), 
 +      - voltage $\rightarrow$ current (transconductance, $\frac{I_{\rm out}}{U_{\rm in}}$), 
 +    which underpins current sensing, photodiode readout, and controlled current sources.
 </callout> </callout>
  
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 This "trick" via $A_{\rm V}=\frac{1}{k}$ is no longer possible for some of the following circuits. Accordingly, a possible solution via network analysis is to be derived here as well. This "trick" via $A_{\rm V}=\frac{1}{k}$ is no longer possible for some of the following circuits. Accordingly, a possible solution via network analysis is to be derived here as well.
- 
-{{page>uebung_3.4.1&nofooter}} 
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 From the [[3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]] another circuit can be derived, which can be seen in <imgref pic1>. Here, both the green part of the circuit and the purple part correspond to an inverting amplifier. From the [[3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]] another circuit can be derived, which can be seen in <imgref pic1>. Here, both the green part of the circuit and the purple part correspond to an inverting amplifier.
  
-How can $U_\rm O$ be calculated in this circuit? To do this, it is first important to understand what is being sought (compare [[3_opamp_basic_circuits_i#steps_to_the_goal|steps to the goal]]). The goal is to find the relationship between output and input signals: $U_{\rm O}(U_{\rm I1}, U_{\rm I2})$. Different ways to get there were explained in [[electrical_engineering_1:network_analysis|Electrical engineering 1: Network analysis]]. Here we will outline a different way.+How can $U_\rm O$ be calculated in this circuit? To do this, it is first important to understand what is being sought. The goal is to find the relationship between output and input signals: $U_{\rm O}(U_{\rm I1}, U_{\rm I2})$. Different ways to get there, were explained in [[Block07]] and [[Block08]]. Here we will outline a different way.
  
 In the case of a circuit with several sources, superposition is a suitable method, in particular the superposition of the effect of all sources in the circuit. For superposition, it must be ensured that the system behaves linearly. The circuit consists of ohmic resistors and the operational amplifier. These two components give twice the output value when the input value is doubled - they behave linearly. For superposition, the effect of the two visible voltage sources $U_{\rm I1}$ and $U_{\rm I2}$ must be analyzed in the present circuit. \\ In the case of a circuit with several sources, superposition is a suitable method, in particular the superposition of the effect of all sources in the circuit. For superposition, it must be ensured that the system behaves linearly. The circuit consists of ohmic resistors and the operational amplifier. These two components give twice the output value when the input value is doubled - they behave linearly. For superposition, the effect of the two visible voltage sources $U_{\rm I1}$ and $U_{\rm I2}$ must be analyzed in the present circuit. \\
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 </WRAP> </WRAP>
  
-The **Inverting Summing Amplifier** (also called: Summing Amplifier or Voltage Adder) can be extended to any number of inputs. The simulation above shows the superposition of several inputs. Depending on the resistances at the different inputs, a different current flows into the circuit.+The **Inverting Summing Amplifier** (also called: Summing Amplifier or Voltage Adder) can be extended to any number of inputs. \\ The simulation above shows the superposition of several inputs. Depending on the resistances at the different inputs, a different current flows into the circuit.
  
 This circuit was used in analog [[https://en.wikipedia.org/wiki/Electronic_mixer|audio mixers]]. This allows a combination of several signals with different gains (by the input resistors $R_i$ with $i=1, ..., n$). Furthermore, the overall gain can be changed by $R_0$. A big advantage of this circuit is also that the summation at node $\rm K1$ is done on potential $U_\rm D$. This means that capacitive interference concerning the ground potential (and therefore the case) is virtually non-existent. This circuit was used in analog [[https://en.wikipedia.org/wiki/Electronic_mixer|audio mixers]]. This allows a combination of several signals with different gains (by the input resistors $R_i$ with $i=1, ..., n$). Furthermore, the overall gain can be changed by $R_0$. A big advantage of this circuit is also that the summation at node $\rm K1$ is done on potential $U_\rm D$. This means that capacitive interference concerning the ground potential (and therefore the case) is virtually non-existent.
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 The result is: \\ The result is: \\
 $       U_{\rm O} = U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - R_3 \cdot \frac{U_{\rm I1} - U_{\rm p}}{R_1} $ \\ $       U_{\rm O} = U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - R_3 \cdot \frac{U_{\rm I1} - U_{\rm p}}{R_1} $ \\
-$       U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - U_{\rm I1} \cdot \frac{R_3}{R_1} +  U_{\rm I2} \cdot (\frac{R_3}{R_1}\cdot \frac{R_4}{R_2 + R_4})$+$       U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - U_{\rm I1} \cdot \frac{R_3}{R_1} +  U_{\rm I2} \cdot (\frac{R_3}{R_1}\cdot \frac{R_4}{R_2 + R_4})$ \\ 
 $\boxed{U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4}  \frac{R_1 + R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1}}$ $\boxed{U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4}  \frac{R_1 + R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1}}$
  
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 Two simplifications should be considered here: Two simplifications should be considered here:
-  - If $R_1 = R_2$ and $R_3 = R_4$ are chosen, the equation further simplifies to:  \\ (nbsp) $\boxed{U_{\rm O} = U_{\rm I2}\cdot \frac{R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1} = \frac{R_3}{R_1}\cdot(U_{\rm I2}-U_{\rm I1})}$. \\ This variant can be found in various measurement circuits. \\ \\+  - If $R_1 = R_2$ and $R_3 = R_4$ are chosen, the equation further simplifies to:  \\ (nbsp) $\boxed{U_{\rm O} = \frac{R_3}{R_1}\cdot(U_{\rm I2}-U_{\rm I1})}$. \\ This variant can be found in various measurement circuits. \\ \\
   - Alternatively, if $R_1 = R_3$ and $R_2 = R_4$ is chosen, the result is: \\ (nbsp) $\boxed{U_{\rm O}= U_{\rm I2}-U_{\rm I1}}$ \\ This would also result in case 1. if $R_1 = R_2 = R_3 = R_4$ is chosen.   - Alternatively, if $R_1 = R_3$ and $R_2 = R_4$ is chosen, the result is: \\ (nbsp) $\boxed{U_{\rm O}= U_{\rm I2}-U_{\rm I1}}$ \\ This would also result in case 1. if $R_1 = R_2 = R_3 = R_4$ is chosen.
  
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
-===== 4.3 Instrumentation Amplifier ===== 
  
-This type of amplifier shall be analyzed in the following exercise. \\ \\ +==== Current-Voltage-Converter ====
-An example of a pure instrumentation amplifier is the components {{circuit_design:ina818.pdf}}.  +
- +
-Often instrumentation amplifiers are used in programmable gain amplifiers (PGA). PGAs allow the manipulation of the amplification factor by a digital interface.  +
- +
- +
-<WRAP><panel type="default">  +
-<imgcaption pic3| Instrumentation Amplifier> +
-</imgcaption> +
-\\ {{drawio>Instrumentenverstärker.svg}} +
-</panel></WRAP> +
- +
- +
-~~PAGEBREAK~~ ~~CLEARFIX~~ +
- +
-{{page>uebung_4.3.1&nofooter}} +
- +
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-gKkH56heEBsZ4R74a7e3nHX88eVPx9Pc9DCvG9j1fS9qFjIh32gdcuG8Wx0kXIYMCPE8zwvWMIBvQxr3A2NUOggAPYQ-D0YQSFYBZJAomM+BI1V5lYWpQUGUxBzokiVXXahmEBP9WNo9p6KKP4imgZj9HEgYhJEgFyWHTRUio2SuKUsl2LVMkoCcLBEy4wwklBHBkCNeZVK3djFVIq4cDSdjTGEkiHDkZgW2s3SQAASX4EioRAX4WCgCAZI4gBVESyCKKh1k3ByQAi5yUQ4jAeIoGiwCwLycD4IA noborder}} +
-</WRAP> +
- +
-~~PAGEBREAK~~ ~~CLEARFIX~~ +
-===== 4.4 Current-Voltage-Converter =====+
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-==== Application - Programmable Gain Amplifier ===+==== Applications ==== 
 +=== Programmable Gain Amplifier ===
  
 Often in applications an analog signal is too small to process (e.g. to digitalize it afterward). \\ Often in applications an analog signal is too small to process (e.g. to digitalize it afterward). \\
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 </WRAP> </WRAP>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Common pitfalls ===== ===== Common pitfalls =====
-  * ...+  * Mixing up open-loop and closed-loop gain: 
 +      - open-loop: $U_{\rm O}=A_{\rm D}\,U_{\rm D}$, 
 +      - closed-loop: $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}$ is set mainly by resistor ratios. 
 +  * Forgetting the conditions for the “golden rules”: $U_{\rm p}\approx U_{\rm m}$ only holds when the op-amp is in negative feedback and not saturated. 
 +  * Sign errors: 
 +      - inverting amplifier has a minus sign $A_{\rm V}=-\frac{R_2}{R_1}$, 
 +      - transimpedance example gives $U_{\rm O}=-R_1 I_{\rm in}$ (direction conventions matter). 
 +  * Misusing “virtual ground”: 
 +      - the inverting input node can be near $0~\rm V$, but it is **not** physically connected to ground and cannot source/sink arbitrary current. 
 +  * Superposition mistakes: 
 +      - when “turning off” a voltage source, replace it by a short; when “turning off” a current source, replace it by an open. 
 +      - ensure linear operation (no clipping, no saturation). 
 +  * Ignoring practical limits: 
 +      - too small $R_1+R_2$ loads the op-amp output (current limit), 
 +      - too large resistors make bias currents and offsets more visible, 
 +      - finite supply rails limit $U_{\rm O}$ and can break the ideal assumptions.
  
 ===== Exercises ===== ===== Exercises =====
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 {{page>uebung_3.5.1&nofooter}} {{page>uebung_3.5.1&nofooter}}
 {{page>uebung_3.5.2&nofooter}} {{page>uebung_3.5.2&nofooter}}
-{{page>uebung_3.5.3&nofooter}} 
 {{page>uebung_3.5.4&nofooter}} {{page>uebung_3.5.4&nofooter}}
 {{page>uebung_3.5.5&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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