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--- electrical_engineering_and_electronics_1:block22 [2025/12/13 22:57] – mexleadmin
+++ electrical_engineering_and_electronics_1:block22 [2025/12/14 23:36] (aktuell) – mexleadmin
@@ Zeile 4: / Zeile 4: @@
 <callout>
 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>
@@ Zeile 19: / Zeile 28: @@
 ===== 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 =====
 <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>
@@ Zeile 410: / Zeile 466: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== 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 =====
@@ Zeile 419: / 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>
 ===== Embedded resources =====