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--- electrical_engineering_and_electronics_1:block22 [2025/12/14 23:26] – mexleadmin
+++ electrical_engineering_and_electronics_1:block22 [2026/01/10 10:01] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
 ====== Block 22 — Negative-feedback Op-Amp Circuits ======
-===== Learning objectives =====
+===== 22.0 Intro =====
+==== 22.0.1 Learning objectives ====
 <callout>
 After this 90-minute block, you can
@@ Zeile 18: / Zeile 20: @@
 </callout>
-===== Preparation at Home =====
+==== 22.0.2 Preparation at Home ====
 Well, again
@@ Zeile 27: / Zeile 29: @@
   * ...
-===== 90-minute plan =====
+==== 22.0.3 90-minute plan ====
   - 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) $U_{\rm D}=U_{\rm p}-U_{\rm m}\rightarrow 0$.
@@ Zeile 68: / Zeile 70: @@
     - Outlook: differential amplifier as subtraction / common-mode rejection; application circuits (PGA, instrumentation concepts).
+==== 22.0.4  Conceptual overview ====
-===== 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.
@@ Zeile 87: / Zeile 88: @@
 </callout>
-===== Core content =====
+===== 22.1 Core content =====
 <WRAP>
 <callout type="info" icon="true">
-==== Introductory Example ====
+==== 22.1.1 Introductory Example ====
 In various applications, currents must be measured. In an electric motor, for example, the torque is caused by the current flowing through the motor. A motor control and a simple overcurrent shutdown are based on the knowledge of the current. For further processing, a voltage must be generated from the current. The simplest current-to-voltage converter is the ohmic resistor. A sufficiently large voltage as required by a microcontroller, for example, cannot be achieved with this. So not only does the current have to be converted, but also the generated potential difference has to be amplified.
@@ Zeile 104: / Zeile 105: @@
 </callout></WRAP>
-==== Voltage follower ====
+==== 22.1.2 Voltage follower ====
@@ Zeile 156: / Zeile 157: @@
  </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Non-inverting amplifier ====
+==== 22.1.3 Non-inverting amplifier ====
 So far, the entire output voltage has been negative-feedback. Now only a part of the voltage is to be fed back. \\ To do this, the output voltage can be reduced using a voltage divider $R_1+R_2$. The circuit for this can be seen in <imgref pic5>.
@@ Zeile 213: / Zeile 214: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Inverting Amplifier ====
+==== 22.1.4 Inverting Amplifier ====
 The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). \\
@@ Zeile 304: / Zeile 305: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Inverting Summing Amplifier ====
+==== 22.1.5 Inverting Summing Amplifier ====
 <WRAP><panel type="default">
@@ Zeile 339: / Zeile 340: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Differential Amplifier / Subtractor ====
+==== 22.1.6 Differential Amplifier / Subtractor ====
 <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjA7CAMB00LAThat6AcNYGYJJwDZCIMkAWQjHSCQkAVgCZHHpGBTAWjDACgAhiCYYsTJuRDkcLcZKxgGIHkrAJ17LoVgNy5BhggSM0Qk0IMlWnXsLkk0GaQxMI0VRuh8A7sNHgwWX8maCwvX2lZCSkZAJYvACc-BUCYll548HU+JOR01JEUzLUEHJAZdgqpBHKQmCzSgHM02vZI8qJ6xOTW8CQWKvYSrwB5FrkpJHoJ9i8AN3ARXpNe9nS8VlnsJD4AJVrennpB8BBCdiUtuAYfHqrC4XJZ26r7-0Hb9qqnyrrwvoGdUCWA+TxwUjA0yejD05WgknYGAA+kwkdAkU8kMi1Gj4OoJOQwE8zAwkWAyXwweBYRI1kY4QjqSi0Ri1KiQrjyeTUUSkThbnk4osijA9gcMspIUKhmcLvUcLAiYp+hYcAYkPhwdc+AtgUKVhK1uANpcurAdlSiZI9a50v16pBmejMciOXByeieRA+TqQCtXoCtiwIDRNmadgsDalYobhCAQ2Aw7NzXwAEbKchYFbicq6UXpzoGnNqhF8AAeUmtGHovGmBikwnYAEkAHYAFw4CS4AB1uy2AGadjgtgDGHG7AGdey2APYtydtgAW46nfYAMgBLFsccsZxMhGuSJixI-sAAmG-7-dgk+nAGUN40WwIADZ8IA noborder}}
@@ Zeile 390: / Zeile 391: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Current-Voltage-Converter ====
+==== 22.1.7 Current-Voltage-Converter ====
 <WRAP><panel type="default">
@@ Zeile 418: / Zeile 419: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Voltage-to-Current Converter ====
+==== 22.1.8 Voltage-to-Current Converter ====
 <WRAP><panel type="default">
@@ Zeile 444: / Zeile 445: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Applications ====
+===== 22.2 Applications =====
-=== Programmable Gain Amplifier ===
+=== 22.2.1 Programmable Gain Amplifier ===
 Often in applications an analog signal is too small to process (e.g. to digitalize it afterward). \\
@@ Zeile 465: / Zeile 466: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-===== Common pitfalls =====
+===== 22.3 Common pitfalls =====
   * Mixing up open-loop and closed-loop gain:
       - open-loop: $U_{\rm O}=A_{\rm D}\,U_{\rm D}$,
@@ Zeile 483: / Zeile 484: @@
       - finite supply rails limit $U_{\rm O}$ and can break the ideal assumptions.
-===== Exercises =====
+===== 22.4 Exercises =====
 ==== Worked examples ====
@@ Zeile 490: / Zeile 491: @@
 {{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 =====