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electrical_engineering_and_electronics_1:block21 [2025/12/13 19:19] mexleadminelectrical_engineering_and_electronics_1:block21 [2025/12/14 22:26] (aktuell) mexleadmin
Zeile 4: Zeile 4:
 <callout> <callout>
 After this 90-minute block, you can After this 90-minute block, you can
-  * ...+  * explain what an operational amplifier (op-amp) is **as a black-box voltage amplifier** with two inputs (inverting / non-inverting) and one output. 
 +  * correctly label and use the voltages \(U_{\rm p}\), \(U_{\rm m}\) and the **differential voltage** \(U_{\rm D}\). 
 +  * state and apply the **basic equation** of the (idealized) op-amp. 
 +  * state and use the **golden rules** (ideal op-amp model) 
 +  * distinguish **open-loop gain** \(A_{\rm D}=U_{\rm O}/U_{\rm D}\) from **closed-loop / circuit voltage gain** \(A_{\rm V}=U_{\rm O}/U_{\rm I}\). 
 +  * explain what **feedback** is and clearly differentiate **negative feedback** (stabilizing) from **positive feedback** (reinforcing / potentially unstable). 
 +  * describe key **non-ideal** limitations of real op-amps at the qualitative level (finite gain, finite input resistance & bias currents, limited output swing and output current, nonzero output resistance). 
 +  * explain the difference between **bipolar** and **unipolar** op-amp power supply and what this implies for the possible output voltage range.
 </callout> </callout>
  
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 ===== 90-minute plan ===== ===== 90-minute plan =====
-  - Warm-up (min):  +  - Warm-up (10 min): 
-    - ....  +    - Hook: audio amplifier clipping example (undistorted vs overdriven waveform/spectrum) → why “ideal amplification” is not automatic. 
-  - Core concepts & derivations (min): +    - Recall: what does “amplify a voltage” mean? What would an ideal voltage amplifier look like (voltmeter at input, voltage source at output)? 
-    - ... +  - Core concepts & derivations (55–60 min): 
-  - Practice (min): ... +    - Op-amp as a black box + symbols (10–15 min) 
-  - Wrap-up (min): Summary box; common pitfalls checklist.+      - Triangle symbol(s), inverting/non-inverting inputs, output, supply rails. 
 +      - Differential voltage definition \(U_{\rm D}=U_{\rm p}-U_{\rm m}\). 
 +    - Ideal op-amp model (15 min) 
 +      - Basic equation \(U_{\rm O}=A_{\rm D}U_{\rm D}\). 
 +      - Golden rules; interpret each rule physically (input ≈ voltmeter, output ≈ ideal source). 
 +    - Real op-amp limits (10–15 min) 
 +      - Output saturation (rails / headroom), finite \(A_{\rm D}\), small input currents, limited output current. 
 +      - Unipolar vs bipolar supply: output range and operating point. 
 +    - Feedback concept (15 min) 
 +      - Meaning of feedback; block diagram vs circuit diagram. 
 +      - Sign convention: positive vs negative feedback. 
 +      - Big idea: with negative feedback and large \(A_{\rm D}\), the **closed-loop gain** becomes mostly set by the feedback network (introduce \(k\) and the result \(A_{\rm V}\approx 1/k\) as the motivating target; details can be finished in later blocks if needed)
 +  - Practice (15–20 min): 
 +    - Quick symbol + sign drills: identify \(U_{\rm p}\), \(U_{\rm m}\), \(U_{\rm D}\), and predict the direction of \(U_{\rm O}\) change. 
 +    - “Golden rules” micro-exercises: 
 +      - Decide when you may set \(U_{\rm p}\approx U_{\rm m}\) and \(I_{\rm p}\approx I_{\rm m}\approx 0\). 
 +    - Feedback classification: 
 +      - Given a block diagram with \(kU_{\rm O}\) fed back, classify as positive/negative feedback and state the qualitative consequence (stabilize vs runaway/oscillate)
 +  - Wrap-up (min): 
 +    - Summary box: basic equation, golden rules, open-loop vs closed-loop gain, feedback sign. 
 +    - Common pitfalls checklist (below).
  
 ===== Conceptual overview ===== ===== Conceptual overview =====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
-  - ...+  - Think of an op-amp as a **differential voltage sensor + powerful output stage**: 
 +      - it measures the difference \(U_{\rm D}=U_{\rm p}-U_{\rm m}\), 
 +      - then tries to produce \(U_{\rm O}=A_{\rm D}U_{\rm D}\). 
 + 
 +  - The “magic” of op-amp circuits comes from **negative feedback**: 
 +      - with large \(A_{\rm D}\), the circuit forces \(U_{\rm D}\) to be (almost) zero in normal operation, 
 +      - so you can treat \(U_{\rm p}\approx U_{\rm m}\) and \(I_{\rm p}\approx I_{\rm m}\approx 0\) as powerful design rules, 
 +      - and the **external feedback network** determines the closed-loop behavior (gain, impedance, linearity). 
 + 
 +  - Open-loop vs closed-loop is the key separation: 
 +      - **open-loop gain** \(A_{\rm D}\) is huge but poorly controlled, 
 +      - **closed-loop gain** \(A_{\rm V}\) is what we design to be stable, predictable, and useful. 
 + 
 +  - Reality check: 
 +      - real op-amps are limited by supply rails, maximum output current, finite speed, and nonzero input/output resistances. 
 +      - choosing unipolar vs bipolar supply changes what “zero” and “negative output” even mean in the circuit.
 </callout> </callout>
  
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   * **Input resistance**: \\ The input side is drawn in the figure. \\ The resistance given here is the differential resistance $R_{\rm D}$ where the differential voltage $U_{\rm D}$ drops. \\ Therefore, $R_{\rm D} \rightarrow \infty$ is valid. Thus, the input currents $I_\rm p \rightarrow 0$ and $I_\rm m \rightarrow 0$. \\ \\   * **Input resistance**: \\ The input side is drawn in the figure. \\ The resistance given here is the differential resistance $R_{\rm D}$ where the differential voltage $U_{\rm D}$ drops. \\ Therefore, $R_{\rm D} \rightarrow \infty$ is valid. Thus, the input currents $I_\rm p \rightarrow 0$ and $I_\rm m \rightarrow 0$. \\ \\
   * **Output resistance**, **Basic equation**: \\ On the output side, $R_\rm O=0$ gives that $U_{\rm O} = A_{\rm D}\cdot U_\rm D$. \\ This is the basic equation of the amplifier circuit. \\ Ideally, the op-amp amplifies linearly, as indicated in the equation. Specifically, for a differential voltage of $0 ~\rm V$, the output voltage is $0 ~\rm V$. \\ \\   * **Output resistance**, **Basic equation**: \\ On the output side, $R_\rm O=0$ gives that $U_{\rm O} = A_{\rm D}\cdot U_\rm D$. \\ This is the basic equation of the amplifier circuit. \\ Ideally, the op-amp amplifies linearly, as indicated in the equation. Specifically, for a differential voltage of $0 ~\rm V$, the output voltage is $0 ~\rm V$. \\ \\
-  * **Voltage Gain**: \\ $A_\rm D$ is very large. \\ Ideally, the following applies: $A_ \rm D \rightarrow \infty$+  * **Differential Gain**: \\ $A_\rm D$ is very large. \\ Ideally, the following applies: $A_ \rm D \rightarrow \infty$
  
 <WRAP><panel type="default"> <imgcaption picA|Operational amplifier equivalent circuit> </imgcaption> {{drawio>OPV_Ersatzschaltbild.svg}} </panel></WRAP> <WRAP><panel type="default"> <imgcaption picA|Operational amplifier equivalent circuit> </imgcaption> {{drawio>OPV_Ersatzschaltbild.svg}} </panel></WRAP>
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 <WRAP column 100%> <panel type="danger" title="Remember: basic equation and golden rules"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> <WRAP column 100%> <panel type="danger" title="Remember: basic equation and golden rules"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
  
-  - The output voltage depends on the differential voltage via the differential gain: $U_{\rm O} = A_{\rm D} \cdot U_\rm D$ This is the **basic equation**  of the amplifier circuit.+  - The output voltage depends on the differential voltage via the differential gain: $U_{\rm O} = A_{\rm D} \cdot U_\rm D$. \\ This is the **basic equation**  of the amplifier circuit.
   - The **golden rules** of the ideal amplifier are:   - The **golden rules** of the ideal amplifier are:
       - The differential gain goes to infinity: $A_\rm D \rightarrow \infty$       - The differential gain goes to infinity: $A_\rm D \rightarrow \infty$
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 </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~ </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-In <imgref pic001> a simulation of an ideal amplifier is shown. The input source specifies the voltage to be amplified. The amplifier with an amplification factor of 100 has the connections for input and output voltage are drawn in. On the right side, a resistor is provided as load; this can be varied via a switch. \\ \\+In <imgref pic001> a simulation of an ideal amplifier is shown. The input source specifies the voltage to be amplified. \\ The idealized amplifier with an differential gain of 400'000 has the connections for input and output voltage are drawn in. On the right side, a resistor is provided as load; this can be varied via a switch. \\ \\
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
 <WRAP><well> <WRAP><well>
 <imgcaption pic001|Ideal Amplifier></imgcaption> \\ <imgcaption pic001|Ideal Amplifier></imgcaption> \\
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 </well></WRAP> </well></WRAP>
  
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   - Ideally, no current flows into the amplifier on the input side.   - Ideally, no current flows into the amplifier on the input side.
   - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. The amplifier thus tries to maintain the desired voltage.   - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. The amplifier thus tries to maintain the desired voltage.
-  - On the output side of the amplifier, the current can flow in either direction. \\ The amplifier adjusts the current so that the amplified voltage $U_A=\pm 2.5~\rm V$ can be measured at the output.+  - On the output side of the amplifier, the current can flow in either direction. \\ The amplifier adjusts the current so that the amplified voltage $U_A=\pm 4.0~\rm V$ can be measured at the output.
 \\  \\ 
  
Zeile 156: Zeile 198:
       * The output voltage can only follow the input voltage as far as the power supply allows. \\ In real operational amplifiers, only so-called **rail-to-rail** operational amplifiers can exploit the range down to a few $100 ~\rm mV$ to $U_\rm S$. \\ Other operational amplifiers have an **output limit**, which is $1 ... 2 ~\rm V$ below the supply voltage.       * The output voltage can only follow the input voltage as far as the power supply allows. \\ In real operational amplifiers, only so-called **rail-to-rail** operational amplifiers can exploit the range down to a few $100 ~\rm mV$ to $U_\rm S$. \\ Other operational amplifiers have an **output limit**, which is $1 ... 2 ~\rm V$ below the supply voltage.
       * If the supply voltages are not symmetrical ($U_{\rm sm} \neq -U_{\rm sp}$), then the characteristic also shifts.       * If the supply voltages are not symmetrical ($U_{\rm sm} \neq -U_{\rm sp}$), then the characteristic also shifts.
-      * The ideal operational amplifier produces the same output voltage $U_{\rm O}=A_{\rm D} \cdot U_{\rm D}$ as long as $U_{\rm D} = U_{\rm p} - U_{\rm m}$ is the same.\\  For the real operational amplifier with fixed $A_{\rm D}$, output voltage $U_{\rm O1}$ for $U_{\rm D1}=5 ~\rm V - 4.9 ~V$ is different from $U_{\rm O2}$ for $U_{\rm D1}=0.1 ~\rm V - 0 ~V$. \\ \\+      * The ideal operational amplifier produces the same output voltage $U_{\rm O}=A_{\rm D} \cdot U_{\rm D}$ as long as $U_{\rm D} = U_{\rm p} - U_{\rm m}$ is the same.\\  For the real operational amplifier with fixed $A_{\rm D}$, output voltage $U_{\rm O1}$ for $U_{\rm D1}=5 ~\rm V - 4.9 ~V$ is different from $U_{\rm O2}$ for $U_{\rm D2}=0.1 ~\rm V - 0 ~V$. \\ \\
   * Differential gain  $\boldsymbol{A_\rm D}$: \\ The differential gain is usually between $A_\rm D = 20'000 ... 400'000$. \\ \\    * Differential gain  $\boldsymbol{A_\rm D}$: \\ The differential gain is usually between $A_\rm D = 20'000 ... 400'000$. \\ \\ 
   * Input resistance $\boldsymbol{R_\rm D}$: \\ For real operational amplifiers, the input resistance $R_\rm I > 1 M\Omega$ and the input current $|I_\rm p|$ or $|I_ \rm m|$ is less than $1 ~\rm µA$. \\ \\   * Input resistance $\boldsymbol{R_\rm D}$: \\ For real operational amplifiers, the input resistance $R_\rm I > 1 M\Omega$ and the input current $|I_\rm p|$ or $|I_ \rm m|$ is less than $1 ~\rm µA$. \\ \\
Zeile 162: Zeile 204:
  
  
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-The simulation shows a **(simulated) real amplifier**. The input source has a high internal resistance. This means it has a high impedance and can only supply a small amount of current. The amplifier with a gain of 100 has - besides the connections for input and output voltage - also connections for the supply voltage drawn in. On the right side, a resistor is provided as load; this can be varied via a switch.+The simulation shows a **(simulated) real amplifier**. The input source has a high internal resistance. This means it has a high impedance and can only supply a small amount of current. \\ The amplifier with a differential gain of 400'000 has - besides the connections for input and output voltage - also connections for the supply voltage drawn in. \\ On the right side, a resistor is provided as load; this can be varied via a switch.
  ~~PAGEBREAK~~ ~~CLEARFIX~~  ~~PAGEBREAK~~ ~~CLEARFIX~~
  
 In the simulation some properties of an amplifier can be seen: In the simulation some properties of an amplifier can be seen:
   - On the input side, a small current flows into the amplifier.   - On the input side, a small current flows into the amplifier.
-  - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. The amplifier thus tries to maintain the desired voltage.+  - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. \\ The amplifier thus tries to maintain the desired voltage.
   - The amplifier can output current as well as absorb current. \\ The current on the output side flows in and out of the amplifier through the supply voltage connections.   - The amplifier can output current as well as absorb current. \\ The current on the output side flows in and out of the amplifier through the supply voltage connections.
-  - The simulation is based on a real amplifier. This has a small deviation from the expected value $U_{\rm O}=\pm 2.5~\rm V$ at the output voltage. Part of the deviation will be described later in this chapter.+  - The simulation is based on a real amplifier. This has a small deviation from the expected value $U_{\rm O}=\pm 4.0~\rm V$ at the output voltage. 
  
 \\ \\ \\ \\
-=== Power supply of the operational amplifier ===+=== Voltage Supply of the Operational Amplifier === 
 + 
 +The op-amp needs an additional voltage supply to be able to actively output more power. \\ 
 +This two supplies are also called **rails**. 
 +In general, the rails are drawn on top and on below the triangular shape of the op-amp.
  
 For the voltage supply of the operational amplifier, a distinction is made between unipolar and bipolar: For the voltage supply of the operational amplifier, a distinction is made between unipolar and bipolar:
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-The op-amps in the simulation replicate real op-amps in some respects: The voltage gain is $A_\rm D = 100'000$. \\ The transfer characteristics $U_{\rm O}(U_\rm D)$ show proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{\rm sp}|=|U_{\rm sm}|=5 ~\rm V$ (not shown). \\ The modulation limits and the voltage gain can be changed in the simulation via "Edit component" (double click).+The op-amps in the simulation replicate real op-amps in some respects: The differential gain is $A_\rm D = 100'000$. \\ The transfer characteristics $U_{\rm O}(U_\rm D)$ show proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{\rm sp}|=|U_{\rm sm}|=5 ~\rm V$ (not shown). \\ The modulation limits and the differential gain can be changed in the simulation via "Edit component" (double click).
  
 <WRAP><imgcaption pic10|unipolar and bipolar supply> </imgcaption> \\ <WRAP><imgcaption pic10|unipolar and bipolar supply> </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjA7CAMB00OgFgJxvRzA2GsUA4UwloUsBmcpAVmomp3KjGhGteoFMBaMMAKABM0KGRBJ84JEnGSZAVQD6AeQHDmjQSG4SQ5LQsUAHfgHcQ+VrrG7o-AE4hBELeSyTnB-THB2A5k4u4t6e4rYw-CTk4ChaxDL6rIKCMqzQiviK6QAegvgsYIKKLFmwYIq8ikWZRenp5KXQ1FlCIoGu7k4kep2GyoL8AEbayanaYG6+UHbZI+SsE9G85NTgkzJgkgCSAHZGAK4ALgA6AM4AygCWfjsAhgA2QzGrK1DkYKtIWKn8s2A2+ASwlkUA2kgAMgB7W4AEzOkJ2Z2URwOh1aGhA320uixhgAtvxbuAcLpLLIQCg2NpVixEAhcHSmCgIBAUKRaKNBEwfLTEOjxrELKxeFhWGT8fzeILvCLWN5DCYAG4CrRkqWuLQLcRpeCITD6zA+HXUSLQJYpBbScaFJwpHnNTI5PIFWq66AfQTUfAQQjQLCe0rlSrVRqIVAGg1YLLFZoNOBNFpRHqSMA4UUeO0LZosFLR3L5aAwopIPAR-U8VPFCsQCrlIpR+Nhsv6hvR9KCWDNGaY6BxJDRLDEKQk8CSeS-TEfcDUMCY5zTmRgkAAEQn4eJbDyMXEo5ASIEswklJw7FnlKX48PgPJqBkM53mxAijX+Gim0XKDf3ofkmfs3Yb5EGw7wxIuu7cNQABqa4MGwJI+mw567lBkRcnoEALMCg5xMCdjqBuSCptichERK5jivSNj0nhbREeQ146JI9FgUoJj4VRlqLvSfQDH8YgTBqCwrD+IBQrC8KIqcyKHKiTz-KsLxTnwmr8qEREWgRIBKKo-ABKEZKhGIaQOO0G76WkvimoBBhiIkbAQDgcrpI6ij5i6xR1GUtZVBkPn1PUjRdqhgEcMJdnUMJdhIGh-yzps0T6HFr4REmbhMdQWjpnBFkfIoSB5s6hZFCUcBBnWigNm6yDNtg0bzNGHZBbM3BYDg3CUOA9DaPMO52gAQpcRiQvctz2GckHDYctx+JwZznPsRhGPcACeE4tZS+hxN++ifF0Wk7INw2jeNk3TbNFwLUtq3RdE4S6D1UVoeEGW+ClT2SFYGzGTd5IoJ8khYKsj23ZIf2bhYQPBb9nwyPgkP-jgs74HIInHEYKH-im9L4IjZoiRB0GY+IL1w8Tu2Pmc0CE3BegRRY6UjhT6PQT9ujSKCi7csDN5IFAL39tMUNs7zN6QG9IO0wYCT+uL5IrFoL1uCprPpQYTH4CpRKkdjViUVSCx0tRVVMiybJ-RlKRctMvh0lD3AafF1oLMl3NMIusGDiSsHdl8cqChIrhg0uGNsG0GV88JEVWOBIcSHzwKoPHt27n+sjqx+6ugina6fno0Why8NAiZeocJS9EWuLQqNSbxk7nt+FCzqg0ePqusyivBx6PgHxcTq+coTBYkzvCOdpnPw9HTiAABis4WY2CA1YavAgAAsrc2QnfcU0zfwQA 1000,500 noborder}} +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00OgFgJxvRzA2GsUA4UwloUsBmcpAVmomp3KjGhGteoFMBaMMAKABM0KGRBJ84JEnGSZAVQD6AeQHDmjQSG4SQ5LQsUAHfgHcQ+VrrG7o-AE4hBELeSyTnB-THB2A5k4u4t6e4rYw-CTk4ChaxDL6rIKCMqzQiviK6QAegvgsYIKKLFmwYIq8ikWZRenp5KXQ1FlCIoGu7k4kep2GyoL8AEbayanaYG6+UHbZI+SsE9G85NTgkzJgkgCSAHZGAK4ALgA6AM4AygCWfjsAhgA2QzGrK1DkYKtIWKn8s2A2+ASwlkUA2kgAMgB7W4AEzOkJ2Z2URwOh1aGhA320uixhgAtvxbuAcLpLLIQCg2NpVixEAhcHSmCgIBAUKRaKNBEwfLTEOjxrELKxeFhWGT8fzeILvCLWN5DCYAG4CrRkqWuLQLcRpeCITD6zA+HXUSLQJYpBbScaFJwpHnNTI5PIFWq66AfQTUfAQQjQLCe0rlSrVRqIVAGg1YLLFZoNOBNFpRHqSMA4UUeO0LZosFLR3L5aAwopIPAR-U8VPFCsQCrlIpR+Nhsv6hvR9KCWDNGaY6BxJDRLDEKQk8CSeS-TEfcDUMCY5zTmRgkAAEQn4eJbDyMXEo5ASIEswklJw7FnlKX48PgPJqBkM53mxAijX+Gim0XKDf3ofkmfs3Yb5EGw7wxIuu7cNQABqa4MGwJI+mw567lBkRcnoEALMCg5xMCdjqBuSCptichERK5jivSNj0nhbREeQ146JI9FgUoJj4VRlqLvSfQDH8YgTBqCwrD+IBQrC8KIqcyKHKiTz-KsLxTnwmr8qEREWgRIBKKo-ABKEZKhGIaQOO0G76WkvimoBBhiIkbAQDgcrpI6ij5i6xR1GUtZVBkPn1PUjRdqhgEcMJdnUMJdhIGh-yzps0T6HFr4REmbhMdQWjpnBFkfIoSB5s6hZFCUcBBnWigNm6yDNtg0bzNGHZBbM3BYDg3CUOA9DaPMO52gAQpcRiQvctz2GckHDYctx+JwZznPsRhGPcACeE4tZS+hxN++ifF0Wk7INw2jeNk3TbNFwLUtq3RdE4S6D1UVoeEGW+ClT2SFYGzGTd5IoJ8khYKsj23ZIf2bhYQPBb9nwyPgkP-jgs74HIInHEYKH-im9L4IjZoiRB0GY+IL1w8Tu2Pmc0CE3BegRRY6UjhT6PQT9ujSKCi7csDN5IFAL39tMUNs7zN6QG9IO0wYCT+uL5IrFoL1uCprPpQYTH4CpRKkdjViUVSCx0tRVVMiybJ-RlKRctMvh0lD3AafF1oLMl3NMIusGDiSsHdl8cqChIrhg0uGNsG0GV88JEVWOBIcSHzwKoPHt27n+sjqx+6ugina6fno0Why8NAiZeocJS9EWuLQqNSbxk7nt+FCzqg0ePqusyivBx6PgHxcTq+coTBYkzvCOdpnPw9HTiAABis4WY2CA1YavAgAAsrc2QnfcU0zfwQA 1000,500 noborder}} 
 </WRAP> </WRAP>
  
Zeile 206: Zeile 252:
 The block diagram does not claim to conserve energy or charge but serves to provide an overview of the effects and interrelationships. Thus Kirchhoff's rules are usually not applicable there. \\ The block diagram does not claim to conserve energy or charge but serves to provide an overview of the effects and interrelationships. Thus Kirchhoff's rules are usually not applicable there. \\
  
-<imgref pic4> shows a block diagram of a feedback amplifier consisting of an ideal voltage amplifier with gain $A_\rm D$ drawn in the center. +<imgref pic4> shows a block diagram of a feedback amplifier consisting of an ideal voltage amplifier with differential gain $A_\rm D$ drawn in the center. 
 The output voltage $U_\rm O$, reduced by the factor $k$, is fed back via a feedback element.  The output voltage $U_\rm O$, reduced by the factor $k$, is fed back via a feedback element. 
 The circle symbol with the arithmetic symbols (in the block diagram on the left) shows how the incoming values must be offset against each other.  The circle symbol with the arithmetic symbols (in the block diagram on the left) shows how the incoming values must be offset against each other. 
Zeile 217: Zeile 263:
 </panel></WRAP> </panel></WRAP>
  
-The advantage of a real amplifier in negative feedback is that the gain $A_\rm V$ of the whole system depends only negligibly on the gain factor $A_\rm D$ of the real amplifier if $A_\rm D$ is very large (see also task 1.3.2).  +<WRAP><imgcaption pic10|unipolar and bipolar supply> </imgcaption> \\ 
-In this case, the gain $ A_{\rm V}=\frac {1}{k}$To avoid oscillation of the whole system, the amplifier must contain a delay element. +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQJhAZngFgGwSwDiyzCQFYloSM0sQTbpaBTAWjDACgAlEVjKsLA1YBOFAIYSQgmPGhgYCuCXYB3EEiT9pqfOGlJ2sEAGUArgCMALgCcAhgGNLAfQBuchShSoQAHQAO4B486grygMXAaAreAGoAwtHGAKIANgC2PgDOYJlImaiZEj45RRkAFLbM5gCU7BhgQmxi0sx4DJA08mZWdo6u7tDFZRXVhiAAYoyMACbmDgDWffIMnpGZIPyrGWtiIZlzkZJxCSnpGfkZGINt0JoluZm2AFTmtfXreOp4H1hUqDqBE2ms3sCzc8nqtwyuQyTxeRgAgqk-MkAJYAMxRjGsURC3nhTgAIpEliAAKqEkLyFDkgDypksICg8hi8SSaTymUuGRIBR81k4jAyKIylgA9tYTpkeRkIHzYrY-A4UWKJeyZbzrPLFfZleLJRcNQKhSK9WroVQBsUHs9eZlsgaCrJ8o6EMI3cI8O6PdAMKhWrKXXISgMHVkWAGMsJYO7PTGEBAwHUWMJg7JsryMpasq82h0kB9hB1pPJEcj0ZjsSHobCYHb1oHznA3a7hK7WkhC6hpXAEPaGFywOHA4hm9Bm2BhBBoDo8L7A33a9m1ImtMtNHoJKp1OvxNvfuQYFuNPu2nw0AfoFvWEXcxvD2oMB9UAeWgxn5u1K-z0JWnvD9xf1IGhH3UEgaEkaQe36RRYGUDAkG8L9yA+ZhWz-S94MQwDoBQqd0NqBDwHzb9wDPVA3gwwjiCfA8VzQVBN0w887wEcASEYwjnxYjo8MvcweGoqRV3UVAoEoxDCzvVhiHAXiCIkm8BJksB2PvASz13aSmk3AAPATCMocB3BIKAqHXGk-CxWxLBRUUADtbGSTJS1RDEsXYPSsC8MhwC4rs6C0cZJhmeYPJ4PAxFEnhVMTfs9BMUxUjCjBhHAURwDYNKAvilzy3cvTmCgUhUpoJAvPWIiQDmHwAEcwpIbwyoYIh1CwbxAp8AALervHqZrwWnCrWK6sLeBAN14ohIaUHxAlaq4EBf0kkDJIg65EEkCRYPYUVAgtdYxw+Og4BJHZUHYIA 1000,500 noborder}}  
 +</WRAP> 
 + 
 +There is a big advantage of a real amplifier in negative feedback: \\ The voltage gain $A_\rm V$ of the whole system depends in this case only negligibly on the differential gain $A_\rm D$ (assuming $A_\rm D$ is very large). \\ \\ 
 +In this case, the voltage gain is: 
 + 
 +$$A_{\rm V}=\frac {1}{k + \frac {1}{A_{\rm D}}} $
 + 
 +$$\boxed{ A_{\rm V}=\frac {1}{k} \quad \Bigg|_{A_{\rm D} \rightarrow \infty} }$$ 
 + 
 +To avoid oscillation of the whole system, the amplifier must contain a delay element. \\
 This is present in the real amplifier in such a way that the output voltage $U_\rm O$ cannot change infinitely fast. [(Note2>That a voltage change can only take place in a finitely long time is also true for the input voltage. However, this cannot be influenced by the amplifier, but is externally specified.)]. This is present in the real amplifier in such a way that the output voltage $U_\rm O$ cannot change infinitely fast. [(Note2>That a voltage change can only take place in a finitely long time is also true for the input voltage. However, this cannot be influenced by the amplifier, but is externally specified.)].
  
Zeile 240: Zeile 296:
 <panel type="danger" title="Note: Gain factors"> <panel type="danger" title="Note: Gain factors">
 <WRAP group><WRAP column 3%>{{fa>exclamation?32}}</WRAP><WRAP column 85%> <WRAP group><WRAP column 3%>{{fa>exclamation?32}}</WRAP><WRAP column 85%>
-The **differential gain** or **open-loop gain** $\boldsymbol{A_\rm D}$ (German: Differenzverstärkung) refers only to the input and output voltage of the inner amplifier: $A_{\rm D}=\frac{U_\rm O}{U_\rm D}$. +The **differential gain** or **open-loop gain** $\boldsymbol{A_\rm D}$ (German: Differenzverstärkung) refers only to the input and output voltage of the inner amplifier:  
-This acts only without external feedback. It is also called open-loop gain. \\ \\+$$A_{\rm D}=\frac{U_\rm O}{U_\rm D}$$ 
 +This gain is the amplification without external feedback. \\ It is also called open-loop gain. \\ \\
  
-The **voltage gain** $\boldsymbol{A_\rm V}$ refers to the input and output voltage of the whole circuit with feedback: $A_{\rm V}=\frac{U_\rm O}{U_\rm I}$. \\ It is also called closed-loop gain. \\ \\+The **voltage gain** $\boldsymbol{A_\rm V}$ refers to the input and output voltage of the whole circuit with feedback:  
 +$$A_{\rm V}=\frac{U_\rm O}{U_\rm I}$$  
 +It is also called closed-loop gain. \\ \\
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
 </WRAP> </WRAP>
Zeile 252: Zeile 311:
  
 ===== Common pitfalls ===== ===== Common pitfalls =====
-  * ...+  * **Mixing up the inputs:** confusing the inverting input $U_{\rm m}$ (minus) with the non-inverting input $U_{\rm p}$ (plus)A wrong sign flips the whole behavior. 
 +  * **Wrong differential voltage:** forgetting that $U_{\rm D}$ = $U_{\rm p}$ - $U_{\rm m}$. 
 +  * **Using the golden rules outside their valid context:** 
 +      - $U_{\rm p} \approx $U_{\rm m}$ is only justified when the op-amp is in **linear operation** with **negative feedback** and not saturated. 
 +      - $I_{\rm p} \approx $I_{\rm m} \approx 0$  is an idealization; real input bias currents may matter in high-impedance circuits. 
 +  * **Assuming unlimited output voltage:** the output is limited by the **supply rails** (and headroom). Once saturated, linear equations break. 
 +  * **Confusing open-loop and closed-loop gain:** $A_{\rm D}$ (open-loop) is huge and device-dependent; $A_{\rm V}$ (closed-loop) is what the feedback network sets. 
 +  * **Ignoring supply type:** unipolar supply does **not** allow negative output voltages (without a mid-supply reference). Many textbook sketches silently assume bipolar rails. 
 +  * **Assuming unlimited output current:** real op-amps have output current limits; too-small load resistance causes clipping/distortion. 
 +  * **Treating block diagrams like circuit diagrams:** block diagrams show cause–effect; Kirchhoff’s laws do not automatically apply inside blocks. 
 +  * **Misclassifying feedback sign:** feeding output to the inverting input is typically **negative feedback**, while to the non-inverting input is typically **positive feedback** (depending on the network). 
 + 
 + 
 + 
 +===== Learning Questions ===== 
 + 
 +  * Explain the difference between the unipolar and bipolar power supply of an opamp. 
 +  * Draw a sketch for bipolar and one for unipolar power supply. 
 +  * What are the advantages and disadvantages of unipolar and bipolar supply in opamp? 
 +  * What are the golden rules? 
 +  * What is the basic equation of the opamp?
  
 ===== Exercises ===== ===== Exercises =====
-==== Worked examples ==== 
  
-...+<panel type="info" title="Exercise 1.3.2 Calculations for negative feedback"> 
 +<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 + 
 +<WRAP right><panel type="default">  
 +<imgcaption pic1_3_2_rueckkopplung2|Block diagram of a feedback amplifier> 
 +</imgcaption> 
 +{{drawio>BlockschaltbildRueckkopplung.svg}} 
 +</panel></WRAP> 
 + 
 +For the principle of negative feedback, the adjacent block diagram was given in the script. Here $A_\rm D$ is the so-called differential gain, i.e. the gain of the difference between the input voltage $U_\rm I$ and the feedback voltage. 
 +  - Derive the voltage gain $A_\rm V$ as a function of the differential gain $A_\rm D$ and the feedback factor $k$. Note that $A_{\rm V} = {{U_\rm O}\over{U_\rm I}} = f(A_{\rm D}, k)$ and give the derivation. 
 +  - What is the voltage gain $A_\rm V$ for an ideal differential gain ($A_\rm D \rightarrow \infty $)? 
 +  - Find the voltage gain $A_\rm V$ for feedback $k = 0.001$ with differential gain $A_{\rm D1} = 100'000$ and $A_{\rm D2} = 200'000$. \\ Real differential amplifiers, more precisely operational amplifiers, are considered in more detail in Chapter 3. Two operational amplifiers of the same type can have noticeably different values in the differential gain, e.g., due to specimen scattering, aging, or temperature drift. \\ Looking at the result from $A_{\rm D1}$ and $A_{\rm D2}$, what can be said about such a variation of a large differential gain value by, say, $50~\%$? 
 +  - State how the voltage gain behaves for the following feedback parameter $k$ with an ideal differential gain and correctly assign the following statements (some are not needed, some are needed more than once): \\ (A) Positive feedback, \\ (B) Negative feedback, \\ (C) Damping, \\ (D) gain, \\ (E) voltage gain equals open-loop gain, \\ (F) $U_{\rm O} = U_{\rm I}$, \\ (G) $U_{\rm O} = - U_{\rm I}$, \\ (H) gain equal 0. \\ \\ 
 +    - $k < -0$  
 +    - $k = 0$  
 +    - $0 < k < 1$  
 +    - $k = 1$  
 +    - $k > 1$  
 + 
 +</WRAP></WRAP></panel> 
 + 
 +<panel type="info" title="Exercise 21.1 Op-amp basics: symbols and signs"> 
 +  * Given an operational amplifier symbol, label the following quantities: 
 +      - non-inverting input voltage $U_{\rm p}$, 
 +      - inverting input voltage $U_{\rm m}$, 
 +      - output voltage $U_{\rm O}$, 
 +      - (if present) the supply voltages $U_{\rm sp}$ and $U_{\rm sm}$. 
 + 
 +  * For each case below, state whether the output voltage $U_{\rm O}$ initially moves **upwards** or **downwards** (assume linear operation): 
 +      - $U_{\rm p}$ increases slightly over $U_{\rm m}$. 
 +      - $U_{\rm m}$ increases slightly over $U_{\rm p}$. 
 +      - $U_{\rm p} = U_{\rm m}$. 
 + 
 +  * Compute the differential voltage: $U_{\rm D}$ for $U_{\rm p} = 2.1\,\rm V$ and $U_{\rm m} = 2.0\,\rm V$. 
 + 
 +  * Using a differential gain of $A_{\rm D} = 200{'}000$, compute the **ideal** output voltage $U_{\rm O}$. 
 +  * Explain briefly why this output voltage cannot be realized in practice when the op-amp is powered from supply rails of $\pm 5\,\rm V$. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.2 Differential vs single-ended thinking"> 
 +An op-amp has $A_{\rm D}=150{'}000$ and is powered from $\pm 12\,\rm V$. 
 +  - Compute $U_{\rm O}$ for $U_{\rm p}=1.002\,\rm V$ and $U_{\rm m}=1.000\,\rm V$ (ideal equation). 
 +  - Decide whether the result is physically possible. 
 +  - Explain why even very small differences between $U_{\rm p}$ and $U_{\rm m}$ are sufficient to drive the output into saturation in open-loop operation. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.3 Unipolar supply and output biasing"> 
 +An op-amp operates from a unipolar supply $0\,\rm V$ to $9\,\rm V$. 
 +  - What output voltage corresponds to “zero differential input” in a typical unipolar configuration? 
 +  - Why is this value often chosen close to $U_{\rm S}/2$? 
 +  -  Describe one practical consequence if the output is biased too close to one supply rail. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.4 Unipolar supply and virtual ground intuition"> 
 +An op-amp uses a unipolar supply $0\,\rm V \dots 10\,\rm V$. \\ 
 +If you want to amplify a small sinus signal centered around $0\,\rm V$, why is it a problem to connect it directly to an input? 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.5 Classify feedback (fast diagnosis)"> 
 +  * For each statement, mark **true/false** and correct the false ones: 
 +      -  Feeding back a fraction of the output to the inverting input always creates negative feedback. 
 +      -  With negative feedback and large $A_{\rm D}$, the op-amp tends to keep $U_{\rm D}$ close to 0. 
 +      -  Positive feedback generally stabilizes the operating point and improves linearity. 
 +      -  If the output is saturated at a rail, $U_{\rm p} \approx U_{\rm m}$ must still be true. 
 +  * For each configuration below, classify the feedback as positive or negative (assume resistive feedback networks): 
 +      -  Output fed through a divider to $U_{\rm m}$, $U_{\rm p}$ driven by the input source. 
 +      -  Output fed through a divider to $U_{\rm p}$, $U_{\rm m}$ driven by the input source. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.6 Saturation and clipping reasoning"> 
 +An op-amp is powered from $\pm 5\,\rm V$ (bipolar). The output swing is limited to about $\pm 4\,\rm V$. 
 +  - If $U_{\rm D}=+50\,\mu\rm V$ and $A_{\rm D}=200{,}000$, compute the ideal $U_{\rm O}$. Is saturation expected? 
 +  - Repeat for $U_{\rm D}=+10\,\rm mV$. 
 +  - Explain in one sentence why clipping produces distortion in audio signals. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.7 Input bias currents (qualitative + estimate)"> 
 +A sensor with source resistance $R_{\rm S}=1\,\rm M\Omega$ drives the non-inverting input. \\ 
 +The real op-amp dows not only show an internal resistance, but also a small current source on the input pins. \\ 
 +This input bias current is in this exercise $I_{\rm B}=200\,\rm nA$. 
 +   - Estimate the voltage error at the input caused by $I_{\rm B}$ flowing through $R_{\rm S}$. 
 +   - Explain when such an error matters and when it is negligible. 
 +</panel> 
 + 
 +<panel type="info" title="Exercise 21.8 Output current limit and load selection"> 
 +A real op-amp can supply at most $I_{\rm O,max}=20\,\rm mA$. \\  
 +It is intended to drive a load resistor $R_{\rm L}$ from an output voltage of $U_{\rm O}=3\,\rm V$. 
 +  - What is the minimum $R_{\rm L}$ to avoid exceeding the output current limit? 
 +  - If $R_{\rm L}$ is smaller than this value, what happens to the output waveform for a sine input? 
 + 
 +Bonus: If the op-amp can also sink $20\,\rm mA$, does that change your answer to (a)? 
 +</panel> 
 + 
  
 ===== Embedded resources ===== ===== Embedded resources =====