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--- circuit_design:3_opamp_basic_circuits_i [2021/11/21 14:28]
slinn [Bearbeiten - Panel]
+++ circuit_design:3_opamp_basic_circuits_i [2023/09/19 22:16] (aktuell)
mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 3. Basic circuits of operational amplifiers I ======
+====== 3 Basic circuits of operational amplifiers I ======
 <callout>.
   * [[https://www.arrow.com/en/research-and-events/articles/fundamentals-of-op-amp-circuits|Operational amplifier basic circuits]] provide a nice introduction.
-  * Textbook and Workbook on [[https://www.amazon.de/Operational-Amplifier-Circuits-Analysis-Design/dp/0750694688|Operational Amplifier Circuits: Analysis and Design by John C. C. Nelson]].
+   * [[https://link.springer.com/book/10.1007%2F978-3-319-28127-8| Operational Amplifiers (Johan Huijsing)]] [viewable via the university network]
   * On Youtube the following videos are recommended: [[https://www.youtube.com/watch?v=lJDjWZqhpVc&list=PLzUN9-WgjT3PcvDFD5cI9COE9E53CrcgA&ab_channel=KhanAcademy|What is an operational amplifier?]], [[https://www.youtube.com/watch?v=_Ut-nQ535iE&list=PLzUN9-WgjT3PcvDFD5cI9COE9E53CrcgA&index=3&ab_channel=KhanAcademy|Inverting operation amplifier circuit]], [[https://www.youtube.com/watch?v=_Ut-nQ535iE&list=PLzUN9-WgjT3PcvDFD5cI9COE9E53CrcgA&index=3&ab_channel=KhanAcademy|Non-inverting operation amplifier circuit]].
@@ Zeile 16: / Zeile 16: @@
 {{:elektronische_schaltungstechnik:hallo.mp3|hallo.mp3}}
  \\
-{{drawio>hallo}} </panel></WRAP>
+{{drawio>hallo.svg}} </panel></WRAP>
 <WRAP><panel type="default">
@@ Zeile 24: / Zeile 24: @@
 {{:elektronische_schaltungstechnik:hallo_verzerrt.mp3|hallo_verzerrt.mp3}}
-{{drawio>hallo_verzerrt}}
+{{drawio>hallo_verzerrt.svg}}
 </panel></WRAP>
@@ Zeile 30: / Zeile 31: @@
 === Introductory example ===
-Acoustic amplifiers, such as those found in mobile phones, laptops or hi-fi systems, often exhibit an unpleasant characteristic when heavily amplified: the previously undistorted signal is no longer passed on as usual, but [[https://en.wikipedia.org/wiki/Total_harmonic_distortion#Definitions_and_examples|clatters]]. It is distorted in such a way that it no longer sounds pleasant.
+Acoustic amplifiers, such as those found in mobile phones, laptops, or hi-fi systems, often exhibit an unpleasant characteristic when heavily amplified: the previously undistorted signal is no longer passed on as usual, but [[https://en.wikipedia.org/wiki/Total_harmonic_distortion#Definitions_and_examples|clatters]]. It is distorted in such a way that it no longer sounds pleasant.
 For this purpose, you will find an acoustic example with pictures in <imgref pic11> and <imgref pic12> respectively. The bottom of each image shows the time course of the voltage output to a loudspeaker (x-axis: time, y-axis: frequency). The upper picture has three dimensions: It shows in color intensity which frequencies are used at which time. The frequencies in grey areas are not used. If a frequency is shown in red at one point in time, it has a relatively large amplitude.
@@ Zeile 37: / Zeile 38: @@
-The signal distortion is due to the design of the amplifier, which can only output a maximum possible voltage and otherwise [[https://en.wikipedia.org/wiki/Clipping_(audio)|clipping]]. The structure of an acoustic amplifier is similar to that of a feedback operational amplifier, as seen in the simulation.
+The signal distortion is due to the design of the amplifier, which can only output the maximum possible voltage and otherwise [[https://en.wikipedia.org/wiki/Clipping_(audio)|clipping]]. The structure of an acoustic amplifier is similar to that of a feedback operational amplifier, as seen in the simulation.
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.0005+0.22070718156067046+55+10+50%0A207+80+256+80+240+4+U_O%0A207+-64+240+-64+224+4+U_I%0AR+-64+240+-96+240+0+1+40+5+0+0+0.5%0Ar+176+304+176+352+0+1000%0Ag+176+352+176+368+0%0A403+-64+176+16+208+0+1_1_0_4098_5_0.1_0_2_1_3%0A403+128+160+224+224+0+0_1_0_4098_20_0.1_0_2_0_3%0Ax+-146+402+-24+405+4+18+Eingangssignal%0Ax+138+399+243+402+4+18+Lautsprecher%0Aw+0+272+-16+272+0%0Ag+80+352+80+368+0%0Ar+80+304+80+256+0+190000%0Ar+80+304+80+352+0+10000%0Aw+-64+240+0+240+0%0Aa+0+256+80+256+9+15+-15+1000000+-0.7500024664848447+-1.243449435405756+100000%0Aw+80+304+-16+304+0%0Aw+-16+304+-16+272+0%0Al+176+272+176+304+0+1+-0.014999892735530778%0Aw+80+256+176+256+0%0Aw+176+256+176+272+0%0A noborder}} </WRAP>
 Acoustic amplifiers are usually constructed like operational amplifiers, which will be considered in this chapter. </callout></WRAP>
+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.0005+0.22070718156067046+55+10+50%0A207+80+256+80+240+4+U_O%0A207+-64+240+-64+224+4+U_I%0AR+-64+240+-96+240+0+1+40+5+0+0+0.5%0Ar+176+304+176+352+0+1000%0Ag+176+352+176+368+0%0A403+-64+176+16+208+0+1_1_0_4098_5_0.1_0_2_1_3%0A403+128+160+224+224+0+0_1_0_4098_20_0.1_0_2_0_3%0Ax+-146+402+-24+405+4+18+Eingangssignal%0Ax+138+399+243+402+4+18+Lautsprecher%0Aw+0+272+-16+272+0%0Ag+80+352+80+368+0%0Ar+80+304+80+256+0+190000%0Ar+80+304+80+352+0+10000%0Aw+-64+240+0+240+0%0Aa+0+256+80+256+9+15+-15+1000000+-0.7500024664848447+-1.243449435405756+100000%0Aw+80+304+-16+304+0%0Aw+-16+304+-16+272+0%0Al+176+272+176+304+0+1+-0.014999892735530778%0Aw+80+256+176+256+0%0Aw+176+256+176+272+0%0A noborder}} </WRAP>
 <WRAP><callout>
@@ Zeile 61: / Zeile 62: @@
 <WRAP><panel type="default">
-<imgcaption pic0|operatingamplifier>
+<imgcaption pic0|Operating Amplifier?>
 </imgcaption>
 {{drawio>OP_Verstärker}} </panel></WRAP>
@@ Zeile 67: / Zeile 68: @@
 === Historical outline ===
-To understand the origin of the operational amplifier, a brief outline of microprocessor history will be presented here. The first microprocessor was commercially distributed in 1971 by INTEL as [[https://en.wikipedia.org/wiki/Intel_4004|Intel 4004]], which could only represent a 4-bit number space with one word and had a clock frequency of 500 kHz. This was already extended after half a year as [[https://en.wikipedia.org/wiki/Intel_8008|8008]] on a 8-Bit Word, so that calculations could be accomplished substantially faster. From this second generation the [[https://en.wikipedia.org/wiki/X86|8086]] was developed in 1978, which had been the basis of many PCs and other electronic components with a 16-bit word for a very long time. In 1985 this processor was extended as 80386 to a 32-Bit Word. From this time also the perhaps well-known question originates whether software should be installed as x86 or x64 code. Processors with 64-bit Word have been available since 1999, but they have not yet penetrated all niches.
+To understand the origin of the operational amplifier, a brief outline of microprocessor history will be presented here. The first microprocessor was commercially distributed in 1971 by INTEL as [[https://en.wikipedia.org/wiki/Intel_4004|Intel 4004]], which could only represent a 4-bit number space with one word and had a clock frequency of 500 kHz. This was already extended after half a year as [[https://en.wikipedia.org/wiki/Intel_8008|8008]] on an 8-Bit word so that calculations could be accomplished substantially faster. From this second generation, the [[https://en.wikipedia.org/wiki/X86|8086]] was developed in 1978, which had been the basis of many PCs and other electronic components with a 16-bit word for a very long time. In 1985 this processor was extended from 80386 to a 32-Bit Word. From this time also the perhaps well-known question originates whether software should be installed as x86 or x64 code. Processors with 64-bit Word have been available since 1999, but they have not yet penetrated all niches.
-With this historical outline in mind, the question arises as to how Apollo 11 was able to control both the first lunar mission and the lunar module, and how the orbit could be calculated so accurately that this would not pose a risk. Development of the [[https://en.wikipedia.org/wiki/Apollo_Guidance_Computer|Apollo Gudance Computer]] was started in 1961, 10 years before the first commercial microprocessor. This processor, with about 1MHz frequency and 16-bit word width, could provide performance as early as 1966 that was not commercially available until 15 years later. Nevertheless, these computers could not solve the differential equation systems necessary for trajectory planning in a reasonable amount of time. This was only possible by combining the well-known digital processor and the [[https://en.wikipedia.org/wiki/Analog_computer|analog computers]] that had been in use since the 1920s. Analog computers are only slightly reminiscent of the classical image of the computer. Although they also consist of a network of components, they do not have any software. Rather, various properties of electronic components are used for computation (e.g., the exponential behavior of the Shockley equation of the diode). For the Apollo missions, the [[http://www.analogmuseum.org/library/eai380_short.pdf|EAI 380 hybrid computer]] was used. This and similar hybrid computers continued to be used into the 1990s to solve systems of differential equations for aerospace, construction, and (under a slightly different guise) as synthesizers in the music industry. The interested reader is referred to the video [[https://www.youtube.com/watch?v=LF35eXfCMIQ&ab_channel=Alex|Analog computers]], which explains why they are superior to even the most modern computers.
+With this historical outline in mind, the question arises as to how Apollo 11 was able to control both the first lunar mission and the lunar module, and how the orbit could be calculated so accurately that this would not pose a risk. Development of the [[https://en.wikipedia.org/wiki/Apollo_Guidance_Computer|Apollo Guidance Computer]] was started in 1961, 10 years before the first commercial microprocessor. This processor, with about 1MHz frequency and 16-bit word width, could provide performance as early as 1966 that was not commercially available until 15 years later. Nevertheless, these computers could not solve the differential equation systems necessary for trajectory planning in a reasonable amount of time. This was only possible by combining the well-known digital processor and the [[https://en.wikipedia.org/wiki/Analog_computer|analog computers]] that had been in use since the 1920s. Analog computers are only slightly reminiscent of the classical image of the computer. Although they also consist of a network of components, they do not have any software. Rather, various properties of electronic components are used for computation (e.g., the exponential behavior of the Shockley equation of the diode). For the Apollo missions, the [[http://www.analogmuseum.org/library/eai380_short.pdf|EAI 380 hybrid computer]] was used. This and similar hybrid computers continued to be used into the 1990s to solve systems of differential equations for aerospace, construction, and (under a slightly different guise) as synthesizers in the music industry. The interested reader is referred to the video [[https://www.youtube.com/watch?v=LF35eXfCMIQ&ab_channel=Alex|Analog computers]], which explains why they are superior to even the most modern computers.
 <WRAP> <imgcaption picZ|Operational Amplifier Module (OPA) in a PIC microcontroller manufactured by Microchip (c) Microchip>. {{  :circuit_design:opa_8bitpic.jpg?200|opa_8bitpic.jpg}}</imgcaption> </WRAP>
-The core component of analog computers is the operational amplifier. With this it is possible to build and combine adders, subtractors, multipliers, exponentials, integrators and differentiators. The "calculation" happens almost instantaneously after setting the desired input voltages.
+The core component of analog computers is the operational amplifier. With this, it is possible to build and combine adders, subtractors, multipliers, exponentials, integrators, and differentiators. The "calculation" happens almost instantaneously after setting the desired input voltages.
-The operational amplifier has outlasted this time and nowadays offers a starting point for a wide variety of signal conditioning, such as active filters, amplifier stages, input/output resistance matching and various (integrated) control loops. Several microcontrollers have op-amps built right in and can be configured by suitable code. In <imgref picZ> a PIC microprocessor is shown, which contains an integrated operational amplifier.
+The operational amplifier has outlasted this time and nowadays offers a starting point for a wide variety of signal conditioning, such as active filters, amplifier stages, input/output resistance matching, and various (integrated) control loops. Several microcontrollers have op-amps built right in and can be configured by suitable code. In <imgref picZ> a PIC microprocessor is shown, which contains an integrated operational amplifier.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 81: / Zeile 82: @@
 ===== 3.1 Circuit symbols and basic circuitry =====
-This chapter deals with operational amplifiers or in general with measuring amplifiers. One application for this is the measurement of voltages, currents and resistances. These must be determined very precisely in some applications, for example for an accurate temperature measurement. In this case, amplification of the measurement signals is useful and necessary.
+This chapter deals with operational amplifiers or in general with measuring amplifiers. One application for this is the measurement of voltages, currents, and resistances. These must be determined very precisely in some applications, for example for accurate temperature measurement. In this case, amplification of the measurement signals is useful and necessary.
-This amplification is done by measuring amplifiers. Measuring amplifiers have to fulfil some characteristics:
+This amplification is done by measuring amplifiers. Measuring amplifiers have to fulfill some characteristics:
   * Measurement amplifiers should not have any feedback effect on the measured variable. An operational amplifier should have the highest possible input resistance. This prevents the voltage to be amplified from collapsing.
-  * Measuring amplifiers should have a high sensitivity. An operational amplifier should have a large differential gain $A_D$.
+  * Measuring amplifiers should have a high sensitivity. An operational amplifier should have a large differential gain $A_\rm D$.
-  * Measuring amplifiers should show a defined transmission behaviour, i.e. the output signal should be clearly related to the input signal. An operational amplifier concretely should show a linear relationship.
+  * Measuring amplifiers should show a defined transmission behavior, i.e. the output signal should be clearly related to the input signal. An operational amplifier concretely should show a linear relationship.
-  * Measuring amplifiers should show a good dynamic behaviour. The output signal of an operational amplifier should follow the input signal without any time delay.
+  * Measuring amplifiers should show good dynamic behavior. The output signal of an operational amplifier should follow the input signal without any time delay.
-  * Measurement amplifiers should produce an "impressed output signal". This means that the components at the amplifier output cannot change the output signal. An operational amplifier, specifically, should be able to maintain the desired output signal with the necessary current to do so. Since the current $I_O$ can become very large (by electronic standards), this means that an operational amplifier must have a low output resistance $R_O =\frac{U_O}{I_O}$.
+  * Measurement amplifiers should produce an "impressed output signal". This means that the components at the amplifier output cannot change the output signal. An operational amplifier, specifically, should be able to maintain the desired output signal with the necessary current to do so. Since the current $I_\rm O$ can become very large (by electronic standards), this means that an operational amplifier must have a low output resistance $R_{\rm O} =\frac{U_{\rm O}}{I_{\rm O}}$.
-<WRAP><panel type="default"> <imgcaption pic1|Equivalent circuit diagram of an amplifier> </imgcaption> {{drawio>Ersatzschaltbild_eines_Verstärkers}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption pic1|Equivalent circuit diagram of an amplifier> </imgcaption> {{drawio>Ersatzschaltbild_eines_Verstärkers.svg}}</panel></WRAP>
-In general, a measurement amplifier is constructed as in <imgref pic1>. This has already been described in chapter [[:circuit_design:1_amplifier_basics#amplifier_-_a_black_box_is_going_to_be_specified|1 Basics to amplifiers]]. In the following, only operational amplifiers will be considered. An operational amplifier is a measurement amplifier, which is often used in electrical engineering.
+In general, a measurement amplifier is constructed as in <imgref pic1>. This has already been described in chapter [[:circuit_design:1_amplifier_basics#amplifier_-_a_black_box_is_going_to_be_specified|Amplifier Basics]]. In the following, only operational amplifiers will be considered. An operational amplifier is a measurement amplifier, which is often used in electrical engineering.
-The circuit symbol of the amplifier is an isosceles triangle, at the apex of which the output signal originates and in the base of which the input signal enters. In <imgref pic3> you can see different switching symbols:
+The circuit symbol of the amplifier is an isosceles triangle, at the apex of which the output signal originates and at the base of which the input signal enters. In <imgref pic3> you can see different circuit symbols:
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<WRAP><panel type="default"> <imgcaption pic3|Switching symbols of amplifiers> </imgcaption> {{drawio>Schaltzeichen_OPV}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption pic3|Circuit Symbols of Amplifiers> </imgcaption> {{drawio>Schaltzeichen_OPV.svg}} </panel></WRAP>
-  * Switching symbol (1): In block diagrams (not to be confused with circuit diagrams, see [[:circuit_design:https:wiki.mexle.org:circuit_design:1_amplifier_basics|chapter 1]]) this switching symbol is used for general amplifiers. The input signal enters an input and exits through an input. This symbol will not be found again until Chapter 5.
+  * Circuit symbol (1): In block diagrams (not to be confused with circuit diagrams, see [[https://wiki.mexle.org/circuit_design/1_amplifier_basics|chapter 1]]) this circuit symbol is used for general amplifiers. The input signal enters an input and exits through an input. This symbol will not be found again until Chapter 5.
-  * Switching symbol (2): According to DIN EN 60617, this circuit diagram is to be used for operational amplifiers. It indicates with the infinity sign the ideally infinitely high amplification. in the following this symbol is not used, because it is not used in all international circuits and tools.
+  * Circuit symbol (2): According to DIN EN 60617, this circuit diagram is to be used for operational amplifiers. It indicates with the infinity sign the ideally infinitely high amplification. in the following this symbol is not used, because it is not used in all international circuits and tools.
-  * Switching symbol (3): The switching symbol (3) is the most commonly used symbol for an operational amplifier. On the left is the **inverting input**  with voltage $U_m$ (__m__inus) and the **non-inverting input**  with $U_p$ (__p__lus). The output with voltage $U_O$ is shown on the right.
+  * Circuit symbol (3): The circuit symbol (3) is the most commonly used symbol for an operational amplifier. On the left is the **inverting input**  with voltage $U_{\rm m}$ (__m__inus) and the **non-inverting input**  with $U_{\rm p}$ (__p__lus). The output with voltage $U_{\rm O}$ is shown on the right.
-  * Switching symbol (4): The switching symbol (4) is additionally drawn with the supply voltages $U_{sp}$ (__s__upply __p__lus) and $U_{sm}$ (__s__upply __m__inus). Power is provided from the supply for the output voltage of the operational amplifier.
+  * Circuit symbol (4): The circuit symbol (4) is additionally drawn with the supply voltages $U_{\rm sp}$ (__s__upply __p__lus) and $U_{\rm sm}$ (__s__upply __m__inus). Power is provided from the supply for the output voltage of the operational amplifier.
-  * Switching symbol (5) and (6): these symbols show __**no**__  operational amplifier. These symbols show the NOT gate and the tri-state gate. Both of these components have already been discussed in [[:fundamentals_of_digital_engineering:binaer_logic#applications_of_binaer_logic|fundamentals_of_digital_engineering]]. Unfortunately, the representation of these digital components in various circuits is not unlike the operational amplifier. An example of this is the transceivers[(_ckgedit_QUOT___trans__mitter and re__ceiver__", meaning transmitter-receiver, or interface adapter)] {{:circuit_design:sp3485cn-ltr.pdf|SP3481 or SP3485}}. If digital input values are considered, assume that the circuit symbol does not represent an operational amplifier.
+  * Circuit symbols (5) and (6): these symbols show __**no**__  operational amplifier. These symbols show the NOT gate and the tri-state gate. Both of these components have already been discussed in [[https://wiki.mexle.org/introduction_to_digital_systems/start|fundamentals of digital engineering]]. Unfortunately, the representation of these digital components in various circuits is not unlike the operational amplifier. An example of this is the transceivers[(_ckgedit_QUOT___trans__mitter and re__ceiver__", meaning transmitter-receiver, or interface adapter)] {{:circuit_design:sp3485cn-ltr.pdf|SP3481 or SP3485}}. If digital input values are considered, assuming that the circuit symbol does not represent an operational amplifier.
 <WRAP column 100%> <panel type="danger" title="Notice: opamp input"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
-<WRAP><panel type="default"> <imgcaption pic2|Voltages at the operational amplifier> </imgcaption> {{drawio>Spannungen_am_OPV}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption pic2|Voltages at the operational amplifier> </imgcaption> {{drawio>Spannungen_am_OPV.svg}} </panel></WRAP>
-The inputs of the operational amplifier are designated as **inverting input**  $U_m$ and **non-inverting input**  $U_p$.
+The inputs of the operational amplifier are designated as **inverting input**  $U_\rm m$ and **non-inverting input**  $U_\rm p$.
-The voltage $U_D = U_p - U_m$ is called the differential voltage (see <imgref pic2>).
+The voltage $U_{\rm D} = U_{\rm p} - U_{\rm m}$ is called the differential voltage (see <imgref pic2>).
 </WRAP></WRAP></panel> </WRAP>
@@ Zeile 121: / Zeile 122: @@
 ===== 3.2 Basic Equation / Golden Rules =====
-The operational amplifier is a voltage amplifier. Thus it follows from chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#idealized_amplifier_base_types|idealized amplifier basic types]] that for the ideal case the input resistance must be infinite and the output resistance $R_O=0$.
+The operational amplifier is a voltage amplifier. Thus it follows from the chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#idealized_amplifier_base_types|idealized amplifier basic types]] that for the ideal case, the input resistance must be infinite and the output resistance $R_{\rm O}=0$.
 The <imgref picA> shows an ideal voltage amplifier. This is characterized by the following properties:
-  * **input resistance**: The input side is drawn in the figure. The resistance given here is the differential resistance $R_D$ where the differential voltage $U_D$ drops. Therefore $R_D \rightarrow \infty$ is valid. Thus, the input currents $I_p \rightarrow 0$ and $I_m \rightarrow 0$.
+  * **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, $R_O=0$ gives that $U_O = A_D\cdot U_D$. 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 $0V$, the output voltage is $0V$.
+  * **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**: From the chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]] it is known that $A_D$ must be very large. Ideally, the following applies: $A_D \rightarrow \infty$
+  * **Voltage Gain**: From the chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]] it is known that $A_\rm D$ must be 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}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption picA|Operational amplifier equivalent circuit> </imgcaption> {{drawio>OPV_Ersatzschaltbild.svg}} </panel></WRAP>
 <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: $U_O = A_D \cdot U_D$ 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: $A_D \rightarrow \infty$
+      - The differential gain goes to infinity: $A_\rm D \rightarrow \infty$
-      - The input resistance goes to infinity: $R_D \rightarrow \infty$
+      - The input resistance goes to infinity: $R_\rm D \rightarrow \infty$
-      - The output resistance is 0: $R_O = 0$
+      - The output resistance is 0: $R_\rm O = 0$
 </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 144: / Zeile 145: @@
 These rules have different limits in the real amplifier:
-  * $\boldsymbol{U_O = A_D \cdot U_D}$:
+  * $\boldsymbol{U_{\rm O} = A_{\rm D} \cdot U_{\rm D}}$:
-      * 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 $100mV$ to $U_S$. Other operational amplifiers have an **output limit**, which is $1...2V$ 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 ($U_{sm} \neq -U_{sp}$), 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 $U_O=A_D \cdot U_D$ as long as $U_D = U_p - U_m$ is the same. For the real operational amplifier with fixed $A_D$, output voltage $U_{O1}$ for $U_{D1}=5V - 4.9V$ is different from $U_{O2}$ for $U_{D1}=0.1V - 0V$.
+      * 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 $U_{\rm D1}=0.1 ~\rm V - 0 ~V$.
-  * $\boldsymbol{A_D}$: The differential gain is usually between $A_D = 20'000 ... 400'000$.
+  * $\boldsymbol{A_\rm D}$: The differential gain is usually between $A_\rm D = 20'000 ... 400'000$.
-  * $\boldsymbol{R_D}$: For real operational amplifiers, the input resistance $R_I > 1 M\Omega$ and the input current $|I_p|$ or $|I_m|$ is less than $1 \mu A$.
+  * $\boldsymbol{R_\rm D}$: 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 $1 ~\rm µA$.
-  * $\boldsymbol{R_O}$: In real operational amplifiers, the output resistance $R_O$ is usually a few $\Omega$ and limited by a maximum current (in the range of a few tens of $mA$ to a few $A$)
+  * $\boldsymbol{R_\rm O}$: In real operational amplifiers, the output resistance $R_\rm O$ is usually a few $\Omega$ and limited by a maximum current (in the range of a few tens of $\rm mA$ to a few $\rm A$)
-The op-amp in the simulation replicates a real op-amp in some respects: The voltage gain is $A_D = 100'000$. The transfer characteristic $U_O(U_D)$ shows proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{sp}|=|U_{sm}|=15V$ (not shown). The modulation limits and the voltage gain can be changed in the simulation via "Edit component" (double click).
+The op-amp in the simulation replicates a real op-amp in some respects: The voltage gain is $A_\rm D = 100'000$. The transfer characteristic $U_{\rm O}(U_\rm D)$ shows proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{\rm sp}|=|U_{\rm sm}|=15 ~\rm V$ (not shown). The modulation limits and the voltage gain can be changed in the simulation via "Edit component" (double click).
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00OgFgJxvRzA2GsUA4UwloUsBmcpAVmomp3KjGhGteoFMBaMMAKABM0KFiQhBDcGEESp4gKoB9AIJCRIXjkElN+VjtaKlAB34B3EEnxycY2zH4AnCTom7Bb8shjho-AHNXcW9xTxDhX38ScnAkcWIwiFlwqKV8JWglEgIlYUzYMEy84vJ1KFTDcBQU3WM1ACNNYRtU3kjKMEcAD2byFBByKV54weoBhJsASQA7EwBXABcAHQBnAGUASwCZgEMAG34m4grBAcEsKEocf16wbzHtahDxq3AbABkAe12AEzWvjM1gB5JYLRblTRgbTJPQGWHGGb8XZQp44awOAZgahQnEsRAIXAE2h8LB8aj4QTJejPHGsfGISFaVihVEGfA2RFM6EGYY8iQIkDKMwANzZgx8zIc9KsrDgiH6mCV6BuUVg1H4MWabkSmjIwV82PSxRyGXlCDJ0CG2KwkgKRSygiURrKWu4+uu0lkSHIsnp1Gd0B0JrOWSQeGVyp40Od0YgxSdWVufiMpC9VhkbzANgU-Du9wzzEkVkikxAABE8+AcZQkoMmFmbGpejWdVcqY2hVWulQsT3qN73iAlFXYjlwGPM2WR3cJtm2FAkIIumXeNQAGpV7i+sY2FlkzvYzdAA noborder}} </WRAP>
+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00OgFgJxvRzA2GsUA4UwloUsBmcpAVmomp3KjGhGteoFMBaMMAKABM0KFiQhBDcGEESp4gKoB9APJCRIXjkElN+VjtaKlAB34B3EEnxycY2zH4AnCTom7Bb8shjho-AHNXcW9xTxDhX38ScnAkcWIwiFlwqKV8JWglEgIlFkzYMEylQWLydShUw3AUFN1jNQAjTWEbVN5IyjBHAA8W8hQQcileeKHqQYSbAEkAOxMAVwAXAB0AZwBlAEsA2YBDABt+ZuJKwUHBLChKHH8+sG9x7WoQiatwGwAZAHs9gBN1t9ZusVMtFksKpowNpknoDLDjABbfh7KHPHDWByDaiaHEsRAIXAEgj4LDUSA0eLQQQQW5+AmQrSsUJogz4GxIxnQgwjbkSBEgZRmABurKGPiZDlY3V0cGJmAVGCiuGo-BiLTciU0ZGCvjA1HSxRyGTlCG85PoEHwMgNcCKWVK+qU5XV3B1N2ksiQ5Fk0oNLB0xR61ugf1KSDwisVPGheRjEGKpVNCDAUcVJuKDsy-HuhJynqsMneYBsChz0liSEg4EkVkiUxAABFy-qhlR+W2oA2AILlnE+hLXGnF0stoaoaTjL0fEBKcuV0jgStFhtz+6TEtWLuCGUz7jUABq5e4PvGNl4zKwu83h7VglibsY+ASxCGz8cQA noborder}} </WRAP>
 === Power supply of the operational amplifier ===
@@ Zeile 162: / Zeile 162: @@
 For the voltage supply of the operational amplifier, a distinction is made between unipolar and bipolar:
-With the **bipolar voltage supply**, the same voltage with different sign is applied to both supply terminals in terms of magnitude (<imgref pic10> (1)). This allows the output voltage $U_O$ of the amplifier to point in either direction (<imgref pic10> (2)). The power supply must be designed in such a way that it can provide both voltages. A differential voltage of $U_D=0$ also results in an output voltage of $U_O=0$.
+With the **bipolar voltage supply**, the same voltage levels but with different signs are applied to both supply terminals in terms of magnitude (<imgref pic10> (1)). This allows the output voltage $U_{\rm O}$ of the amplifier to point in either direction (<imgref pic10> (2)). The power supply must be designed in such a way that it can provide both voltages. A differential voltage of $U_{\rm D}=0$ also results in an output voltage of $U_{\rm O}=0$.
-With the **unipolar power supply**, the negative supply terminal is connected to ground (<imgref pic10> (3)). As a result, the output voltage $U_O$ of the amplifier can only take on non-negative values (<imgref pic10> (4)). In this case, the supply can be provided by a single voltage source (e.g., a battery). With a differential voltage of $U_D=0$, the output voltage is $U_O=\frac{1}{2}\cdot U_{sp}$.
+With the **unipolar power supply**, the negative supply terminal is connected to the ground (<imgref pic10> (3)). As a result, the output voltage $U_{\rm O}$ of the amplifier can only take on non-negative values (<imgref pic10> (4)). In this case, the supply can be provided by a single voltage source (e.g., a battery). With a differential voltage of $U_\rm D=0$, the output voltage is $U_{\rm O}=\frac{1}{2}\cdot U_{\rm sp}$.
-<WRAP><panel type="default"> <imgcaption pic10|unipolar and bipolar supply> </imgcaption> {{drawio>unipolare_und_bipolare_Versorgung}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption pic10|unipolar and bipolar supply> </imgcaption> {{drawio>unipolare_und_bipolare_Versorgung.svg}} </panel></WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 173: / Zeile 173: @@
-In the chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]] it was described that an amplifier with high open-loop gain can be "tamed" by feeding back a part of the output signal with a negative sign. In the simplest case, the output signal could be fed directly to the negative input of the operational amplifier. The input signal $U_I$ of the entire circuit is applied to the positive input. In <imgref pic4> this circuit is shown.
+In the chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]] it was described that an amplifier with high open-loop gain can be "tamed" by feeding back a part of the output signal with a negative sign. In the simplest case, the output signal could be fed directly to the negative input of the operational amplifier. The input signal $U_\rm I$ of the entire circuit is applied to the positive input. In <imgref pic4> this circuit is shown.
-<WRAP><panel type="default"> <imgcaption pic4|tension follower> </imgcaption> {{drawio>Spannungsfolger_Schaltung}} </panel></WRAP>
+<WRAP><panel type="default"> <imgcaption pic4|voltage follower> </imgcaption> {{drawio>Spannungsfolger_Schaltung.svg}} </panel></WRAP>
 Using this circuit, the procedure for solving amplifier circuits is now to be illustrated.
-  - The aim is always to create a relation between output voltage $U_O$ and input voltage $U_I$. <fc #800080>Thus, the goal here is the voltage gain $A_V=\frac{U_O}{U_I}$.</fc>
+  - The aim is always to create a relation between output voltage $U_\rm O$ and input voltage $U_ \rm I$. <fc #800080>Thus, the goal here is the voltage gain $A_{\rm V}=\frac{U_{\rm O}}{U_\rm I}$.</fc>
-  - Before calculating, it should be checked how many equations describe the system and thus have to be set up. This can be determined by the **number of variables**. This is done by counting through the currents and voltages of the circuit. _fc #800080>In this case, there are 3 currents and 3 voltages. So the **number of equations**  needed is 6.</fc>
+  - Before calculating, it should be checked how many equations describe the system and thus have to be set up. This can be determined by the **number of variables**. This is done by counting through the currents and voltages of the circuit. <fc #800080>In this case, there are 3 currents and 3 voltages. So the **number of equations**  needed is 6.</fc>
   - Now **equations are set up**  that can be used. These are:
-      - **Basic equation**: (1) $U_O = U_D \cdot A_D$
+      - **Basic equation**: (1) $U_{\rm O} = U_{\rm D} \cdot A_\rm D$
-      - **Golden rules**: $R_D \rightarrow \infty$ so that (2+3) $I_p = I_m = 0$, $A_D \rightarrow \infty$, $R_O = 0$
+      - **Golden rules**: $R_\rm D \rightarrow \infty$ so that (2+3) $I_{\rm p} = I_{\rm m} = 0$, $A_{\rm D} \rightarrow \infty$, $R_\rm O = 0$
-      - Consideration of the existing **loops**: <fc #800080>in this example, there is only one loop (4) $-U_I + U_D + U_O =0$.</fc> **__Caution__**  : loops can __not__  enter the amplifier through an input and exit through the output! Also to be noted is the direction of $U_D$.
+      - Consideration of the existing **loops**: <fc #800080>in this example, there is only one loop (4) $-U_{\rm I} + U_{\rm D} + U_{\rm O} = 0$.</fc> **__Caution__**: loops can __not__  enter the amplifier through input and exit through the output! Also to be noted is the direction of $U_{\rm D}$.
-      - Consideration of the existing **nodes**: <fc #800080>in this example, there is only one node (5) $I_o = I_m$</fc>.
+      - Consideration of the existing **nodes**: <fc #800080>in this example, there is only one node (5) $I_{\rm o} = I_m$</fc>.
-  - <fc #800080>There appears to be a missing equation. However, this is not correct, because there is still an equation hidden in the objective: (0) $A_V=\frac{U_O}{U_I}$</fc>
+  - <fc #800080>There appears to be a missing equation. However, this is not correct, because there is still an equation hidden in the objective: (0) $A_{\rm V}=\frac{U_{\rm O}}{U_\rm I}$</fc>
   - Now, to **solve the equations**, the equations must be cleverly inserted into each other in such a way that there are no dependencies on the variables left at the end.
-The calculation is done here once in detail (clicking on arrow to the right "►" leads to the next step, [[:circuit_design:rechnung_spannungsfolger|alternative representation]]): {{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_spannungsfolger?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=122x104# left noborder}} ~~PAGEBREAK~~ ~~CLEARFIX~~
+The calculation is done here once in detail (clicking on the arrow to the right "►" leads to the next step, [[:circuit_design:rechnung_spannungsfolger|alternative representation]]):
+{{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_spannungsfolger?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
-So the voltage gain is $A_V=1$. This would also have been seen from chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]]. There it was derived that for $A_D\rightarrow\infty$ the voltage gain just results from $k$: $A_V=\frac{1}{k}$. Since the entire output voltage is fed back here, $k=1$ and thus also $A_V=1$.
-The output voltage $U_O$ is therefore equal to the input voltage $U_I$. This is where the name "voltage follower" comes from. Now one could assume, that this amplifier is of little help, because also a direct connection would deliver $U_O=U_I$. But the important thing here is: because of the operational amplifier, there is __no feedback__  from U_O to U_I. This means that a resistor on the output side will not load the input side. In the simulation , the "Resistance" slider (on the right) can be used to change the load resistance. This changes the current flow, but not the voltage.
+~~PAGEBREAK~~ ~~CLEARFIX~~
+So the voltage gain is $A_\rm V = 1$. This would also have been seen in chapter [[https://wiki.mexle.org/circuit_design/1_amplifier_basics#feedback|feedback]]. There it was derived that for $A_\rm D \rightarrow \infty$ the voltage gain just results from $k$: $A_{\rm V}=\frac{1}{k}$. Since the entire output voltage is fed back here, $k=1$ and thus also $A_\rm V=1$.
+The output voltage $U_\rm O$ is therefore equal to the input voltage $U_\rm I$. This is where the name "voltage follower" comes from. Now one could assume, that this amplifier is of little help because also a direct connection would deliver $U_{\rm O}=U_{\rm I}$. But the important thing here is: because of the operational amplifier, there is __no feedback__  from U_O to U_I. This means that a resistor on the output side will not load the input side. In the simulation, the "Resistance" slider (on the right) can be used to change the load resistance. This changes the current flow, but not the voltage.
 <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjAnCAMB00IKw1gJldA7FsAOMiAbNIdgCyEiLJjRV2ICmAtGGAFAaYiaoipEQESgMpkQAVQD6AeXYBDEMzK5+ggMx9RIVQSV7aCIyDJpUBCJjJtEmdetzdT6MIXXDUmRGS-QB4IwR2AHcQTTVKcIxVaBCwrWhVZVVomDjk-kSlFX4yOljQ5l5+YuZcOk8+WK59TBFSyC1i8WkASU4sJXKSvmZCcUqTSSl2gCVa+t7UwbowEwYYRbhEdjz1JVQ88FQksB3MmJBCKTApaCk8iFwpRDPYU-PUKWP1Veh1vdn99AGs-JOzhdoFcbncHlInudXgAjDZbH76DBhPSxAAeG3cYXU3FYZHE6i24jwIFaADsAA4AVwALgAdADOAGUAJYAc1J8gANuxYWwRP5UOpIq40ui2Nx7NxNnMCXQiaoADIAe3kABMGUrSQyZDSqdT2GArOAIFotq4KsU5oZ8mhcOgQKNGPTmfTqfJSQBjRgdbjmiLgRBNPgtWTsVnGoMR-i4Q6xbFylKCYraOb5OI5bRRQRx7BKZOlDOWpbsIA noborder}} </WRAP>
-This behaviour can also be explained in another way: The input signal usually come from a voltage source, which can only produce low currents. That means the input signals are high impedance ($\text{high impedance}=\frac{\text{voltage}}{\text{low current}}$). However, a load of arbitrary impedance can be applied to the output. That is, to keep the output signal constant, a large current must be provided depending on the load. As the output resistance of the amplifier approaches 0, the signal is actually low impedance ($\text{low impedance}=\frac{\text{voltage}}{\text{(likely) large current}}$). This is where the second name of the circuit "**impedance converter**" comes from.
+This behavior can also be explained in another way: The input signal usually comes from a voltage source, which can only produce low currents. That means the input signals are high impedance ($\text{high impedance}=\frac{\text{voltage}}{\text{low current}}$). However, a load of arbitrary impedance can be applied to the output. That is, to keep the output signal constant, a large current must be provided depending on the load. As the output resistance of the amplifier approaches 0, the signal is low impedance ($\text{low impedance}=\frac{\text{voltage}}{\text{(likely) large current}}$). This is where the second name of the circuit "**impedance converter**" comes from.
 ~~PAGEBREAK~~ ~~CLEARFIX~~ <wrap #steps_to_goal />
-<WRAP column 100%> <panel type="danger" title="Remember: steps to the goal"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
+<WRAP column 100%>
+<panel type="danger" title="Remember: steps to the goal"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
 To solve tasks, the following procedure helps:
   - Where to? Clarification of the goal (here: always the relation between output and input signal)
-  - What to? Clarification of what is needed (here: always equations. Number of equations can be determined by number of variables)
+  - What to? Clarification of what is needed (here: always equations. The number of needed equations can be determined by the number of variables)
-  - With what? Clarification of what is already available (here: known equations: Stress amplification equation, basic equation, golden rules, loop/node theorem, relationships of stresses and currents of components).
+  - With what? Clarification of what is already available (here: known equations: voltage amplification equation, basic equation, golden rules, loop/node theorem, relationships of voltages and currents of components).
-  - Go. Work out the solution (here:inserting the equations) It helps to rearrange the equation so that $1/A_D$ appears without a prefactor. It is valid: $1/A_D \xrightarrow{A_D \rightarrow \infty} 0$
+  - Go. Work out the solution (here: inserting the equations) It helps to rearrange the equation so that $1/A_\rm D$ appears without a prefactor. It is valid: $1/A_{\rm D} \xrightarrow{A_{\rm D} \rightarrow \infty} 0$
-</WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
+</WRAP></WRAP></panel>
+ </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== 3.4 Non-inverting amplifier =====
@@ Zeile 217: / Zeile 224: @@
 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>.
-By considering the feedback, the result can be quickly derived here as well: only $\frac{R_2}{R_1+R_2}\cdot U_O$ is fed back from the output voltage $U_O$. So the feedback factor is $k=\frac{R_2}{R_1+R_2}$ and thus the voltage gain becomes $A_V=\frac{R_1+R_2}{R_2}$.
+By considering the feedback, the result can be quickly derived here as well: only $\frac{R_2}{R_1+R_2}\cdot U_{\rm O}$ is fed back from the output voltage $U_{\rm O}$. So the feedback factor is $k=\frac{R_2}{R_1+R_2}$ and thus the voltage gain becomes $A_{\rm V}=\frac{R_1+R_2}{R_2}$.
-This "trick" via $A_V=\frac{1}{k}$ 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}}
@@ Zeile 225: / Zeile 232: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<WRAP>
+<WRAP right>
-<panel type="default"> \\ <imgcaption pic5|Non-inverting amplifier> \\ </imgcaption> \\  \\ <button size="xs" type="link" collapse="Nichtinvertierender_Verstärker_Schaltung">{{icon>undo}}</button> \\  \\ <collapse id="Nichtinvertierender_Verstärker_Schaltung" collapsed="true">{{drawio>Nichtinvertierender_Verstärker_Schaltung}}</collapse> \\ <collapse id="Nichtinvertierender_Verstärker_Schaltung" collapsed="false">{{drawio>Nichtinvertierender_Verstärker_Schaltung2}}</collapse> \\ </panel>
+<panel type="default"> \\ <imgcaption pic5|Non-inverting amplifier> \\ </imgcaption> \\  \\ <button size="xs" type="link" collapse="Nichtinvertierender_Verstärker_Schaltung">{{icon>undo}}</button> \\  \\ <collapse id="Nichtinvertierender_Verstärker_Schaltung" collapsed="true">{{drawio>Nichtinvertierender_Verstärker_Schaltung.svg}}</collapse> \\ <collapse id="Nichtinvertierender_Verstärker_Schaltung" collapsed="false">{{drawio>Nichtinvertierender_Verstärker_Schaltung2.svg}}</collapse> \\ </panel>
 </WRAP>
 ^Step^Description^Implementation|
-|1|What is wanted?|<button size="xs" type="link" collapse="Step_1_1">{{icon>eye}}</button><collapse id="Step_1_1" collapsed="true">$A_V = \frac{U_O}{U_I}=?$</collapse>|
+|1|What is wanted?|<button size="xs" type="link" collapse="Step_1_1">{{icon>eye}}</button><collapse id="Step_1_1" collapsed="true">$A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=?$</collapse>|
-|2|Counting the variables \\ $->$ Number of equations needed|<button size="xs" type="link" collapse="Step_1_2">{{icon>eye}}</button><collapse id="Step_1_2" collapsed="true">5 voltages + 5 currents \\ $->$ Number of equations needed: 10</collapse>|
+|2|Counting the variables \\ $\rightarrow$ Number of equations needed|<button size="xs" type="link" collapse="Step_1_2">{{icon>eye}}</button><collapse id="Step_1_2" collapsed="true">5 voltages + 5 currents \\ $\rightarrow$ Number of equations needed: 10</collapse>|
-|3|Setting up the equations|<button size="xs" type="link" collapse="Step_1_3">{{icon>eye}} always usable equations</button><collapse id="Step_1_3" collapsed="true">(1) Basic equation: $U_O = A_D \cdot U_D$ \\ <fc #ffa500> Golden rules: \\ $R_D \rightarrow \infty$ so that (2+3) $I_p \rightarrow 0$ and $I_m \rightarrow 0$ \\ $R_O = 0$ \\ $A_D \rightarrow \infty$ </fc></collapse><button size="xs" type="link" collapse="Step_1_4">{{icon>eye}} Loops and nodes</button> (see <button size="xs" type="link" collapse="Non-inverting_amplifier_circuit">{{icon>undo}}</button>)<collapse id="Step_1_4" collapsed="true">(4) Loop I: $-U_I + U_D + U_2 = 0$ \\ (5) Loop II: $-U_2 -U_1 + U_O = 0$ \\ (6) Node I: $I_o = I_1$ \\ (7) Node II / voltage divider: $I_1 - I_2 - I_m = 0$</collapse><button size="xs" type="link" collapse="Step_1_5">{{icon>eye}} $U, I$ relationships across components</button><collapse id="Step_1_5" collapsed="true">(8) Resistor $R_1= \frac{U_1}{I_1}$ \\ (9) Resistor $R_2= \frac{U_2}{I_2}$</collapse>|
+|3|Setting up the equations|<button size="xs" type="link" collapse="Step_1_3">{{icon>eye}} always usable equations</button><collapse id="Step_1_3" collapsed="true">(1) Basic equation: $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$ \\ <fc #ffa500> Golden rules: \\ $R_\rm D \rightarrow \infty$ so that (2+3) $I_\rm p \rightarrow 0$ and $I_\rm m \rightarrow 0$ \\ $R_{\rm O} = 0$ \\ $A_\rm D \rightarrow \infty$ </fc></collapse><button size="xs" type="link" collapse="Step_1_4">{{icon>eye}} Loops and nodes</button> (see <button size="xs" type="link" collapse="Non-inverting_amplifier_circuit">{{icon>undo}}</button>)<collapse id="Step_1_4" collapsed="true">(4) Loop I: $-U_{\rm I} + U_{\rm D} + U_2 = 0$ \\ (5) Loop II: $-U_2 -U_1 + U_{\rm O} = 0$ \\ (6) Node I: $I_{\rm o} = I_1$ \\ (7) Node II / voltage divider: $I_1 - I_2 - I_\rm m = 0$</collapse><button size="xs" type="link" collapse="Step_1_5">{{icon>eye}} $U, I$ relationships across components</button><collapse id="Step_1_5" collapsed="true">(8) Resistor $R_1= \frac{U_1}{I_1}$ \\ (9) Resistor $R_2= \frac{U_2}{I_2}$</collapse>|
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-The calculation is done here again in detail (clicking right arrow "►" leads to the next step, [[:circuit_design:rechnung_nichtinvertierender_verstaerker|alternative representation]]): {{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_nichtinvertierender_verstaerker?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 noborder}} ~~PAGEBREAK~~ ~~CLEARFIX~~
+The calculation is done here again in detail (clicking the right arrow "►" leads to the next step, [[:circuit_design:rechnung_nichtinvertierender_verstaerker|alternative representation]]):
+{{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_nichtinvertierender_verstaerker?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
+~~PAGEBREAK~~ ~~CLEARFIX~~
 <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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 noborder}} </WRAP>
-So the voltage gain of the non-inverting amplifier is $A_V=\frac{R_1+R_2}{R_2}$ or $A_V=1+\frac{R_1}{R_2}$. Thus, the numerical value $A_V$ can only become larger than 1. This is shown again in the simulation. In real circuits, the resistors $R_1$ and $R_2$ will be in the range between a few $100 \Omega$ and a few $M\Omega$. If the sum of the resistors is too small, the operational amplifier will be heavily loaded. However, the output current must not exceed the maximum current. If the sum of the resistors is too large, the current $I_1=I_2$ can come into the range of the current $I_m$, which is present in the real operational amplifier.
+So the voltage gain of the non-inverting amplifier is $A_{\rm V}=\frac{R_1+R_2}{R_2}$ or $A_{\rm V}=1+\frac{R_1}{R_2}$. Thus, the numerical value $A_{\rm V}$ can only become larger than 1. This is shown again in the simulation. In real circuits, the resistors $R_1$ and $R_2$ will be in the range between a few $100 ~\Omega$ and a few $\rm M\Omega$. If the sum of the resistors is too small, the operational amplifier will be heavily loaded. However, the output current must not exceed the maximum current. If the sum of the resistors is too large, the current $I_1 = I_2$ can come into the range of the current $I_\rm m$, which is present in the real operational amplifier.
-The __**input and output resistance of the entire circuit**__  should also be considered here. Both resistors are marked here with a superscript 0 to distinguish them from the input and output resistance of the operational amplifier. The input resistance $R_{I}^0$ is given by $R_{I}^0=\frac{U_I}{I_I}$ with $I_I=I_p$. Thus, for the ideal operational amplifier, it is also true that the input resistance $R_{I}^0=\frac{U_I}{I_p} \rightarrow \infty$ becomes when $I_p \rightarrow 0$.
+The __**input and output resistance of the entire circuit**__  should also be considered here. Both resistors are marked here with a superscript 0 to distinguish them from the input and output resistance of the operational amplifier. The input resistance $R_{\rm I}^0$ is given by $R_{\rm I}^0=\frac{U_\rm I}{I_\rm I}$ with $I_{\rm I}=I_\rm p$. Thus, for the ideal operational amplifier, it is also true that the input resistance $R_{\rm I}^0=\frac{U_\rm I}{I_\rm p} \rightarrow \infty$ becomes when $I_\rm p \rightarrow 0$.
-In the **real case**  it is important in how far the total input resistance depends on the input resistance of the operational amplifier $R_{I}^0(R_D)$. This can be derived as follows: (clicking on right arrow "►" leads to the next step, [[:circuit_design:rechnung_nichtinvertierender_verstaerker_eingangswiderstand|alternative representation]]):
+In the **real case**  it is important in how far the total input resistance depends on the input resistance of the operational amplifier $R_{\rm I}^0(R_\rm D)$. This can be derived as follows: (clicking on the right arrow "►" leads to the next step, [[:circuit_design:rechnung_nichtinvertierender_verstaerker_eingangswiderstand|alternative representation]]):
-{{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_nichtinvertierender_verstaerker_eingangswiderstand?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 noborder}} ~~PAGEBREAK~~ ~~CLEARFIX~~
+{{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_nichtinvertierender_verstaerker_eingangswiderstand?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
+~~PAGEBREAK~~ ~~CLEARFIX~~
-So it can be assumed simplistically, that the input resistance of the whole circuit is many times higher than the input resistance of the operational amplifier. The output resistance $R_O^0$ of the whole circuit with real operational amplifiers shall only be sketched: In this case, the output resistance $R_O$ of the operational amplifier is in parallel with $R_1 + R_2$. Thus the output resistance $R_O^0$ will be somewhat smaller than $R_O$.
+So it can be assumed simplistically, that the input resistance of the whole circuit is many times higher than the input resistance of the operational amplifier. The output resistance $R_\rm O^0$ of the whole circuit with real operational amplifiers shall only be sketched: In this case, the output resistance $R_\rm O$ of the operational amplifier is in parallel with $R_1 + R_2$. Thus the output resistance $R_\rm O^0$ will be somewhat smaller than $R_\rm O$.
 <WRAP column 100%> <panel type="danger" title="Notice: non-inverting amplifier"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
@@ Zeile 254: / Zeile 268: @@
 For the __non-inverting amplifier__, the following holds:
-  * The input voltage $U_I$ is at the __non-inverting input__  of the operational amplifier.
+  * The input voltage $U_\rm I$ is at the __non-inverting input__  of the operational amplifier.
   * The feedback is done by a voltage divider $R_1 + R_2$
-  * The voltage gain is $A_V=\frac{R_1+R_2}{R_2}$ or $A_V=1+\frac{R_1}{R_2}$ and is always greater than 1.
+  * The voltage gain is $A_{\rm V}=\frac{R_1+R_2}{R_2}$ or $A_{\rm V}=1+\frac{R_1}{R_2}$ and is always greater than 1.
-  * Both input and output resistances of the overall circuit are smaller than these for the (real) operational amplifier used.
+  * Both input and output resistances of the overall circuit are smaller than those for the (real) operational amplifier used.
 </WRAP></WRAP></panel> </WRAP>
@@ Zeile 265: / Zeile 279: @@
 ===== 3.5 Inverting Amplifier =====
-<WRAP><panel type="default"> <imgcaption pic6|Inverting Amplifier> </imgcaption> {{drawio>Invertierender_Verstärker_Schaltung}} </panel></WRAP>
+The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_\rm I$, $\rm GND$, and $U_\rm O$. These terminals can be rearranged - while keeping the output terminal $U_\rm O$.
-<WRAP><imgcaption pic8|Converting non-inverting amplifier to inverting amplifier>  (here: $U_E=U_I$ and $U_A = U_O$){{:circuit_design:inv_2_ninv.gif}}</imgcaption></WRAP>
+<WRAP><panel type="default"> <imgcaption pic6|Inverting Amplifier> </imgcaption> {{drawio>Invertierender_Verstärker_Schaltung.svg}} </panel></WRAP>
-The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_I$, $GND$ and $U_O$. These terminals can be rearranged - while keeping the output terminal $U_O$. Thus the voltage divider $R_1 + R_2$ is no longer between $U_O$ and $GND$, but between $U_O$ and $U_O$, see <imgref pic6>.
+Thus the voltage divider $R_1 + R_2$ is no longer between $U_\rm O$ and $\rm GND$, but between $U_\rm O$ and $U_\rm O$, see <imgref pic6>.
 In this circuit, the resistor $R_2$ is also called the negative feedback resistor.
+<WRAP><imgcaption pic8|Converting non-inverting amplifier to inverting amplifier. U_E is the input voltage (Eingangsspannung), U_A is the output voltage (Ausgangspannung).>  (here: $U_{\rm E}=U_{\rm I}$ and $U_{\rm A} = U_{\rm O}$){{:circuit_design:inv_2_ninv.gif}}</imgcaption></WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-RwUjOsAuvSuUwA0uw7qbSurHmpxC22EA noborder}} </WRAP>
+Before the voltage gain is determined, the node $\rm K1$ in <imgref pic6> is to be considered first. This is just larger than the ground potential by the voltage $U_\rm D$; thus, it lies on the potential difference $U_\rm D$. For a feedback amplifier with finite voltage supply, $U_\rm O$ can only be finite, and thus $U_{\rm D}= U_{\rm O} / A_{\rm D} \rightarrow 0$ (cf. [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i#basic_equationgolden_rules|basic_equation of the operational amplifier]]), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $\rm K1$ is __always__  at ground potential in the ideal operational amplifier. This property is called **virtual ground**  because there is no direct short to ground. The op-amp regulates its output voltage $U_\rm O$ in such a way that the voltage divider sets a potential of $0~\rm V$ at node $\rm K1$. This can also be seen in the simulation by the voltage curve at $\rm K1$.
-Before the voltage gain is determined, the node $K1$ in <imgref pic6> is to be considered first. This is just larger than the ground potential by the voltage $U_D$; thus, it lies on the potential difference $U_D$. For a feedback amplifier with finite voltage supply, $U_O$ can only be finite and thus $U_D= U_O / A_D \rightarrow 0$ (cf. [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i#basic_equationgolden_rules|basic_equation of the operational amplifier]]), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $K1$ is __always__  at ground potential in the ideal operational amplifier. This property is called **virtual ground**  because there is no direct short to ground. The op-amp regulates its output voltage $U_O$ in such a way that the voltage divider sets a potential of $0V$ at node $K1$. This can also be seen in the simulation by the voltage curve at $K1$.
-<WRAP column 100%> <panel type="danger" title="Notice: virtual ground"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> For the ideal feedback amplifier, $U_D \rightarrow 0$ holds. This means that the same voltage is always present at both inputs. if one of the two voltages is fixed, for example by connecting ground potential or even by a fixed voltage source, this property is called **virtual ground**. </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
+The following diagram shows again the interactive simulation. \\
+$R_{\rm 1}$ and $R_{\rm 2}$ can be manipulated by the sliders. Hit ''Run / STOP'' to run
-<WRAP right><panel type="default"> <imgcaption pic9|Voltage divider in inverting amplifier> </imgcaption> {{drawio>Spannungsteiler_im_invertierenden_Verstärker}} </panel></WRAP>
+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5BOJyWoVaAmTkDseBGADgIFYA2Sc-AFnJFNJAMgddIFMBaAggKBy4QRVpgrNMREGPo0QAVQD6AeQF4QPStNyYNBXPUw6QcpQEk1QruTmYarLjSl3WpxRYBKerUd08Eur5QJibswTBUCCRg+nC4CLi4MVIwpHwATtJ20vZZcmCQcqwscHwA5nnghZVgOOF89hBc2PnkUjwEujROwQSKNIqQigAeJJCdiiyD0H08ipjz0KSDk4pg05BiK-IAOgDOFo3M3eBt2l09rEMDQ6MFExFwfY-jinMLmEsrQ+svm8tDXZ7VSCSouZgBHKuBQqPgAIw04JaejqMSYkD4w0RYHQYDAVgIJzAuTkxBAZgAdgAHACuABd9gBlACWZQpAEMADbw45CbDoTDUcBgegYrEEArgcRifKkdCkqQAGQA9uyACb7ZUU-bKem0ul8ADuwSCmnOUHK3nNZoKRQtxpE0nEjpwKQywlEkBS0i9JvgGMyoly1lsuWK-qNfu94Ix7JNztEzuYTB4TBK-ow-tIgpwNhFmxoxLovX9GKOjikIqkju63uI-RWoxYBHIkyGL1mfQWHbQvaQRBW9d+sHgW0B+wA0vxQZ72i2IV0QFO+GApPoQAAxAi9H3wPQgDz8VfMehbnc4Pc8A+YPhAA noborder}} </WRAP>
-For the determination of the voltage gain, the consideration of the feedback $A_V=\frac{1}{k}$ seems to be of little use at first. Instead, however, the determination via network analysis is possible. <imgref pic6> shows a possible variant to choose the loops for this purpose. However, network analysis is not to be done here, but is given as Exercise 3.5.1 below.
+<WRAP column 100%> <panel type="danger" title="Notice: virtual ground"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> For the ideal feedback amplifier, $U_\rm D \rightarrow 0$ holds. This means that the same voltage is always present at both inputs. if one of the two voltages is fixed, for example by connecting ground potential or even by a fixed voltage source, this property is called **virtual ground**. </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~
+<WRAP right><panel type="default"> <imgcaption pic9|Voltage divider in inverting amplifier> </imgcaption>{{drawio>Spannungsteiler_im_invertierenden_Verstärker.svg}} </panel></WRAP>
+For the determination of the voltage gain, the consideration of the feedback $A_{\rm V}=\frac{1}{k}$ seems to be of little use at first. Instead, however, the determination via network analysis is possible. <imgref pic6> shows a possible variant to choose the loops for this purpose. However, network analysis is not to be done here, but is given in Exercise 3.5.1 below.
 Instead, two other ways of derivation will be shown here to bring further approaches closer. For the first derivation, the **voltage divider**  $\boldsymbol{R_1 + R_2}$ is considered. For the unloaded voltage divider, the general rule is:
-\begin{align*} U_2 = U_G \cdot \frac{R_2}{R_1 + R_2} \end{align*}
+\begin{align*} U_2 = U_{12} \cdot \frac{R_2}{R_1 + R_2} \end{align*}
 This equation is now to be adapted for concrete use. First <imgref pic6>, the voltages of the voltage divider can be read as given in <imgref pic9>. From this, using the general voltage divider formula:
@@ Zeile 293: / Zeile 312: @@
 \begin{align*} U_2 = ( U_I - U_O ) \cdot \frac{R_2}{R_1 + R_2} \end{align*}
-With the virtual masses at node $K1$ in <imgref pic9>, it holds that $U_2$ points away from the (virtual) mass and thus equals $U_O$ in magnitude. By the same reasoning, $U_I = U_1$ holds. Thus it follows:
+With the virtual mass at node $\rm K1$ in <imgref pic9>, it holds that $U_2$ points away from the (virtual) mass and thus $U_2 = U_\rm O$. Similarly, $U_{\rm I} = U_1$ holds. Thus it follows:
-\begin{align*} - U_O = ( U_I - U_O ) \cdot \frac{R_2}{R_1 + R_2} \end{align*}
+\begin{align*}
+- U_{\rm O} = ( U_{\rm I} - U_{\rm O} ) \cdot \frac{R_2}{R_1 + R_2}
+\end{align*}
 And from that:
-<WRAP left 70%> \begin{align*} - U_O &= U_I \cdot \frac{R_2}{R_1 + R_2} - U_O \cdot \frac{R_2}{R_1 + R_2} \\ - U_O + U_O \cdot \frac{R_2}{R_1 + R_2} &= U_I \cdot \frac{R_2}{R_1 + R_2} \\ U_O \cdot (1 - \frac{R_2}{R_1 + R_2}) &= U_I \cdot \frac{R_2}{R_1 + R_2} \\ \frac{U_O}{U_I} &= \frac{\frac{R_2}{R_1 + R_2}}{1 - \frac{R_2}{R_1 + R_2}} \\ \frac{U_O}{U_I} &= \frac{R_2}{R_1 + R_2 - R_2} = \frac{-R_2}{R_1} \\ \\ \boxed{A_V = - \frac{R_2}{R_1}} \end{align*} </WRAP>
+<WRAP left 70%> \begin{align*}
+- U_O                                       &= U_I \cdot \frac{R_2}{R_1 + R_2} - U_O \cdot \frac{R_2}{R_1 + R_2} \\
+- U_O + U_O \cdot  \frac{R_2}{R_1 + R_2}    &= U_I \cdot \frac{R_2}{R_1 + R_2} \\
+  U_O       \cdot (\frac{R_2}{R_1 + R_2}-1) &= U_I \cdot \frac{R_2}{R_1 + R_2} \\
+                   \frac{U_O}{U_I}          &= \frac{\frac{R_2}{R_1 + R_2}}{\frac{R_2}{R_1 + R_2}-1} \\
+                   \frac{U_O}{U_I}          &=       \frac{R_2}{R_2 - (R_1 + R_2)} = \frac{R_2}{-R_1} \\ \\
+\boxed{                                  A_V =     - \frac{R_2}{R_1}}
+\end{align*} </WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<WRAP><panel type="default"> <imgcaption pic7|Inverting Amplifier - Animation> </imgcaption>
-{{url>https://www.geogebra.org/material/iframe/id/hhxhcqbp/width/600/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 400,500 noborder}} </panel></WRAP>
 For the second derivation, the current flow through the resistors $R_1$ and $R_2$ of the unloaded voltage divider is to be considered. These two currents $I_1$ and $I_2$ are just equal. Thus:
@@ Zeile 315: / Zeile 339: @@
 \begin{align*} \frac{U_1}{R_1}=\frac{U_2}{R_2} \end{align*}
-This can also be converted into a "seesaw" or mechanical analogue via **like triangles**. In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. [(Note1> To complete the mechanical analogue of the setup, one can assume that there is an external "force source". This always acts in such a way that it always lands on the height reference surface at the point corresponding to the virtual mass)].
+This can also be converted into a "seesaw" or mechanical analog via **like triangles**. In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. [(Note1> To complete the mechanical analogue of the setup, one can assume that there is an external "force source". This always acts in such a way that it always lands on the height reference surface at the point corresponding to the virtual mass)].
-Now, if a certain height (voltage $U_I$) is set, a certain height on the right side (voltage $U_O$) is obtained via force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in <imgref pic7> above. In the figure, all points marked in red (<fc #ff0000>{{fa>circle?10}}</fc>) can be manipulated. Accordingly, the input voltage $U_I = U_{in}$ is adjustable and automatically results in a voltage $U_O=U_{out}$. In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed.
+<WRAP><panel type="default"> <imgcaption pic7|Inverting Amplifier - Animation> </imgcaption>
+{{url>https://www.geogebra.org/material/iframe/id/hhxhcqbp/width/600/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 400,500 noborder}} </panel></WRAP>
-The **input resistance of the entire circuit**  $R_I^0=\frac{U_I}{I_I}$ is easily obtained by considering the input side: since $K1$ is at $0V$, $U_1 = U_I$. The complete current flowing into the input passes through resistor $R_1$. So it is then true that he input resistance is $R_I = R_1$. At the **output resistance of the whole circuit**  $R_O^0$ there is again a parallel connection between the output resistance of the operational amplifier $R_O$ and the resistor $R_2$. So the output resistance will be slightly smaller than the output resistance of the operational amplifier $R_O$.
+Now, if a certain height (voltage $U_\rm I$) is set, a certain height on the right side (voltage $U_\rm O$) is obtained via the force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in <imgref pic7> above. In the figure, all points marked in red (<fc #ff0000>{{fa>circle?10}}</fc>) can be manipulated. Accordingly, the input voltage $U_{\rm I} = U_{\rm in}$ is adjustable and automatically results in a voltage $U_{\rm O}=U_{\rm out}$. In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed.
+The **input resistance of the entire circuit**  $R_{\rm I}^0=\frac{U_{\rm I}}{I_{\rm I}}$ is easily obtained by considering the input side: since $\rm K1$ is at $0 ~\rm V$, $U_1 = U_\rm I$. The complete current flowing into the input passes through resistor $R_1$. So it is then true that the input resistance is $R_{\rm I} = R_1$. At the **output resistance of the whole circuit** $R_{\rm O}^0$, there is again a parallel connection between the output resistance of the operational amplifier $R_\rm O$ and the resistor $R_2$. So the output resistance will be slightly smaller than the output resistance of the operational amplifier $R_\rm O$.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 327: / Zeile 354: @@
 In the case of the __inverting amplifier__:
-  * The input voltage $U_I$ is at the __inverting input__  of the operational amplifier.
+  * The input voltage $U_\rm I$ is at the __inverting input__  of the operational amplifier.
   * The feedback is done by a voltage divider of $R_1$ and $R_2$.
-  * The voltage gain is $A_V= - \frac{R_2}{R_2}$ and is always less than or greater than 0. However, the magnitude of the voltage gain can be greater than or less than 1.
+  * The voltage gain is $A_{\rm V}= - \frac{R_2}{R_2}$ and is always less than or greater than 0. However, the magnitude of the voltage gain can be greater than or less than 1.
   * The input resistance of the whole circuit is defined by $R_1$ and is usually smaller than the input resistance of the used (real) operational amplifier. \\ The output resistance is smaller than that of the used (real) operational amplifier.
@@ Zeile 335: / Zeile 362: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
 ====== Exercises ======
@@ Zeile 346: / Zeile 374: @@
 {{page>uebung_3.5.3&nofooter}}
 {{page>uebung_3.5.4&nofooter}}
+{{page>uebung_3.5.5&nofooter}}
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 351: / Zeile 380: @@
 ====== Learning Questions ======
-  * Explain the difference between unipolar and bipolar power supply of an OPV.
+  * Explain the difference between the unipolar and bipolar power supply of an OPV.
   * Draw a sketch for bipolar and one for unipolar power supply.
   * What are the advantages and disadvantages of unipolar and bipolar supply in OPV?