====== Block 22 — Negative-feedback Op-Amp Circuits ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can
  * explain why negative feedback “tames” an op-amp with very large open-loop gain $A_{\rm D}$, and use the ideal assumptions ($A_{\rm D}\rightarrow\infty$, $R_{\rm D}\rightarrow\infty$, $R_{\rm O}\rightarrow 0$) to analyze circuits.
  * apply a systematic equation-based workflow (goal $\rightarrow$ variables $\rightarrow$ equations $\rightarrow$ solving) to derive transfer functions such as $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}$ for basic feedback circuits.
  * recognize and use the concept of **virtual ground** (more generally: $U_{\rm D}\rightarrow 0$) at the inverting input in negative-feedback circuits.
  * apply the **superposition method** to linear op-amp circuits with multiple independent sources (e.g. the inverting summing amplifier) and compute $U_{\rm O}(U_{\rm I1},U_{\rm I2},\dots)$.
  * identify the circuit topologies and transfer functions of
      - the voltage follower ($A_{\rm V}=1$),
      - the non-inverting amplifier ($A_{\rm V}=1+\frac{R_1}{R_2}$),
      - the inverting amplifier ($A_{\rm V}=-\frac{R_2}{R_1}$),
      - the inverting summing amplifier ($U_{\rm O}=-\sum_i \frac{R_0}{R_i}U_{{\rm I}i}$),
      - the current-to-voltage converter (transimpedance), $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm in}}=-R_1$,
      - the voltage-to-current converter (transconductance), $S=\frac{I_{\rm out}}{U_{\rm in}}$.
  * name typical applications of these circuits (buffer/impedance converter, programmable gain amplifier, summing/mixing, differential measurement, current sensing, photodiode readout, voltage-controlled current source).
</callout>

===== Preparation at Home =====

Well, again 
  * read through the present chapter and write down anything you did not understand.
  * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting). 

For checking your understanding please do the following exercises:
  * ...

===== 90-minute plan =====
  - Warm-up (10 min):
    - Quick recall: ideal op-amp model and “golden rules” in negative feedback: \\ $I_{\rm p}\approx 0$, $I_{\rm m}\approx 0$, and (with feedback) $U_{\rm D}=U_{\rm p}-U_{\rm m}\rightarrow 0$.
    - One-minute concept check: why a voltage follower is useful although $A_{\rm V}=1$ (buffering / impedance conversion).
  - Core concepts & derivations (55 min):
    - Introductory motivation: current sensing (10 min)
      - Why a shunt resistor alone may be insufficient; need for current-to-voltage conversion plus amplification.
      - Link to the idea of “transimpedance” as gain: $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm in}}$.
    - Voltage follower (10 min)
      - Circuit idea: $U_{\rm O}$ fed back directly to the inverting input.
      - Derive $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}=1$ using $U_{\rm D}\rightarrow 0$.
      - Interpretation: decouples input from load (high input resistance, low output resistance) → impedance converter / buffer.

    - Non-inverting amplifier (10 min)
      - Feedback factor from divider: $k=\frac{R_2}{R_1+R_2}$.
      - Transfer function: $A_{\rm V}=\frac{1}{k}=1+\frac{R_1}{R_2}$.
      - Design notes: practical resistor range, loading, bias currents (qualitative).

    - Inverting amplifier + virtual ground (15 min)
      - Virtual ground at node $\rm K1$ (inverting input): in ideal feedback $U_{\rm D}\rightarrow 0$, so node is held near ground without a physical short.
      - Derive $A_{\rm V}=-\frac{R_2}{R_1}$ (via divider approach or equal currents through $R_1$ and $R_2$).
      - Input resistance insight: $R_{\rm I}^0\approx R_1$ (ideal), and output resistance reduced by feedback.

    - Superposition with the inverting summing amplifier (10 min)
      - Linearity check: resistors + (ideal) op-amp behavior under negative feedback.
      - Apply superposition: analyze each source separately (others set to zero), then sum: $ U_{\rm O}=-\left(\frac{R_0}{R_1}U_{\rm I1}+\frac{R_0}{R_2}U_{\rm I2}+\cdots\right). $
      - Application link: mixers, weighted sums, DAC concept.

  - Practice (20 min):
    - Mini-problems (individually → pair check):
      - Choose $R_1,R_2$ for a non-inverting amplifier with target gain $A_{\rm V}$.
      - Inverting amplifier: given $R_1,R_2$, compute $U_{\rm O}(t)$ for a supplied $U_{\rm I}(t)$; mark phase inversion.
      - Summing amplifier: compute $U_{\rm O}$ for two sources; verify via superposition.
      - Transimpedance amplifier: compute $U_{\rm O}$ for a given $I_{\rm in}$ and $R_1$; interpret sign.
      - Transconductance circuit: given $U_{\rm in}$, estimate $I_{\rm out}$ (qualitative if full derivation is later).

  - Wrap-up (5 min):
    - Summary box: “Feedback enforces $U_{\rm D}\rightarrow 0$; resistors set the gain.”
    - Common pitfalls checklist (see below).
    - Outlook: differential amplifier as subtraction / common-mode rejection; application circuits (PGA, instrumentation concepts).


===== Conceptual overview =====
<callout icon="fa fa-lightbulb-o" color="blue">
  * Negative feedback turns a very large (and imperfect) op-amp gain $A_{\rm D}$ into predictable closed-loop behavior: the circuit “chooses” $U_{\rm O}$ so that the differential input voltage $U_{\rm D}=U_{\rm p}-U_{\rm m}$ becomes (almost) zero.
  * In the ideal feedback limit ($A_{\rm D}\rightarrow\infty$) we can treat
      - $I_{\rm p}=I_{\rm m}=0$ (no input currents),
      - $U_{\rm p}\approx U_{\rm m}$ (virtual short between inputs),
      - and the output as an ideal voltage source ($R_{\rm O}\approx 0$).
  * The voltage follower is the simplest example: $A_{\rm V}=1$ but with a huge practical effect—high input impedance and low output impedance (buffering).
  * Resistors in the feedback path set **ratios**, and these ratios define gains:
      - non-inverting: $A_{\rm V}=1+\frac{R_1}{R_2}$,
      - inverting: $A_{\rm V}=-\frac{R_2}{R_1}$ (with virtual ground at the summing node).
  * Multi-source op-amp circuits remain linear, so superposition works: a summing amplifier is just a controlled way of forming weighted sums.
  * Negative-feedback op-amp circuits are not only “voltage amplifiers”: they also realize signal conversions:
      - current $\rightarrow$ voltage (transimpedance, $\frac{U_{\rm O}}{I_{\rm in}}$),
      - voltage $\rightarrow$ current (transconductance, $\frac{I_{\rm out}}{U_{\rm in}}$),
    which underpins current sensing, photodiode readout, and controlled current sources.
</callout>

===== Core content =====

<WRAP>
<callout type="info" icon="true">
==== Introductory Example ====

In various applications, currents must be measured. In an electric motor, for example, the torque is caused by the current flowing through the motor. A motor control and a simple overcurrent shutdown are based on the knowledge of the current. For further processing, a voltage must be generated from the current. The simplest current-to-voltage converter is the ohmic resistor. A sufficiently large voltage as required by a microcontroller, for example, cannot be achieved with this. So not only does the current have to be converted, but also the generated potential difference has to be amplified.

One such current sense amplifier is the [[http://www.ti.com/lit/ds/symlink/ina240.pdf|INA 240]] device. This is installed as shown below. In the simulation, a real current source feeds the electrotechnical image of a DC motor on the left (in the example: inductance with $L_{\rm L}=10~\rm mH$ and internal resistance $R_{\rm L}=1~\Omega$). The current flowing from the motor is conducted through a measuring resistor ($R_{~\rm M}=0.01~\Omega$) which is noticeably smaller than the internal resistance of the motor. Thus, most of the power acts in the motor and the current is only marginally affected by the sense resistor. The simulation above shows the inner workings of the current measuring amplifier.

The following explains ways in which such circuits can be understood.

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</WRAP>

</callout></WRAP>

==== Voltage follower ====


In the [[Block21]] 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|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 $U_\rm O$ 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> \\ \\
  - Now **equations are set up**  that can be used. These are:
      - **Basic equation**: (1) $U_{\rm O} = U_{\rm D} \cdot A_\rm D$
      - **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) $-U_{\rm I} + U_{\rm D} + U_{\rm O} = 0$.</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) $I_{\rm o} = I_m$</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) $A_{\rm V}=\frac{U_{\rm O}}{U_\rm I}$</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 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}}


~~PAGEBREAK~~ ~~CLEARFIX~~


So the voltage gain is $A_\rm V = 1$. This would also have been seen in chapter [[Block21]]. 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 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%>

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. The number of needed equations can be determined by the number of variables) \\ \\
  - With what? \\ Clarification of what is already available (here: known equations: voltage gain 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 $1/A_\rm D$ appears without a prefactor. It is valid: $1/A_{\rm D} \xrightarrow{A_{\rm D} \rightarrow \infty} 0$

</WRAP></WRAP></panel>
 </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~

==== Non-inverting amplifier ====

So far, the entire output voltage has been negative-feedback. Now only a part of the voltage is to be fed back. \\ To do this, the output voltage can be reduced using a voltage divider $R_1+R_2$. The circuit for this can be seen in <imgref pic5>.

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 $A_{\rm V}=\frac{1}{k}$ is no longer possible for some of the following circuits. Accordingly, a possible solution via network analysis is to be derived here as well.

~~PAGEBREAK~~ ~~CLEARFIX~~

<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.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">$A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=?$</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: $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$ \\ <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 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~~

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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 $R_{\rm I}^0$ 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 $R_{\rm I}^0(R_\rm D)$. \\ 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=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 $R_\rm O^0$ 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%>

For the __non-inverting amplifier__, the following holds:

  * 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 $A_{\rm V}=\frac{R_1+R_2}{R_2}$ 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 those for the (real) operational amplifier used.

</WRAP></WRAP></panel> </WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Inverting Amplifier ====

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 voltage divider): $U_\rm I$, $\rm GND$, and $U_\rm O$. \\ 
These terminals can be rearranged - while keeping the output terminal $U_\rm O$. 

<WRAP><panel type="default"> <imgcaption pic6|Inverting Amplifier> </imgcaption> {{drawio>Invertierender_Verstärker_Schaltung.svg}} </panel></WRAP>

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. \\ 
$U_\rm E$ is the input voltage (Eingangsspannung), $U_\rm A$ 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~~

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 $U_{\rm D}= U_{\rm O} / A_{\rm D} \rightarrow 0$, 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 $\rm K1$.


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>{{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>

<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_{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:

\begin{align*} U_2 = ( U_I - U_O ) \cdot \frac{R_2}{R_1 + R_2} \end{align*}

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_{\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 (\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~~

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:

\begin{align*} I_\boxed{}=\frac{U_\boxed{}}{R_\boxed{}}=const. \quad \text{with} \: \boxed{}=\{1,2\} \end{align*}

respectively

\begin{align*} \frac{U_1}{R_1}=\frac{U_2}{R_2} \end{align*}

This can also be converted into a "seesaw" or mechanical analog via **similar triangles** (see {{wp>Similarity_(geometry)#Similar_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)].

<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>

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~~

<WRAP column 100%> <panel type="danger" title="Notice: Inverting Amplifier"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

In the case of the __inverting 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 $A_{\rm V}= - \frac{R_2}{R_2}$ 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.

</WRAP></WRAP></panel> </WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Inverting Summing Amplifier ====

<WRAP><panel type="default"> 
<imgcaption pic1|Inverting Summing Amplifier >
</imgcaption>
\\ {{drawio>Umkehraddierer.svg}}
</panel></WRAP>

From the [[3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]] another circuit can be derived, which can be seen in <imgref pic1>. Here, both the green part of the circuit and the purple part correspond to an inverting amplifier.

How can $U_\rm O$ be calculated in this circuit? To do this, it is first important to understand what is being sought. The goal is to find the relationship between output and input signals: $U_{\rm O}(U_{\rm I1}, U_{\rm I2})$. Different ways to get there, were explained in [[Block07]] and [[Block08]]. Here we will outline a different way.

In the case of a circuit with several sources, superposition is a suitable method, in particular the superposition of the effect of all sources in the circuit. For superposition, it must be ensured that the system behaves linearly. The circuit consists of ohmic resistors and the operational amplifier. These two components give twice the output value when the input value is doubled - they behave linearly. For superposition, the effect of the two visible voltage sources $U_{\rm I1}$ and $U_{\rm I2}$ must be analyzed in the present circuit. \\
In **case 1** the voltage source $U_{\rm I1}$ must be considered - the voltage source $U_{\rm I2}$ must be short-circuited for this purpose. The equivalent circuit formed corresponds to an inverting amplifier across $R_2$ and $R_0$. However, there is an additional resistor $R_1$ between the inputs of the operational amplifier. What is the influence of this resistor? The differential voltage $U_\rm D$ between the inputs of the operational amplifier approaches 0. Thus, the following also applies to the current through $R_1$: $I_{1(1)} \rightarrow 0$. Thus the circuit in case 1 is exactly an inverting amplifier. For case 1, $A_{V(1)} = \frac{U_{O(1)}}{U_{\rm I1}} = - \frac{R_0}{R_1}$ and thus: $U_{O(1)}= - \frac{R_0}{R_1} \cdot U_{\rm I1}$. \\
Using the same procedure, **case 2** for considering the voltage source $U_2$ gives: $U_{\rm O(2)}= - \frac{R_0}{R_2} \cdot U_{\rm I2}$. \\
In superposition, the effect results from the **addition of partial effects**:

$\boxed{U_{\rm O} = \sum_i U_{\rm O(i)} = - (\frac{R_0}{R_2} \cdot U_{I2} + \frac{R_0}{R_1} \cdot U_{I1})}$.

Also, considering the node set for $\rm K1$ in <imgref pic1> gives the same result.

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</WRAP>

The **Inverting Summing Amplifier** (also called: Summing Amplifier or Voltage Adder) can be extended to any number of inputs. \\ The simulation above shows the superposition of several inputs. Depending on the resistances at the different inputs, a different current flows into the circuit.

This circuit was used in analog [[https://en.wikipedia.org/wiki/Electronic_mixer|audio mixers]]. This allows a combination of several signals with different gains (by the input resistors $R_i$ with $i=1, ..., n$). Furthermore, the overall gain can be changed by $R_0$. A big advantage of this circuit is also that the summation at node $\rm K1$ is done on potential $U_\rm D$. This means that capacitive interference concerning the ground potential (and therefore the case) is virtually non-existent.

A very similar concept allows the construction of a [[elektronische_schaltungstechnik/8_weiterfuehrendes#digital-analog-wander_dac|Digital-Analog Converter, DAC]].

~~PAGEBREAK~~ ~~CLEARFIX~~

{{page>uebung_4.1.1&nofooter}}

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Differential Amplifier / Subtractor ====

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</WRAP>

In addition to the (reverse) adder, there is also a circuit for subtracting two input values. This circuit became the core of the introductory example. But also in the simulation below this circuit is shown in another example: In this case, a [[https://en.wikipedia.org/wiki/Differential_signalling|differential input signal]] is shown on the left. Differential means that the signal on one line is __, not__ transmitted concerning a reference voltage (usually ground potential) on a second line. Instead, the signal is transmitted to both lines in opposite directions. If a disturbance acts equally on both lines (which is often the case when lines are close to each other), the effect of the disturbance can be eliminated by forming the difference.

How can the relationship $U_{\rm O}(U_{\rm I1}, U_{\rm I2})$ between output and input signals be determined for this circuit?

<WRAP>
<panel type="default"> 
<imgcaption pic2|Differential Amplifier>
</imgcaption>
\\ {{drawio>Differenzverstärker.svg}}
</panel>
</WRAP>

Again, various network analysis concepts could be used to look at the circuit (e.g. superposition or mesh and node sets). Again, another possibility is to split the circuit as color-coded in the <imgref pic2>. \\
The __green part__ shows a voltage divider $R2 + R4$. Since the input resistance of the operational amplifier is very large, this voltage divider is unloaded. The voltage at node $\rm K2$ or at the noninverting input $U_\rm p$ is just given by the voltage divider: $U_{\rm p} = U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4}$. \\
The __violet part__ corresponds to an inverting amplifier, but the voltage at the node $\rm K1$ or at the inverting input $U_\rm m$ is just equal to $U_\rm p$ due to the feedback, since $U_\rm D \rightarrow \infty$. Thus, the current flowing into node $\rm K1$ via $R_1$ results from $I_1=\frac{U_{\rm I1} - U_\rm p}{R_1}$. The output voltage is given by $U_{\rm O} = U_{\rm p} - U_3$, where the voltage $U_3$ is given by the resistance $R_3$ and the current through $R_3$. The current through $R_3$ is just the same as the current through $R_1$, i.e. $I_1$.

The result is: \\
$       U_{\rm O} = U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - R_3 \cdot \frac{U_{\rm I1} - U_{\rm p}}{R_1} $ \\
$       U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4} - U_{\rm I1} \cdot \frac{R_3}{R_1} +  U_{\rm I2} \cdot (\frac{R_3}{R_1}\cdot \frac{R_4}{R_2 + R_4})$ \\ 
$\boxed{U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4}  \frac{R_1 + R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1}}$

~~PAGEBREAK~~ ~~CLEARFIX~~

<WRAP><panel type="default"> 
<imgcaption pic3|Differential Amplifier - Animation>
</imgcaption> 

<collapse id="foo" collapsed="true"><well>
{{url>https://www.geogebra.org/material/iframe/id/msxjcgz4/width/450/height/300/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 450,300 noborder}}
</well></collapse>

<collapse id="foo" collapsed="false">
<button type="warning" collapse="foo">Please click to see the animation!</button>
</collapse>
</panel></WRAP>

Two simplifications should be considered here:
  - If $R_1 = R_2$ and $R_3 = R_4$ are chosen, the equation further simplifies to:  \\ (nbsp) $\boxed{U_{\rm O} = \frac{R_3}{R_1}\cdot(U_{\rm I2}-U_{\rm I1})}$. \\ This variant can be found in various measurement circuits. \\ \\
  - Alternatively, if $R_1 = R_3$ and $R_2 = R_4$ is chosen, the result is: \\ (nbsp) $\boxed{U_{\rm O}= U_{\rm I2}-U_{\rm I1}}$ \\ This would also result in case 1. if $R_1 = R_2 = R_3 = R_4$ is chosen.

The animation shows how the 2nd case would result in similar triangles. The connection of the two "seesaws" at the point $\rm K_1 K_2$ is caused by the operational amplifier, through which the voltage $U_\rm p$ and $U_\rm m$ converge to $U_\rm D \rightarrow 0$.

A big advantage of this circuit is that even very large voltages can be used as input voltage, if $R_1 \gg R_3$ and $R_2 \gg R_4$ are chosen. This would divide the input voltages down and display a fraction of the difference as the result. The main drawback of the circuit is that the gain/attenuation depends on more than one resistor. This makes a quick choice of gain difficult.

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Current-Voltage-Converter ====

<WRAP><panel type="default"> 
<imgcaption pic4| Current-Voltage-Converter>
</imgcaption>
\\ {{drawio>Strom-Spannungs-Wandler.svg}}
</panel></WRAP>

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.0005+0.03528660814588489+57+5+50%0Aa+384+224+480+224+8+15+-15+1000000+0.00009999900000999991+0+100000%0Ag+384+240+384+272+0%0Ar+384+160+480+160+0+1000%0Aw+480+160+480+224+0%0Aw+384+160+384+208+0%0A370+320+208+384+208+1+0%0Ai+272+208+320+208+0+0.01%0Aw+272+208+272+256+0%0Ag+272+256+272+272+0%0Ax+258+167+332+170+4+14+Stromquelle%0A368+480+224+528+224+0+0%0A38+6+0+-0.01+0.01+Strom%5Csder%5CsStromquelle%0A noborder}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~

In <imgref pic4> one can see the circuit of a current-voltage converter. The current-to-voltage converter changes its __output voltage__ based on an __input current__. This circuit is also called a [[https://en.wikipedia.org/wiki/Transimpedance_amplifier|transimpedance amplifier]] because here the transfer resistance - that is, the trans-impedance - represents the gain. Generally, the gain was expressed as 
$$A={ {\rm output} \over {\rm input} }$$. 

In the case of the current-to-voltage converter, the gain is defined as:

$$R = {{U_{\rm out}} \over I_{\rm in}} ={{U_{\rm o}} \over I_{\rm I}} = - R_1$$

$R_1$ is the resistor used in the circuit.

In the simulation, the slider on the right ("Current of current source") can be varied. This changes the input current and thus the output voltage.
 
This circuit can be used, for example, to read a [[https://en.wikipedia.org/wiki/Photodiode|photodiode in volt-free circuit]] ([[https://en.wikipedia.org/wiki/Transimpedance_amplifier|further explanation]] and integrated circuit {{elektronische_schaltungstechnik:tsl250r.pdf}}).


~~PAGEBREAK~~ ~~CLEARFIX~~
==== Voltage-to-Current Converter ====

<WRAP><panel type="default"> 
<imgcaption pic5| Voltage-to-Current Converter>
</imgcaption>
\\ {{drawio>Spannungs-Strom-Wandler.svg}}
</panel></WRAP>

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</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~

Next, consider the voltage-to-current converter. With this, an __output current__ is set proportional to an __input voltage__.

Here, the general gain $$A={ {\rm output} \over {\rm input} }$$ to

$$S = {{I_{\rm out}} \over U_{\rm in}} = {{I_{\rm o}} \over U_{\rm I}}$$

The quantity $S$ is called the transfer conductance.

This circuit can be used, for example, to generate a voltage-regulated current source. \\
In practical applications, often specialized amplifiers, called {{wp>Operational Transconductance Amplifier}} (transconductance from __trans__mission __conductance__), are used.

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Applications ====
=== Programmable Gain Amplifier ===

Often in applications an analog signal is too small to process (e.g. to digitalize it afterward). \\
To amplify it an OpAmp can be used. However, for a wide input range, it might be beneficial to have an adjustable scale.

This can be done with a simple non-inverting amplifier combined with a resistor network as seen in the next simulation. \\
In this case, a so-called **single-ended** input is used. This means the input voltage is always referred to the ground. 

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l3AWAnC1b0DYq2QJgRgrgBwDsGpArApEsSJaQyAhJQKYC0AjNwFABDELkpYkWMJVwhxMkN0ogei7vHjZEKSnGqQa3Ghiwwd8tXD4AnFsXrcwWBLflSoM81Zt2M9J3eIIbkhqnn7CcF7C3MbuFtZhuGDSCU5uqhYA7pEKyc7RkFB8WWH52T6FxXkB2UjSkEU1ufS4EfWVzdGRuPgVXT0JSYUA5l2pCaJubeCusgidslNhs3kkvSUO2auLeRgFrNKl23a1LIPcJ4uD3YEJPVNJduTCE9xPRzJYlBuyuA0iWMQCi16IDhHwAG7hOyuYEgUEFPYI2CQRQI7CUPgjWE5KHyN6FUqSA5PMDVV5YbgaSCU9LmeB8Uo0AoGG5nBCBSkmalmOn0xkRWH7XGc5E03n04oRFngMns3qkjlyhXyXZrAWQejK2H3aqw5XnOoNIXcVYlLYNB54iSucmFFpMBzQ6T61yBADK-CyjpczvKhyNnRNe0D5vtLBDezOLRYIDd9TDcwp1WNydji0DycDFz4YbAG3wBQcQJoMbjfAAMuB8yWC8Ia-IQAAzAQAGwAzuxmPVK3msLhOrX+32G83253UTnIA7850i1FHLH+AAjKsc1QsShIYQYQ0ADx9IFIEBa0ieHOkAEEAHatgD2QwAOm2AJJXvj7r4ArebsTJeSXm8W3vJ8AHkAFcABd3xVSlBynOsIHPEAABEAEshlQiDWyfV8-heJ5-nkE01mcWZ5mMPggA noborder}}
</WRAP>

\\ \\
When the signal is not referred to the ground, the following circuit based on an instrumentation amplifier can be used. \\
In this case, the input signal is **differential**. Referred to the ground the input signal (here the difference of $5 ~\rm mV$) can have an offset voltage with regard to the ground. \\
An example of this setup is the [[https://www.ti.com/lit/ds/symlink/ina351.pdf|INA 351]].

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgAnEPKsPG7lTRVekMGPZceIFGEKCZ2FFBDFxkTgpQ08WwgNFi2AdwVgM86XwMmtcO7o2npaXS7lRbLpVpoivQn6+jgHgFuDy5vJODhEy+J6mHtpULkEaAG4gLMTKuAI5yvoqVOmwkEgiUNAIbHJUTMVgudk+zcoQMHAQ6vB9GvXZHoPDhLzV3cj9fXVjQ-Io2nGLAp3lPdMzpiyRc0xtczH7HXtt4UfJS8cyS0c+K63KrolxYDQCVu8vTFYtjastDRochMHQRXTXOS6AQAZXY2zBUOyiPOtl42Gi4C+QMg5HRu3AGNoIBhOLx8V4C3ssJiQnsdJCwOyLgQ8iYrJkHJpbAAMsyhD4WGAnvcVAAzACGABsAM4Mahk-mKZTshY+WHsABG2TUqnolBkWE8AA9sigELo5MQzQlzMpVsoAIIAO2lAHsAOYAHRlAElnWxTToILJrQgaCHFsThSAXe7vTKAPIAVwALoGhgbZBAWJABLIBA6QAARACWHtLqelPv9bAl4BQukbTYSze44CQLCQvXgE0gpAHA76NFIjlgG02dYbbgZCmtYE7C6mfQmb0Ho5HpE3Y7E3c2MWewstjenLyPp-PxGitnPmA6J5SZ4fQWej6OvxVVhjMS-n94395JVzTZVgng5ZQqElWV5UqKclF4Fo-AQ5RLUXPd+j7QdKDEPAcDwfRii6fsB2wlBsBwDA3iafdNHg2JpEMDQuDottXzwRxl00CMFnYkAWN4xiuI8aQ6IYzjTBYhJ+JCCS0DCeRRP8UxuIUFT2heNTEPsdSLhafIhhPfSUFsQoQGKFgT2KYyuGeFxbRPKhsPgTRXxfVsBIofpQjeD5-wgtgslM7AEBVdT9KqMoxBKapag9IY9JC+K8kIDiNFNJhtAEbBzDNMB6DootEzFMU5XTbZzyaSybHKwy81PcLQjvU8mpiS9LCoK8XmkTrfy6oQMBQoRCLYAAHWgmkQqzePoJTaGSXiaCmmS5siSaAX8oA noborder}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~
===== Common pitfalls =====
  * Mixing up open-loop and closed-loop gain:
      - open-loop: $U_{\rm O}=A_{\rm D}\,U_{\rm D}$,
      - closed-loop: $A_{\rm V}=\frac{U_{\rm O}}{U_{\rm I}}$ is set mainly by resistor ratios.
  * Forgetting the conditions for the “golden rules”: $U_{\rm p}\approx U_{\rm m}$ only holds when the op-amp is in negative feedback and not saturated.
  * Sign errors:
      - inverting amplifier has a minus sign $A_{\rm V}=-\frac{R_2}{R_1}$,
      - transimpedance example gives $U_{\rm O}=-R_1 I_{\rm in}$ (direction conventions matter).
  * Misusing “virtual ground”:
      - the inverting input node can be near $0~\rm V$, but it is **not** physically connected to ground and cannot source/sink arbitrary current.
  * Superposition mistakes:
      - when “turning off” a voltage source, replace it by a short; when “turning off” a current source, replace it by an open.
      - ensure linear operation (no clipping, no saturation).
  * Ignoring practical limits:
      - too small $R_1+R_2$ loads the op-amp output (current limit),
      - too large resistors make bias currents and offsets more visible,
      - finite supply rails limit $U_{\rm O}$ and can break the ideal assumptions.

===== Exercises =====
==== Worked examples ====

{{page>uebung_3.5.1&nofooter}}
{{page>uebung_3.5.2&nofooter}}
{{page>uebung_3.5.4&nofooter}}
{{page>uebung_3.5.5&nofooter}}

<panel type="info" title="Task 22.1 Voltage follower as impedance converter"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A voltage follower is built with an ideal op-amp. \\  
The input is a voltage source $U_{\rm I}=2.0~\rm V$ with internal resistance $R_{\rm S}=10~\rm k\Omega$.   \\
The output drives a load resistor $R_{\rm L}$ which is varied between $100~\Omega$ and $100~\rm k\Omega$.

  - Determine the input current drawn from the source for $R_{\rm L}=100~\Omega$ and for $R_{\rm L}=100~\rm k\Omega$.
  - Explain briefly why the load does not “pull down” the source voltage in this circuit.

<button size="xs" type="link" collapse="Loesung_22_1_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_1_Tipps" collapsed="true">
  * The input voltage source sees (ideally) infinite input resistance.
</collapse>

<button size="xs" type="link" collapse="Loesung_22_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_1_Endergebnis" collapsed="true">
  * Input current from the source: $I_{\rm S}\approx 0$ for both load values (ideal model).
</collapse>

</WRAP></WRAP></panel>


<panel type="info" title="Task 22.2 Non-inverting amplifier design"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A non-inverting amplifier should have a voltage gain of $A_{\rm V}=11$.

  - Choose resistor values $R_1$ and $R_2$ in the kOhm-range.
  - If $U_{\rm I}=0.25~\rm V$, compute $U_{\rm O}$ (ideal op-amp, no saturation).
  - What happens to $U_{\rm O}$ if the op-amp supply rails are $\pm 2.5~\rm V$?

<button size="xs" type="link" collapse="Loesung_22_2_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_2_Tipps" collapsed="true">
  * Rearrange $1+\frac{R_1}{R_2}=11$ to a resistor ratio.
  * Check the computed $U_{\rm O}$ against the supply rails (clipping).
</collapse>

<button size="xs" type="link" collapse="Loesung_22_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_2_Endergebnis" collapsed="true">
  * One possible choice: $R_2=10~\rm k\Omega$, $R_1=100~\rm k\Omega$.
  * Ideal: $U_{\rm O}=11\cdot 0.25~\rm V=2.75~\rm V$.
  * With $\pm 2.5~\rm V$ rails: $U_{\rm O}$ clips near $+2.5~\rm V$ (model-dependent headroom).
</collapse>

</WRAP></WRAP></panel>


<panel type="info" title="Task 22.3 Inverting amplifier and virtual ground"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

An inverting amplifier is built with $R_1=2.2~\rm k\Omega$ and $R_2=22~\rm k\Omega$.  
The non-inverting input is connected to ground.

  - Compute the closed-loop gain $A_{\rm V}$.
  - For an input $U_{\rm I}(t)=0.30~\rm V~\sin(2\pi\cdot 1\,{\rm kHz}\cdot t)$, determine $U_{\rm O}(t)$.
  - State the potential at the inverting input node (the summing node) in the ideal negative-feedback case.

<button size="xs" type="link" collapse="Loesung_22_3_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_3_Tipps" collapsed="true">
  * Use $A_{\rm V}=-\frac{R_2}{R_1}$ for the inverting amplifier.
  * Virtual ground means $U_{\rm m}\approx U_{\rm p}=0~\rm V$ (not a physical short).
</collapse>

<button size="xs" type="link" collapse="Loesung_22_3_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_3_Endergebnis" collapsed="true">
  * $A_{\rm V}=-\frac{22~\rm k\Omega}{2.2~\rm k\Omega}=-10$.
  * $U_{\rm O}(t)=-3.0~\rm V~\sin(2\pi\cdot 1\,{\rm kHz}\cdot t)$.
  * Summing node potential: approximately $0~\rm V$ (virtual ground).
</collapse>

</WRAP></WRAP></panel>


<panel type="info" title="Task 22.4 Inverting summing amplifier via superposition"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

An inverting summing amplifier has $R_0=10~\rm k\Omega$, $R_1=10~\rm k\Omega$, and $R_2=20~\rm k\Omega$.  
Two inputs are applied: $U_{\rm I1}=+1.0~\rm V$ and $U_{\rm I2}=-0.5~\rm V$.

  - Use superposition to compute $U_{\rm O}$.
  - Compute the same result by writing the sum directly as a weighted sum.
  - Explain briefly why the resistor between the op-amp inputs carries (ideally) no current.

<button size="xs" type="link" collapse="Loesung_22_4_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_4_Tipps" collapsed="true">
  * For each input alone: treat the circuit as an inverting amplifier with gain $-\frac{R_0}{R_i}$.
  * Superposition: set the other voltage source to $0$ (replace by a short).
  * With feedback: $U_{\rm D}\rightarrow 0$ implies negligible current through a resistor between inputs.
</collapse>

<button size="xs" type="link" collapse="Loesung_22_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_4_Endergebnis" collapsed="true">
  * $U_{\rm O(1)}=-\frac{R_0}{R_1}U_{\rm I1}=-\frac{10}{10}\cdot 1.0~\rm V=-1.0~\rm V$.
  * $U_{\rm O(2)}=-\frac{R_0}{R_2}U_{\rm I2}=-\frac{10}{20}\cdot(-0.5~\rm V)=+0.25~\rm V$.
  * $U_{\rm O}=U_{\rm O(1)}+U_{\rm O(2)}=-0.75~\rm V$.
</collapse>

</WRAP></WRAP></panel>


<panel type="info" title="Task 22.5 Current-to-voltage converter (transimpedance amplifier)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A photodiode is modeled as an ideal current source delivering $I_{\rm I}$ into a transimpedance amplifier.  \\
The feedback resistor is $R_1=220~\rm k\Omega$ and the non-inverting input is grounded.

  - Determine the transfer resistance $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm I}}$.
  - Compute $U_{\rm O}$ for $I_{\rm I}=+2.0~\rm \mu A$.
  - What sign does $U_{\rm O}$ have for a positive $I_{\rm I}$ (according to the circuit convention)?

<button size="xs" type="link" collapse="Loesung_22_5_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_5_Tipps" collapsed="true">
  * For the current-to-voltage converter in the given convention: $U_{\rm O}=-R_1 I_{\rm I}$.
  * Keep units consistent: $\rm \mu A$ and $\rm k\Omega$.
</collapse>

<button size="xs" type="link" collapse="Loesung_22_5_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_5_Endergebnis" collapsed="true">
  * $R_{\rm T}=\frac{U_{\rm O}}{I_{\rm I}}=-R_1=-220~\rm k\Omega$.
  * $U_{\rm O}=-(220~\rm k\Omega)\cdot (2.0~\rm \mu A)=-0.44~\rm V$.
  * For $I_{\rm I}>0$: $U_{\rm O}<0$ (with this sign convention).
</collapse>

</WRAP></WRAP></panel>


<panel type="info" title="Task 22.6 Voltage-to-current converter (transconductance)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A voltage-to-current converter should generate an output current proportional to an input voltage, with transfer conductance
\[
S=2.0~\rm mA/V.
\]
Assume the circuit uses a single resistor $R$ to set the current (ideal op-amp behavior), such that approximately $I_{\rm O}\approx \frac{U_{\rm I}}{R}$.

  - Determine $R$ for the desired $S$.
  - Compute $I_{\rm O}$ for $U_{\rm I}=0.6~\rm V$.
  - Briefly name one application of such a circuit.

<button size="xs" type="link" collapse="Loesung_22_6_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_22_6_Tipps" collapsed="true">
  * If $I_{\rm O}\approx \frac{U_{\rm I}}{R}$, then $S\approx \frac{1}{R}$.
  * Convert $2.0~\rm mA/V$ into $\rm A/V$ before inverting.
</collapse>

<button size="xs" type="link" collapse="Loesung_22_6_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_22_6_Endergebnis" collapsed="true">
  * $S=2.0~\rm mA/V=2.0\times 10^{-3}~\rm A/V \Rightarrow R=\frac{1}{S}=500~\Omega$.
  * $I_{\rm O}=S\,U_{\rm I}=(2.0~\rm mA/V)\cdot 0.6~\rm V=1.2~\rm mA$.
  * Example application: voltage-controlled current source (e.g. driving an actuator/LED current or biasing a sensor).
</collapse>

</WRAP></WRAP></panel>


===== Embedded resources =====
<WRAP column half>
Non-inverting operation amplifier circuit \\
{{youtube>_Ut-nQ535iE}}
</WRAP>


\\

~~PAGEBREAK~~ ~~CLEARFIX~~