====== 4 Basic Circuits II ======

<callout>
  * Also recommended for basic circuits II is the [[https://www.ablic.com/en/semicon/products/analog/opamp/intro/]]
  * [[https://link.springer.com/book/10.1007%2F978-3-319-28127-8| Operational Amplifiers (Johan Huijsing)]] [viewable via the university network]
</callout>

<WRAP>
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=== 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>
=== Objectives for Basic Circuits II ===
After this lesson, you should:
  - Be able to apply the superposition method to operational amplifier circuits.
  - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantages, and disadvantages).
  - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.
  - Be able to name applications for the summing inverter, voltage-to-current converter, and current-to-voltage converter.
</callout></WRAP>

===== 4.1 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 (compare [[3_opamp_basic_circuits_i#steps_to_the_goal|steps to the goal]]). 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 [[electrical_engineering_1:network_analysis|Electrical engineering 1: Network analysis]]. 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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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]].

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===== 4.2 Differential Amplifier / Subtractor =====

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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>
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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})$
$\boxed{U_{\rm O}=  U_{\rm I2}\cdot \frac{R_4}{R_2 + R_4}  \frac{R_1 + R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1}}$

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<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}}
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<collapse id="foo" collapsed="false">
<button type="warning" collapse="foo">Please click to see the animation!</button>
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Two simplifications should be considered here:
  - If $R_1 = R_2$ and $R_3 = R_4$ are chosen, the equation further simplifies to:  \\ (nbsp) $\boxed{U_{\rm O} = U_{\rm I2}\cdot \frac{R_3}{R_1} - U_{\rm I1} \cdot \frac{R_3}{R_1} = \frac{R_3}{R_1}\cdot(U_{\rm I2}-U_{\rm I1})}$. \\ 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~~
===== 4.3 Instrumentation Amplifier =====

This type of amplifier shall be analyzed in the following exercise. \\ \\
An example of a pure instrumentation amplifier is the components {{circuit_design:ina818.pdf}}. 

Often instrumentation amplifiers are used in programmable gain amplifiers (PGA). PGAs allow the manipulation of the amplification factor by a digital interface. 


<WRAP><panel type="default"> 
<imgcaption pic3| Instrumentation Amplifier>
</imgcaption>
\\ {{drawio>Instrumentenverstärker.svg}}
</panel></WRAP>


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===== 4.4 Current-Voltage-Converter =====

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

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~~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}}).


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===== 4.5 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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~~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. 

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\\ \\
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]].

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====== Exercises ======

{{page>Übungsblatt6&nofooter}}


====== Further Reading ======
  * [[https://en.wikipedia.org/wiki/Low-dropout_regulator| Low Dropout Controller]]

====== Learning Questions ======
  * State applications for the reverse adder.
  * Explain the working of a current-to-voltage converter.
  * Name 3 applications for an operational amplifier.