theory:sensor_technology:st3_measurement_technology

*Most measurements are done using electronic tools. In the electronic world, we distinguish analysis in the amplitude, time and frequency domain. Each domain has its own vocabulary, toolset, mathematical models and application areas. In this chapter, such basic concepts of electronics are discussed.*

According to Coulomb's law, an electric charge $q_{A}$ will encounter a force when facing another charge $q_{B}$. This force is equal to

\begin{equation} \vec{F}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{A}q_{B}}{r^{2}}\vec{e_{r}} \label{eq:Coulomb} \end{equation}

with $r$ the distance between the two charges and $\epsilon_{0}$ the permittivity of vacuum being $8.85\cdot 10^{-12} F/m$. Charges are expressed in $Coulombs$, where the elementary charge (the charge of a single electron) equals $1.602214\cdot 10^{-19} C$. The unity vector $\vec{e_{r}}$, pointing from source $q_{A}$ to source $q_{B}$, defines force $\vec{F}$ as a vector.

Another interpretation of Coulomb's law is to say that charge $q_{A}$ is subject to a force when it is in the *electric field* of charge $q_{B}$. This electric field can be defined by normalizing to the charge $q_{A}$:
\begin{equation}
\vec{E}=\frac{\vec{F}}{q_{A}}.
\label{eq:ElectricField}
\end{equation}

Once there is a field, we can define *electric potential* $U$ as
\begin{equation}
\vec{U}_{12}=-\int_{1}^{2}\vec{E}\cdot d\vec{s}
\label{eq:ElectricPotential}
\end{equation}
which is the effort it takes to move a charge along a line from $1$ to $2$ in steps $d\vec{s}$. The dot-vectorproduct is used to include only the component of the trajectory that is parallel to the vector $\vec{E}$. In a homogenous field with strength $E$, and discarding vector directions, the potential difference $U$ over a distance $x$ becomes
\begin{equation}
U=-Ex.
\label{eq:ElectricPotential2}
\end{equation}

A consequence of Gauss law (briefly mentioned in the chapter about Magnetic Sensors as part of the Maxwell equations), is that there is no electric field inside an electric conductor. As a result, equation \eqref{eq:ElectricPotential2} shows there is no potential difference over an electric conductor. This means that in an electric wire (normally copper) there is no voltage drop, so we can interconnect electronic components with copper wires to supply them with supply voltages and signals.

Potential differences are expressed with the unit of $Volts$. As a result, a postential difference is also referred to as *the voltage*.

Charged particles, like electrons in a copper wire, will move under the influence of an electric potential difference. *Electric current* is defined as the change of charge as a function of time, so
\begin{equation}
I=\frac{dq}{dt}
\label{eq:ElectricCurrent}
\end{equation}
with $I$ the current in $Amperes$. The product of voltage and current has the meaning of *electric power*:
\begin{equation}
P=U \times I
\label{eq:ElectricPower}
\end{equation}
which has the unit of $Watts$.

In electronic circuits there are several ways of drawing sources of voltage and current. In figure 1 we can see:

- [a] The European symbol of an ideal voltage source
- [b] The American symbol of an ideal voltage source
- [c] The symbol of a voltage source that is an electrochemical battery
- [d] The European symbol of an ideal current source
- [e] The American symbol of an ideal current source.

There is a separate symbol for a battery to emphasize sources that are constant, like a power supply. With that, it clearly distinguishes from the basic voltage source that insinuates it is an input of a circuit: a voltage source that may be variable or dependent. Nevetheless, both provide a constant and ideal voltage.

Constant voltage and current sources as shown in figure 1 are said to give *Direct Current (DC)*: all charge is forced in a single direction. There are many time varying source signals possible, but one has special interest: a sinusoidal signal. Such signals, either current or potential, are referred to as *Alternating Current (AC)*.

The circuit symbol for a sinusoidal source is given in figure 2. The sinusoidal signal itself can be described by the following equation: \begin{equation} U_{Sine}\left( t \right )=U_{Offset} + \hat{U} \sin ( \omega t + \phi) \label{eq:SinusoidalSignal} \end{equation} which can be drawn as shown in figure figure 3.

The relation between the characteristics in figure 3 and equation \eqref{eq:SinusoidalSignal} is explained by the four terms amplitude, phase, period and offset.

The period and phase describe the time behaviour of the signal. The period can be expressed in the *radian frequency* $\omega$ of the equation as

- Period = $\frac{1}{f}$ = $\frac{2\pi}{\omega}$

which also introduces $f$ for the *frequency* in the unit of $Hz$ = $s^{-1}$.

The signal repeats every $2 /pi$ period of the sine equation. An offset in radian frequency, results in a shift in time of the signal:

- Phase = $-\phi$

This expression of phase is not completely correct in relation to figure 3, where phase is expressed as a time-shift. The expression of phase as defined in equation \eqref{eq:SinusoidalSignal} is more common: a shift in radians. We can say that a sine wave shifted by a phase of $2 \pi$ becomes equal to the original wave. However, on an oscilloscope we will see a shift in time.

While the period and phase describe the time behaviour of the signal, the offset and amplitude define the vertical variation in $Volts$ (or $Amperes$ for a current source). The offset, also *bias*, of the signal is

- Offset = $U_{Offset}$

which is the average of the signal. Where the sine-wave has the “zero-crossings”. It is also correct to say that the offset is the *DC-component* while the amplitude is the *AC-component*.

The amplitude is represented in the equation by

- Amplitude = $\hat{U}$

which is how much the signal varies above and under the offset. In practical measurements, we measure the top-top value: the difference between the highest peak en the lowest peak. By understanding the sine function, we can see that is equal to two times the amplitude: \begin{equation} U_{pp}=2 \hat{U} \label{eq:VPeakPeak} \end{equation}

In power AC electronics, there is normally a large amplitude and zero offset. In practical electronic circuits for signal processing, we are dealing with relatively large offsets (“DC operational points”) and small signals (“AC modulations”). We will later see that we can calculate the circuit separately for the AC and DC point. This is because of the *Superposition principle*.
\begin{equation}
U\left( t \right )=U_{Offset} + u \left ( t \right )
\label{eq:ACDC}
\end{equation}
With this way of expressing the equation for the signal, we can clearly distinguish between DC “large signal” behaviour (capital font) and AC “small signal” behaviour (small font). So $U$ is the quasi-static DC operational point and $u$ is the small signal which only has an amplitude, phase and frequency.

There is another way of expressing the signal strenth besides the amplitude and the peak-peak value: the root-mean-square or *RMS* value. For a periodic signal $U(t)$ the RMS value $U_{RMS}$ is defined as

\begin{equation} U_{RMS}=\sqrt{\frac{1}{T}\int_{0}^{T} U^{2} \left ( t \right ) dt} \label{eq:RMS} \end{equation}

with $T$ the period. This equation describes nothing more than to take the square of all instantaneaus signal levels, average them, and take the root to come back to the dimension of $Volts$. It is logical not to take the average, because that would reduce to the offset voltage. The RMS value has more a meaning of avarage power of the signal because power is also the square of potentials: $P = U \cdot I = U^{2}/R$.

When calculating the RMS voltage of a sine wave without an offset we find: \begin{equation} \begin{aligned} U_{RMS}&=\sqrt{\frac{1}{T}\int_{0}^{T} \left ( U_{Sine} \left ( t \right ) \right )^{2} dt} \\ &=\sqrt{\frac{1}{T}\int_{0}^{T} \left ( \hat{U} \sin(\omega t) \right )^{2} dt}\\ &=\sqrt{\frac{1}{T}\int_{0}^{T} \hat{U}^{2} \sin^{2}(\omega t) dt}\\ &=\sqrt{\frac{1}{T}\int_{0}^{T} \hat{U}^{2} \left ( \tfrac{1}{2}- \tfrac{1}{2}\cos(2 \omega t) \right ) dt}\\ &=\sqrt{\frac{1}{T}\int_{0}^{T} \tfrac{1}{2} \hat{U}^{2} dt - \frac{1}{T} \int_{0}^{T} \tfrac{1}{2}\cos(2 \omega t) dt}\\ &=\sqrt{\frac{1}{T} \left [ \tfrac{1}{2} \hat{U}^{2}t \right ]_{0}^{T} - \frac{1}{T} \left [ \tfrac{1}{4 \omega}\sin(2 \omega t) \right ]_{0}^{T} }\\ &=\sqrt{\tfrac{1}{2} \hat{U}^{2} - 0 }\\ &=\tfrac{1}{2}\sqrt{2} \cdot \hat{U} \end{aligned} \label{eq:RMSsine} \end{equation} So: the RMS value of a sine-shaped signal which has no offset is $\tfrac{1}{2}\sqrt{2}\approx 0.707$ times the amplitude $\hat{U}$. The $230 V$ value of our power supply system is an RMS value. This means that the amplitude of the power supply is actually $325 V$.

Note that voltages (potential differences) are measured *over* a component and currents are measured *through* a wire or component. This is a direct consequence of the definition: voltage is a potential *difference* and current is the amount of charge conducted *through* a wire. How to measure is indicated in figure 4. where a simple circuit of a voltage source connected to a light bulb is taken.

The current meter should not affect the circuit, and should therefore have a very low resistance (a concept which is introduced in a section below). A voltmeter should also not affect the circuit and must therefore have a very high resistance.

In figure 5 [a] a lamp is drawn which is connected to a constant voltage $U$: the voltage of the lamp is constant with $U_{L}=U$. As a result, there is a constant current $I_{L}$ flowing through the lamp. For many electronic components, there appears to be a constant relationship between the voltage and current as shown in figure 6 [a]. Such components are said to be *resistive* or *Ohmic*. It is *Ohm's law* that defines the relationship

\begin{equation} R=\frac{U}{I} \label{eq:OhmsLaw} \end{equation}

more commonly written as

\begin{equation} U=I \cdot R \label{eq:OhmsLaw2} \end{equation}

with $R$ the *resistance* of the component. A unified resistive element is drawn in figure 5 [b]. The unit of resistance is $Ohm$ $[\Omega]$.

In practical elements, the linear relation of equation \eqref{eq:OhmsLaw2} and figure 6 [a] is not observed over the full range. The [b] figure of figure 6 shows the $I-U$ curve of a *non-linear* component: current is not proportional to the potential, but to the square or power of the potential. In this case, we can not speak of a single resistor value. However, what we can do is to define the resistance at a single point or over a small regime. In this case, we define a *differential resistance* $r$ which is

\begin{equation} r=\frac{\delta U}{\delta I}=\frac{u}{i} \label{eq:DifferentialResistance} \end{equation}

also indicated in figure 6 [b]. A small font is used for the same reasons as mentioned with equation \eqref{eq:ACDC}: this is about the small variations on top of a large signal setting-point.

It may be confusing that for a nonlinear line, the large signal operational point resistance $R$ is not equal to the small signal differential resistance $r$. The large signal operational point resistance $R$ representes the ratio between the voltage $U$ needed to reach that operational point and the current $I$ that is drawn in that point. The small signal differential resistance $r$ represents the resistance experienced to put a small voltage on top of the operational point. We can compare this to walking up a mountain. If we are at $1000m$ above sea level on a mountain, but the sea is at $100km$ distance, we have to look down at an angle of $1\%$ to see the sea. This is the operational point. But to make one step of $1m$ to walk up the mountain, it may take an increase of $10\%$ depending on the steepnes of the mountain at that operational point. This gives a much higher differential resistance. When solving electronic problems and designing circuits, we deal with the AC and DC component exactly in that way: small signal and large signal term of equation \eqref{eq:ACDC} may experience different resistances.

Resistors can be placed in series or in parallel as shown in figure 7. The question is how we can express the total equivalent resistance in the resistor values $R_{1}$, $R_{2}$ and $R_{3}$. First, the series circuit of figure [a] can be evaluated by

\begin{equation} U_{Series}=U_{R_{1}}+U_{R_{2}}+U_{R_{3}} \end{equation}

because voltages are potential differences between the nodes of the resistors. Next, we can replace the voltages by the Ohm's laws of the individual resistors:

\begin{equation} U_{Series}=I_{R_{1}}R_{1}+I_{R_{2}}R_{2}+I_{R_{3}}R_{3}\\ \end{equation}

which is equal to

\begin{equation} U_{Series}=\left (R_{1}+R_{2}+R_{3}\right ) I_{Series} \end{equation}

because the currents through the resistors must be equal. As a result, we find

\begin{equation} R_{Series}=R_{1}+R_{2}+R_{3} =\sum_{i} R_{i} \label{eq:SeriesResistors} \end{equation}

for the series equivalent resistance of the resistors.

Similar, we can find for the parallel circuit of figure [b] that

\begin{equation} I_{Parallel}=I_{R_{1}}+I_{R_{2}}+I_{R_{3}} \end{equation}

because currents must add-up since these are just flows of electrons which can not be lost. Next, we can replace the currents by the Ohm's laws of the individual resistors:

\begin{equation} I_{Parallel}=\frac{I_{R_{1}}}{R_{1}}+\frac{I_{R_{2}}}{R_{2}}+\frac{I_{R_{3}}}{R_{3}}\\ \end{equation}

which is equal to

\begin{equation} I_{Parallel}=\left (\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\right ) U_{Parallel} \end{equation}

because the voltages over the resistors must be equal. As a result, we find

\begin{equation} \frac{1}{R_{Parallel}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} =\sum_{i} \frac{1}{R_{i}} \label{eq:ParallelResistors} \end{equation}

for the parallel equivalent resistance of the resistors.

As an example, we will calculate the resistance the source of figure 8 will “see”. When that resistance is knows, we can calculate the current that is drawn form the source.

First, combine resistor $R_2$ and $R_3$:

\begin{equation} \begin{aligned} \frac{1}{R_{23}}&=\frac{1}{R_2}+\frac{1}{R_2}\\ R_{23}&=\frac{R_2 R_3}{R_2 + R_3}\\ &=\frac{R_2 R_3}{R_2 + R_3}\\ &=\frac{30\Omega \cdot 300\Omega}{30\Omega+300\Omega}\\ &\approx 27.3 \Omega \end{aligned} \end{equation}

where the way of expressing the equivalent parallel resistance as the quationt of the product and the sum of the individual resosances is quite common. Next, calculate the series resistance of $R_1$ and $R_{23}$:

\begin{equation} \begin{aligned} R_{123}&=R_1+R_{23}\\ &\approx 227.3 \Omega \end{aligned} \end{equation}

which will give a source current of $10V/227\Omega = 44.0mA$.

The configuration of figure 9 is commonly seen and is referred to as a *Voltage divider*.

To calculate the output voltage $U_{out}$, we calculate the current through $R_2$ first:

\begin{equation} I_{R_{2}}=\frac{U_{in}}{R_1+R_2} \end{equation}

This current gives the output voltage $U_{out}$

\begin{equation} U_{out}=I_{R_{2}}R_2=\frac{U_{in}R_2}{R_1+R_2} \end{equation} so, the overall equation of the voltage divider is:

\begin{equation} U_{out}=U_{in} \frac{R_2}{R_1+R_2}. \label{eq:VoltageDivider} \end{equation}

This is an important relation because many configurations consist of this. The equation \eqref{eq:VoltageDivider} is only true if no (or a very small) current is drawn from $U_{out}$, because we assumed $I_{R_1}=I_{R_2}$.

In the previous section, we have used some rules regarding voltages and currents for resistors in series and in parallel. For example, we used that two resistors in series will have the same current through them because no electrons are lost. Two resistors in parallel will have the same voltage accross them. These rules are generalized in the *Kirchhoff's laws*. Before introducing them, some terms and definitions have to be introduced.

When we consider a circuit like the one of figure 10 as a *network* of electronic components, the following terms can be introduced:

- A
*component*is an electronic element which has a specific current-voltage relationship between the pins - A
*node*is the connection point of components - A
*branch*is the line unsplitted between two nodes - A
*loop*is any path we can walk from any node back to the same node.

Now the Kirchhoff laws say:

*Kirchhoffs current law*: The sum of all currents flowing into a node equals zero ($\Sigma$I_{node}=0) → no charge is lost in a node*Kirchhoffs voltage law*: The sum of voltages along a loop equals zero ($\Sigma$U_{loop}=0) → no energy is los when travelling along a closed loop.

We can use either law for solving the network completely: meaning, to determine all node-voltages and branch currents. A network can be solved by three steps:

- Write down all component equations: the eqations determining the $U$ - $I$ characteristics of the components. This is Ohms law for a resistor, $U_{x}=Constant$ for a voltage source, etc.
- Write down the network equations: the equations that follow from either one of the Kirchhoff laws
- Solve the set of linear equations. In case of Kirchhoffs voltage law, the values to be solved are the loop currents, in case of the current law, these are the node potentials.

This procedure is explained with the example of figure 11. The question is to solve all network node-voltages and branch currents.

With Kirchhoffs current law we will solve the independent node-voltages first, and then we are able to solve any branch-current. The structural procedure is:

- Give all nodes a name. Here are three remarks to make:
- One node must be grounded to define it to $0V$. Because voltages are defined as potential differences, we must create one absolute reference. It does not matter which one is grounded, so the most logical one can be taken. For example the one that is the lowest in the drawing or one negative pole of a source
- Every fixed voltage source defines one dependent node-voltage because if one side is defined, the other side is defined as well by the voltage of the source
- A current source does not add a new node because the voltage accross it is zero

- Give all branch currents a name and direction. The direction can be chosen arbitrary: we will find a minus sign when the real current appears to be in the opposite direction
- Write down all component equations
- Write down the Kirchhoff current laws for all independent nodes
- substitute the component equations in the Kirchhoff current laws and solve the set of equations. When there are N independent nodes, there will be N equations that can be solved.

In figure 12 this leaves two independent nodes ($a$ and $b$) and five branch currents $I_1$ to $I_5$. The component equations are:

\begin{equation} \begin{aligned} U_{R_{1}}&=\left ( U_{Source}-U_a \right ) &= I_1 R_1\\ U_{R_{2}}&=\left ( U_a-0V \right ) &= I_2 R_2\\ U_{R_{3}}&=\left ( U_a-U_b \right ) &= I_3 R_3\\ U_{R_{4}}&=\left ( U_{Source}-U_b \right ) &= I_4 R_4\\ U_{R_{5}}&=\left ( U_b-0V \right ) &= I_5 R_5 \end{aligned} \label{eq:ComponentEquations} \end{equation}

and the Kirchhoff current equations are:

\begin{equation} \begin{aligned} \text{Node a:} \quad I_1 - I_2 -I_3 &= 0\\ \text{Node b:} \quad I_4 - I_5 +I_3 &= 0 \end{aligned} \end{equation}

Substitution gives:

\begin{equation} \begin{aligned} \text{Node a:} \quad \frac{U_{Source}-U_a}{R_1} - \frac{U_a-0V}{R_2} - \frac{U_a-U_b}{R_3} &= 0\\ \text{Node b:} \quad \frac{U_{Source}-U_b}{R_4} - \frac{U_b-0V}{R_5} +\frac{U_a-U_b}{R_3} &= 0 \end{aligned} \end{equation}

which can be solved by:

\begin{equation} \begin{aligned} \text{Node a:} \quad \frac{U_{Source}}{R_1} -\frac{U_a}{R_1} - \frac{U_a}{R_2} - \frac{U_a}{R_3} +\frac{U_b}{R_3} &= 0\\ \text{Node b:} \quad \frac{U_{Source}}{R_4} - \frac{U_b}{R_4} - \frac{U_b}{R_5} +\frac{U_a}{R_3} - \frac{U_b}{R_3} &= 0 \end{aligned} \end{equation}

\begin{equation} \begin{aligned} \text{Node a:} \quad \left ( \frac{1}{R_1} \right )U_{Source} - \left ( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right )U_a + \left( \frac{1}{R_3}\right )U_b &= 0\\ \text{Node b:} \quad \left ( \frac{1}{R_4} \right ) U_{Source} - \left( \frac{1}{R_4} + \frac{1}{R_5} + \frac{1}{R_3}\right) U_b + \left ( \frac{1}{R_3} \right )U_a &= 0 \end{aligned} \end{equation} These are two linear equations with two variables $U_a$ and $U_b$. By substituting all values from figure 11 we find $U_a=6.1277V$ and $U_b=7.6596V$. Once these are known, we can calculate any branch current with the component equations \eqref{eq:ComponentEquations}. For example, the current $I_3$ through resistor $R_3$ can be found using

\begin{equation} \begin{aligned} U_{R_{3}}&=\left ( U_a-U_b \right ) = I_3 R_3\\ I_3 &=\left ( U_a-U_b \right )/ R_3\\ &=\left ( 6.1276V-7.6596V \right )/ 6k\Omega\\ &\approx -0.2553mA \end{aligned} \end{equation}

which has a minus sign because apparently, the current is in the opposite direction as assumed in figure 12.

Now we solve the same circuit with Kirchhoffs voltage law. With Kirchhoffs voltage law we will solve the loop currents first, and then we are able to calculate the node-voltages. The structural procedure is:

- Give all loops a name and direction. Here are two remarks to make:
- Only the loops containing a single maze are needed
- The loop direction can be chosen arbitrary: we will find a minus sign when the real current appears to be in the other direction

- Write down all component equations
- Write down the Kirchhoff voltage laws for all chosen loops
- substitute the component equations in the Kirchhoff voltage laws and solve the set of equations.

In figure 13 there are three open mazes that are defined as the independent current-loops $\alpha$, $\beta$ and $\gamma$. We can fill-in the component equations directly in the three Kirchhoff voltage laws:

\begin{equation} \begin{aligned} \text{Loop } \alpha \text{:} & \quad U_{Source} + \left ( I_\beta - I_\alpha \right ) R_1 + \left ( I_\gamma - I_\alpha \right ) R_2 &= 0\\ \text{Loop } \beta \text{:} & \quad \left ( I_\alpha - I_\beta \right ) R_1 + \left ( - I_\beta \right ) R_4 + \left ( I_\gamma - I_\beta \right ) R_3 &= 0\\ \text{Loop } \gamma \text{:} & \quad \left ( I_\alpha - I_3 \right ) R_2 + \left ( I_\beta - I_\gamma \right ) R_3+ \left ( - I_\gamma \right ) R_5 &= 0 \end{aligned} \label{eq:KirchhoffVoltageExample} \end{equation}

What is important here, is the sign of the currents and sourcesL

- When stepping along a voltage source from $-$ to $+$ there is a positive voltage step (the potential increases)
- When stepping over a passive element like a resistor, there is a voltage drop if we step in the direction of the current.

For example, when stepping along loop $\alpha$ and passing by resistor $R_1$, the current $I_\alpha$ gives a voltage drop: $-I_\alpha R_1$. Along the same loop and same resistor, current $I_\beta$ gives a voltage increase of $+I_\beta R_1$. As a result, the first term in loop $\alpha$ of equation \eqref{eq:KirchhoffVoltageExample} is equal to $+(I_\beta-I_\alpha) R_1 [Volt]$.

Manipulation of equation \eqref{eq:KirchhoffVoltageExample} gives:

\begin{equation} \begin{aligned} \text{Loop } \alpha \text{:} & \quad U_{Source} - \left ( R_1+ R_2 \right ) I_\alpha + R_1 I_\beta + R_2 I_\gamma &= 0\\ \text{Loop } \beta \text{:} & \quad R_1 I_\alpha- \left ( R_1 + R_3 + R_4 \right ) I_\beta + R_3 I_\gamma &= 0\\ \text{Loop } \gamma \text{:} & \quad R_2 I_\alpha + R_3 I_\beta - \left ( R_2 + R_3 + R_5 \right )I_\gamma &= 0 \end{aligned} \end{equation}

which can be solved for $I_\alpha$, $I_\beta$ and $I_\gamma$. By substituting all values from figure 11 we find $I_\alpha=8.043mA$, $I_\beta=2.170mA$ and $I_\gamma=1.915mA$. Once these are known, we can calculate any branch current (by adding the loop-currents involved) and any node voltage with the component equations \eqref{eq:ComponentEquations}. For example, the current $I_3$ through resistor $R_3$ can be found using $I_3=I_\gamma-I_\beta \approx -2.553mA$, which is equal to the result of Kirchhoffs current law. To calculate a node voltage, for example for node $a$ of figure 12 with respect to ground, we can calculate the voltage over $R_2$:

\begin{equation} \begin{aligned} U_a&=U_{R_{2}}\\ &= \left ( I_\alpha-I_\gamma \right ) R_2\\ &\approx 6.1277V \end{aligned} \end{equation}

which is also equal to the result of Kirchhoffs current law..

The *superposition* theorem claims that:

*The voltage over, or the current through, a branch in a network is equal to the sum of all contributions of the sources (current and voltage) in the network*.

In other words, we can calculate the voltage on a certain node by setting all sources to zero one by one and adding the contributions. When doing this, we have to realize that:

- setting a voltage source to zero means that it is replaced by a short-circuit ($U=0$)
- setting a current source to zero means that is removed to make an open-loop ($I=0$).

This is best explained by an example. Consider figure 14: a network with one current source and one voltage source. If we are interested in the voltage over $R_2$, we can used Kirchhoff (thereis one independent node for Kirchhoffs current law or two independent current loops for Kirchhoffs voltage law). The alternative is to use the superposition theorem.

First, set all sources to zero, except for $U_1$. This is shown in figure 15[a]: the current source is set to zero (left out). The partial voltage $U^{'}_{R_2}$ accross $R_2$ is

\begin{equation} U^{'}_{R_2}=U_1\frac{R_2}{R_1 + R_2} \end{equation}

because this is a voltage divider.

Next, set all sources to zero, except for $I_1$. This is shown in figure 15[b]: the voltage source is set to zero (short-circuit). The partial voltage $U^{''}_{R_2}$ accross $R_2$ is

\begin{equation} U^{''}_{R_2}=I_1 \frac{R_1 R_2}{R_1+R_2} \end{equation}

where we use that the voltage over $R_2$ equals the voltage over $R_2$ parallel to $R_1$.

In total we find

\begin{equation} \begin{aligned} U_{R_2}&=U^{'}_{R_2}+U^{''}_{R_2}\\ &=U_1\frac{R_2}{R_1 + R_2}+I_1 \frac{R_1 R_2}{R_1+R_2}\\ &=\frac{R_2}{R_1 + R_2} \left ( U_1+I_1 R_1 \right )\\ &=\frac{7\Omega}{3\Omega + 7\Omega} \left ( 10V+6A \cdot 3\Omega \right )\\ &\approx 19.6V \end{aligned} \end{equation}

The theorems of Norton and Thévenin are helpful to model a complex source as two ideal components. The theorems are:

- Thévenin: Any linear network of current sources, voltage sources and resistors, measured between two pins, is equivalent to a single ideal voltage source with a resistor in series.
- Norton: Any linear network of current sources, voltage sources and resistors, measured between two pins, is equivalent to a single ideal current source with a resistor in parallel.

The equivalent circuits are given in figure 16. In these theorems, “equivalent” means that they behave the same with respect to voltage and current. For example, if we load the circuit of figure 16[a] with a $1k\Omega$ resistor, we will observe a certain voltage over the resistor and a certain current through the resistor. This voltage and current is the same of the complex network of which these are the equivalent.

This is once more explained with figure 17: the figures [a], [b] and [c] whould behave the same for any load resistance $R_L$.

The question is, how to determmine the equivalent resistance and the Thévenin equivalent source (or Norton source)? There are in fact two methods. The first one is illustrated by figure 18. First we define the short-circuit current and the open-circuit voltage. The *Short-Circuit current* $I_{SC}$ can be determined by

\begin{equation} \begin{aligned} U_{out}\left ( R_L=0\Omega \right)&=0V\\ I_{out}\left ( R_L=0\Omega \right)&=I_{SC}=\frac{U_T}{R_T}=I_N \end{aligned} \label{eq:UI_ShortCircuit} \end{equation}

and the *Open-Circuit voltage* $U_{OC}$ is

\begin{equation} \begin{aligned} U_{out}\left (R_L=\infty \Omega \right )&=U_{OC}=U_T=I_N R_N\\ I_{out}\left (R_L=\infty \Omega \right )&=0A \end{aligned} \label{eq:UI_OpenCircuit} \end{equation}

By combining the equations \eqref{eq:UI_ShortCircuit} and \eqref{eq:UI_OpenCircuit} for $I_{SC}$ and $U_{OC}$ we can conclude that

\begin{equation} R_T=R_N \end{equation}

so the Norton and Thévenin equivalent resistors are the same. If both the Norton and Thévenin theorems are true, they must be interchangeable. So that is the case because $R_T=R_N$ and $U_T=R_T I_N$.

In practice, the method above can not be done: short-circuiting a circuit may damage the circuit, or may at least bring the circuit in a non-linear regime. What is normally done, is that the circuit is measured first open loop (with a high-resistive voltmeter) and next the output voltage is measured with a known load-resistance $R_L$. In that case, the equivalent components can still be calculated.

The second method is not applicable at all in practice, but can only be used to calculate the equivalent components from a known circuit:

- The open-circuit voltage is equal to $U_T$, just as with the previous method
- The equivalent resistance $R_T$ or $R_N$ is equal to the resistance of the circuit when all sources are switched to zero.

The Thévenin equivalent circuit as shown in figure 16[a] has a consequence which may not be intuitive. In case the circuit is unloaded ($U_L=\infty\Omega$), there is no voltage drop over the resistor, and $U_{out}=U_T$. This can be understood by Ohms law: when there is no current through a resistor, there is no voltage drop.

Many sensors are resistors where the resistance is dependent on a certain environmental value like temperature or light. For example, a thermistor is a temperature dependent resistor. With such a sensor, the voltage divider of figure 9 can be used to convert the resistance into a voltage. This is needed because most meters are essentailly voltmeters and not resistance meters.

In figure 19 a voltage divider is prepared to convert a change in resitance into a voltage. This is done by using a constant supply voltage $U_s$. The output voltage $U_{out}$ can now be calculated with the equation for a voltage divider:

\begin{equation} U_{out}(T)=\frac{R_2}{R_1(T)+R_2} U_s. \label{eq:VoltageDividerResistiveSensor} \end{equation}

The response of equation \eqref{eq:VoltageDividerResistiveSensor} is plotted in figure 20[a]. There are a few shortcomings of this response:

- The output is not linear. However, over a short range it may be sufficient.
- There is an offset-voltage on top of which there is only a small T-dependent variation.

The second shortcoming may be a big issue. In some typical applications (like Pt100 temperature sensors) we see less than 1% variation. For other applications (like an LDR to detect only light and dark) there may be enough signal. Also electronic amplifiers (“differential amplifiers” and “instrumentation amplifiers”) may not perform well when the offset signal is more than a hundred times the signal of interest. In addition, the small variation of the output signal can not be amplified because that would amplify the huge offset as well.

A solution is to use a *Wheatstone Bridge* as shown in figure 21. This is in fact a voltage divider with the sensor, compared to a second voltage divider without a sensor. The resistor values $R_2$, $R_3$ and $R_4$ are normally chosen equal, and equal to the nominal value of $R_1(T)$.

Now the output signal $U_{out}(T)$ is shifted to a level around $0V$: the offset is removed. Now we can amplify the signal without any problems. The equation describing the output voltage is

\begin{equation} \begin{aligned} U_{out}(T) &= \left( \frac{R_2}{R_1(T)+R_2} -\frac{1}{2} \right )U_s \\ &= -\frac{1}{2} \frac{ R_1(T)-R_2 }{R_1(T)+R_2} U_s \end{aligned} \label{eq:OutputWheatstoneBridge} \end{equation}

where $R_3$ and $R_4$ are not visible because they are equal to each other. The sensitivity around the nominal value of the resistive sensor is

\begin{equation} \frac{\partial U_{out}}{\partial R_1}=-\frac{U_s}{4R_1}. \label{eq:SensitivityWheatstoneBridge} \end{equation}

The response is shown in figure 20[b]. We can see that the curve shape has not changed: it is still non-linear. In some applications we can replace $R_4$ with a temperature sensor as well: in that case, the response becomes an s-shape which is quite linear around $R_1(T)=0$.

Basic electric sensing elements are either a volt-meter or current-meter. For example, the *Analog to Digital Converter (ADC)* input of a microcontroller is normally a voltmeter. The more classical *moving coil meter* is in essence a current meter.

The ideal voltage meter and current meter of figure 4 can be expanded to practical meters in a similar way as the Norton and Thévenin equivalents are adding resistive elements. In figure 22 we can see that practical meters have an internal resistance $R_i$:

- In case of a voltmeter (figure [a]) this resistor should be modelled in series to the meter,
- In case of an current meter (figure [b]) this resistor should be modelled in parallel to the meter.

To affect the measured circuit minimally, the $R_i$ must be low for a current meter and high for a voltmeter.

One of the basic building blocks is the classical *moving coil meter* of figure 23. Internally, such a meter is an inductive coil in a permanent magnet. Depending on the applied electric current, the coil experiences a force in the permanent magnet. This force is counterbalanced by a mechanical spring. This system makes the position of the needle proportional to the applied current: so in essence, this meter is a current meter. Based on the used copper wire of the coil, such a meter has a constant internal resistance $R_i$.

To use this meter as either a voltmeter or current meter over a custom range, we have to add a series resistor or parallel resistor respectively.

In figure 24, the current meter will become a custom-range voltmeter when the appropriate series resistor $R_{series}$ is chosen.

For example, if we want to configure the currentmeter for a range of $0V..10V$, and the meter is a $100\mu A$ meter (full range is $100\mu A$) with an internal resistance of $R_i=20\Omega$, we can find:

\begin{equation} \begin{aligned} R_{series}&=\frac{U_{R_{series}}}{I_{R_{series}}}\\ &=\frac{U_{Total}-U_{R_{i}}}{I_{Total}}\\ &=\frac{U_{Total}-I_{R_{i}}R_{i}}{I_{Total}}\\ &=\frac{U_{Total}}{I_{Total}}-R_{i}\\ &=\frac{10V}{20\mu A}-20\Omega\approx 500 k\Omega \end{aligned} \label{eq:SeriesResistorVoltmeterFromAmmeter} \end{equation} which is based on all full scale values for $U$ and $I$. So, the final configuration has a relatively high internal resistance as should be the case for voltmeters.

In figure 25, the current meter will become a custom-range practical amperemeter when the appropriate parallel resistor is chosen. Such a parallel resistor is called a *shunt resistor*, so we prefer to use the symbol $R_{shunt}$ for this parallel resistor.

For example, if we want to configure the currentmeter for a range of $0V..100mA$, and the meter is a $100\mu A$ meter (full range is $100\mu A$) with an internal resistance of $R_i=20\Omega$, we can find:

\begin{equation} \begin{aligned} R_{shunt}&=\frac{U_{R_{shunt}}}{I_{R_{shunt}}}\\ &=\frac{U_{Total}}{I_{Total}-I_{R_{i}}}\\ &=\frac{I_{R_{i}}R_{i}}{I_{Total}-I_{R_{i}}}\\ &=\frac{20\mu A \cdot 20\Omega}{100mA-20\mu A} \approx 4m\Omega \end{aligned} \label{eq:ParallelResistorAmmmeterFromAmmeter} \end{equation} which has a relatively low internal resistance as should be the case for current meters.

So far, all circuits only had resistors and sources and where therefore constant or quasi-static. When including time-dependent components like capacitors and inductors, the mathematics to evaluate them have to be expanded with differential equations. This requires new mathematical tools to solve differential equations and to transform to the frequency domain. This will be dealt with in later courses, but this chapter gives a preview.

A metal plate carrying a potential $U$ can induce a charge $Q$ on a second plate:

\begin{equation} Q=C \cdot U \label{eq:UQC} \end{equation}

where the proportional factor $C$ is called the *capacitance*. Capacitance has the unit of $Farad [F]$ with $1F=1 Coulomb/Volt$. From an electronic circuit perspective, it is more interesting to take the derivative with time of the charge:

\begin{equation} I=\frac{\partial Q}{\partial t}=C \frac{\partial U}{\partial t} \label{eq:UIcapacitor} \end{equation}

which says that the current $I$ through a capacitor is proportioanl to the change of voltage $U$. As a result, a capacitor is a component that does not pass DC voltages ($dU/dt=0$). However, an AC voltage is passed because the first derivative of a sine wave is a cosine wave. So we can see a capacitor as a component that passes fast changes in voltage, but blocks DC components. We call this *high-pass behaviour*.

Take for example the simple RC Network of figure 26.

To capture such a system in a mathematical equation, we first write down the component equations. In this case, that is Ohm's law for the resistor and the voltage-current relation for the capacitor. Next, we need the network equations. These follow from Kirchhoff's current and voltage laws. \begin{equation} \begin{matrix} \begin{aligned} I_{C}&=C\frac{\partial U_{C}}{\partial t} \\ U_{R}&=I_{R}R \\ I_{C}&=I_{R} \\ U_{out}&=U_{C}=U_{in}-U_{R} \end{aligned} \end{matrix} \label{eq:RC_Network} \end{equation} After combining the equations, we find \begin{equation} U_{out}+RC\frac{\partial U_{out}}{\partial t}=U_{in} \label{eq:RC_DiffEquation} \end{equation}

which can be solved analytically for a step-input at t = 0 sec as

\begin{equation} U_{out}\left ( t \right ) = U_{in} \left ( 1-e^{-\frac{t}{RC}} \right ). \label{eq:RC_RC_AnSolution} \end{equation}

Laplace transforms and Fourier transforms give easier mathematics than differential equations. As an example we look at the same RC network as used in figure 26. After introducing the complex number $i$ all sinusoidal signals become simple linear functions. In electronics, we prefer to use the symbol $j$ for the complex number $i$ to avoid confusion with the small-signal current $i$.

After Fourier transform, the complex component equations become

\begin{equation} \begin{matrix} U_C=\frac{1}{j \omega C} I_C\\ U_R=I_R R \end{matrix} \label{eq:RC_NetworkImpedanceComponentEq} \end{equation} We could combine the component equations with the network equations (from Kirchhoff), but because all components have become simple impedances, we can also work with simple voltage dividers. Using a voltage divider, we can see that \begin{equation} U_{out}\left ( j\omega \right ) = \frac{Z_{C}\left ( j\omega \right )}{Z_{C}\left ( j\omega \right )+Z_{R}\left ( j\omega \right )}U_{in}\left ( j\omega \right ) \label{eq:RC_ImpedanceVoltageDivider} \end{equation}

which can be substituted easily with the component equations in impedance form

\begin{equation} U_{out}\left ( j\omega \right ) = \frac{1}{1+j\omega RC}U_{in}\left ( j\omega \right ). \label{eq:RC_AnZSolution} \end{equation}

The result is an expression of the frequency transfer function of the RC circuit as a filter. It has a shape as shown in figure 28.

- Chapter 1: Measurement Theory
- Chapter 2: Measurement Errors
- Chapter 3: Electronic Measurements
- Chapter 4: Sensor-Actuator Systems ← Next
- Chapter 5: Modelling
- Chapter 6: Modelling: The Accelerometer - example of a second order system
- Chapter 7: Modelling: Scaling - why small things appear to be stiffer
- Chapter 8: Modelling: Lumped Element Models
- Chapter 9: Modelling: Finite Element Models
- Chapter 11: Modelling: Systems Theory
- Chapter 12: Modelling: Numerical Integration
- Chapter 13: Signal Conditioning and Sensor Read-out
- Chapter 14: Resistive Sensors
- Chapter 15: Capacitive Sensors
- Chapter 16: Magnetic Sensors
- Chapter 17: Optical Sensors
- Chapter 18: Actuators - an example of an electrodynamic motor
- Chapter 19: Actuator principles for small speakers
- Chapter 20: ADC and DAC
- Chapter 21: Bus Interfaces - SPI, I
^{2}C, IO-Link, Ethernet based - Appendix A: Systematic unit conversion
- Appendix B: Circuits, Graphs, Tables, Pictures and Code
- Appendix C: Common Mode Rejection Ratio (CMRR)
- Appendix D: A Schmitt Trigger for sensor level detection

theory/sensor_technology/st3_measurement_technology.txt · Last modified: 2018/11/16 11:36 by glangereis