# 5 Filter Circuits

• Also recommended for basic circuits II is the nice introduction and textbook found here.

#### Introductory example

Various applications work in harsh environments, where a clear digital signal becomes a noisy signal at the receiver (e.g. sensors in the engine compartment or industrial environments, satellite communication). In the simulation above, the left scope shows the original signal. The second scope shows the noisy signal.

One possibility to process such noisy signals is the use of filters. Filters have already been described in Electrical Engineering 1. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequently optimize the use of the signal, active filters can be applied. These are often built up by operational amplifiers.

Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{O1}$ after the first filter stage already shows significantly less noise. In the signal $U_{O2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply.

With the switch (left in the simulation) a frequency-variable test signal (Chirp or sweep) can be fed in. This clearly shows that the filter produces a smaller amplitude at the filter output from high-frequency oscillations at the input.

This chapter will explain the basics of active high and low pass filters from operational amplifiers.

#### Objectives for Basic Circuits II

After this lesson, you should:

1. Be able to apply the superposition method to operational amplifier circuits.
2. Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantages, and disadvantages).
3. Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.
4. Be able to name applications for the inverse integrator, voltage-to-current converter, and current-to-voltage converter.

To be able to analyze filter circuits, various possibilities for representing the numerical values should be explained beforehand.

Decibels (dB's) for Engineers - A Tutorial

The Decibel scale is an auxiliary unit of measure that facilitates handling with ratios (e.g. $U_2/U_1$). These ratios are called level in engineering (in German: Pegel). The level makes it possible to refer to a reference quantity. In electronic circuitry, the decibel is used as a dimensionless unit for current or voltage ratios. In the future, this will be particularly interesting for the amplification $A_{\rm V} = \frac{U_\rm O}{U_\rm I}$ and factors.

The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation:

$\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 {\rm dB} \cdot \log_{10} A_\rm V}$     resp.      $A_{\rm C}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{I_2}{I_1}\right)$

#### Technical level in dB

Name Symbol Formula Reference value for 0dB
Voltage level $\rm dBV$ $20{\rm dB} \cdot \log_{10}(V/V_{\rm ref})$ $\rm 0dBV \widehat{=} 1V$
Power level $\rm dBm$ $10{\rm dB} \cdot \log_{10}(P/P_{\rm ref})$ $\rm 0dBm \widehat{=} 1mW$
Power level $\rm dBW$ $10{\rm dB} \cdot \log_{10}(P/P_{\rm ref})$ $\rm 0dBW \widehat{=} 1W$
Full scale level $\rm dBFS$ $20{\rm dB} \cdot \log_{10}(V/V_{\rm max})$ $\rm 0dBFS \widehat{=} V_{max}$
Sound pressure level $\rm dBA$ $20{\rm dB} \cdot \log_{10}(p/p_{\rm ref})$ $\rm 0dBA \widehat{=} 20 µPa$

Note that this equation changes somewhat for power quantities, i.e. ratios of $P$.
Since $P \sim U^2$ or $U \sim P^\frac{1}{2}$, we get:
$A_P^{\rm dB}= 20 ~{\rm dB} \cdot \log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$
$=\color{blue}{20 ~{\rm dB} \cdot \frac{1}{2}} \cdot \log_{10}\left(\frac{P_2}{P_1}\right) =\color{blue}{10 {\rm dB}} \cdot \log_{10}\left(\frac{P_2}{P_1}\right)$

The table above shows various dB-levels which are frequently used in engineering. In the following, only the voltage level is used and indicated with the symbol $\rm dB$.

#### Simple examples of voltage levels in dB

linear factor level [$\rm dB$]
$\times 0.01$ $-40 ~{\rm dB}$
$\times 0.1$ $-20 ~{\rm dB}$
$\times 1$ $0 ~{\rm dB}$
$\times 2$ $\approx +6~{\rm dB}$
$\times 10$ $+20 ~{\rm dB}$
$\times 100$ $+40 ~{\rm dB}$

By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{\rm dB}$ in $\rm dB$. Occasionally, the superscript $\boxed{}^{\rm dB}$ is omitted below. Instead, the level is denoted by the unit after the numerical value.

Examples:

1. For $A_{\rm V}= \color{green}{1}$ we get

$A_{\rm V}^{\rm dB}(\color{green}{1}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{1})} \quad = 20 {\rm dB} \cdot 0 \quad \ \boldsymbol{= 0 {\rm dB}}$
Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$.

2. For $A_{\rm V}= \color{green}{0.01}$ we get

$A_{\rm V}^{\rm dB}(\color{green}{0.01}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{0.01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{0.01})} = 20 {\rm dB} \cdot (-2) \boldsymbol{= -40 {\rm dB}}$
Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, so even $\color{blue}{x}=-2$.

3. For $A_{\rm V}= \color{green}{2}$, we get

$A_{\rm V}^{\rm dB}(\color{green}{2}) = 20 {\rm dB} \cdot \underbrace{\log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{2})} \quad\approx 20 {\rm dB} \cdot 0.30103 \boldsymbol{\approx 6 {\rm dB}}$
Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$.

#### Use of the dB measure

The decibel offers some advantages, which are used in the filter elements considered below:

• handier numerical values: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_{\rm V} = 10000000 \rightarrow A_{\rm V}^{\rm dB}= 140{\rm dB}$. This also results in less „zero counting“.
• Relationship to Sensory Perceptions: Sensory perceptions such as brightness and loudness have an almost exponential effect. That is, any tenfold increase in the underlying physical quantity (number of photons or sound pressure) does not have ten times the effect but seems to have an additive effect.
• easier math: The logarithm in the defining equation turns any multiplication of linear factors into addition of levels: $A_{\rm V}^{\rm dB}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1) + 20~{\rm dB} \cdot \log_{10}(A_2) = A_{\rm V}^{\rm dB}(A_1) + A_{\rm V}^{\rm dB}(A_2)$

Abb. 1: Example: filters in series

Especially the last point of the calculation should be considered again. In Abbildung 1 several amplifiers connected in series can be seen with exemplary voltage gain values.
The total gain here is the product of the individual gains: $A_{\rm V, eq}=\prod A_{i} = A_1 \cdot A_2 \cdot A_3$.
The determination of the total gain was rather laborious before the time of the pocket calculator due to the multiplications. For the levels, the result is an addition: $A_{\rm V,eq}^{\rm dB}=\sum A_i^{\rm dB} = A_1^{\rm dB} + A_2^{\rm dB} + A_3^{\rm dB}$.
Here this would be: $A_{\rm V,eq}^{\rm dB} = \sum A^{\rm dB} = 88~{\rm dB} + (-58~{\rm dB}) + 14~{\rm dB} = 44~{\rm dB}$.

#### Remember: dB measure

For current and voltage levels:

1. A linear factor of $\color{green}{\times 10}$ results in level $+ 20~{\rm dB}$.
2. A linear factor of $\color{green}{\times 2}$ results in a level of $+ 6~{\rm dB}$.
3. The linear value $A_{\rm V} = 1$ corresponds to $0 ~{\rm dB}$.

For systems connected in series, to determine the amplification

1. multiply the linear measure $A_\rm V$ and
2. add the level $A_{\rm V}^{\rm dB}$.

#### More difficult examples of voltage levels in dB

With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20~{\rm dB}$ and $\color{green}{\times 2} \rightarrow + 6~{\rm dB}$ the linear values can easily be determined from a level in $dB$ without a calculator.

Examples:

1. $A_{\rm V}^{\rm dB}=58~{\rm dB}$
with interpolation points: $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$
This becomes linear $\qquad \qquad \qquad \qquad \qquad \qquad \quad \ A_{\rm V} = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$

2. $A_{\rm V}^{\rm dB}=56~{\rm dB}$
with interpolation points: $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$
This becomes linear $\qquad \qquad \qquad \qquad \qquad \qquad \quad\ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad 2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$
or alternatively $qquad \qquad \qquad \qquad \qquad \qquad A_{\rm V}^{\rm dB} = 20~{\rm dB} + 36~{\rm dB} \rightarrow A_{\rm V}= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$

3. $A_{\rm V}^{\rm dB}=55~{\rm dB}$
with interpolation points: $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB} + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$
This becomes linear $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ A_{\rm V} = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3} \qquad \cdot \qquad 2^\color{teal}{-\frac{1}{2}} \approx 100 \cdot 8 \cdot 0.707 = 560$

The value $-3~{\rm dB}$ will still be used in the following examples.

Abb. 2: Representation of complex numbers

The Bode diagram aims to show the transmission behavior of systems clearly and concisely.

#### Preliminary consideration: complex numbers

A complex number can always be reduced to two real number values. For the exact definition of these numerical values, there are different possibilities (Abbildung 2):

1. Definition over real part $\Re(\underline{A}_{\rm V})=A_{\rm V} \cdot \cos(\varphi)$ and imaginary part $\Im(\underline{A}_{\rm V})=A_{\rm V} \cdot \sin(\varphi)$ in $\underline{A}_{\rm V}= \Re(\underline{A}_{\rm V}) + {\rm j} \cdot \Im(\underline{A}_{\rm V})$
2. Definition over absolute value $A_{\rm V} = |\underline{A}_{\rm V}|$ and phase $\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right)$ in $\underline{A}_{\rm V}=A_{\rm V} \cdot {\rm e}^{{\rm j} \varphi}$

The 2nd definition is more appropriate when considering frequency-dependent voltage gain since it allows the „time shift“ (phase) to be separated from the gain.

#### Developing the Bode Diagram

To better understand the frequency dependence of the voltage gain, it can be plotted as a function of frequency $f$ as $|A_{\rm V}(f)|$. It is useful to represent the voltage gain as level $|A_{\rm V}^{\rm dB}(f)|$. The simulation above shows the $|A_{\rm V}^{\rm dB}(f)|$ curve for a low pass filter in the lower half of the image. The curve starts at $0~{\rm dB}$ on the left and drops to about $-45~{\rm dB}$. If the mouse is dragged over the diagram, the gain value in ${\rm dB}$ can be displayed for each frequency. Clicking on the curve displays current flow and voltage ratios. Only clicking on the leftmost frequency range results in a situation where $U_{\rm O}$ assumes noticeably high voltages (visible via the color of the line).

In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve. However, if the frequency axis is also logarithmized, a different picture results. This is done via Options » Linear Scale. Now the initially flat course of the voltage gain for low frequencies and the drop for higher frequencies becomes very clear. Also, the cut-off frequency can be read at the point of the „bending“.

The phase can be made visible via Options » Show Phase.

#### Description of the Bode diagram

Abb. 3: Principle image of the Bode diagram

The bottom diagram allows the representation of a complex-valued, frequency-dependent quantity in logarithmic form. It is also referred to as the frequency response and is divided into (cf. Abbildung 3):

• the amplitude response which represents amplitude in double logarithmic form (dB level is a logarithmic representation); and
• the phase response which represents the phase single logarithmically.

A jump in frequency by a factor of $\times 10$ is called a decade (abbreviated Dec.). The amplitude response of various functions will be briefly discussed ($\mathcal{C}$ is an arbitrary frequency-independent factor):

1. $|A_{\rm V}(f)| = \mathcal{C} \cdot f$:
If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain.
This results in an increase of $+20~{\rm dB}$ per decade.
2. $|A_{\rm V}(f)| = \mathcal{C} / f$:
If a function is considered that is reciprocal to $f$, then a tenfold increase in frequency results in a decrease in voltage gain to one-tenth.
This results in a drop of $-20~{\rm dB}$ per decade (cf. Abbildung 3 at high frequencies).
3. $|A_{\rm V}(f)| = \mathcal{C} / f^n$:
Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/10^n$th.
This results in a drop of $-20~{\rm dB} \cdot n$ per decade.

As an alternative to the actual course, $|A_{\rm V}(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments.

#### Remember: Bode diagram

The bottom diagram (= frequency response) consists of:

1. Amplitude response: amplitude in dB over logarithmic frequency (i.e. double logarithmic representation).
2. Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation)

This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$. In particular, this is true for $A(\omega) \sim \omega$, i.e., a slope of $+ ~\rm 20dB/decade$, and for $A(\omega) \sim \frac{1}{\omega}$, i.e., a slope of $-20~\rm dB/decade$

Abb. 4: Filter circuits

Up to now, operational amplifier circuits have been considered in which negative feedback took place via ohmic resistors. In the following, operational amplifier circuits with storing components ($C$, $L$) will be analyzed.

In electronics, inductors are rarely used for this purpose. This has several reasons:

1. Inductors are possible in integrated circuits but are somewhat more difficult to calculate as such an element.
2. Inductors require a current source as current storage. The internal resistance results in a continuous power loss.

Instead of inductors, capacitors are used in microelectronics and filter technology for sensor signals. The passive circuits formed are correspondingly also called $R$-$C$-links and were analyzed in Elektrotechnik 2.

The following basic circuit is a modified, inverting amplifier (Abbildung 4) in which one or both ohmic resistors are replaced by (complex-valued) impedances.

For the first circuit, only the part between output voltage $U_{\rm O}$ and virtual ground should be replaced by a capacitor (Abbildung 5).

The first active filter circuit can be seen in the simulation above. The superimposed square-wave voltage sources result in a step function as input voltage. This generates a current via the RC element. If we now look at the output voltage in comparison to the input voltage, we can see that:

1. for each constant input value $U_{\rm I} \neq 0$ an output value with a fixed slope results and
2. for each positive input value $U_{\rm I} > 0$ a negative slope results, and for a negative input value a positive slope results.

The circuit thus created is called an inverse integrator or inverting integrator.

If you look at the circuit, you can see that the node $\rm K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly.

Abb. 5: Inverse Integrator

Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in Abbildung 5 for this purpose. The transfer function $U_{\rm O} = f(U_{\rm I})$ is now to be determined.

$A_{\rm V} = ? \quad \rightarrow \quad U_{\rm O} = f(U_{\rm I})$

#### given equations

Given the following equations:

 I. Basic equation $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$ II. Mesh 1 $-U_{\rm I} + U_R - U_{\rm D} = 0$ III. Mesh 2 $U_{\rm D} + U_C + U_{\rm O} = 0$ IV. Node $I_R = I_C$ V. Capacity C $C = { Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C {\rm d}t+ Q_0(t_0))$ VI. Resistance R $R = { U_R \over I_R }$

#### Calculation

The calculation is performed once in detail here (clicking on the right arrow „►“ leads to the next step, alternative representation):

Abb. 6: Example of a signal-time curve of the Inverting Integrator

Using an example, the signal-time curve at the inverting integrator shall be explained.

1. Let $R=5 ~\rm k\Omega$, $C=1 ~\rm µF$, and the input voltage waveform $U_{\rm I}$ shown in Abbildung 6 be given.
2. We are looking for the output voltage $U_{\rm O}$.

Solution:

1. Over the given values of $R$ and $C$, the time constant $\tau$ is determined.
2. With the inverse integrator the input value is integrated and inverted. For the given course of the input voltage, the calculation of interpolation points is sufficient.
3. With the formula derived in 5.1.1 $U_{\rm O}$ can be composed section by section:

The calculation is performed once in detail here (clicking on the right arrow „►“ leads to the next step, alternative representation):

#### Notice: signal-time curve of the inverting integrator

If a constant input voltage $U_{\rm I}$ is applied to the inverting integrator, the output voltage $U_{\rm O}$ just equals $-U_{\rm I}$ after the time constant $\tau$ (see Abbildung 6, light gray arrow).

To be able to determine absolute value and phase, purely sinusoidal input and output variables are to be considered first. The following function is used as input voltage $U_{\rm I}$:

$U_{\rm I}(t)= \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$

This definition of the input voltage can now be substituted into the above equation for $U_{\rm O}$:

The calculation is performed once in detail here (clicking on the right arrow „►“ leads to the next step, alternative representation):

Abb. 7: Sketch of the bottom diagram from the inverting integrator

The absolute value $|A_{\rm V}|$ is given by the amplitude ratio of $\hat{U}_{\rm O} \over \hat{U}_{\rm I}$: $$|A_{\rm V}|={\hat{U}_{\rm O} \over \hat{U}_{\rm I}} = {1 \over {\omega \cdot R\cdot C}}$$

The phase can be determined from the „time offset“ of the peak input voltage $U_{\rm I} = \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$ and output voltage $U_{\rm O} = { {\hat{U}_{\rm I} } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$ can be determined. The phase is given by considering the trigonometric functions and the sign:

$U_{\rm I} = + \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$
$U_{\rm O} = + \hat{U}_{\rm O} \cdot \cos(\omega \cdot t) = + \hat{U}_{\rm O} \cdot \sin(\omega \cdot t + 90°)$
$\rightarrow \varphi = 90°$

#### Consideration of extreme frequencies

To be able to sketch the course in the bottom diagram, the behavior of the transfer function $U_{\rm O}=f(U_{\rm I})$ in the extreme cases for low (${\omega}\rightarrow 0$) and high frequencies (${\omega}\rightarrow \infty$) shall be considered. For absolute value $|A_{\rm V}|$ and phase $\varphi$ we obtain:

$|A_{\rm V}({\omega}\rightarrow 0 \ \; ) | \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$
$|A_{\rm V}({\omega}\rightarrow \infty) | \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow\infty}\quad 0$

$\varphi = +90° \qquad \forall \ \omega$

From these boundary conditions, the frequency response can already be sketched, see Abbildung 7.

In the previous chapters, it could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the inverting integrator. Now the complex calculation will be considered as a method that simplifies the analysis of such circuits. For the complex calculation, the resistances and capacitances are replaced by complex impedances:

$U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$
$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ with $\quad \underline{Z}_C= \frac{1}{{\rm j} \cdot \omega \cdot C}$

However, this consideration can only be implemented under certain boundary conditions:

1. sinusoidal quantities: complex current or voltage pointers (see Introduction in AC circuits) can only represent sinusoidal quantities.
2. Steady state: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since the switch-on. This takes out disturbances that are generated by switching on.

Abb. 8: Circuit of the inverting integrator with complex impedances

This can now be used to calculate the circuit (Abbildung 8):

$\underline{Z}_1=R$

$\underline{Z}_2=\frac{1}{{\rm j} \cdot \omega \cdot C} = \frac{\rm -j}{\omega \cdot C}$

From the basic circuit of the inverting amplifier, its voltage gain is known:

$A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=-\frac{R_2}{R_1}$

This results in the complex:

$\underline{A}_{\rm V} = \frac{\underline{U}_{\rm O}}{\underline{U}_{\rm I}}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{\rm j}{\omega \cdot R \cdot C}$

#### Plausibility check via extreme frequency consideration

From the formulaFrom the circuit
$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow 0} \infty$

For $\omega \rightarrow 0$, the capacitor behaves like a high impedance resistor
Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of $\underline{U}_{\rm O} \rightarrow \infty$.

$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow \infty} 0$

For $\omega \rightarrow \infty$ the capacitor behaves like a short circuit
Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of $\underline{U}_{\rm O} \rightarrow 0$.

#### Absolute Value and Phase

Abb. 9: Course of the arc tangent

The absolute value $A_{\rm V}$ is given by:

$|\underline{A}_{\rm V}|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$

Specifically, for a absolute value of $|\underline{A}_{\rm V}(0~{\rm dB})|$ at $0~{\rm dB}$, the result is:

$|\underline{A}_{\rm V}(0~{\rm dB})|\overset{!}{=} 1 \widehat{=} 0~{\rm dB} \rightarrow \omega(0~{\rm dB}) = \frac{1}{R \cdot C}$

The phase $\varphi$ is calculated via

$\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = \arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = \arctan \left( \infty \right) = +90°$

The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (Abbildung 9) for $x \rightarrow \infty$.

Abb. 10: Bode diagram of the inverting integrator

The frequency response is to be illustrated using an example.

1. Let $R=1 ~\rm k\Omega$, $C=16 ~\rm nF$ be given.
2. We are looking for the Bode diagram

Solution

1. Determining the time constant:
$\tau = R \cdot C = 16 ~\rm µs$

2. Determining the frequency $f$ for $|\underline{A}_{\rm V}(0~{\rm dB})|$: $\omega(0~{\rm dB})= \frac{1}{\tau} = 2\pi \cdot f(0~{\rm dB})$
This gives $f(0~{\rm dB})$ via:
$f(0~{\rm dB})=\frac{1}{2\pi} \cdot \frac{1}{16 ~\rm µs} \approx 10 ~\rm kHz$

3. Consideration of the slope:
$|\underline{A}_{\rm V}|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$
From this, a tenfold increase in $f$ results in one-tenth the absolute value $|\underline{A}_{\rm V}|$, i.e., a slope of $-20~{\rm dB}$ per decade

4. From this information, the full Bode diagram can be determined (Abbildung 10)

Abb. 11: Circuit of the active Low Pass Filter

Another circuit can be derived from the inverting integrator. For this purpose, the hitherto purely capacitive value of the impedance between the output side and the virtual ground is to be supplemented by a resistive component. This circuit can be seen in Abbildung 11. In the following, this circuit

• first be considered practically with a simulation,
• then a picture of the system's effect will be drawn up without detailed calculation, and finally
• be checked by a circuit analysis with complex calculations.

#### Low Pass Filter in Simulation

In the simulation above, the circuit from Abbildung 11 is shown again. In addition, two switches $\rm S1$ and $\rm S2$ are built into the circuit by which the various feedback paths can be disabled:

1. If only switch $\rm S1$ is closed, the circuit is an inverting amplifier.
2. If only switch $\rm S2$ is closed, the circuit is an inverting integrator.

In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in ${\rm dB}$, or the phase in degrees.

#### Exercise 5.2.1. circuit analysis in simulation

1. First set up an inverting amplifier and read off the gain and phase (click on the Bode diagram).
2. Now change the circuit to an inverting integrator and read off the gain and phase there as well.
3. Now both switches should be closed.
1. In which frequency ranges does the inverting amplifier or the inverting integrator work approximately?
2. Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1~\rm k\Omega$ resistor and the capacitor? What is the value of the gain and phase here?
3. After reading through the following analyses, the gain and phase at the „kink point“ can be determined. Do these deviate from your measurement?

#### Extreme frequency consideration

From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of the absolute value of voltage gain and phase can be analyzed.

1. $\omega \rightarrow 0$:
1. The capacitor acts like an open circuit
2. Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$
3. Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially
4. Thus the circuit behaves like an inverting amplifier: $|\underline{A}_{\rm V}|=|-\frac{R_2}{R_1}|$
5. For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier.
2. $\omega \rightarrow \infty$:
1. The capacitor acts like a short circuit
2. Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$
3. Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts
4. Thus the circuit behaves like a inverting integrator: $|\underline{A}_{\rm V}|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$ and $\varphi=+ 90°$

From this, it can be seen that

• for low frequencies a constant gain is expected and
• for high frequencies a drop is known from the inverting integrator.

#### expected Bode diagram

Abb. 12: Expected bottom diagram of Low Pass Filter

A Bode diagram can be estimated from the extreme frequency consideration.

Frequency Response:

• For low frequencies, the filter behaves like an inverting amplifier with $|\underline{A}_{\rm V}| \xrightarrow{f \rightarrow 0} |-\frac{R_2}{R_1} |$
• For higher frequencies, the filter behaves like an inverting integrator with $|\underline{A}_{\rm V}| \xrightarrow{f \rightarrow \infty} |-\frac{1}{\omega \cdot R_1 \cdot C}|$
• There is a frequency where both situations seem to occur simultaneously

Phase response:

• For low frequencies, the filter behaves like an inverting amplifier with $\varphi=\pm 180°$.
• For higher frequencies the filter behaves like an inverting integrator with $\varphi=+90°$
• There is a frequency where both situations seem to occur simultaneously

For the intermediate area, there must be a transition between the two extremal situations.

One problem still seems to be that for the inverting amplifier, it is not clear whether the phase is now $+180°$ or $-180°$. In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ {\rm d}t$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just a $90°$ phase shift from the inverting amplifier to high frequencies at low frequencies.

From this knowledge, we get an expected Bode diagram as seen in Abbildung 12.

#### Notice: filters in the Bode diagram

The following rules apply to filters:

• for each energy storage element in the circuit, the order of the filter increases by 1.
• for each energy storage element in the circuit ($C$, $L$) there is an amplitude change of $-20~\rm dB/dec$
• each energy storage element in the circuit ($C$, $L$) results in a phase change of $-90°$
• The phase response is monotonically decreasing.

#### RC element and cut-off frequency

In the circuit, the parallel circuit $R_2$ and $C$ behave like a passive RC element. That is, it affects the frequency response. At a certain frequency, the circuit behaves just in such a way that the current runs half over $R_2$ and $C$, thus acting „half“ like an inverting amplifier and „half“ like an inverting integrator. This frequency is the cut-off frequency $f_{c}$:

For this:
$|\underline{X}_C|=R_2$
$\frac{1}{\omega_{c} \cdot C}=R_2 \rightarrow \omega_{c} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{c}$
$\boxed{f_{c} = \frac{1}{2\pi \cdot R_2 \cdot C}}$

Now the circuit is to be analyzed again using complex calculations. The impedances are again understood as complex numbers. The starting point is again the voltage gain of the (complex) inverting amplifier. The impedances from Abbildung 11 are taken into account:

$\underline{A}_{\rm V}=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{{\rm j} \cdot \omega \cdot C}}{R_2 + \frac{1}{{\rm j} \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$
$\boxed{\underline{A}_{\rm V}= - \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}}$

#### Calculation of Absolute Value and Phase

For the calculation of the absolute value $A_{\rm V}$ a „trick“ can be used. In principle, the absolute value can always be determined by multiplication with the conjugate complex value. But here it is easier to calculate the absolute value of the fraction via the absolute value of the numerator and the absolute value of the denominator:

$|\underline{A}_{\rm V}| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{|\mathcal{c}|}$

This results in the absolute value:

$\boxed{|\underline{A}_{\rm V}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$

For the phase, $\varphi$ real value $\Re(\underline{A}_{\rm V})$ and imaginary value $\Im(\underline{A}_{\rm V})$ must be determined by multiplication with the conjugate complex value.

$\varphi = \arctan(\frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})})$

But here, too, there is a „trick“:

$\underline{A}_{\rm V}= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}} \cdot \frac{1 - {\rm j} \omega \cdot R_2 \cdot C}{\color{blue}{1 - {\rm j} \omega \cdot R_2 \cdot C}}$

After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$ because all factors of the constant are real:

$\underline{A}_{\rm V}= \color{blue}{\mathcal{C}} \cdot (1 - {\rm j} \omega \cdot R_2 \cdot C)$

Thus, the phase $\varphi = \arctan\left(\frac{\color{teal}{\Im(\underline{A}_{\rm V})}}{\color{brown}{\Re(\underline{A}_{\rm V})}}\right)$ is obtained as.

$\underline{A}_{\rm V}= \mathcal{C} \cdot (\color{brown}{1} + {\rm j} \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$

$\boxed{\varphi = \arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$

#### Consideration of extreme frequencies

For the absolute value we get

1. for $\omega \rightarrow 0$:
$|\underline{A}_{\rm V}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 }}$,     since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \gg 1$
So the absolute value of the gain tends towards $|\underline{A}_{\rm V}| = \frac{R_2}{R_1}$.
The effect is similar to the inverting amplifier

2. at $\omega \rightarrow \infty$:
$|\underline{A}_{\rm V}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{\color{blue}{(\omega \cdot R_2 \cdot C)^2}}}$,     since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \ll 1$
So the absolute value of gain tends towards $|\underline{A}_{\rm V}| = \frac{1}{\color{blue}{\omega \cdot} R_1 \color{blue}{\cdot C}}$.
This resembles the inverse integrator

Abb. 13: arc tangent

For finding the phase $\color{red}{\varphi} = \arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (Abbildung 13). For the extremal values $\omega$ of results in:

1. at $\omega \rightarrow 0$: the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$.
2. at $\omega \rightarrow \infty$: $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$.

In the diagram, the argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuous because also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the upper branch: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in Abbildung 13. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram.

This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$.

#### Calculation of the cut-off frequency

The cut-off frequency can also be understood as the transition from the inverting amplifier to the inverting integrator. In the Bode diagram (Abbildung 12), the cut-off frequency can be found at the intersection of the straight lines for the inverting amplifier and the inverting integrator.

Thus, for the cut-off frequency $f_{c}$ we get

$\frac{R_2}{R_1} = \frac{1}{\omega_{c} R_1 \cdot C}$
$\omega_{c} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{c}$

At the cut-off frequency, the result is an absolute value of:

$|\underline{A}_{V,c}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega_{c} \cdot R_2 \cdot C)^2}}= \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C)^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + 1^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{2}}$

$\boxed{|\underline{A}_{V,c}| = \frac{1}{2} \sqrt{2} \cdot \frac{R_2}{R_1} = -3~{\rm dB} + |\underline{A}^{\rm dB}_{\rm V}(\omega \rightarrow 0)|}$

The phase at the cut-off frequency is:

$\varphi_{c} = \arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = \arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = \arctan\left(-1 \right)$

$\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$

Because of the $-3~{\rm dB}$ attenuation of the low-frequency gain at the cut-off frequency, it is also called the $-3~{\rm dB}$ cut-off frequency.

Abb. 14: Circuit of the inverting Differentiator

Abb. 15: Bode diagram of the inverting differentiator

In Abbildung 14 an inverting differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped.

In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is resulting in the output. The derivative at the reversal points („peaks“) of the signal cannot be determined (see Differentiability of the sum function). This leads to problems in the calculation during the simulation and can be seen as overshoot or „deflection“ at $U_{\rm O}$. To reduce this, a small resistor (relative to the feedback resistor) is inserted after the capacitor.

In the following, only the results will be discussed without calculation.

Circuit analysis via differential equation yields:

$\boxed{U_{\rm O} = - R \cdot C \frac{\rm d}{{\rm d}t}U_{\rm I}}$

With complex calculation, the transfer function becomes:

$\boxed{\underline{A}_{\rm V}=-{\rm j} \cdot \omega \cdot R \cdot C}$

From this, the Bode diagram shown in Abbildung 15 can be determined.

#### Exercise 5.3.1 Inverting Differentiator

For the inverting differentiator shown in Abbildung 14, derive the complex voltage gain, and its absolute value and phase using complex calculus as shown for the inverting integrator. In doing so, implement the following steps:

1. Circuit analysis using differential equation
2. Determination of absolute value and phase from differential equation (incl. consideration of extreme cases)
3. Example of a signal-time-curve with $R = 10 ~\rm k\Omega$ and $C = 2 ~\rm µF$ and $U_{\rm I}$ - as shown in the diagram.
4. Circuit analysis using a complex calculation
5. Consideration of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$
6. Frequency response (Bode plot) for a circuit with $R = 10 ~\rm k\Omega$ and $C = 16 ~\rm nF$.

A high pass filter can be created from the inverting differentiator, if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (Abbildung 16). The simulation above shows this high pass. A reverse integrator forms from this with switch $\rm S1$ closed and switch $\rm S2$ open. When the switch is inverted, an inverting amplifier is formed. In the simulation, clicking on a frequency point again shows the distribution of the currents.

Abb. 16: Circuit of the High Pass Filter

With complex calculation this results in: $\boxed{\underline{A}_{\rm V} = - \frac{R_2}{R_1} \cdot \frac{{\rm j} \cdot \omega \cdot R_1 \cdot C}{1 + {\rm j} \cdot \omega \cdot R_1 \cdot C}}$

From this, the Bode diagram shown in Abbildung 17 can be determined.

Abb. 17: Bode diagram of the High Pass Filter

#### Exercise 5.4.1 1st order high pass

In previous chapter the gain $A_{\rm V}$ of the 1st order low pass filter was derived based on its circuit. In the same way, now the gain for a high pass filter (cf. Abbildung 16) shall be derived.

1. Behavior of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$.
2. Expected Bode diagram
3. RC element and cut-off frequency
4. Circuit analysis with complex calculation
5. Calculation of absolute value and phase

Abb. 18: Overview high pass Filter / low pass Filter

# Exercises

#### Exercise 5.0.1. Converting linear Factors to dB

Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points $20 ~{\rm dB}$ ≙ factor 10 and $6 ~{\rm dB}$ ≙ factor 2.

Solve without a calculator (note: $\sqrt{2}\approx 1.414$, ${1\over\sqrt{2}}\approx 0.707$).

As an example, the calculation is sketched for the value $10 ~{\rm dB}$.

levelover interpolation points in $\rm dB$over interpolation points linearlinear factor
$10 ~{\rm dB}$$5 \cdot 6 ~{\rm dB} - 20 ~{\rm dB}$$2^5 \cdot {1\over 10}$$3,2$
$2 ~{\rm dB}$
$4 ~{\rm dB}$
$6 ~{\rm dB}$
$8 ~{\rm dB}$
$12 ~{\rm dB}$
$14 ~{\rm dB}$
$16 ~{\rm dB}$
$18 ~{\rm dB}$
$15 ~{\rm dB}$
$79 ~{\rm dB}$
$128 ~{\rm dB}$

#### Exercise 5.0.2. Series of Amplifiers

The following image shows a series of amplifiers, which shall be analyzed.

The amplification factors are:

• $A_{\rm V1} =80$
• $A_{\rm V2} =0.0125$
• $A_{\rm V3} =250'000$

Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$).

1. Rearrange the given linear factors as exponents of $2$ and $10$, e.g. $2^6 \cdot 10^7$
2. Use the exponent values to transfer it into $\rm dB$, in this example: $6 \cdot 6~{\rm dB} + 7 \cdot 20~{\rm dB}$
3. Calculate the $\rm dB$ value, in this example: $36~{\rm dB} + 140~{\rm dB} = 176~{\rm dB}$

Amp. Name $\rightarrow$ lin. Factor $\rightarrow$ re-arrange $\rightarrow$ as exponents of $2$ and $10$ $\rightarrow$ transfer into $\rm dB$ $\rightarrow$ result in $\rm dB$
$A_{\rm V1}$ $80$ $8 \cdot 10$ $2^3 \cdot 10^1$ $3 \cdot 6~{\rm dB} + 1 \cdot 20~{\rm dB}$ $38~{\rm dB}$
$A_{\rm V2}$ $0.0125$ $0.125\cdot 0.1$ $2^{-3} \cdot 10^{-1}$ $-3 \cdot 6~{\rm dB} + (-1) \cdot 20~{\rm dB}$ $-38~{\rm dB}$
$A_{\rm V3}$ $250'000$ $0.25 \cdot 1'000'000$ $2^{-2} \cdot 10^{6}$ $-2 \cdot 6~{\rm dB} + 6 \cdot 20~{\rm dB}$ $108~{\rm dB}$

$A_{\rm V} = 250'000$
$A_{\rm V}^{\rm dB} = 108~{\rm dB}$

#### Exercise 5.1.1 Analysis of Circuits in Tina TI

The following circuit shall be given with $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ and a sinusoidal input voltage $U_{\rm I} = 1 ~\rm V$ with $f = 1 ~\rm kHz$

As described in the course, the Bode diagram can be displayed in Tina TI via Analysis > AC Analysis > AC Transfer Characteristic.
In the following, frequencies from 100 Hz to 1 GHz are relevant.

1. Simulate this circuit in Tina TI
- with an ideal operational amplifier
- using the operational amplifiers uA776, LM301A and LM318.
1. Attach the Bode diagram.
2. Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$.
2. What happens if instead of $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ the same time constant is implemented with $R= 10 ~\rm M\Omega$, $C = 1.6 ~\rm nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier.
3. Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 ~\rm k\Omega$ and use the LM318 op-amp.
1. Attach the Bode diagram.
2. What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup?
3. Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of $1~{\rm dB}$)? Use zoom and/or cursor to determine.
4. Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to the ground. The inverting input should have the above sinusoidal input voltage.
1. Attach the Bode diagram.
2. What is the cut-off frequency?
3. How many $\rm dB$ per decade does the amplitude response drop at high frequencies?

# Learning questions

• What affects the rise time of an amplifier circuit?
• What is the difference between an ideal and a real operational amplifier?

# References

References to the media used