3 Basic circuits of operational amplifiers I

3.0 Introduction

Historical outline

To understand the origin of the operational amplifier, a brief outline of microprocessor history will be presented here. The first microprocessor was commercially distributed in 1971 by INTEL as Intel 4004, which could only represent a 4-bit number space with one word and had a clock frequency of 500 kHz. This was already extended after half a year as 8008 on an 8-Bit word so that calculations could be accomplished substantially faster. From this second generation, the 8086 was developed in 1978, which had been the basis of many PCs and other electronic components with a 16-bit word for a very long time. In 1985 this processor was extended from 80386 to a 32-Bit Word. From this time also the perhaps well-known question originates whether software should be installed as x86 or x64 code. Processors with 64-bit Word have been available since 1999, but they have not yet penetrated all niches.

With this historical outline in mind, the question arises as to how Apollo 11 was able to control both the first lunar mission and the lunar module, and how the orbit could be calculated so accurately that this would not pose a risk. Development of the Apollo Guidance Computer was started in 1961, 10 years before the first commercial microprocessor. This processor, with about 1MHz frequency and 16-bit word width, could provide performance as early as 1966 that was not commercially available until 15 years later. Nevertheless, these computers could not solve the differential equation systems necessary for trajectory planning in a reasonable amount of time. This was only possible by combining the well-known digital processor and the analog computers that had been in use since the 1920s. Analog computers are only slightly reminiscent of the classical image of the computer. Although they also consist of a network of components, they do not have any software. Rather, various properties of electronic components are used for computation (e.g., the exponential behavior of the Shockley equation of the diode). For the Apollo missions, the EAI 380 hybrid computer was used. This and similar hybrid computers continued to be used into the 1990s to solve systems of differential equations for aerospace, construction, and (under a slightly different guise) as synthesizers in the music industry. The interested reader is referred to the video Analog computers, which explains why they are superior to even the most modern computers.

The core component of analog computers is the operational amplifier. With this, it is possible to build and combine adders, subtractors, multipliers, exponentials, integrators, and differentiators. The „calculation“ happens almost instantaneously after setting the desired input voltages.

The operational amplifier has outlasted this time and nowadays offers a starting point for a wide variety of signal conditioning, such as active filters, amplifier stages, input/output resistance matching, and various (integrated) control loops. Several microcontrollers have op-amps built right in and can be configured by suitable code. In Abbildung 4 a PIC microprocessor is shown, which contains an integrated operational amplifier.

3.1 Circuit symbols and basic circuitry

This chapter deals with operational amplifiers or in general with measuring amplifiers. One application for this is the measurement of voltages, currents, and resistances. These must be determined very precisely in some applications, for example for accurate temperature measurement. In this case, amplification of the measurement signals is useful and necessary.

This amplification is done by measuring amplifiers. Measuring amplifiers have to fulfill some characteristics:

• Measurement amplifiers should not have any feedback effect on the measured variable. An operational amplifier should have the highest possible input resistance. This prevents the voltage to be amplified from collapsing.
• Measuring amplifiers should have a high sensitivity. An operational amplifier should have a large differential gain $A_\rm D$.
• Measuring amplifiers should show a defined transmission behavior, i.e. the output signal should be clearly related to the input signal. An operational amplifier concretely should show a linear relationship.
• Measuring amplifiers should show good dynamic behavior. The output signal of an operational amplifier should follow the input signal without any time delay.
• Measurement amplifiers should produce an „impressed output signal“. This means that the components at the amplifier output cannot change the output signal. An operational amplifier, specifically, should be able to maintain the desired output signal with the necessary current to do so. Since the current $I_\rm O$ can become very large (by electronic standards), this means that an operational amplifier must have a low output resistance $R_{\rm O} =\frac{U_{\rm O}}{I_{\rm O}}$.

In general, a measurement amplifier is constructed as in Abbildung 5. This has already been described in chapter Amplifier Basics. In the following, only operational amplifiers will be considered. An operational amplifier is a measurement amplifier, which is often used in electrical engineering.

The circuit symbol of the amplifier is an isosceles triangle, at the apex of which the output signal originates and at the base of which the input signal enters. In Abbildung 6 you can see different circuit symbols:

• Circuit symbol (1): In block diagrams (not to be confused with circuit diagrams, see chapter 1) this circuit symbol is used for general amplifiers. The input signal enters an input and exits through an input. This symbol will not be found again until Chapter 5.
• Circuit symbol (2): According to DIN EN 60617, this circuit diagram is to be used for operational amplifiers. It indicates with the infinity sign the ideally infinitely high amplification. in the following this symbol is not used, because it is not used in all international circuits and tools.
• Circuit symbol (3): The circuit symbol (3) is the most commonly used symbol for an operational amplifier. On the left is the inverting input with voltage $U_{\rm m}$ (minus) and the non-inverting input with $U_{\rm p}$ (plus). The output with voltage $U_{\rm O}$ is shown on the right.
• Circuit symbol (4): The circuit symbol (4) is additionally drawn with the supply voltages $U_{\rm sp}$ (supply plus) and $U_{\rm sm}$ (supply minus). Power is provided from the supply for the output voltage of the operational amplifier.
• Circuit symbols (5) and (6): these symbols show no operational amplifier. These symbols show the NOT gate and the tri-state gate. Both of these components have already been discussed in fundamentals of digital engineering. Unfortunately, the representation of these digital components in various circuits is not unlike the operational amplifier. An example of this is the transceivers¹⁾ SP3481 or SP3485. If digital input values are considered, assuming that the circuit symbol does not represent an operational amplifier.

3.2 Basic Equation / Golden Rules

The operational amplifier is a voltage amplifier. Thus it follows from the chapter idealized amplifier basic types that for the ideal case, the input resistance must be infinite and the output resistance $R_{\rm O}=0$.

The Abbildung 8 shows an ideal voltage amplifier. This is characterized by the following properties:

• input resistance: The input side is drawn in the figure. The resistance given here is the differential resistance $R_{\rm D}$ where the differential voltage $U_{\rm D}$ drops. Therefore $R_{\rm D} \rightarrow \infty$ is valid. Thus, the input currents $I_\rm p \rightarrow 0$ and $I_\rm m \rightarrow 0$.
• Output resistance, Basic equation: On the output side, $R_\rm O=0$ gives that $U_{\rm O} = A_{\rm D}\cdot U_\rm D$. This is the basic equation of the amplifier circuit. Ideally, the op-amp amplifies linearly, as indicated in the equation. Specifically, for a differential voltage of $0 ~\rm V$, the output voltage is $0 ~\rm V$.
• Voltage Gain: From the chapter feedback it is known that $A_\rm D$ must be very large. Ideally, the following applies: $A_ \rm D \rightarrow \infty$

These rules have different limits in the real amplifier:

• $\boldsymbol{U_{\rm O} = A_{\rm D} \cdot U_{\rm D}}$:
  • The output voltage can only follow the input voltage as far as the power supply allows. In real operational amplifiers, only so-called rail-to-rail operational amplifiers can exploit the range down to a few $100 ~\rm mV$ to $U_\rm S$. Other operational amplifiers have an output limit, which is $1 ... 2 ~\rm V$ below the supply voltage.
  • If the supply voltages are not symmetrical ($U_{\rm sm} \neq -U_{\rm sp}$), then the characteristic also shifts.
  • The ideal operational amplifier produces the same output voltage $U_{\rm O}=A_{\rm D} \cdot U_{\rm D}$ as long as $U_{\rm D} = U_{\rm p} - U_{\rm m}$ is the same. For the real operational amplifier with fixed $A_{\rm D}$, output voltage $U_{\rm O1}$ for $U_{\rm D1}=5 ~\rm V - 4.9 ~V$ is different from $U_{\rm O2}$ for $U_{\rm D1}=0.1 ~\rm V - 0 ~V$.
• $\boldsymbol{A_\rm D}$: The differential gain is usually between $A_\rm D = 20'000 ... 400'000$.
• $\boldsymbol{R_\rm D}$: For real operational amplifiers, the input resistance $R_\rm I > 1 M\Omega$ and the input current $|I_\rm p|$ or $|I_ \rm m|$ is less than $1 ~\rm µA$.
• $\boldsymbol{R_\rm O}$: In real operational amplifiers, the output resistance $R_\rm O$ is usually a few $\Omega$ and limited by a maximum current (in the range of a few tens of $\rm mA$ to a few $\rm A$)

The op-amp in the simulation replicates a real op-amp in some respects: The voltage gain is $A_\rm D = 100'000$. The transfer characteristic $U_{\rm O}(U_\rm D)$ shows proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{\rm sp}|=|U_{\rm sm}|=15 ~\rm V$ (not shown). The modulation limits and the voltage gain can be changed in the simulation via „Edit component“ (double click).

Power supply of the operational amplifier

For the voltage supply of the operational amplifier, a distinction is made between unipolar and bipolar:

With the bipolar voltage supply, the same voltage levels but with different signs are applied to both supply terminals in terms of magnitude (Abbildung 9 (1)). This allows the output voltage $U_{\rm O}$ of the amplifier to point in either direction (Abbildung 9 (2)). The power supply must be designed in such a way that it can provide both voltages. A differential voltage of $U_{\rm D}=0$ also results in an output voltage of $U_{\rm O}=0$.

With the unipolar power supply, the negative supply terminal is connected to the ground (Abbildung 9 (3)). As a result, the output voltage $U_{\rm O}$ of the amplifier can only take on non-negative values (Abbildung 9 (4)). In this case, the supply can be provided by a single voltage source (e.g., a battery). With a differential voltage of $U_\rm D=0$, the output voltage is $U_{\rm O}=\frac{1}{2}\cdot U_{\rm sp}$.

3.3 Voltage follower

In the chapter feedback it was described that an amplifier with high open-loop gain can be „tamed“ by feeding back a part of the output signal with a negative sign. In the simplest case, the output signal could be fed directly to the negative input of the operational amplifier. The input signal $U_\rm I$ of the entire circuit is applied to the positive input. In Abbildung 10 this circuit is shown.

Using this circuit, the procedure for solving amplifier circuits is now to be illustrated.

1. The aim is always to create a relation between output voltage $U_\rm O$ and input voltage $U_ \rm I$. Thus, the goal here is the voltage gain $A_{\rm V}=\frac{U_{\rm O}}{U_\rm I}$.
2. Before calculating, it should be checked how many equations describe the system and thus have to be set up. This can be determined by the number of variables. This is done by counting through the currents and voltages of the circuit. In this case, there are 3 currents and 3 voltages. So the number of equations needed is 6.
3. Now equations are set up that can be used. These are:
   1. Basic equation: (1) $U_{\rm O} = U_{\rm D} \cdot A_\rm D$
   2. Golden rules: $R_\rm D \rightarrow \infty$ so that (2+3) $I_{\rm p} = I_{\rm m} = 0$, $A_{\rm D} \rightarrow \infty$, $R_\rm O = 0$
   3. Consideration of the existing loops: in this example, there is only one loop (4) $-U_{\rm I} + U_{\rm D} + U_{\rm O} = 0$. Caution: loops can not enter the amplifier through input and exit through the output! Also to be noted is the direction of $U_{\rm D}$.
   4. Consideration of the existing nodes: in this example, there is only one node (5) $I_{\rm o} = I_m$.
4. There appears to be a missing equation. However, this is not correct, because there is still an equation hidden in the objective: (0) $A_{\rm V}=\frac{U_{\rm O}}{U_\rm I}$
5. Now, to solve the equations, the equations must be cleverly inserted into each other in such a way that there are no dependencies on the variables left at the end.

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

So the voltage gain is $A_\rm V = 1$. This would also have been seen in chapter feedback. There it was derived that for $A_\rm D \rightarrow \infty$ the voltage gain just results from $k$: $A_{\rm V}=\frac{1}{k}$. Since the entire output voltage is fed back here, $k=1$ and thus also $A_\rm V=1$.

The output voltage $U_\rm O$ is therefore equal to the input voltage $U_\rm I$. This is where the name „voltage follower“ comes from. Now one could assume, that this amplifier is of little help because also a direct connection would deliver $U_{\rm O}=U_{\rm I}$. But the important thing here is: because of the operational amplifier, there is no feedback from U_O to U_I. This means that a resistor on the output side will not load the input side. In the simulation, the „Resistance“ slider (on the right) can be used to change the load resistance. This changes the current flow, but not the voltage.

This behavior can also be explained in another way: The input signal usually comes from a voltage source, which can only produce low currents. That means the input signals are high impedance ($\text{high impedance}=\frac{\text{voltage}}{\text{low current}}$). However, a load of arbitrary impedance can be applied to the output. That is, to keep the output signal constant, a large current must be provided depending on the load. As the output resistance of the amplifier approaches 0, the signal is low impedance ($\text{low impedance}=\frac{\text{voltage}}{\text{(likely) large current}}$). This is where the second name of the circuit „impedance converter“ comes from.

3.4 Non-inverting amplifier

So far, the entire output voltage has been negative-feedback. Now only a part of the voltage is to be fed back. To do this, the output voltage can be reduced using a voltage divider $R_1+R_2$. The circuit for this can be seen in Abbildung 11.

By considering the feedback, the result can be quickly derived here as well: only $\frac{R_2}{R_1+R_2}\cdot U_{\rm O}$ is fed back from the output voltage $U_{\rm O}$. So the feedback factor is $k=\frac{R_2}{R_1+R_2}$ and thus the voltage gain becomes $A_{\rm V}=\frac{R_1+R_2}{R_2}$.

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

Abb. 11: Non-inverting amplifier

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

So the voltage gain of the non-inverting amplifier is $A_{\rm V}=\frac{R_1+R_2}{R_2}$ or $A_{\rm V}=1+\frac{R_1}{R_2}$. Thus, the numerical value $A_{\rm V}$ can only become larger than 1. This is shown again in the simulation. In real circuits, the resistors $R_1$ and $R_2$ will be in the range between a few $100 ~\Omega$ and a few $\rm M\Omega$. If the sum of the resistors is too small, the operational amplifier will be heavily loaded. However, the output current must not exceed the maximum current. If the sum of the resistors is too large, the current $I_1 = I_2$ can come into the range of the current $I_\rm m$, which is present in the real operational amplifier.

The input and output resistance of the entire circuit should also be considered here. Both resistors are marked here with a superscript 0 to distinguish them from the input and output resistance of the operational amplifier. The input resistance $R_{\rm I}^0$ is given by $R_{\rm I}^0=\frac{U_\rm I}{I_\rm I}$ with $I_{\rm I}=I_\rm p$. Thus, for the ideal operational amplifier, it is also true that the input resistance $R_{\rm I}^0=\frac{U_\rm I}{I_\rm p} \rightarrow \infty$ becomes when $I_\rm p \rightarrow 0$.

In the real case it is important in how far the total input resistance depends on the input resistance of the operational amplifier $R_{\rm I}^0(R_\rm D)$. This can be derived as follows: (clicking on the right arrow „►“ leads to the next step, alternative representation):

So it can be assumed simplistically, that the input resistance of the whole circuit is many times higher than the input resistance of the operational amplifier. The output resistance $R_\rm O^0$ of the whole circuit with real operational amplifiers shall only be sketched: In this case, the output resistance $R_\rm O$ of the operational amplifier is in parallel with $R_1 + R_2$. Thus the output resistance $R_\rm O^0$ will be somewhat smaller than $R_\rm O$.

3.5 Inverting Amplifier

The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see Abbildung 13). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_\rm I$, $\rm GND$, and $U_\rm O$. These terminals can be rearranged - while keeping the output terminal $U_\rm O$.

Thus the voltage divider $R_1 + R_2$ is no longer between $U_\rm O$ and $\rm GND$, but between $U_\rm O$ and $U_\rm O$, see Abbildung 12. In this circuit, the resistor $R_2$ is also called the negative feedback resistor.

Before the voltage gain is determined, the node $\rm K1$ in Abbildung 12 is to be considered first. This is just larger than the ground potential by the voltage $U_\rm D$; thus, it lies on the potential difference $U_\rm D$. For a feedback amplifier with finite voltage supply, $U_\rm O$ can only be finite, and thus $U_{\rm D}= U_{\rm O} / A_{\rm D} \rightarrow 0$ (cf. basic_equation of the operational amplifier), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $\rm K1$ is always at ground potential in the ideal operational amplifier. This property is called virtual ground because there is no direct short to ground. The op-amp regulates its output voltage $U_\rm O$ in such a way that the voltage divider sets a potential of $0~\rm V$ at node $\rm K1$. This can also be seen in the simulation by the voltage curve at $\rm K1$.

The following diagram shows again the interactive simulation.
$R_{\rm 1}$ and $R_{\rm 2}$ can be manipulated by the sliders. Hit Run / STOP to run

For the determination of the voltage gain, the consideration of the feedback $A_{\rm V}=\frac{1}{k}$ seems to be of little use at first. Instead, however, the determination via network analysis is possible. Abbildung 12 shows a possible variant to choose the loops for this purpose. However, network analysis is not to be done here, but is given in Exercise 3.5.1 below.

Instead, two other ways of derivation will be shown here to bring further approaches closer. For the first derivation, the voltage divider $\boldsymbol{R_1 + R_2}$ is considered. For the unloaded voltage divider, the general rule is:

\begin{align*} U_2 = U_{12} \cdot \frac{R_2}{R_1 + R_2} \end{align*}

This equation is now to be adapted for concrete use. First Abbildung 12, the voltages of the voltage divider can be read as given in Abbildung 14. From this, using the general voltage divider formula:

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

With the virtual mass at node $\rm K1$ in Abbildung 14, it holds that $U_2$ points away from the (virtual) mass and thus $U_2 = U_\rm O$. Similarly, $U_{\rm I} = U_1$ holds. Thus it follows:

\begin{align*} - U_{\rm O} = ( U_{\rm I} - U_{\rm O} ) \cdot \frac{R_2}{R_1 + R_2} \end{align*}

And from that:

For the second derivation, the current flow through the resistors $R_1$ and $R_2$ of the unloaded voltage divider is to be considered. These two currents $I_1$ and $I_2$ are just equal. Thus:

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

respectively

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

This can also be converted into a „seesaw“ or mechanical analog via like triangles. In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. ²⁾.

Now, if a certain height (voltage $U_\rm I$) is set, a certain height on the right side (voltage $U_\rm O$) is obtained via the force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in Abbildung 15 above. In the figure, all points marked in red () can be manipulated. Accordingly, the input voltage $U_{\rm I} = U_{\rm in}$ is adjustable and automatically results in a voltage $U_{\rm O}=U_{\rm out}$. In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed.

The input resistance of the entire circuit $R_{\rm I}^0=\frac{U_{\rm I}}{I_{\rm I}}$ is easily obtained by considering the input side: since $\rm K1$ is at $0 ~\rm V$, $U_1 = U_\rm I$. The complete current flowing into the input passes through resistor $R_1$. So it is then true that the input resistance is $R_{\rm I} = R_1$. At the output resistance of the whole circuit $R_{\rm O}^0$, there is again a parallel connection between the output resistance of the operational amplifier $R_\rm O$ and the resistor $R_2$. So the output resistance will be slightly smaller than the output resistance of the operational amplifier $R_\rm O$.

Exercises

120ms, 0
160ms, 0.1m
200ms, 0
260ms, 0
270ms,-0.2m
320ms, 1m
340ms,-0.3m
350ms, 0
440ms, 0
520ms, 0.1m
600ms, 0
820ms, 0
860ms, 0.1m
900ms, 0
960ms, 0
970ms, -0.2m
1020ms, 1m
1040ms,-0.3m
1050ms, 0
1140ms, 0
1220ms, 0.1m
1300ms, 0
1400ms, 0

Learning Questions

• Explain the difference between the unipolar and bipolar power supply of an OPV.
• Draw a sketch for bipolar and one for unipolar power supply.
• What are the advantages and disadvantages of unipolar and bipolar supply in OPV?
• What is the meaning of „virtual ground“? Under what circumstances is it found in opamp circuits?
• What are the golden rules in negative feedback amplifier circuits?
• What is the basic equation in negative feedback amplifier circuits?
• What are impedance transformers used for?
• How does the input voltage relate to the output voltage in an impedance converter and why?
• What does „voltage follower“ mean in the context of operational amplifiers and what are the characteristics of the voltage follower?

References to the media used

Element	License	Link
Abbildung 4: Image with internal operational amplifier	(c) Microchip	http://ww1.microchip.com/downloads/en/appnotes/90003132a.pdf