How an operational amplifier works. Operational Amplifiers for Beginners

Operational amplifier (op-amp) An Operational Amplifier (OpAmp), popularly known as an opamp, is a direct current amplifier (OCA) with a very high gain. The phrase "dc amplifier" does not mean that an op-amp can only amplify direct current. This means starting from a frequency of zero Hertz, and this is direct current.

The term "operational" has been strengthened for a long time, since the first samples of the op-amp were used for various mathematical operations such as integration, differentiation, summation, and so on. The gain of the op-amp depends on its type, purpose, structure and can exceed 1 million!

Operational amplifier circuit

In the diagrams, the operational amplifier is indicated as follows:

or so

Most often, op amps on the diagrams are indicated without power leads.

An input with a plus sign is called a non-inverting input, and an input with a minus sign is called an inverting input. Do not confuse these two signs with power polarity! They DO NOT say that it is mandatory to apply a signal with negative polarity to the inverting input, but for a NOT inverting signal with positive polarity, and further you will understand why.

Power supply for operational amplifiers

If the power outputs are not specified, then it is considered that the op-amp is supplied with a bipolar supply +E and -E Volts. It is also labeled +U and -U, V CC and V EE , Vc and V E . Most often it is +15 and -15 Volts. Bipolar nutrition is also called bipolar nutrition. How to understand this - bipolar power?

Let's imagine a battery

I think you all know that the battery has a "plus" and a "minus". In this case, the “minus” of the batteries is taken as zero, and the batteries are already counted relative to zero. In our case, the battery voltage is 1.5 volts.

And let's take another such battery and connect them in series:

So, we will have a total voltage of 3 Volts, if we take the minus of the first battery as zero.

But what if we take the minus of the second battery to zero and measure all the voltages relative to it?

Here we just got a bipolar power supply.

Ideal and real model of the operational amplifier

In order to understand the essence of the operation of the OS, consider it ideal And real models.

1) an ideal op-amp is infinitely large.

In real op-amps, the value of the input resistance depends on the purpose of the op-amp (universal, video, precision, etc.), the type of transistors used and the input stage circuitry, and can range from hundreds of ohms to tens of megohms. A typical value for a general purpose op amp is a few MΩ.

2) The second rule follows from the first rule. Since the input impedance of an ideal op-amp is infinitely large, the input will be zero.

In fact, this assumption is quite true for op-amps with inputs, whose input currents can be less than picoamps. But there is also an op-amp with an input. Here, the input current can already be tens of microamperes.

3) The output impedance of an ideal op amp is zero.

This means that the voltage at the output of the op-amp will not change when the load current changes. In real general purpose op amps, it is tens of ohms (usually 50 ohms).
In addition, the output impedance depends on the frequency of the signal.

4) The gain in an ideal op-amp is infinitely large. In reality, it is limited by the internal circuitry of the op-amp, and the output voltage is limited by the supply voltage.

5) Since the gain is infinitely large, therefore, the voltage difference between the inputs of an ideal op-amp is zero. Otherwise, even if the potential of one input is greater or less than at least by the charge of one electron, then the output will be an infinitely large potential.

6) The gain in an ideal op-amp does not depend on the frequency of the signal and is constant at all frequencies. In real op amps, this condition is satisfied only for low frequencies up to any cutoff frequency, which is individual for each op-amp. Typically, the cutoff frequency is taken as a 3 dB gain drop, or down to 0.7 of the gain at zero frequency (DC).

The circuit of the simplest op-amp on transistors looks something like this:

The principle of operation of the operational amplifier

Let's take a look at how OS works.

The principle of operation of the OU is very simple. It compares two voltages and at the output it already produces a negative or positive supply potential. It all depends on which input the potential is greater. If the potential at the non-inverting input U1 is greater than at the inverting U2, then the output will be + Upit, if the potential at the inverting input U2 is greater than at the non-inverting U1, then the output will be -Upit. That's the whole principle ;-).

Let's explore this principle in the Proteus simulator. To do this, we choose the simplest and most common operational amplifier LM358 (analogues 1040UD1, 1053UD2, 1401UD5) and assemble a primitive circuit showing the principle of operation

Let's apply 2 Volts to the NON-inverting input, and 1 Volt to the inverting input. Since the potential is greater at the non-inverting input, therefore, we should get + Upit at the output. We got 13.5 volts, which is close to this value.

But why not 15 volts? The internal circuitry of the op-amp is to blame for everything. The maximum value of the op amp may not always be equal to the positive or negative supply voltage. It can deviate from 0.5 to 1.5 Volts, depending on the type of op-amp.

But, as they say, the family is not without its freaks, and therefore op-amps have long appeared on the market, which can produce an acceptable supply voltage at the output, that is, in our case, these are values ​​\u200b\u200bclose to +15 and -15 Volts. Such a feature is called Rail-to-Rail, which is literally translated from English. “from rail to rail”, and in the language of electronics “from one power rail to another”.

Let's now apply more potential to the inverting input than to the non-inverting one. We supply 2 Volts to the inverting one, and 1 Volt to the non-inverting one:

As you can see, in this moment the output “lay down” on -Upit, since the potential at the inverting input was greater than at the non-inverting one.

In order not to download the Proteus software package once again, you can simulate the operation of an ideal op-amp online using the Falstad program. To do this, select the Circuits-Op-Amps->OpAmp tab. As a result, the following diagram will appear on your screen:

On the right control panel, you will see sliders for adding voltage to the op-amp inputs and you can already visually see what happens at the output of the op-amp when the voltage at the inputs changes.

So, we have considered the case when the voltage at the inputs can vary. But what happens if they are equal? What will Proteus show us in this case? Hmm, showed + Upit.

And what will Falstad show? Zero Volt.

Who to believe? Nobody! In real life, it is impossible to do this in order to drive absolutely equal voltages to two inputs. Therefore, such a state of the op-amp will be unstable and the output values ​​\u200b\u200bcan take on either -E Volts or +E Volts.

Let's apply a sinusoidal signal with an amplitude of 1 volt and a frequency of 1 kilohertz to the non-inverting input, and put the inverting signal to the ground, that is, to zero.

Let's see what we have on the virtual oscilloscope:

What can be said in this case? When the sinusoidal signal is in the negative region, we have -Upit at the output of the op-amp, and when the sinusoidal signal is in the positive region, then we have +Upit at the output. Also pay attention to the fact that the voltage at the output of the op-amp cannot change its value abruptly. Therefore, in the op-amp there is such a parameter as the slew rate of the output voltage V Uout .

This parameter shows how quickly the output voltage of the op-amp can change when working in pulse circuits. Measured in Volt/sec. Well, how did you figure out what more value this parameter, the better the op amp behaves in pulse circuits. For LM358, this parameter is 0.6 V/µs.

Featuring Jeer

I often remember my first acquaintance with an operational amplifier (op-amp). I always knew that these mysterious triangles on the diagrams would be useful to me in life. However, the long sleepless nights spent studying their working principle did not lead to anything. There are many articles on this topic, but it seems to me that the very basics are not obvious. I'll try to approach a little from the other side and dispel terrible secrets OU.

Let's try to figure out what "operations" our operational amplifier amplifies.

Problem: There is a signal source, such as a signal from a microphone or a guitar pickup. If the microphone is connected directly to the headphones, then most likely you will not hear anything, at best it will be a barely perceptible sound.

Imagine, instead of a microphone, a person who is trying to lift a heavy slab, of course, he cannot do it, just like the microphone is not able to shake the speaker. But if this person uses little force to operate the crane, then he will be able to lift any load within the capacity of the crane. Those. crane in this case amplifier. An analogue of the lifting capacity of a crane is the power of the amplifier. The meaning of the gain should be clear from the picture. The frequency and waveform remains the same, only the amplitude changes.

Now we know that in order to hear the sound from the speakers, an amplifier is needed. While we do not know how it works and what is inside it, however, we already know that there must be legs to which the signal is applied, which we want to amplify Uin, as well as legs from which the amplified signal Uout is removed.

The question is to what voltage can the signal be amplified? You will say: “I want to amplify 220V to 1000000V”, but this is not possible, why? Because, the original signal is amplified by an external source. The external source will be the supply voltage of the op-amp. Similarly, a crane cannot lift a load above its height (we agree that it cannot :)). Therefore, the voltage at the output of the op-amp cannot exceed the supply voltage. In reality, it is even slightly less than the supply voltage. For example, for LM324, the supply voltage is from 3 to 32V.

Now we know that the op-amp needs external power, let's draw these legs

By the way, we are used to the fact that our power supply is unipolar + 5V and ground. Here is a subtle point, if you want to amplify a signal that has negative values,

then you need to connect to -Upit, namely the source of negative voltage, and not the ground. If you connect the ground, it turns out that there is no voltage source and the “lower” (negative) part of the signal will not be amplified, i.e. part of the signal will be “cut off”, more on this in the example.

Similarly, if the signal is amplified more than the supply voltage, then in those places where the signal exceeds the supply voltage, the signal will be “cut off”, i.e. instead of a sinusoid, we will see something like this

The main question remains, how to set the gain? Very simple - a voltage divider. But first, let's move on to more real notation. Any op-amp has at least 5 legs - 2 power, which was mentioned above, an inverted input (-), a non-inverted input (+) and an output.

Therefore, depending on which input the original signal is applied to, two types of switching are distinguished: non-inverting amplifier

The gain, which is equal to K=(R4/R3)+1. In this case, K=4. In this case, the output waveform does not change.

And inverting, with gain K=-(R2/R1). For this scheme K=3. The output signal will be in phase with the input.

Let's move from words to deeds. A meander with a frequency of 1 kHz was taken as the initial signal. The signal has both positive and negative values ​​(the middle of the screen is 0). Signal amplitude 50mV.

I connect the op-amp (L324) according to the scheme of a non-inverting amplifier. The food is unipolar. At the output of the op-amp, a signal of the same shape, but with a larger amplitude. It is probably not entirely clear why the signal is of such amplitude and why it has shifted upwards.

Let's try to figure it out. The amplitude of the original signal is 50mV, R4=30k, R3=10k, we substitute into the formula 50*(30/10+1)=200mV, very similar to what is seen on the oscilloscope. Why did the signal move up? Recall the disadvantage of unipolar power, anything below 0 cannot be amplified, so the signal is cut off at 0.

Now imagine that if a negative voltage source were connected to the power pin, say -5V, then the signal amplitude would double !!! Therefore, the volume would also increase significantly.

Actually, this is a small preface, before starting to study the OU, all of the above is just a drop in the ocean, if you liked it, write, we will gradually master other applications of the OU, incl. and practical plans.

There are many important topics in an electronics course. Today we will try to deal with operational amplifiers.
Start over. An operational amplifier is such a “thing” that allows you to operate in every possible way with analog signals. The simplest and most basic are amplification, attenuation, addition, subtraction, and many others (for example, differentiation or logarithm). The vast majority of operations on operational amplifiers (hereinafter referred to as op-amps) are performed using positive and negative feedback.
In this article, we will consider a certain "ideal" OS, because there is no point in switching to a specific model. The ideal means that the input resistance will tend to infinity (hence the input current will tend to zero), and the output resistance, on the contrary, will tend to zero (which means that the load should not affect the output voltage). Also, any ideal op amp should amplify signals of any frequency. Well, and most importantly, the gain with no feedback should also tend to infinity.

Get to the point

The operational amplifier on the circuits is very often indicated by an equilateral triangle. On the left are the inputs, which are marked "-" and "+", on the right - the output. Voltage can be applied to any of the inputs, one of which changes the polarity of the voltage (that's why it was called inverting), the other does not change (it is logical to assume that it is called non-inverting). The power supply of the op-amp, most often, is bipolar. Normally, positive and negative supply voltages have the same value (but different sign!).
In the simplest case, you can connect voltage sources directly to the inputs of the op-amp. And then the output voltage will be calculated by the formula:
, where is the voltage at the non-inverting input, is the voltage at the inverting input, is the voltage at the output, and is the gain without feedback.
Let's look at the ideal op-amp from the point of view of Proteus.


I suggest you play with it. A voltage of 1V was applied to the non-inverting input. On the inverting 3V. We use the "ideal" OS. So, we get: But here we have a limiter, because we won't be able to amplify the signal above our supply voltage. Thus, the output will still get -15V. Outcome:


Let's change the gain (for you to believe me). Let the Voltage Gain parameter become equal to two. The same problem is clearly solved.

The real application of the op-amp on the example of inverting and non-inverting amplifiers

There are two such major rules:
I. The output of the op-amp tends to make the differential voltage (the difference between the voltage at the inverting and non-inverting inputs) equal to zero.
II. The op-amp inputs draw no current.
The first rule is implemented through feedback. Those. voltage is transferred from the output to the input in such a way that the potential difference becomes zero.
These are, so to speak, "sacred canons" in the topic of OU.
And now, more specifically. Inverting amplifier looks exactly like this (pay attention to how the inputs are located):


Based on the first "canon" we get the proportion:
, and having a little "conjure" with the formula, we derive the value for the gain of the inverting op-amp:

The screenshot above needs no comments. Just plug everything in and check it out.

Next stage - non-inverting amplifier.
Everything is also simple here. The voltage is applied directly to the non-inverting input. Feedback is applied to the inverting input. The voltage at the inverting input will be:
, but applying the first rule, it can be argued that

And again, "grand" knowledge in the field of higher mathematics allows us to proceed to the formula:
Here is a complete screenshot that you can double-check if you want:

Finally, I will give a couple of interesting circuits so that you do not get the impression that op-amps can only amplify voltage.

Voltage follower (buffer amplifier). The principle of operation is the same as that of a transistor repeater. Used in high load circuits. Also, with its help, you can solve the problem of impedance matching if there are unwanted voltage dividers in the circuit. The scheme is simple to genius:

summing amplifier. It can be used if you want to add (subtract) several signals. For clarity - a diagram (again, pay attention to the location of the inputs):


Also, pay attention to the fact that R1 = R2 = R3 = R4, and R5 = R6. The calculation formula in this case will be: (familiar, isn't it?)
Thus, we see that the voltage values ​​that are applied to the non-inverting input "acquire" a plus sign. On the inverting - minus.

Conclusion

Operational amplifier circuits are extremely diverse. In more complex cases, you can find active filter circuits, ADCs and storage sampling devices, power amplifiers, current-to-voltage converters, and many many other circuits.

List of sources

A short list of sources that will help you quickly get used to both the OS and electronics in general:
Wikipedia
P. Horowitz, W. Hill. "The Art of Circuitry"
B. Baker. "What a digital designer needs to know about analog electronics"
Lecture notes on electronics (preferably your own)
UPD.: Thank you UFO for invitation

There are many important topics in an electronics course. Today we will try to deal with operational amplifiers.
Start over. An operational amplifier is such a “thing” that allows you to operate in every possible way with analog signals. The simplest and most basic are amplification, attenuation, addition, subtraction, and many others (for example, differentiation or logarithm). The vast majority of operations on operational amplifiers (hereinafter referred to as op-amps) are performed using positive and negative feedback.
In this article, we will consider a certain "ideal" OS, because there is no point in switching to a specific model. The ideal means that the input resistance will tend to infinity (hence the input current will tend to zero), and the output resistance, on the contrary, will tend to zero (which means that the load should not affect the output voltage). Also, any ideal op amp should amplify signals of any frequency. Well, and most importantly, the gain with no feedback should also tend to infinity.

Get to the point

The operational amplifier on the circuits is very often indicated by an equilateral triangle. On the left are the inputs, which are marked "-" and "+", on the right - the output. Voltage can be applied to any of the inputs, one of which changes the polarity of the voltage (that's why it was called inverting), the other does not change (it is logical to assume that it is called non-inverting). The power supply of the op-amp, most often, is bipolar. Normally, positive and negative supply voltages have the same value (but different sign!).
In the simplest case, you can connect voltage sources directly to the inputs of the op-amp. And then the output voltage will be calculated by the formula:
, where is the voltage at the non-inverting input, is the voltage at the inverting input, is the voltage at the output, and is the gain without feedback.
Let's look at the ideal op-amp from the point of view of Proteus.


I suggest you play with it. A voltage of 1V was applied to the non-inverting input. On the inverting 3V. We use the "ideal" OS. So, we get: But here we have a limiter, because we won't be able to amplify the signal above our supply voltage. Thus, the output will still get -15V. Outcome:

Let's change the gain (for you to believe me). Let the Voltage Gain parameter become equal to two. The same problem is clearly solved.

The real application of the op-amp on the example of inverting and non-inverting amplifiers

There are two such major rules:
I. The output of the op-amp tends to make the differential voltage (the difference between the voltage at the inverting and non-inverting inputs) equal to zero.
II. The op-amp inputs draw no current.
The first rule is implemented through feedback. Those. voltage is transferred from the output to the input in such a way that the potential difference becomes zero.
These are, so to speak, "sacred canons" in the topic of OU.
And now, more specifically. Inverting amplifier looks exactly like this (pay attention to how the inputs are located):


Based on the first "canon" we get the proportion:
, and having a little "conjure" with the formula, we derive the value for the gain of the inverting op-amp:

The screenshot above needs no comments. Just plug everything in and check it out.

Next stage - non-inverting amplifier.
Everything is also simple here. The voltage is applied directly to the non-inverting input. Feedback is applied to the inverting input. The voltage at the inverting input will be:
, but applying the first rule, it can be argued that

And again, "grand" knowledge in the field of higher mathematics allows us to proceed to the formula:
Here is a complete screenshot that you can double-check if you want:

Finally, I will give a couple of interesting circuits so that you do not get the impression that op-amps can only amplify voltage.

Voltage follower (buffer amplifier). The principle of operation is the same as that of a transistor repeater. Used in high load circuits. Also, with its help, you can solve the problem of impedance matching if there are unwanted voltage dividers in the circuit. The scheme is simple to genius:

summing amplifier. It can be used if you want to add (subtract) several signals. For clarity - a diagram (again, pay attention to the location of the inputs):


Also, pay attention to the fact that R1 = R2 = R3 = R4, and R5 = R6. The calculation formula in this case will be: (familiar, isn't it?)
Thus, we see that the voltage values ​​that are applied to the non-inverting input "acquire" a plus sign. On the inverting - minus.

Conclusion

Operational amplifier circuits are extremely diverse. In more complex cases, you can find active filter circuits, ADCs and storage sampling devices, power amplifiers, current-to-voltage converters, and many many other circuits.

List of sources

A short list of sources that will help you quickly get used to both the OS and electronics in general:
Wikipedia
P. Horowitz, W. Hill. "The Art of Circuitry"
B. Baker. "What a digital designer needs to know about analog electronics"
Lecture notes on electronics (preferably your own)

Tags: operational amplifiers, op amps, electronics, for beginners

) we will work with OP97 and AD620. Consider the AD620 first. In the datasheet for it, it is indicated as follows:

Fig.2b

AD620 is an instrumental op amp. The word instrumental speaks of its better characteristics compared to a conventional op-amp. The amplified signal is fed to the inputs +IN and respectively -IN. The gain of this amplifier is set using a resistor connected to the Rg inputs (there are two of them, respectively - No. 1 and No. 8). Which resistor, which gain corresponds to - look in the datasheet. The power supply of the AD620 op amp is bipolar. This means that it has power leads, which are referred to as +Vs and -Vs. And now if we connect, for example, a 5V battery to them (moreover, the minus of the batteries must be connected to -Vs, and the plus, respectively, to +Vs), and we apply a potential difference to the signal inputs +IN and, respectively -IN, which must be amplified, then a signal amplified by K times (where K is the gain set by Rg - see above) we can remove from this device by connecting to the OUTPUT pin and the point of the circuit we are assembling, with a potential of 2.5V relatively battery minus. The point with a potential of 2.5V relative to the minus of the battery is called the zero point. This is the very zero relative to which the potential (amplified signal) is measured at the OUTPUT pin of the amplifier. This point can be obtained using an ordinary resistive divider of the form Fig. 3b.

fig.3b

Thus, the simplest circuit connection of this op-amp is as follows:

So plus batteries relative to the zero point has a potential equal to +2.5V, and the minus of the battery relative to the zero point has a potential of -2.5V (see Fig. 3b). That is, the potential of the zero point is exactly in the middle between the plus and minus of the battery. Hence the name this method power supply - bipolar supply (since it turns out that we applied minus 2.5V to the -Vs output of the amplifier relative to the zero point, and to + Vs we applied plus 2.5V relative to the zero point).
It should also be noted that the potentials applied to the +IN and –IN inputs of the amplifier relative to the zero of the circuit must have a value within the same limits as the potentials of the power source. That is, if we applied –2.5V and +2.5V to –Vs and +Vs, respectively, then –IN and +IN cannot be supplied, for example, 230V and 230.1V, respectively. In this example, the potential difference 230.1–230=0.1V, although small, will not be amplified. According to the datasheet, it is necessary to find out the acceptable range of potentials at the inputs of the corresponding op-amp. For example, for the AD620, in accordance with its datasheet Input Voltage Range (input voltage range), when power is applied to –2.5V and 2.5V to –Vs and +Vs, the voltage relative to zero on –IN or +IN should be no more than Vs–1.2V = 2.5–1.2 = 1.3V and not less than –Vs+1.9V = –2.5+1.9 = –0.6V. This means that if, for example, 0.2V and 0.3V are applied to –IN and +IN, respectively, then now the potential difference between –IN and +IN to the same 0.1V can already be amplified. In the electrocardiograph circuit (see Fig. 5), in order for the potentials from the human body supplied to the input of the amplifier to be within the same limits as the potentials of the power source, the zero point of the power source is connected using the so-called reference electrode to the patient's right leg ( such a connection is also called the “right leg driver” - right leg driver). As a result, the potentials on the human body will fluctuate within the zero point of the power supply of the amplifier, which means they will fall into the range Vs–1.2V, –Vs+1.9V.
There is also the following important feature. The output voltage of the amplifier must be measured relative to OUTPUT and the neutral wire of the circuit, however, in practice, sometimes some op-amps add their shift to the output signal by one or another constant value. Therefore, in order to remove this constant value from such op-amps and, as a result, measurements relative to the neutral wire of the circuit would be correct, a REF output (the so-called reference input) is usually provided, to which it is necessary to apply the zero potential of the circuit. Moreover, zero potential must be supplied to the REF output from a source with a minimum output impedance, otherwise the supply of zero potential to REF will not achieve the desired effect. Thus, the zero potential is usually applied to the REF input through the so-called follower op-amp, which, as you know, has an output resistance close to zero, while the input, on the contrary, tends to a huge value. A zero potential is applied to the input of the repeater, the gain of which is equal to one, the zero potential is removed from the output of the repeater and fed to REF. The op-amp connected according to the repeater circuit looks like this:

fig.5b

Then the amplifier switching circuit with REF will look like this:

fig.6b

In our electrocardiograph circuit, the repeater that produces the reference voltage for the AD620 amplifiers is built on the basis of OP97 (see Fig. 8) - here the zero potential is applied to the positive input of OP97, and from the output of OP97 the reference zero potential is fed to the REF outputs of the AD620 amplifiers specially designed for this . OP97 is also bipolar.
In addition to op amps with bipolar power, there are also so-called unipolar ones, for example, TLC272. For such amplifiers, the output voltage is measured not relative to the zero point, but relative to the minus of the battery, and, accordingly, the outputs for powering such an op-amp are designated as GND (minus batteries here) and VDD (plus here).
Well, perhaps that's all. This information is enough to understand what to apply where and what, where to measure with amplifiers on our electrocardiograph circuit.

You can also see more about operational amplifiers here:

p.s. For those who are interested in explaining the concepts of mathematics, physics, technology, which is called "on the fingers", we can recommend this book and, in particular, the chapters from its sections "Mathematics", "Physics", "Technology" (the book itself or individual chapters from it you can purchase).