1 Overview
Starting today, let’s talk about op amps. Many brothers have also made this request before. It just so happens that I want to take a deeper look at op amps recently, so I'll take this opportunity to do it step by step.
The operational amplifier is a relatively complicated device compared to resistors, capacitors, transistors, MOS tubes and other devices. It is also commonly used in circuits and has many problems. Obviously, one article cannot explain the operational amplifier at all. Therefore, I may There are many issues to write. I never make a plan for the specific number of issues and what content to write. I just do whatever I want. Brothers can also ask if they have any needs. I can arrange it according to the situation.
Now that the background introduction is over, let’s get started.
2 The first step, ideal operational amplifier
First of all, the first question, why should I say ideal op amp?
Because generally speaking, when we understand something, we first look at it as an ideal. This is the simplest and easiest to understand.
When we get an unfamiliar circuit, we must first know what the circuit is used for, right?
At this time, we do not need to consider the non-ideal characteristics of the devices in the circuit, such as temperature drift, leakage, parasitic inductance, parasitic capacitance, etc. Let's treat it as ideal first, and then look at what functions this circuit achieves. This is generally how operational amplifier circuits are analyzed.
After we know what this circuit is used for, and then look at which characteristics of the device will cause the circuit to fail, or not work as expected, it is time to consider non-ideal characteristics.
Therefore, the purpose of our understanding of the ideal op amp is to quickly analyze the working principle of the circuit and what functions it implements at the beginning.
Secondly, what are the characteristics of an ideal operational amplifier?
The ideal operational amplifier mainly has the following three points:
1. Infinite gain
2. Infinite input impedance
3. The output impedance is 0
So where do these three characteristics come from?
2.1. Infinite gain
It is easy to understand that the gain is infinite, because the gain of general op amps is very large. For example, the open-loop gain of Ti's uA741 is about 105dB. How many times is it calculated?
20log(Av)=105dB , calculated Av=10^5.25=177828, which is about 180,000 times. Compared with the dozens of times of amplification in our general circuits, this is huge.
2.2. Infinite input impedance
The input impedance of an ideal op amp is infinite. Let’s look at the actual op amp. Let’s take Ti’s uA741 as an example, as shown below: As you can see, the
input impedance is still relatively large, with a typical value of 2MΩ. In fact, this chip is in the op amp. The impedance is quite small. For example, another chip from TI, the LM358, has a larger input impedance, with a differential input impedance of 10MΩ and a common mode input impedance of 4GΩ.
In short, the input impedance of the op amp is relatively large. Therefore, when we analyze the working principle of the circuit in principle, we regard the input impedance of the op amp as infinite.
2.3. The output impedance is 0
The output impedance of an ideal op amp can be regarded as 0.
Generally, we do not use op amps to directly drive high-power loads, and the output impedance of op amps is generally tens of Ω or hundreds of Ω. Compared with the subsequent stage, the output impedance can be ignored, so the output impedance of the op amp can be It can be seen that infinitesimal is 0, which is convenient for analysis, and the result will not be much different.
For example, the output impedance of ti's uA741 is 75Ω and
the output impedance of ti's LM358 is 300Ω
. Some people may think that this impedance is not very low. How can it be ignored?
This actually depends on the application. If the equivalent input impedance of the circuit at the back end of the op amp is relatively high, then it can naturally be ignored. If the input impedance of the load itself is only tens or hundreds of Ω, then naturally it cannot be completely ignored.
The above are the three biggest characteristics of an ideal op amp. After we get a circuit, we can generally analyze the basic functions of the op amp circuit by using these three characteristics.
But you may ask this question: We usually use the "virtual short" and "virtual open" of the op amp to analyze the circuit. These three characteristics mentioned above are not at all, so what is going on?
The answer is actually very simple, that is, the "virtual short" and "virtual break" of the op amp are derived based on the above three characteristics of the ideal op amp .
3 virtual break and virtual shortness
3.1. Virtual break
"Virtual break" is relatively simpler than "virtual short".
It was said earlier that the input impedance of an ideal op amp is infinite. To interpret it, if a voltage is applied to the input terminal of the op amp, then the current flowing in and out of the input pin of the op amp is 0. Since the impedance is infinite, there is naturally no current, then It is equivalent to an open circuit, that is, a broken circuit. However, this is different from a complete circuit break, because the op amp still senses the voltage at the input end, so it is not really a circuit break, so it is called a "virtual break".
It can be seen that "virtual break" has nothing to do with what kind of circuit the op amp is connected to. As long as it is an integrated op amp, it can be analyzed using virtual break (strictly speaking, there is still current at the input end of the actual op amp, but Quite small. If the external resistance is too large, causing the resistor current to approach or exceed the tiny current at the input end of the op amp, the "virtual break" will still fail).
Compared with the basically no-threshold use of "virtual break", there is a threshold for the use of "virtual short" of op amps.
3.2. Virtual shortfall
Let’s talk about the conclusion first. There are two conditions for using virtual short.
a. The circuit is a negative feedback circuit
b. The op amp works in the linear amplification zone
To understand these two points, we only need to know where "virtual shortness" comes from
First of all, what does virtual short mean?
We know that the op amp has two input terminals, the non-inverting terminal and the inverting terminal. "Virtual short" means that the voltages at the non-inverting terminal and the inverting terminal are the same, just like a short circuit. So how does it achieve this? The three characteristics of the ideal op amp we mentioned earlier don’t have these?
Let’s take the circuit in the picture below as an example to see why u+ = u- in the end?
Assume that at the beginning, the voltage everywhere is 0, and suddenly u1 changes from 0 to 2.5V in an instant. Because uo is 0V at the beginning, according to the "virtual break", u- has no current flowing into the amplifier, so u- is uo in R1 and R2 The partial voltage on is still 0V.
When u+ is 2.5V instantaneously, u- is 0V, u+>u-, and the amplifier will amplify in the direction of voltage increase, that is, the uo voltage will begin to increase.
When uo increases to 1V, u- is still the partial voltage of uo on R1 and R2, which is 0.5V. At this time u+=2.5V, u-=0.5V, u+>u-, the amplifier continues to amplify the voltage in the positive direction, so uo continues to increase.
The question is, how high will uo increase before it stops? It is easy to think that as long as u+>u-, because we are now discussing an ideal operational amplifier, the amplification factor is infinite, so uo will increase (the amplifier is a device that always amplifies the input voltage Au times, that is, the total satisfy:
Uo=Au*（u+ - u-））
Only when uo increases to 5V, the u- voltage is uo's divided voltage in R1 and R2 is exactly 2.5V, u- is equal to u+. At this time, the amplifier reaches balance and no longer amplifies, that is, the stable state is now.
Then why must the stable state be u- = u+, but not u- > u+?
We can also assume that if uo accidentally exceeds 5V, then u- will be greater than 2.5V, and u- will be greater than u+. At this time, the amplifier will amplify the output voltage in the opposite direction, that is, reduce it, and the final voltage will still be It will approach 5V.
Therefore, no matter what the initial state voltage of the circuit is, the final output will be stable at 5V, and u+ = u-, because once u+ is not equal to u-, then under infinite amplification, the output will inevitably change, and eventually it will lead to u+ = u-.
The above is a bit convoluted. Let’s think about it carefully. Is the logic like this: when u+ is not equal to u-, the output will change, and this change will be sent back to the input, which is u- in the picture, which will lead to u+ and u- The difference becomes smaller, and the difference becomes smaller, which means that the input signal becomes smaller (the input of the op amp is u+ - u-, which is the difference).
In other words, the output signal sends itself to the input terminal through resistors R2 and R1, reducing the input signal. Isn't this negative feedback?
In short, for the above negative feedback circuit, the final result is: u+ = u-.
What would happen if we reversed the non-inverting and inverting ends of the op amp?
Similarly, when the input suddenly changes to 2.5V, because uo is still 0V initially, then u+ is also 0V. At this time, u+ < u-, therefore, the output will decrease and become negative. When the output decreases, according to the voltage division relationship, u+ also decreases, which is also negative. That is to say, u+ is much smaller than u-, that is, the difference between u+ and u- is larger. A larger difference means that the input signal has become larger (the input of the op amp is u+ - u-, which is the difference).
In other words, the output signal sends itself to the input terminal through resistors R2 and R1, strengthening the input signal. Isn't this positive feedback?
It is easy to think that the final steady state is as low as uo can output. If it is powered by a single power supply, then uo=0V, then u+=0V, and u-=2.5V. Obviously, u+ is not equal to u-, In other words, the "virtual short" is not satisfied
. From the above, it can be seen that the "virtual short" must be negative feedback, but there is also a condition. The amplifier needs to work in the linear amplification region. Why do you say that?
Taking the previous negative feedback example as an example, when the input is 2.5V, the output is 5V, but what if the power supply is only 3.3V?
Obviously, the output cannot exceed 5V. At this time, the amplifier is working in the saturation zone. The amplifier can only amplify to 3.3V. Therefore, the output will not reach 5V in the end. Then u- will naturally not reach 2.5V. At this time, u+ will also not be able to reach 2.5V. It is not equal to u-, and it is not satisfied with virtual shortness.
Therefore, to satisfy the virtual short circuit, the amplifier also needs to work in the linear amplification region.
question
Here we can go deeper. Previously we regarded the op amp as an ideal op amp, that is, the gain is infinite, and finally we get u+ = u-, which is virtual short. If the gain is not infinite, but a finite value, what is the relationship between u+ and u-?
First of all, let’s think about what exactly is an op amp?
In fact, the operational amplifier can be regarded as such a thing, it can always amplify the difference between u+ and u- by AO times. Think about it, is it such a thing? In fact, it doesn't know what circuit is connected to the outside. Anyway, it amplifies the difference between u+ and u- by Aoo times, and then sends it to the output uo.
Therefore, there is naturally such a formula:
uo=(u+ - u-)*Yes
Transform it and you get:
u+ - u- = u/Uo
uo is a limited value. If the power supply is 3.3V, uo will not exceed 3.3. Just assume uo=3.3V. If Auo is one million times, Auo=1000000, then:
u+ - u- = 3.3V/1000000 = 3.3uV
It can be seen that the voltage difference between u+ and u- is only 3.3uV, which is quite small. When we analyze the circuit voltage, we can naturally ignore this voltage difference and regard them as equal, that is, "virtual" short". At the same time, we can also see that the larger the open-loop gain Auo of the op amp, the closer u+ and u- are, and they can be regarded as "virtually short".
summary
The above is the entire content of this section. It briefly summarizes the characteristics of the ideal op amp, as well as the meaning, origin, and conditions of use of virtual open and virtual short.
Characteristics of ideal op amp:
1. Infinite gain
2. Infinite input impedance
3. The output impedance is 0
Conditions for using virtual break:
Basically no threshold (the input impedance from the op amp is very large)
Conditions for using virtual short:
1. Negative feedback
2. Work in the linear amplification zone
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