Posted on 16 Comments

Basic Knowledge of Operational Amplifiers

In this paper, 16 questions about operational amplifier are put forward and answered in order to deepen readers’knowledge of operational amplifier.

In general, there will be a balanced resistance in the reverse/in-phase amplifier circuit. What is the function of this balanced resistance?

  • (1) Provide a suitable static bias for the transistor inside the chip. The circuit inside the chip is usually directly coupled, which can automatically adjust the static operating point. However, if an input pin is directly connected to the power supply or ground, its automatic adjustment function is abnormal, because the transistor inside the chip can not raise the voltage of the ground wire or lower the voltage of the power supply. As a result, the chip can not satisfy the conditions of virtual short and virtual break, and the circuit needs to be analyzed separately.
  • (2) To eliminate the influence of static base current on output voltage, the magnitude should be balanced with the equivalent resistance of external DC path at both input terminals, which is also the reason for its name.

What is the function of a capacitor on the feedback resistance of the operational amplifier in the same proportion?

  • (1) The feedback resistor and capacitor form a high-pass filter, and the local high-frequency amplification is particularly severe.
  • (2) Preventing self-excitation.

What are the consequences if the op-amp in-phase amplifier circuit is not connected with the balanced resistance?

  • (1) Burning the operational amplifier may damage the operational amplifier, and the resistance can play the role of voltage divider.

What role can pull-down resistor play in pulling up capacitor at input of operational amplifier?

  • (1) In order to obtain positive and negative feedback, it depends on the specific connection. For example, if I connect the input voltage signal, the output voltage signal and a line at the output end to the input section, then because of the resistance above, part of the output signal passes through the resistance to get a voltage value, shunting the input voltage, making the input voltage smaller, which is a negative feedback. Because the output signal of the signal source is invariable, the output signal can be corrected by negative feedback.

Operational amplifier is connected to integrator. What is the function of shunt resistance RF at both ends of integrator capacitor?

  • (1) Discharge resistance, used to prevent the output voltage out of control.

Why are resistors and capacitors in series at the input of operational amplifiers?

  • (1) If you are familiar with the internal circuit of an operational amplifier, you will know that any operational amplifier consists of several transistors or MOS transistors. In the absence of external components, the operational amplifier is a comparator. When the voltage at the same phase terminal is high, it will output a level similar to the positive voltage, and vice versa. But such an op-amp does not seem to have much use. Only when the external circuit is connected to form a feedback form can the op-amp have the functions of amplification, flip and so on.

What are the consequences of incorrect balancing resistance in Op-Amp in-phase amplifier circuit?

  • (1) The phase end of the same phase is unbalanced, and there will be output when the input is 0. The total output value of the input signal is a fixed number larger (or smaller) than the theoretical output value. (2) Errors caused by input bias current cannot be eliminated.

What’s the amplification factor and input impedance of an ideal integrated operational amplifier? What’s the voltage between its in-phase and inverted inputs?

  • (1) The amplification factor is infinite, and the input impedance is infinitely small. The voltage between the same input and the inverted input is almost the same (not 0!!!!! For example, 10V at the same end and 9.999 at the opposite end. 999V), just finished the electrician exam, remember!

Why is the open-loop gain of an ideal operational amplifier infinite?

  • (1) The open-loop gain of the actual operational amplifier is over 100,000, which is very large. Therefore, the open-loop gain of the actual operational amplifier theory is assumed to be infinite, and the virtual ground is derived from it.
  • (2) The derivation is only for inverted-phase amplifiers. I see in the book that the open-loop gain of operational amplifiers is infinite, so that when we design circuits, the closed-loop gain can not be limited by the open-loop gain, but only depends on the external components. It sacrifices the large open-loop gain for the stability of the closed-loop gain.
  • (3) The deduced virtual ground is not only for the inverted-phase amplifier when the op-amp is connected by negative feedback, but also for the positive feedback.
  • (4) It is well understood that if the gain is very small, the voltage difference between the two ends of the amplifier is relatively large for an output voltage. If the output voltage is connected to a negative feedback state, the voltage difference between the two ends of the amplifier will be inconsistent, which will lead to the error of amplification.
  • (5) There are two conditions for the realization of “virtual short” of op-amp: 1) the open-loop gain A of op-amp should be large enough; 2) the negative feedback circuit should be used. First, we know that the output voltage Vo of the op-amp is equal to the difference between the positive input voltage and the reverse input voltage Vid multiplied by the open-loop gain A of the op-amp. That is, Vo = Vid * A = VI + – VI -)* A (1) is a limited value because the output voltage of the operational amplifier will not exceed the supply voltage in practice. In this case, if A is large, (VI+ – VI-) must be small; if (VI+ – VI-) is small to a certain extent, then we can actually think of it as 0, then there will be VI+ = VI-, that is, the voltage of the in-phase input of the operational amplifier is equal to the voltage of the inverted input, as if connected together, which we call “virtual short”. Road. Note that they are not really connected, and there is resistance between them, which must be borne in mind. In the above discussion, how can we get the result of “virtual short”? Our starting point is formula (1), which is the characteristic of OP amp. There is no problem. We can rest assured. Then, we make two important assumptions. One is that the output voltage of the op-amp is limited, which is no problem. Of course, the output of the op-amp will not exceed the power supply, so this assumption is absolutely true, so we will not mention it later. The second is that the open-loop gain A of the operational amplifier is very large. The general operational amplifier A usually reaches the sixth, seventh or even higher power of 10. This assumption is generally not a problem. But don’t forget that the actual open-loop gain of the operational amplifier is related to its working state. Without the linear region, A is not necessarily large. Therefore, this second assumption is conditional. Let’s remember this first. So we know that when the open-loop gain A of the OP is very large, the OP can be “virtual short”. But it’s just a possibility, not an automatic one, and no one will believe it if you say that its two input terminals are “virtual short”. “Virtual short” can only be realized in a specific circuit. The conditions for the existence of “virtual short” are: 1) the open-loop gain A of operation should be large enough; 2) the negative feedback circuit should be used. When we understand the condition of “virtual short”, we can easily judge when we can use “virtual short” for circuit analysis. In practice, condition (1) is valid for most op amps, and the key is the working area. If it is a circuit in a book, it can be judged by calculation; if it is a practical circuit, it can be known whether the output voltage of the operational amplifier is reasonable or not. Another situation related to “virtual short” is called “virtual ground”, which is “virtual short” when an input terminal is grounded, not a new situation. I don’t think it’s accurate to say that we need deep negative feedback to use “virtual short” in some books. I think the underlying thought is that the operational amplifier is more likely to work in the linear area in the case of deep negative feedback. But this is not absolute. When the input signal is too large, the operational amplifier with deep negative feedback still saturates. Therefore, the output voltage value should be used to judge the most reliable.

Add the input signal directly to the in-phase input, and the inverted input is grounded by resistance. Why U = U+ = Ui_0? Is it not empty?

Question Supplementary: Constituting virtual short should satisfy certain conditions. That constitutes empty land and also meets certain conditions? What is it? Why?

  • (1) In the same-phase amplifier circuit, the output makes U (+) track U (-) automatically by feedback, so U (+) – U (-) will approach zero. It seems that the two ends are short-circuited, so it is called “virtual short”.
  • (2) Because of the phenomenon of virtual short and the high input resistance of the operation, the current flowing through the two input terminals of the operation amplifier is very small, close to zero. This phenomenon is called “virtual break” (virtual break is derived from virtual short, do not think that the contradiction between the two). (3) virtual ground is in the anti-phase operation amplifier circuit, (+) terminal grounded, (-) connected to the input and feedback network. Because of the existence of virtual short, U (-) and U (+) [potential equals 0], so called (-) false grounding – “virtual ground” (4) on the condition that virtual short is an important feature of closed-loop (in short, feedback) working state of phase amplifier circuit, and virtual ground is an important feature of reverse amplifier circuit in closed-loop working state. Attention should be paid to understanding the short and empty conditions (such as “close to equality”) and to ok.

Why do operational amplifiers work like this way

  1. Always feel that the model of operational amplifier is somewhat strange. First of all, it is “virtual short”, because “virtual short”, when the operational amplifier is connected to an in-phase amplifier, the potential of the two input terminals is the same. If the input waveform is measured, it will be the same, which is like a common-mode signal, in fact, on both input terminals. There are still small differential mode signals, which can not be measured by ordinary instruments. However, in this way, the common mode signals of the two input terminals are artificially increased due to the “virtual short” which is the result of deep negative feedback. This poses a challenge to the performance of operational amplifiers. Why do operational amplifiers work this way?
  • (1) Common-mode signal of in-phase amplifier is much larger than that of inverted-phase amplifier, which requires higher common-mode rejection ratio.
  • (2) My view on “Common-mode signal rejection ability of the same-phase and reverse-phase amplifiers” depends mainly on the symmetry and gain of the differential amplifier inside the op-amp (only inside the op-amp). It is obvious that no operational amplifier provides its common mode rejection ratio with additional structural conditions of external circuits. For single-ended input, the equivalent common mode value is half of the input value, whether in-phase or in-phase. However, since the input impedance of in-phase amplification is usually larger than that of reverse amplification, its anti-interference ability is of course poor. As mentioned above, the inverted terminal voltage is almost zero at the inverted input, so the differential collector voltage has only one tube change. In the same phase input, the voltage at the reverse phase end is equal to the voltage at the same phase end, so the common mode voltage and the input voltage are equal! That is to say, the differential collector voltage of the pair of tubes varies in the same direction besides the two tubes which change in different directions at the same time, which is the common mode output voltage. It is added to the voltage of one of the tubes. As a result, the tube tends to be saturated (or cut-off), fortunately, the common-mode voltage amplification is only one-thousandth of the differential-mode amplification. As mentioned above, it does not mean that the common mode rejection ratio (CMRR) of the differential mode input and common mode input of the amplifier is different. A common mode signal equivalent to the input value should be added to the in-phase input. Therefore, in-phase amplification mode should be used cautiously when the input signal is large.

Why do op amps generally need to be amplified in inverse proportion?

  • The major difference between the inverse input method and the same input method is that the inverse input method is approximately equal to the ground potential because there is no current in the resistance (because the input resistance of the op amp is very large) because a balanced resistance is connected to the ground at the same phase end. It is called “virtual ground”, and the potential at the opposite end is very close to that at the same end, so there is also “virtual ground” at the opposite end. The virtual advantage is that there is no common-mode input signal. Even if the CMRR of this operational amplifier is not high, there is no common-mode output. In-phase input connection method, there is no “virtual ground”. When single-ended input signal is used, common-mode input signal will be generated. Even if the operational amplifier with high CMRR is used, there will still be common-mode output. Therefore, in general, the inverted input connection method will be used as far as possible.

Some op-amps have output even if they do not input any voltage, and the output is not small, so VCC/2 is often used as a reference voltage

  • The output of the op-amp without any input is caused by the asymmetric design structure of the op-amp itself, that is to say, the input offset voltage Vos, which is a very important performance parameter of the op-amp. VCC/2 is commonly used as a reference voltage because the operational amplifier is in a single power supply. At this time, the real reference of the operational amplifier is VCC/2. Therefore, a direct current bias of VCC/2 is often provided at the positive and negative power supply. Operating amplifier selection needs to pay attention to a lot of matters, under not very strict conditions, often need to consider the operating voltage, output current, power consumption, gain bandwidth product, price and so on. Of course, when it is used under special conditions, different factors should be considered.

Why do amplifier circuits consisting of operational amplifiers usually sample inverted input mode?

  • (1) The major difference between the inverse phase input method and the same phase input method is that the inverse phase input method is approximately equal to the ground potential because there is no current in this resistance (because the input resistance of the operational amplifier is very large), which is called “virtual ground”, while the inverse phase input method is the same as the same. The potential at the phase end is very close, so there is a “virtual ground” at the reverse end. The virtual advantage is that there is no common-mode input signal. Even if the CMRR of this operational amplifier is not high, there is no common-mode output. In-phase input connection method, there is no “virtual ground”. When single-ended input signal is used, common-mode input signal will be generated. Even if the operational amplifier with high CMRR is used, there will still be common-mode output. Therefore, in general, the inverted input connection method will be used as far as possible.
  • (2) The positive phase is an oscillator, and the reverse phase can stabilize the amplifier. Accessing negative feedback (3) In principle, it is possible to connect the same phase amplifier circuit. However, in practical application, the amplified signal (i.e. differential mode signal) is usually very small, so attention should be paid to suppressing noise (usually common mode signal). In contrast, the common-mode signal suppression ability of the amplifier circuit is very poor, and the signal that needs to be amplified will be submerged in noise, which is not conducive to post-processing. Therefore, the inverse proportional amplifier circuit with better suppression ability is generally chosen.

Important characteristics of operational amplifiers?

  • (1) If the voltage on both input terminals of the operational amplifier is 0V, the output voltage should also be equal to 0V. But in fact, there is always some voltage at the output end, which is called offset voltage VOS. If the output offset voltage is divided by the noise gain of the circuit, the result is called the input offset voltage or the input reference offset voltage. This feature is usually given in VOS in data tables. VOS is equivalent to a voltage source connected in series with the inverted input of the amplifier. Differential voltage must be applied to the two input terminals of the amplifier to produce 0V output.
  • (2) The input impedance of an ideal operational amplifier is infinite, so there will be no current flowing into the input. However, the actual operational amplifier using a bipolar junction transistor (BJT) in the input stage requires some working current, which is called bias current (IB). Usually there are two bias currents: IB + and IB -, which flow into two input terminals, respectively. IB has a wide range of values. The bias current of special type op amps is as low as 60 fA (large Z passes through an electron every 3 micros), while the bias current of some high-speed op amps can be as high as tens of mA.
  • (3) The power supply voltage range required for the first monolithic operational amplifier to work normally is +15V. Nowadays, due to the improvement of circuit speed and the use of low power supply (such as batteries), the power supply of operational amplifier is developing towards low voltage. Although the voltage specifications of op amps are usually specified as symmetrical bipolar voltages (e.g. +15 V), these voltages do not necessarily require symmetrical voltages or bipolar voltages. For an operational amplifier, as long as the input is biased in the active region (i.e., within the common-mode voltage range), the power supply of +15V is equivalent to + 30V/0V or + 20V/-10V. Operational amplifier has no grounding pin unless the negative voltage rail is grounded in a single power supply application. No device in the op-amp circuit needs to be grounded. The input voltage swing of high-speed circuit is smaller than that of low-speed device. The higher the speed of the device, the smaller the geometry of the device, which means the lower the breakdown voltage. Because the breakdown voltage is low, the device must work at a lower power supply voltage. Nowadays, the breakdown voltage of op-amps is generally about (+7V), so the power supply voltage of high-speed op-amps is generally (+5V), and they can also work under the single supply voltage of +5V. For General Purpose Operational amplifiers, the power supply voltage can be as low as + 1.8V. This type of amplifier is powered by a single power supply, but this does not necessarily mean that a low power supply voltage must be used. The terms single supply voltage and low voltage are two related and independent concepts.

What is working principle of operational amplifier?

The core of operational amplifier is a differential amplifier. Two transistors are connected back to back. Share the current of a transverse current source. One transistor is the forward input of the operational amplifier and the other is the reverse input. The forward input transistor is amplified and sent to a power amplifier circuit to amplify the output. In this way, if the voltage of the forward input rises, the output will naturally increase. If the inverted input voltage rises, the inverted and forward three-stage tubes share a constant current source. If the current of the reverse three-stage transistor is larger, the forward one will be smaller, so the output will be lower. So it’s called reverse input. Of course, there are many other functional components in the circuit, but the core is this.

Advertisements

16 thoughts on “Basic Knowledge of Operational Amplifiers

  1. I always used to study post in news papers but now as I am a user of net
    thus from now I am using net for posts, thanks to web.

  2. Excellent blog here! Also your site loads up very fast!
    What host are you using? Can I get your affiliate link to your host?

    I wish my website loaded up as quickly as yours lol

  3. Good job!

  4. Since voltages are just electric potential differences, I’d like to know what point in the circuit is the reference point for the output of the operational amplifier?

  5. Yeah, sir
    Who can explain with enough mathematics why virtual short concepts do not apply to operational amplifiers in the case of positive feedback?

    1. Hi~ rajesh
      I assume you’re talking about a virtual short circuit between the operational amplifier and the entire opamps input terminal, right? In this case, your statement – in this general form – is incorrect. It is:
      The term “virtual short circuit” applies to amplifier units with large open-loop gain, which can be set to infinity (in mathematical operations/calculations). However, this assumption is correct if OPAMP is dynamically stable and operates only in its linear region. Usually, this is the case of negative feedback. However, there are other applications that use both negative and positive feedback. As long as the negative feedback is dominant (the negative feedback coefficient is greater than the positive feedback coefficient), the circuit will remain stable, and the principle of “virtual short circuit” will continue to apply.
      In addition, some active filter circuits, such as all key topologies, need positive feedback to enhance Q value. These circuits have negative feedback to DC (stable operating point) and positive feedback to some specific frequencies (polar frequency region). Of course, in the case of larger Q value, the circuit is closer to the stability limit than the simple amplifier – but as long as negative feedback is dominant, the active filter circuit is stable and works as required (and applies the virtual short circuit principle). In short, “virtual short” scenarios are not suitable for OPAMP with feedback arrangements that lead to dynamic instability. For example, if we have a pure resistance feedback to a non-inverter terminal whose loop gain is greater than a unit (however, the slight positive feedback that results in a loop gain below a unit is stable!).
      However, there may be an exception in the conversion process: the classical Schmidt flip-flop is a resistance positive feedback circuit. When the circuit is used to construct a square wave oscillator, the opamp output switches between two power supply voltage limits – and in the switching phase, the circuit passes through a linear region between the input and output. In this very short time, the voltage between two OPAMP input nodes is very small (virtual short circuit).

  6. Good day, sir
    I just want to ask a question that we have obtained an electrical characteristic table containing the ideal and actual values of the operational amplifier, but it does not include the maximum output current. What is the maximum output current of the actual operational amplifier? What I’m talking about is actually typical values.

    Thank you very much.
    ramesh

    1. Hi, ramesh
      Most op amps are limited to mA or 10s of mA.

  7. Great blog!
    And I am trying to analyze the building blocks of operational amplifiers.

  8. Grreat post!
    I am designing an electronic sensor to monitor whether the temperature in the refrigerator is above or below a certain value. I use an operational amplifier as a processing unit.

    Regards
    Ritesh Shah

  9. Thank you for such a good passage discussed. I really have a great time understanding it.

  10. I blog quite often and I really thank you for your information. The article has really
    peaked my interest. I’m going to take a note of your blog and keep checking for new information about once per week.
    I subscribed to your Feed as well.

    1. Hi~ wallet
      Thank you for your trust.

  11. Hello. I have checked your 3celectrons.com and i see you’ve got some excellent fans.
    fast.

  12. Ahaa, its fastidious dialogue on the topic of this paragraph at this place at this web site, I have read all that,
    so at this time me also commenting at this place.

  13. Excellent post. I was checking constantly this blog and I am impressed!
    Very useful information particularly the last part 🙂 I care
    for such information a lot. I was looking for this
    particular information for a very long time. Thank you and good
    luck.

Leave a Reply

Your email address will not be published.