A detailed analysis of the gain-phase response vs. The important point is that for both circuits the negative feedback must dominate over the positive in order to maintain sufficient phase and assure stability. If the positive feedback dominated it is likely the circuits would oscillate. Up 0 Down Cancel. Hi Thomas, Thank you very much. I would like to know is I have an arbitrary OP-AMP circuit where output is connected to both positive and negative input of op-amp. In this case how I will check and make sure that opamp is in negative feedback.
Could you please suggest some generalized method so that I can apply it to any opamp circuit. Hi Hari, Yes, there is a practical and easy way to make sure that the circuit has sufficient phase margin to remain stable. I removed the signs of the opamp.
Regards hari. Hello Hari, I am not sure that an analysis that doesn't include the operational amplifier input signs makes sense. Hello Hari, For the complete story regarding dual-feedback stability analysis see the attached PowerPoint.
Hi Thomas, Thank you for your time and concern. May I know how to determine the signs for figure 10 Regards Hari. Hello Hari, Actually, the author provides the solution for the circuit shown in Figure 10 on pages 4 and 5: "In fig 10 ,there are two opamps. Could you please ellobrate the above equation. May I Vid1 is the difference voltage between inputs of first opamp. Could you please tell me how you obtained the value of Vid1.
Hello Hari, We know for the circuit to be stable that the feedback to A1 needs to be negative. If you don't mind could you please explain the above derivation with the help of circuit diagram. May I know why A1 needs to be in negative feedback.
Hello Hari, It appears that I am not communicating effectively regarding your negative feedback questions. You can find the link here: This is Section 4, but be sure to view Section 5 as well. Hi Thomas, I never meant your explanation is not clear. If anything wrong happened from my side,please excuse. I am an average person that's why it's taking time to understand. I will spend this weekend to understand this issue and will get back to you.
I analysed the problem with the support of your material. Could you please check the attached document and tell me my approach is wrong or not. Hello Hari, I am glad to see that you are analyzing the circuit. In this case, the inverting input is connected to the output and the non-inverting input serves as the signal input. Following the rules of op-amp behaviour where the op-amp will try to maintain a 0V difference in voltage across the inverting and non-inverting inputs, we can understand that the output follows the input to maintain this 0V difference, hence the name follower.
If the input to this circuit was 1V, then the output would also be 1V, since the output is directly connected to the inverting input, hence making the voltage difference between the inverting and non-inverting pins 0V. The above figure shows the waveforms of the circuit — the yellow waveform is the input, and the blue waveform is the output.
The output is a replica of the input, so we know the follower works. Note the same vertical scale on both channels. What if we want a gain other than 1? This can be done by adding a voltage divider to the output and connecting the inverting input to the middle of the divider. The non-inverting input serves as the signal input as usual. In this case, both resistors are of equal value.
If the input signal again is 1V, then the op-amp will try to change the output in such a way as to make the inverting input 1V in order to maintain a 0V differential across its input. Run simulation here. Social Media. Your thoughts are wonderful and I fully support them. My point is just that they are based more on a detailed and sometimes formal analysis of the INIC circuit what it does than on the disclosure of its philosophy why it does this.
So I will try to roughly fill that gap with my comment. We can consider this circuit from two perspectives: first - as a circuit with only input and no output a load with negative resistance ; second - as a circuit with input and output an amplifier with mixed feedback.
Negative load. Beginning from the early 90's, I spent a lot of effort to reveal and explain in an easy and intuitive way the first perspective. If you are interested and patient enough, you can familiarize yourself with the resources I created in Web; I described them in detail in two questions asked by me in ResearchGate - What is negative impedance? For those who do not have patience to read all of this, here is a very brief explanation. The circuit behaves as an active load dynamic voltage source with internal resistance R that reverses the current through the resistor R in the original Wikipedia picture and "pushes" it back to the input source.
In this way, it converts the resistor R originally consuming a current into a negative "resistor" -R producing a current. It does this by opposing through the resistor a reverse and higher 2V voltage to the input voltage V. This is the output voltage of the operational amplifier and it is not used here Simply the circuit behaves like a source that attacks back the input source Amplifier with mixed feedback.
According to me, this is the subject of the question asked here. As described in the comments above, this circuit is an amplifier with negative feedback, which is partially neutralized by a weaker positive feedback. But what is the point of that? In general, the positive feedback increases the gain of the imperfect amplifiers and it is used in the past remember the Armstrong's regenerative idea.
But in our case, the op-amp has a huge gain and this is not necessary. Then what is the point of using a positive feedback here? As a result, the resistors R2 and R3 can be low resistive. In this amp application, the op-amp output is the circuit output.
But as above, this amplifier has another output Now I want to explain what is the meaning of the fact that this circuit INIC reverts the current direction and passes the current back through the resistor. We can observe three situations:. There is no benefit from this arrangement, it simply passes a reverse current through the input source really, if it is a rechargable battery, it will be charged. The circuit passes a reverse current through the input source, creates a voltage drop across its Ri in addition to its internal voltage, and thus raises its external voltage.
This is the typical INIC application where it is connected with the input source in parallel to a common load. The INIC adds an additional current to the input current thus helping the input source. The Howland current source is a typical application of this idea.
Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. How are positive and negative feedback of opamps so different?
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