Biswajit - the problem is that your circuit cannot work in reality.

It is not true that the circuit has positive and negative feedback - it has only POSITIVE feedback and, hence, has no stable DC operating point.

Why is there no negative feedback? Because each feedback signal is shorted in the voltage source v1.

In case you want to have positive and negative feedback for a working amplifier, the resistors must be chosen so that we have more negative than positive feedbak (negative feedback must dominate).

Dear Biswajit,

welcome,

what is the value of input voltage at point 5 the inverting input of the op amp. What is the output do you expect and how the output is now different from that expected?

Best wishes

Biswajit - the problem is that your circuit cannot work in reality.

It is not true that the circuit has positive and negative feedback - it has only POSITIVE feedback and, hence, has no stable DC operating point.

Why is there no negative feedback? Because each feedback signal is shorted in the voltage source v1.

In case you want to have positive and negative feedback for a working amplifier, the resistors must be chosen so that we have more negative than positive feedbak (negative feedback must dominate).

If we want have the negative feedback we must add some resistor to the signal voltage source or we must use current signal source.

....at least 5k...

Biswajit - it makes no sense to analyze a circuit which is simply useless (like an opamp with out any supply voltages).

The last picture (simulation results with supply voltages applied) shows what happens due to positive feedback: The output voltage is fixed at a voltage close to one of the supply rails . No operation in the linear amplifier region is possible.

Lutz von Wangenheim sir

"The last picture (simulation results with supply voltages applied) shows what happens due to positive feedback"

Due to positive feedback voltage should be close to 15V (close to positive rail) but here output voltage going only 10V.

If i add 5K to inverting terminal then postive feedback and negative feedback factor will be same in that case which one would be more dominant?

Because you left the power supply voltage floating, so, it is expected to get any voltage according to the solution of the circuit. So, as it is said before it is nonsense to leave the power rails floating or the amplifier is not powered.

If the input voltage is 5V, then the potential of the noninverting input will be equal yo 5V because of the op amp properties of very high DC gain. So the the +ve input terminal will be clamped at 5V. It causes a current flow in R1= 5/1k= 5mA. Since the op amp is considered ideal then this current will continue flowing in the feed back resistance R3, developing a voltage drop = 1kx5mA= 5V. So, the output voltage is the sum of the two voltage drops= 5+5= +10v. this what is calculated by the simulator.

Of course i agree with the colleagues that this is not a practical configuration.

Best wishes

Dear Abdelhalim,

sorry to say, but - with all respect - I cannot agree to your analysis/explanation.

At first, your explanation is based on an output voltage of 10v (supply +- 15V), hence you assume that the opamp is in its LINEAR range. Secondly, you are assuming that the diff. voltage between both input terminals is (nearly) zero and , hence, the gain would be very large.

All these assumptions can`t be true in reality because we have positive feedback.

I have performed corresponding PSpice simulation for the shown circuit with positive dc feedback and - of course - the result is: Vout=-14.5volts.

Therefore, also the simulation performed by the questioner is not correct - possibly he has exchanged - by error - both opamp inputs? In this case, the shown results would be OK (negative DC feedback).

Further comment: It would be interesting to know which simulator engine the questioner has used. I am afraid that it was not one of the "classical" circuit simulators. Why not? In his first simulation with one single voltage source V=5V the presented result is -35 V.

My question: How a reliable simulation program (independent on the question if the circuit makes sense or not) can calculate an output voltage that is larger than any voltage source in the circuit?

Dear Lutz,

welcome,

You do not need to be sorry for do not agree with me.I know you very well and you are very respected colleague.

As for the answer if the amplifier is saturated your answer is right and the amplifier output voltage is at near 15V . I agree completely with you. Then there are some mistakes in the data given in the question.

Have a nice time

Best wishes

Dear Abdelhalim,

thank you for your reply.

Fortunately, the world of analog electronics seems to be healthy - in spite of the fact that the rest of our world thinks that we live in a "digital world", which will and must influence everything in tghe near future.

Do we really? I doubt. But this is another subject worth to be discussed between engineers who are involved in both - analog and digital.

Biswajit Singh:

here is my reply to your question:

"If i add 5K to inverting terminal then postive feedback and negative feedback factor will be same in that case which one would be more dominant?"

At first, positive and negative feedback will be - in reality - never be the same because we always have tolerances which will exclude (theoretically) such a case.

But I know, this was not the core and the background of your question.

Theoretically, when both feedback factors are identical, the circuit does not "know" what to do and - perhaps - moves between several states back and forth.

This is exactly what we want to have when we design an oscillator. If this equality between both feedback factors is true for one single frequency only - we have the classical oscillator structure for producing sinusoidal signals (Examples: WIEN-type oscillator, phase-shift oscillators,..).

By the way: Because of the mentioned tolerance problems often an automatic correction of the negative feedback factor is implemeted - controlled by the output amplitude.

I am pretty sure I am too late , but Biswajit , +Vcc and -Vcc are very much needed !!!!

Again, these days opamp can take one source Vcc but can't do without biasing . No lifeline if they are not there as Prof Zekry has mentioned.

Dear colleagues,

I continue thinking the best way to understand something new is to see something well-known old in it...

1. Bi-stable VINIC. The bizarre circuit that is discussed here (and many times before), is called with the pompous name voltage-inversion negative impedance converter (VINIC). Simply speaking, it inverts the voltage applied to the op-amp inverting input thus behaving as an S-shaped negative resistor. It does this "magic" by introducing a small positive feedback in addition to the existing negative feedback (circuits with only negative feedback can only zero the input voltage). Due to this trick, the circuit draws current from the input source that is bigger than the short-circuit current. As a result, the voltage drop across the internal source resistance exceeds the internal source voltage and the overall voltage across this real voltage source changes its polarity. If you are curious to know exactly how this is going, look carefully at this WB story:

https://en.wikibooks.org/wiki/Circuit_Idea/Linear_Mode_of_Voltage_Inversion_NIC#Negative_input_current,_positive_output_voltage_2

However, to do this "magic" the input voltage source must have some internal resistance or, ideally, be a current source (as Josef said above). When it is a perfect voltage source (as here), the op-amp does not manage to change (reverse) its voltage and goes to the supply rail...

So, the first viewpoint at this circuit is to think of it as of a VINIC acting in a bi-stable mode...

Dear Biswajit Singh,

it is very "unusual" to operate the opamp without powering. The real behaviour will, of course, depend on the device internal structure and technology, but I see no practical reason even to try this.

Looking to the second circuit with +/- 15 volts powering looks very easy to analyze by heart. There is voltage source connected directly to the inverting input, let's say +5V, so this input is permanently fixed exactly to this value. Thus the R2 resistor has no influence to the functionality, it is just a ballast load for the output. The R1 and R2 are just a 1/2 divider from output to the noninverting input, so there will always be half of the output voltage. However, this feedback to the noninverting input is the POSITIVE feedback, so that the output can just jump to the saturation.

2. Op-amp inverting Schmitt trigger. Obviously, as everyone has already noticed, in this mode, the resistor R2 does nothing... and, in fact, this bi-stable VINIC acts simply as an op-amp inverting Schmitt trigger:

https://en.wikipedia.org/wiki/Schmitt_trigger#Inverting_Schmitt_trigger

Output without power supply may be due to some hidden protocol inside the simulator. May be it assumes some default power supply for the opamp when it is absent.

with tools we need to set every parameter by ourselves . Eg: CMRR (I hate this) but without it the signals are not sensed @ INV and Non-INV .

Theoretically an infinitesimal negative feedback is present in the circuit due to the finite output impedance of the voltage source. However it is so small that the effect will be invisible. In a simulated model this (the finite output impedance) may also be absent.

Actually, I can see the possible value in wanting to simulate a circuit that contains an unpowered op-amp. In the real world, we (or our customers) sometimes forget to turn the power on! Interestingly, a circuit sometimes works, or tries to work, because the op-amp is getting power through a 'sneak path', such as an on-chip ESD diode. But, having said that, I wouldn't trust an op-amp model to accurately simulate this situation unless it was designed to do so.

Dear Biswajit

most of the Simulation environment do not affect if you connect power source or do not. I suggest click on opamp and see somewhere that dot model. will get an answer.

Also, if you are using virtual component then for sure you need not connect you power source. But your diagram shows TL series opamp (texas) bu still connecting supply is not the major issue as per my practice.If time permits I will share/show u simulation

"But your diagram shows TL series opamp (texas) bu still connecting supply is not the major issue as per my practice. "

This sounds really surprising to me.

I am using Spice-based simulation software since more than 20 years. However - according to my experience, an opamp model without proper DC supplies cannot work as an amplifier. That`s for sure!

Dear Biswajit

Regarding the first circuit: some simulators offer an "ideal" opamp. This component does not have power pins and is used only to test the idea of a circuit. It seems this is not the case, since you are using a commercial IC (TL072) and thus you must connect the supply pins with the suitable power supplies. The result of the simulation when this is not done is nonsense, since the simulator will try to find a solution to the circuit equations that probably does not represent the real one.

Regarding the second circuit: the opamp must be seen as a high gain amplifier. Even starting with a very small differential voltage between the + pin and the - pin the output voltage still will be very high. R1 and R3 behave as a resistive divider to the output voltage. The voltage at the + pin is thus half of the output voltage. This positive feedback keeps the voltage at the + pin as higher as possible. In a real opamp, this represents almost half of the positive supply (something close to 7 V in this case). This keeps the opamp positively saturated. I think your simulator should present this result, i. e. the output voltage should be very close to the positive supply. I do not know it (SIMetryx or SIMPLIS, correct?) but I suggest you to try another one such as Cadence PSpice Lite in order you can confirm the results. Hope I have helped.

Regards,

Antonio

I also agree to prof. Antonio Carlos da Costa Telles .

I also agree with Prof. Carlos's explanations... and also with all the previous comments... and I even voted for them positively.

But I can not stop to marvel how it is possible to explain in detail something well-known without mentioning its common name? It is like swapping the op-amp inputs in the original Biswajit's circuit above and starting to explain how the op-amp changes its output voltage so as to keep the voltage drop across R1 equal to the input voltage... without mentioning that this is actually a non-inverting amplifier? Or to explain how a network of two resistors in series attenuates the input voltage... without saying that this is a voltage divider? As they say in such cases, you "can't see the forest for the trees"...

We, teachers, can not afford not to make a connection between the unknown new and well-known old... just because we work exclusively with (young) people... and they easily understand it that way. Perhaps this is one of the things we differ from engineers in companies.

Quote Cyril: " ... without mentioning that this is actually a non-inverting amplifier"

What means: "this is"

Cyril, excuse me, but you have made the same error you are criticizing:

Which circuit are you speaking of? The circuit without an effective negative feedback, or ....???

Prof Mechkov , ask a nice Analog question preferably current mirrors ... I want to analyze that :)

Lutz, I mean the "dual" circuit we would get if we exchange the two inputs of the operational amplifier in the original Biswajit's circuit - where the R1-R3 voltage divider is connected between the op-amp output and its inverting input, and the input voltage is applied to the non-inverting input. I have corrected my comment above to clarify it.

Aparna, as far as I understand, you want to see the forest for the trees:)?

OK, see what I was discussing with my students on the blackboard after the lecture about diode circuits this Wednesday. What do you think are the "trees" and what the "forest" here? I think this would be a "nice Analog question" to discuss in this forum:

https://www.researchgate.net/post/How_many_types_of_resistance_can_be_represented_by_a_variable_resistor

Cyril - OK, thank you for clarification.

More than that, I like to comment the following sentence:

"We, teachers, can not afford not to make a connection between the unknown new and well-known old... "

Cyril, don`t you think that such an approach could be somewhat dangerous ? If we always refer to "well-known old" systems/circuits/explantions/approaches,....I am afraid, we could be blind for some really new and unconventional ways to introduce complete new ideas?

Lutz,

My note was about how to use the OLD for the purposes of NEW in circuitry. Such a need arises, first - when we try to understand a new circuit, second - when we try to explain a new circuit to someone. Of course, such a discussion would be particularly interesting for people who teach people and write good circuit stories; for others (who design devices and teachers who do not love teaching:) this is usually considered to be of little importance. If we compare the two mental activities, maybe the second is more noble than the first because by understanding you get something while by explaining you give something...

When understanding: Theoretically, it is possible to have an absolutely new circuit... and then we will have to examine it element by element to understand it. Fortunately, almost always we can discern familiar sub-circuits with well-known behavior... and that helps us to get a big picture of the new circuit.

In our case, we can see that R1, R3 and the op-amp constitute an inverting op-amp Schmitt trigger and R2 does not affect its operation because of the perfect input voltage source. So, the whole circuit acts as a Schmitt trigger.

If the input voltage source is real (having some internal resistance), R2 comes into play. Now, if we are familiar with NIC, we will recognize the circuit of a VNIC. So, the whole circuit now is a VNIC... and (it is quite likely) we know in advance how it works.

Finally, we figure out that the circuit is always VNIC but in the first case (perfect voltage source) it operates only in a bistable mode while in the second case (real voltage source) it can operate also in a linear mode.

When explaining: Now we do the opposite - we show well-known parts constituing the circuit and remind how they work. In this case, we can say that this is just an inverting op-amp Schmitt trigger with an additional but unnecessary resistor. Then we state that if we include a real input voltage source, the circuit will turn into a VNIC.

Finally, we can summarize that the circuit is always VNIC that in the first case (perfect voltage source) operates only in a bistable mode while in the second case (real voltage source) it can operate also in a linear mode.

It seems like a circuit i am looking for,

http://www.ti.com/lit/ds/symlink/lm675.pdf

Figure 15 might explain.

But i don't have the equation for the gain.

Look at Page 60 of AN 29 (attached)... and, for a more intuitive explanation, here:

https://www.circuit-fantasia.com/circuit_stories/understanding_circuits/current_source/widlar_op-amp_current_source/widlar_op-amp_current_source.htm

The idea behind this circuit solution is very interesting...