# Finding out the input impedance in a negative feedback op-amp

• Engineering
PhysicsTruth
Homework Statement:
Find out the input impedance in a LM741 Op-Amp with open loop gain around 10E+6, and with non-inverting terminal receiving variable voltage and inverting terminal connected to output (Negative feedback)
Relevant Equations:
##V_{out} = A_{v} (V_{+}-V_{-})## In this case, there is negative feedback. ##V_{out}## is almost equal to ##V_{in}## in this case. But I really can't understand how to even proceed to find out the input impedance in this case, from the equation I know. Can someone help me out in figuring how should I proceed?

Homework Helper
How about the defnition of input impedance ?

Mentor
Homework Statement:: Find out the input impedance in a LM741 Op-Amp with open loop gain around 10E+6, and with non-inverting terminal receiving variable voltage and inverting terminal connected to output (Negative feedback)
Relevant Equations:: ##V_{out} = A_{v} (V_{+}-V_{-})##

Can someone help me out in figuring how should I proceed?
Take a look at the LM741 datasheet?

alan123hk
how should I proceed?
I believe you need to define the relevant parameters, write down the equation, and then solve the equation to get the answer.

• BvU
Mentor
How about the defnition of input impedance ?
I believe you need to define the relevant parameters, write down the equation, and then solve the equation to get the answer.
Maybe I'm missing something, but they are asking for the input impedance of the LM741 opamp in the follower circuit. There is *nothing* else connected to that + input pin...

• PhysicsTruth and alan123hk
alan123hk
Take a look at the LM741 datasheet?
Yes, although the specifications provided by different LM741 manufacturers may have some differences, we still need to evaluate the actual input impedance according to the parameters provided by the manufacturer.
Maybe I'm missing something, but they are asking for the input impedance of the LM741 opamp in the follower circuit. There is *nothing* else connected to that + input pin...
Since this circuit has only one LM741 and no other additional components, the input impedance of the positive port of the LM741 should only be determined by the internal parameters of the LM741, but we have to consider that negative feedback will affect the voltage gain, input impedance and output impedance.

• PhysicsTruth
Gold Member
One option here is to ask the manufacturer, i.e. read the data sheet, they have typical specs for input resistance and capacitance. I don't really see how the feedback to the other terminal impacts this. But it is a strange question, one that real world EEs normally wouldn't worry about.

In any case, you will need to decide how much detail you want in your models. Beyond this you will need to know the transistor level parameters.

What level electronics class is this for? Analog IC design?

• PhysicsTruth
PhysicsTruth
One option here is to ask the manufacturer, i.e. read the data sheet, they have typical specs for input resistance and capacitance. I don't really see how the feedback to the other terminal impacts this. But it is a strange question, one that real world EEs normally wouldn't worry about.

In any case, you will need to decide how much detail you want in your models. Beyond this you will need to know the transistor level parameters.

What level electronics class is this for? Analog IC design?
The datasheet does show me that the input impedance of the LM741 IC is around ##1M \Omega##. But, in the case of negative feedback in this case, I have read that the output impedance decreases, and thus the input impedance increases. I want to find a way in order to correlate the new input impedance to the original input impedance of LM741. That's exactly where I need help.

PhysicsTruth
Maybe I'm missing something, but they are asking for the input impedance of the LM741 opamp in the follower circuit. There is *nothing* else connected to that + input pin...
Yes, but there is a negative feedback in this case. I had read that the input impedance increases in the case of negative feedback. Isn't it possible in this case to correlate the increased input impedance with the original impedance of LM741 IC?

Gold Member
Yes, but there is a negative feedback in this case. I had read that the input impedance increases in the case of negative feedback. Isn't it possible in this case to correlate the increased input impedance with the original impedance of LM741 IC?
In all but the most complex models of this device, the feedback does not affect the NI input. The differential input stage will do a very good job of isolating the two inputs. Even at the transistor level the effect isn't significant.

Again, we don't understand the depth of your models or analysis, since you're not telling us. But consider a small change in the output that would cause changes to the attached network (let's say current flow, for example). How would that affect the NI input, since there are no connections to it? Feedback isn't such a mysterious thing, it is just analysis of the network. If you want to understand this, you need to be thinking in terms of the algebraic equations that describe how one part of the circuit affects the other parts.

• berkeman
Mentor
Yes, but there is a negative feedback in this case. I had read that the input impedance increases in the case of negative feedback. Isn't it possible in this case to correlate the increased input impedance with the original impedance of LM741 IC?
As @DaveE is saying, there is very little effect inside the IC on the input impedance of the non-inverting "+" input in this non-inverting buffer configuration. The things you have read about negative feedback altering the input impedance apply to other opamp (or general amplifier) configurations.

That's why it's important to understand the fundamentals/basics of these circuits. Sometimes the feedback affects the input impedance, and sometimes it does not. So as a good teaching moment, I'll post schematics of your non-inverting buffer configuration and an inverting opamp configuration, and can you comment about whether negative feedback affects the input impedance?

• DaveE
Mentor  https://www.electronics-tutorials.ws/opamp/opamp_2.html

• DaveE
Gold Member
That triangle everyone draws for amplifiers requires some assumptions among the people discussing the circuit. It's better suited for schematics than analysis. If you replace it with a more explicit circuit model, like the one below, it will make the algebraic analysis easy (easier). Of course, there are many possible models, depending on what circuit behaviors you care about. You may add additional complexity, like frequency dependence, to some parameters, or you may choose to set some values to 0 or ∞, if those are good approximations for you're application. LvW
Yes, but there is a negative feedback in this case. I had read that the input impedance increases in the case of negative feedback. Isn't it possible in this case to correlate the increased input impedance with the original impedance of LM741 IC?
According to general system theory, the input impedance of an amplifier with feedback is increased by a factor k=(1+loop gain). For 100% feedback (as in the present case) the input impedance is drastically increased by a factor k=(1+Aol) with Aol=Open-loop gain.

Gold Member
input impedance of an amplifier
Which amplifier? Yes, this is often the solution, but a bit more rigor would be appropriate. Not all amplifiers are built the same way.

• berkeman
LvW
Which amplifier? Yes, this is often the solution, but a bit more rigor would be appropriate. Not all amplifiers are built the same way.
I think, this applies to ALL amplifiers (V-V, V-I, I-V, I-I) - independent on the way they are built.
For the classical opamp (V-V) it is very easy to calculate the loop gain because of the low output impedance and the very high input impedance. But in many other cases it is not a simple task to find the correct loop gain expression .

Question: You wrote "The differential input stage will do a very good job of isolating the two inputs. "
I dont understand this sentence. The two inputs are isolated? Does this mean that the input impedance at the non-inv. input would be independent on the feedback signal at the inverting input?
In contrary - for each diff amplifier (long-tailed pair), the input impedance at one transistor depends on the signal that does exist at the base of the other transistor.

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Gold Member
I think, this applies to ALL amplifiers (V-V, V-I, I-V, I-I) - independent on the way they are built.
For the classical opamp (V-V) it is very easy to calculate the loop gain because of the low output impedance and the very high input impedance. But in many other cases it is not a simple task to find the correct loop gain expression .

Question: You wrote "The differential input stage will do a very good job of isolating the two inputs. "
I dont understand this sentence. The two inputs are isolated? Does this mean that the input impedance at the non-inv. input would be independent on the feedback signal at the inverting input?
In contrary - for each diff amplifier (long-tailed pair), the input impedance at one transistor depends on the signal that does exist at the base of the other transistor.
What if the opamp were a vintage Philbrick, constructed of vacuum tubes? At low frequencies, and in the absence of a path through a resistor network, would not the impedance at the non-inverting grid (nearly infinite) be fairly independent of the signal at the other grid? It seems that the differential impedance between the non-inverting and inverting inputs is crucial for the connection shown in post #1.

LvW
What if the opamp were a vintage Philbrick, constructed of vacuum tubes? At low frequencies, and in the absence of a path through a resistor network, would not the impedance at the non-inverting grid (nearly infinite) be fairly independent of the signal at the other grid? It seems that the differential impedance between the non-inverting and inverting inputs is crucial for the connection shown in post #1.
I think, we do not need to speak about vacuum tubes. Take, for example, a diff. pair consisting of FETs.
In a differential configuration both source terminals are connected to each other - hence, both devices do not operate independently. Of course, the input impedance at one input terminal is very large (mainly determined by parasitics). Nevertheless, it will be influenced by the signal at the other input node.

Mentor
Question: You wrote "The differential input stage will do a very good job of isolating the two inputs. "
I dont understand this sentence. The two inputs are isolated? Does this mean that the input impedance at the non-inv. input would be independent on the feedback signal at the inverting input?
Correct. They go to different transistors in the input differential transistor amplifier, and at least in the linear region of operation, have very little effect on each other. The inverting opamp configuration that I posted above is a configuration where the negative feedback affects the input impedance. The buffer configuration of the OP is different...

LvW
Correct. They go to different transistors in the input differential transistor amplifier, and at least in the linear region of operation, have very little effect on each other.
Sorry - but I cannot agree. Both transistors in a differential amp (long-tailed pair) share a common node (both emitter terminals are connected). Therefore, the input resistance of T1 depends, of course, on the signal that exists at the input of T2.
As a consequence, we have different input resistances for the 3 most popular operational modes:
* unsymmetrical (one base grounded)
* differential-symmetrical (Vin1=-Vin2)
* common mode (Vin1=Vin2)

Mentor
Therefore, the input resistance of T1 depends, of course, on the signal that exists at the input of T2.
Those two voltages are equal. That's what the negative feedback does.

PhysicsTruth
According to general system theory, the input impedance of an amplifier with feedback is increased by a factor k=(1+loop gain). For 100% feedback (as in the present case) the input impedance is drastically increased by a factor k=(1+Aol) with Aol=Open-loop gain.
I came across this result as well, but for an amplifier in general. So, I tried to deduce a proof for the OpAmp's negative feedback as well, and it did turn out to be true (if my proof actually holds valid!) I would be uploading the proof which I have worked out in a short while.

Gold Member
I think, this applies to ALL amplifiers (V-V, V-I, I-V, I-I) - independent on the way they are built.
For the classical opamp (V-V) it is very easy to calculate the loop gain because of the low output impedance and the very high input impedance. But in many other cases it is not a simple task to find the correct loop gain expression .

Question: You wrote "The differential input stage will do a very good job of isolating the two inputs. "
I dont understand this sentence. The two inputs are isolated? Does this mean that the input impedance at the non-inv. input would be independent on the feedback signal at the inverting input?
In contrary - for each diff amplifier (long-tailed pair), the input impedance at one transistor depends on the signal that does exist at the base of the other transistor.
Yes, I think you're correct about the differential pair. I was being sloppy. In practice, the effect for a follower isn't normally significant enough to care about. Also, as you said, for MOS, the impedance is dominated by a huge input capacitance.

This is one reason for the existence of instrumentation amps; stable, balanced input impedance. I've never need to use one in my work since modern op-amps are so good. OTOH, you could argue that most IAs are really separate amplifiers that are combined downstream.

My real point is that you need to look at the amp implementation details for a question like this. It's not a basic circuit question, like you would see in an inverting configuration. It's a question of how the op-amp operates internally. Seems kind of an odd question for the OP; phrased in a general sense but dependent on subtle details.

• PhysicsTruth
PhysicsTruth
Yes, I think you're correct about the differential pair. I was being sloppy. In practice, the effect for a follower isn't normally significant enough to care about. Also, as you said, for MOS, the impedance is dominated by a huge input capacitance.

This is one reason for the existence of instrumentation amps; stable, balanced input impedance. I've never need to use one in my work since modern op-amps are so good. OTOH, you could argue that most IAs are really separate amplifiers that are combined downstream.

My real point is that you need to look at the amp implementation details for a question like this. It's not a basic circuit question, like you would see in an inverting configuration. It's a question of how the op-amp operates internally. Seems kind of an odd question for the OP; phrased in a general sense but dependent on subtle details.
You are absolutely correct there. Since I was asked to derive/calculate the input impedance in this case, I felt that I had to actually analyze it with the bigger picture at hand, rather going into the subtle details, as I have just been introduced to the practicality of an Operational Amplifier. Going through articles, I came across the input on the Newly Realized Input Impedance with respect to the actual impedance, and I tried to formulate a derivation for the same for an Op-Amp.

Gold Member
Also, (totally off-topic):

I think it's interesting that all of the academic op-amp stuff always uses LM741 as their example. Perhaps the first modern op-amp? But, I started doing design work in 1980, and I have NEVER seen an EE actually choose that part. It was obsolete 40 years ago. Perhaps it will live forever in textbooks though.

alan123hk
For operational amplifiers with limited high gain and infinite input impedance, the following will occur.

If it is used in a non-inverting amplifier configuration, the negative feedback does not affect its input impedance, because a finite number multiplied by infinity is still equal to infinity. Even if the input resistance of Lm741 is only about 1Meg Ohm AC resistance, it may be sufficient for most applications, so there may be no special benefit to further increase it.

On the other hand, if it is used in an inverting amplifier configuration, even if the internal resistance of the operational amplifier is infinite, after adding negative feedback, the input impedance will immediately become equal to the resistance of the external input series resistor, but this resistance cannot be too large, so the input impedance will become relatively low, which may be worrying engineers.

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LvW
I came across the input on the Newly Realized Input Impedance with respect to the actual impedance, and I tried to formulate a derivation for the same for an Op-Amp.
It would be interesting to see your results...
One further comment: In your case (unity gain amp, 100% feedback) the resulting input impedance at the non-inverting input could be - mathematically correct - calculated. Howver, the result would have no practical meaning. Why not?
Assuming rin=1MΩ (without feedback) and Aol=100dB (identical to the loop gain) the formal calculation for the input resistance with feedback would give rin=1E6*1E5=1E11 Ohms (100 GΩ).
Such a value has no practical meaning because all the neglected parasitics (leakage and capacitive effects) would dominate.

LvW
Also, (totally off-topic):

I think it's interesting that all of the academic op-amp stuff always uses LM741 as their example. Perhaps the first modern op-amp? But, I started doing design work in 1980, and I have NEVER seen an EE actually choose that part. It was obsolete 40 years ago. Perhaps it will live forever in textbooks though.
Yes - its surprising. That was - and still is - also my observation.
I started doing opamp design in 1968 and I have used the first follower of the legendary µA709 which was the µA741.

LvW
For operational amplifiers with limited high gain and infinite input impedance, the following will occur.

If it is used in a non-inverting amplifier configuration, the negative feedback does not affect its input impedance, because a finite number multiplied by infinity is still equal to infinity. Even if the input resistance of Lm741 is only about 1Meg Ohm AC resistance, it may be sufficient for most applications, so there may be no special benefit to further increase it.
Just for the sake of accuracy - you are not correct.
The input resistance of a non-inv. opamp configuration will be, of course, increase due to negative feedback. In your reasoning you are using "infinity" which is not realistic.
It is another question if the formal calculation of the enlargement of the input impedance is important - if compared with real conditions (see my separate answer, post#28). But it is simply false to state that "negative feedback does not effect its input impedance".

alan123hk
Just for the sake of accuracy - you are not correct.
The input resistance of a non-inv. opamp configuration will be, of course, increase due to negative feedback. In your reasoning you are using "infinity" which is not realistic.
It is another question if the formal calculation of the enlargement of the input impedance is important - if compared with real conditions (see my separate answer, post#28). But it is simply false to state that "negative feedback does not effect its input impedance".

Of course, infinite impedance does not actually exist, I just tried to use a mathematical model, so I don't think this is incorrect, but it can be said that it does not accurately describe the actual situation. For example, we cannot say that the calculation results obtained by using an ideal voltage source, an ideal current source, or an ideal transformer are incorrect because they are unrealistic. If they are incorrect, then it seems that we should not apply them extensively in textbooks. I just want to use it to describe a situation, that is, a thing is already big, and if we add it, the actual effect will not change.

As for the benefits of increasing the input impedance, this is of course subjective, which will vary according to different people's opinions and application conditions.

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alan123hk
The following nanopower operational amplifier is an interesting example.

Its specifications do not specify input resistance, so users can only assume that its input resistance is close to infinity during the design process.

Of course, it also has input bias current and input capacitance. The bias current is extremely low, about 100fA. The input differential mode capacitance and common mode capacitance are 7pf and 3pf respectively, and these capacitive reactances are almost negligible in the case of extremely low frequency operation. 🐻

Gold Member
Its specifications do not specify input resistance, so users can only assume that its input resistance is close to infinity during the design process.
No! The designer should assume that the input impedance hasn't been characterized. They should assume that:

1) It isn't a required screening in manufacturing.

or

2) The marketing people didn't think it was worth the effort to do the characterization and to publish it in the "typical" section.

There is a difference between unknown, unpublished, zero, and infinite. There are many, many times in my career that I had to deal with uncharacterized or poorly characterized performance specs that mattered to me but not so much to the mass market. This is why the BOM at EVERY reputable manufacturer has approved suppliers for each individual part. Just because the parts have the same (or very similar) part numbers or data sheets, doesn't mean they are the same.

As an aside, one of the things they never teach EEs in school is how data sheets are made, what they mean, and how to use them. They are primarily a marketing document.