Why are inverting op amp amplifiers used in most cases instead of non inverting amplifiers?
In my opinion one of the advantages is that unlike a non-inverting OP-amplifier which has a minimum possible gain of 1, an inverting Op-amp can be used as an attenuater with a gain less than 1 as well. Another advantage is that the input impedance of non-inverting op-amp is infinite while that of Inverting op-amp it is equal to the resistance between the input and the inverting terminal, so a better impedance matching can be obtained.
In my experience its the other way around - non-inverting configurations seem much more common. I myself prefer inverting for the minimization of common-mode distortion in the input stage, but there is a noise drawback amongst other caveats.
Finally i got the answer in vlsi text book. slew rate of inverting ampr is more and cmrr is more compared to non inverting ampr
Inverting op-amps provide more stability to the system than non-inverting op-amp.In case of inverting op-amp negative feedback is used that is always desirable for a stable system.
Speed of the opamp depends inversely on gain bandwidth product. This product is less for inverting compared to non inverting hence output rise time is less for inverting but in non inverting topology this product is same as open loop opamp
i think that now you have got the answer. You are satisfied with my answer
The words OPerational AMPlifier is given to a class of differential input high gain high input impedance amplifier because when two impedance Z1(S) and Z2(S) are are connected around that the voltage transfer function becomes -(Z1(S)/Z2(S)). This is the standard configuration where the input voltage source is applied to the inverting terminal of the op amp through a series impedance Z1 and Z2 is connected as a feedback impedance. The non inverting input terminal is connected to ground. The expression is obtained by two limiting conditions viz., gain tends to infinity and input impedance also tends to infinity. the second condition is necessary to assume input current equal to zero, where the first condition leads to the concept of virtual ground. If you consider "A", the gain of the amplifier to be a very large real number then even if you interchange the inverting and non inverting terminal then also you will get the same result. However for practical op amps this is not the case. as a first order approximation the gain can be expressed as B/(S+a) where B is a very large real positive number and 'a' is between 1~5. With this approximation the op amp can be modeled as v0 = (B/(S+a))*(V1 - V2) where V1 and V2 are the inputs applied to the non inverting and inverting input terminals respectively and V0 is the output voltage. With "A" real the corresponding model was V0 = A*(V1 - V2). If you use the practical op amp model to analyze the standard configuration with the limit B tends to infinity you would get the same result in steady state. However with input terminal interchanged the transfer function would contain an RHS pole that would result instability. This is the reason why inverting terminal is used in op amp configuration. In some application the non inverting terminal is also used along with the inverting terminal but care needs to be taken to ensure that the overall negative feedback is more than the positive feedback.
Quote B. Desai: "Stable due to negative feedback".
Exactly the opposite is true!
An amplifier without any feedback is stable by nature.
Negative feedback has many, many advantages and is used for nearly each practical amplifier - however, the price to be paid is: Dynamic stability is decreased.
General rule: Negative feedback improves DC stability (of the operating point) and decreases the stability margin.
Sorry for the typographical mistake. The correct transfer function is -(Z2(S)/Z1(S)) for the configuration described in my answer posted just now.
I work equally well with inverting and non-inverting designs, so your question seems to me somewhat curious. Lutz's answer, however, is in the right way and allows me to give you a reason why the inverting topology is less prone to be unstable than the non-inverting one, specially when using high speed, broadband Op Amps. To see why this is so, let us consider the capacitive coupling between the input active terminal and the corresponding active output terminal.
These two "terminals" can be viewed as two metallic ones at some distance d..... Two conductors at distance d use to bear an electrical capacitance in our actual world.... Thus, there is a capacitive coupling between the input and the output of the amplifier you are building. This coupling tends to be positive feedback in the non-inverting topology whereas it tends to be negative feedback in the inverting topology (e.g. the "well known" Miller effect.). From this, the inverting topology seems more prone to be succesful for unaware designers than the non-inverting one..... Nature helps those working with negative feedback, although electronic "artists" can "play" both...
Hello Jose-Ignacio,
I am not quite sure if your above explanation really applies.
For my opinion, there is ALWAYS a parasitic capacitance between output and inverting opamp input node - independent on the operating mode (inv. or non-inv.).
On the other hand - the feedback through the capacitance between output and non-inverting input is shorted, either because this node is grounded (inv. operation) or because this node is connected to an (ideal) voltage source (non-inv. operation).
I am aware that I have assumed idealized conditions - however, this reasoning applies in principle also for real circuits, do they not?
hi
I know Negative feedback is responsible to assure a stability to circuit.Inverting amplifier has usually voltage shunt (transresitance)feedback (if RL is connected) or current series (transconductance) (without RL), while non inv is voltage series. now u compare their AC performance parameters as( AV, Ri, Ro by Millers and Negative feedback approach).
Hello Lutz:
I´m going to quote my answer using your own words, namely your sentence:
"For my opinion, there is ALWAYS a parasitic capacitance between output and inverting opamp input node - independent on the operating mode (inv. or non-inv.)."
Let us call C- and C+ respectively to these two capacitances, and let us consider that due to their geometry of conductors in space, both are of similar value.
Let us consider next that we are driving the input of the amplifier with a matched generator of impedance Z=50 ohms +j0 ohms (e.g. through a 50 ohms coaxial cable).
When the inverting configuration is used, C+ is grounded as you said and C- forms with the 50 ohms "viewed towards the generator" a high pass feedback filter with cut-off frequency fc=1/[2pi(50)C-] Hz. At those frequencies f>fc, this filter feeds-back signal to the inverting input with low phase lag, thus being negative feedback because the output signal is sampled and feedback with zero phase (roughly) to the inverting pin.
However, when the non-inverting configuration is used, C- is grounded and now C+ forms with the 50 ohms "viewed towards the generator" a high pass feedback filter with cut-off frequency fc=1/[2pi(50)C+] Hz. At those frequencies f>fc, this filter feeds-back signal to the input with low phase lag, thus being positive feedback because the output signal is sampled and feedback roughly with zero phase to the non-inverting pin.
Therefore, for high speed (broadband) amplifiers, the inverting configuration has negative feedback "by design" (Nature: metallic conductors in space) whereas the non-inverting configuration has positive feedback for the same geometrical reason, being this the idea I was trying to put forward in my first answer following yours.
Moreover, the small phase advance of the aforesaid high-pass, feedback filter to the non-inverting input tends to be counterbalanced by the phase lag expected for the amplifying electronics within the amplifier that will be band-limited by some cut-off frequency. This way, positive feedback with zero phase can appear in the non-inverting stage but NOT in the inverting one.
In summary, considering in the space the two input terminals and the output one, there is a "feedback assimmetry" between inverting and non-inverting topologies.
Hello Jose-Ignacio,
thank you for your reply and your explanation.
You may be surprised - but I completely agree with your analysis.
However, please take into account that my previous answer was based on a certain assumption (explicitly mentioned) : Ideal voltage source (zero input resistance). Thus, I see no conflict between us.
On the other hand - I have the feeling that the discussion is more or less an academical one. I was engaged in analog circuit design since more than 30 years and I must confess that the decision "inverting vs. non-inverting gain" never was governed by the parasitic influences as described in your last contribution.
The main aspect - of course - always was on the system level:
What is needed with respect to the overall performance?
More than that, as you know there always are other unwanted effects (caused by other parasitic capacitances/inductances), which (very?) often overshadow the effects as described by you.
Nevertheless, thanks again. I think your explanation was a big support to clear the situation.
Regards
L.
Correction (since it is not possible to edit a contribution):
Please read in the 5th line: (zero source resistance).
Thank you.
Maybe another reason to prefer the inverting configuration is... just the tradition? The first op-amps were single-ended and the R1-R2 network was the only way to subtract (compare) in a "parallel" way the input and output voltages. It acts as a resistive summer with weighted inputs according to the superposition principle:
http://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#Varying_both_V1_and_V2:_a_resistive_summer
I remember the analog computers from my student years; they had only single-ended-input op-amps.
Regards, Cyril.
Jose-Ignacio, I will note only that the arrangement with a stray capacitance between the op-amp output and the non-inverting input is a good example of a negative capacitor (capacitive INIC). Of course, this is just the other name of the positive feedback... Regards, Cyril.
Another hint: the stray negative capacitance above (C) can neutralize an equivalent input stray capacitance:
http://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Op-amp_implementation
And this is another example how to make two undesired quantities cancel each other.
Hello Cyril,
Thank you for the nice and interesting examples.
However, I have a general question:
The first original question implies (now supported by you: "Maybe another reason to prefer the inverting configuration...") that the inverting configuration would be preferred over the non-inverting topology. Is this really the case?
Can such a claim be confirmed by your experience?
For my opinion - as I have mentioned before - the decision inverting/non-inverting is governed primarily by system aspects.
Example: Some filter topologies need non-inv. and some other need inverting amplifiers.
Since the gain of the operational amplifier is extremely large, to ensure stability it is necessary that the resultant feedback must be negative, and therefore use of inverting terminal is mandatory. However the non inverting terminal can also be used .provided condition stated above is maintained.
Lutz, I second your final conclusion. Here is my story...
The more general problem is (was) to create a voltage follower. We already know from our routine (Harold Black also-:) how to solve it - by "active copying" (formally, "negative feedback"). So, we have to build an NFB voltage follower. For this purpose, we have first to produce some output voltage, e.g. by an amplifier. Then, we have to compare (subtract or sum with an opposite sign) the output with the input voltage and to change the output voltage until it becomes (almost) equal to the input voltage. So, we need a subtractor (a summer). We know two possible ways to do it - by connecting the two voltage sources in series (KVL) or in parallel (KCL). Thus, depending on the used summer , we obtain two possible arrangements - noninverting and inverting; each of them has advantages and disadvantages.
IMO the role of the summer is crucial; it predetermines the other elements and the circuit properties. Thus, if we have chosen a series "summer" (simply a loop), it needs an op-amp with a differential input. Also, the output voltage has to have the same polarity as the input voltage with the purpose the two voltages to be subtracted (compared); in many cases the circuit can be supplied only by one power supply. So, the circuit is noninverting since the summer "needs" it to be noninverting-:)
If we have chosen a "parallel" summer (the two resistors); it needs a bare single-ended op-amp input (inverting). And if we have only op-amps with differential inputs (today's situation), we ground the needless noninverting input-:) Now, the output voltage has to have an opposite polarity regarding the input voltage with the purpose the two voltages to be subtracted (compared); so the circuit obligatory needs a split supply. Again, the circuit is inverting since the summer "wants" it to be inverting-:)
We may observe this duality (series and parallel) in circuitry ever since the Kirchoff's times and it gives this circuit variety...
Regards, Cyril.
Quote: "Since the gain of the operational amplifier is extremely large, to ensure stability it is necessary that the resultant feedback must be negative, and therefore use of inverting terminal is mandatory".
Sorry to say, but this statement is confusing (in fact, it is not true).
It is correct, that negative feedback is necessary - however, not "to ensure" stability but to fix a bias point within the linear operating range.
In contrary, negative feedback REDUCES dynamic stability.
However, these considerations are independent on the choice of the input node (inverting or non-inverting operation).
@Lutz von Wangenheim Consider a classical finite gain inverting amplifier design using an operational amplifier. Normally you use the inverting input for that and connect the non inverting input to ground. If you swap these two input terminals the output of the amplifier would be latched to either of the power supply voltage. This is so because the pole of the resulting configuration will be pushed into the right half of the S-plane. (Possibly location of the poles of a system does determine its stability.) This can be shown very easily by considering operational amplifier as a low pass filter. This can also be verified by fabricating the circuit. However, theoretical simulation may not show this behavior if you consider a real value for the operational amplifier gain..Unfortunately for a practical operational amplifier gain is never a real number but it does follow a low pass characteristics. Please note that this non ideal characteristic is a result of the "parasitic influences".
Hello A. Bandyopadhyay: Excuse me, but it seems that you misunderstood something.
I never have mentioned something like to "swap the input terminals" .
This would lead, of course, to POSITIVE feedback.
In my contribution I was referring to NEGATIVE feedback only!
And it is a well-known fact that such a feedback reduces stability - in contrary to your claim (..."to ensure stability"). And the original question was about the use as inverting/non-inv. amplifier. However, in both cases, of course, only negative feedback is applied.
Regards
LvW
Possibly I did not understand what you mean by stability. I define a system to be stable by the fact that it produces bounded output for the bounded input. Therefore a system will be more stable when its pole has a large negative value [natural response of the system will be of the form exp(-at)]. Most of the practical voltage amplifiers (operational amplifiers are voltage amplifiers) has a low pass characteristics and could you please explain how for such systems "it is a well-known fact that such a feedback reduces stability " I understand by this statement that negative feedback will reduce the negative component of the pole. Since operational amplifiers have very large input impedance therefore I assume negative voltage feedback. MAY BE I am Not understanding your point.
Regards
AKB
Hello again, Prof. Bandyopadhyay:
Quote: "..could you please explain how for such systems "it is a well-known fact that such a feedback reduces stability " I understand by this statement that negative feedback will reduce the negative component of the pole."
Well, let me explain:
An amplifier with lowpass characteristic and without any feedback is always stable (open-loop poles negative-real).
Such an amplifier is equipped with a fixed bias point and the desired gain by applying negative feedback. It is a well known fact (documented in every relevant textbook) that - as a consequence - the stability margin is reduced (phase margin PM=180 deg without feedback). That means, depending on the amount of negative feedback as well as the open-loop pole location, the PM of the amplifier with feedback is smaller than 180 deg. (the closed-loop poles can be even conjugate-complex).
This can be observed by analyzing the step response of the closed-loop, which exhibits peaking or ringing - a typical indication of a reduced stability margin.
Remember: There are opamps, which are unity-gain stable, however, some other types (partly compensated) must NOT be operated with lower gains (heavy feedback), because they become unstable.
In fact - each real opamp with negative feedback will turn into an amplifier with POSITIVE feedback for higher frequencies (observations formulated by H. W. BODE, long time ago), however, in most cases the loop gain for this "critical" frequency range is already below unity. Thus, no instability will occur. But, in any case: Negative feedback reduces dynamic stability properties (smaller gain and phase margin).
Dear Colleagues,
Please allow me to look at the answer from other point of view.
It is the circuit configuration itself which makes the the inverting amplifier op amp more popular than the non inverting op amp.
The non inverting terminal in the inverting amplifier configuration is grounded. This makes the inverting op amp terminal at virtual ground. This basically isolates the input from the output. Practically, there is no common circuit element between them.
Therefore,the input current is determined by the input voltage and the impedance connected between the input and the virtual ground. This current continues to pass in the feedback impedane and the voltage drop across it will be equal to the negative of the output voltage.Therefore due to this virtual ground, one can easily convert current sources into voltage sources and vice verse.One can also implement easily the integration and differentiation. Also ,one can implement the addition process.
The non inverting configuration is called the potentiometric configuration where the input voltage is a fraction of the output voltage as is well known for us.
The main advantage of this configuration is its relatively high input impedance.
In conclusion, the inverting op amp configuration is suitable for performing operations.
Thank you.
Abdelhalim, congratulations! How well you have written it! Yes, it is true that the inverting configuration is more suitable for building various functional circuits implementing mathematical operations. That is why, I have already mentioned it, the inverting single-ended op-amps were so popular in analog computers. I have said the truth about inverting circuits many times; I will say it again (Pisupati had a similar speculation in an old insertion but I didn't know why he erased it later):
All these inverting circuits consist of four elements connected in a loop - two voltage sources (input and output) and two passive elements. The two voltage sources are connected in the same direction so that their voltages are summed. The output voltage source (the op-amp output) and the second passive element neutralize each other. So these two elements disappear (negative + positive impedance = 0) thus creating the illusion that there are only two elements in the circuit - the input voltage source and the first passive element, and only they define the current.
Maybe, you have noted that I have not used the virtual ground concept here and yet I have managed to reveal the phenomenon. The virtual ground is a concept closely related to the real ground concept; "virtual ground" is a point; it implies a potential of a point, not a voltage difference between two points. But if you have not a ground at all? Simply, your circuit is not grounded... My explanation is based on the whole loop, not only on the common point. The power of this explanation is that it is ever true, even in the case of ungrounded circuit; it is more universal... and it shows how to create virtual elements by adding an additional voltage...
I would like to ask you what you mean when saying "This basically isolates the input from the output. Practically, there is no common circuit element between them." IMO exactly the opposite is true for the inverting configuration - the input and the output are connected by the network of two series connected passive elements (a kind of a "bridge") and the same current flows through all these elements. Conversely, in the noninverting configuration there is no connection between the input and output and no current flows. This is a fundamental difference between the two configurations even in the case of a positive feedback (inverting and noninverting op-amp Schmitt trigger). Because of this difference, only inverting circuits can exhibit negative impedance since the output can affect the input current...
About "...This current continues to pass in the feedback impedane and the voltage drop across it will be equal to the negative of the output voltage..." I would reverse the causality:
This current continues to pass in the feedback impedance and "the output voltage will be equal to the negative of the voltage drop across the feedback impedane"
(the first is the voltage drop across the feedback impedance and the second is the op-amp output voltage)
About "...The non inverting configuration is called the potentiometric configuration where the input voltage is a fraction of the output voltage as is well known for us..." I would say that it is not obligatory to apply a fraction of the output voltage. You can apply the whole output voltage (voltage follower) or even the amplified output voltage (active voltage divider). Really, the last sounds guite odd and paradoxical (it means to connect an amplifer instead a voltage divider in the negative feedback).
Also, it is possible to connect an active element, e.g., a bipolar transistor in the negative feedback of the inverting configuration; the so-called "transdiode" is a good example. It also shows how to reverse a transistor (to make the output collector current - an input, and the input base-emitter voltage - an output). But this topic deserves our special atention...
Best regards, Cyril.
Lutz, Abdelhalim, Pisupati, Erik... and all the colleagues here, don't you think that our discussions resemble brainstorming or synectics sessions? What wonderful ideas come into the world in these discussions!
Best regards to all of you, Cyril
Cyril, yes I agree..
In his last contribution Prof. A. Zekry gave some good examples for my opinion (as I have already earlier formulated) that operational requirements dominate the decision inverting/non-inverting configuration (instead of parasitic influences).
Examples: I-V conversion, integrators, active filters with good active sensitivity figures, ...).
I would add resistive sensors (photo-, thermal-, etc.), connected in the feedback loop, R-2R ladders...
All these circuits can be implemented by the noninverting configuration as well with the advantage of the high input impedance. But the problem is that the output voltage is not the "pure" voltage across the second passive element (as it is in the case of the inverting configuration); here it includes the input voltage as well...
Dear Cyril,
Thank you for your complements. And thank you too for the the comments raised on my answer. The concept of virtual ground is highly recommended in the the op amp circuit analysis and we debate on it in a separate discussion. I think that at the end of discussion one is convinced with the logic answers.
I see the inverting op amp from my perspective and you see it from your perspective. BUT when coming to quantitative results we will agree.
There is a difference between isolated and independent. The two voltages are isolated in the sense there is no common element between them assuming virtual ground other wise they physically share the input impedance of the amplifier. Therefore i said practically. The dependence comes from the other main op amp characteristics which is negligible input current. Therefore, the current flowing in the input resistance continues to pass in the output impedance.And this which makes this circuit superior. It is simple but genius circuit..
Best regards.
Zekry
Besides the limitations of the common-mode input range, the non-inverting configuration possesses an inherent error due to the finite CMRR.
Hi Cyril - from your last answer I`ve got the feeling that you would support the statement which is contained in the original question ("inverting amplifiers are used in most cases...").
Is this your experience? I really don`t know ....I have seen no statistics up to now.
Dear Lutz,
These days I was thinking on your question... but I could not find an exhaustive answer... only some partial answers...
First, I think that it depends on the gain of amplifier:
- If we are talking about an amplifier with a gain of 1 (voltage follower/inverter), definetely followers (+1) are much more widespread than inverters (-1).
- In the case of amplifiers with a gain more than 1, I have the feeling that "pure non-inverting amplifiers" are more popular... while applied circuits (including sensors, programmable switches, DACs...) are mainly based on the inverting configuration... because of the miraculous properties of the virtual ground:)
Another way to see which configuration is more popular, is to write in Google the two names, and to count the number of pages:)
- "inverting amplifier" - 339,000 pages
- "non-inverting amplifier" - 142,000 pages
Hi Vasile - yes, I think so (provided there are no hidden" feedback loops).
How could an amplifier without any feedback (that means: Without a loop gain) become self-excited?
Lutz, an open loop amp would be more sensitive to variations in biasing component values. This creates sample to sample variations in gain. Negative feedback can easily remove that problem. Also, open loop amplifiers are prone to changes in gain as they are heated or cooled, also controlled with negative feedback (and other means such as temperature-compensated biasing).
And then there's linearity. Loads are never purely resistive, so they vary over the frequency range. Negative feedback compensates for that variation in the load, so that gain is flatter across the operating spectrum. Negative feedback improves total harmonic distortion, although it may degrade transient intermodulation distortion. There's no free lunch.
And opamps in particular depend on negative feedback, because in principle, they are infinite gain devices. No matter what other arguments there may be, opamps need negative feedback to control their gain and to make them behave in linear fashion across whatever signal spectrum you require. Otherwise, open loop, gain drops relentlessly, as frequency increases.
But I do agree with Yadhunandana's original point. In the case of opamps in particular, the literature seems to favor inverting amps over non-inverting. Not in the case of other amps, but I've noticed this to be the case for opamp circuits. (Although I have no stats to back me up!)
Albert, it seems that you misunderstood my contributions. I didn`t argue against feedback - just the opposite is true. I am aware of the benefits of negative feedback. But this was not the question. In one of my former contributions I wrote:
"Negative feedback has many, many advantages and is used for nearly each practical amplifier - however, the price to be paid is: Dynamic stability is decreased.
General rule: Negative feedback improves DC stability (of the operating point) and decreases the stability margin."
.
Hi Cyril
And what about filters with non-inverting amplifiers? :-)
Sallen-Key for example.
Josef
I agree with Josef that there is something confusing in this question...
Yadhunandana asks us "why inverting amplifiers are used in most cases instead of non-inverting amplifiers"... but we, instead to answer why, questioning this fact:)
I think more correct asked questions would be like "Why in some cases we use an inverting amplifier instead of a non-inverting one?"... or, "When we use an inverting amplifier instead of a non-inverting one?"
...or (if we have the choice): Is there any preference for one of both alternatives?
The resumption of the discussion by Ritesh few days ago and the subsequent reflections awakened my desire to continue delving deeper into the basic ideas behind these circuits... although I have noticed that, for some unexplained (for me) reason, this is not welcome for participants... they leave the discussion... and it stops again...
In the end, I think that these two basic circuits - inverting and noninverting, are great because they represent two great circuit phenomena - "virtual ground" and "bootstrapping"... virtual zero and virtual infinite resistance...
They are dual... so, we can assume they are equally widespread.
https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_adding_V2_to_V1
https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_subtracting_V2_from_V1
Now I want to share my last insight regarding the similarities and differences between the two circuit configurations...
Their common feature is that both maintain a zero voltage between two points (the op-amp inputs)... but they do this in a different way:
- The inverting configuration does this from the side of the input source (the inverting input)... while on the other side (the non-inverting input) the voltage is maintained constant (zero). To do this, it changes the resulting voltage at the inverting input by subtracting (a part of) the op-amp output voltage from the input voltage.
- The noninverting configuration does this from the other side (the inverting input)... while the voltage of the input source (at the non-inverting input) is maintained constant. To do this, it simply changes (part of) the op-amp output voltage applied at the inverting input.
So, although in both circuit configurations there are two equipotential points, only in the second configuration... where the "other voltage" follows the input voltage... and the input source "sees" infinite resistance... the "bootstrapping" effect can be observed...
... the "bootstrapping" effect can be observed...
Cyril - in this context, I wouldn`t use the term "bootstrapping". For my opinion, it is simply the voltage-controlled voltage feedback that is responsible for the increased input resistance.
I rather think that "bootstrapping" is connected with a kind of positive feedback. Otherwise, the relatively large input resistance of the emitter follower would also be the result of "bootstrapping". But we know that it is common practice to further enlarge this input resistance by insertion of a so-called "bootstrap capacitor" which provides positive feedback. THIS really is bootstrapping!
In case of non inverting amplifiers one has to identify the voltage limits in terms of the supply for the op amp. A 15V operated op amp cannot be given >15V at positive input. Inverting amplifiers are current driven and one can also have hundreds of volts as input (provided the current is within limits) and get proper operation.
Inverting amplifiers are current driven and one can also have hundreds of volts as input (provided the current is within limits) and get proper operation.
Example: Supply voltage +-15V.
Input voltage: 150V
Maximum allowable gain: - 0.1.
Question: Can we call such a device "amplifier"?
I expect that you can call it an amplifier even if gain is less than 1. there is also current amplification. If current is 10uA as drawn from the source of 150V, Ri is 15meg. however the resulting 15V with Rf of 1.5 meg can provide current in excess of a few mA.
Hmmm... I was preparing an answer to the Lutz's comment about "bootstrapping"... but I will temporarily stop to say some words about the last comments...
Yes Mani... the ability to apply very high input voltages is a unique property of the inverting configuration... I frequently joke with my students asking them, "Can we apply 1,000V input voltage? Or 1000,000V? And they are quite puzzled... and frightened by such an opportunity...
To explain how it works, I even use the lever analogy of Archimedes:)
But I agree with Lutz that this configuration is a quite strange "attenuating amplifier"... and, in addition, an "inverting attenuator"...
But, as Mani tried to explain, this is an "active attenuator" that can produce bigger output current... Only, I would correct his expalanation above ("however the resulting 15V with Rf of 1.5 meg can provide current in excess of a few mA") since, to show the current amplification here, like in the case of a voltage buffer, we should consider the op-amp output current flowing through an external low-resistive load...
So, "Rf of 1.5 meg" has nothing to do with this "current amplification". I hope it is obvious that "the current of 10uA drawn from the source of 150V, Ri is 15meg" is equal to the current produced by "the resulting 15V with Rf of 1.5 meg".
Lutz,
Are we speaking of the "function" of the circuit
or
are we speaking of the "classification" of the circuit.
"Amplifier" is a very general term .
To your question, I would say "yes" it is an amplifier,
in a general sense,
although it has a negative gain.
Commonly we think of "gain" as producing an increase in signal,
however, in my projects ,
I use Inverting OPA amplifier circuits
having real-time variable gains from 0.005 to 100.0
as wide-range pre-amp stages.
In my application of an OPA ,
I do not want it to oscillate,
nor regulate,
nor current-drive-LED indicators.
I want a wide range amplifier as a Pre-Amp stage.
We can say too that the input resistor (and "virtual zero of the OPA") craetes a voltage to current convertor and the feedback resistor plus OPA creates a current to voltage convertor (inverting).
Glen,
I think the common understanding of amplification is that the output power is greater than the input one... and the ratio between output and input voltages/currents is greater than or equal to one...
Lutz, now about the "bootstrapping"... one of my favorite topics...
On Page 6, you have said:
"Cyril - in this context, I wouldn`t use the term "bootstrapping". For my opinion, it is simply the voltage-controlled voltage feedback that is responsible for the increased input resistance.
I rather think that "bootstrapping" is connected with a kind of positive feedback. Otherwise, the relatively large input resistance of the emitter follower would also be the result of "bootstrapping". But we know that it is common practice to further enlarge this input resistance by insertion of a so-called "bootstrap capacitor" which provides positive feedback. THIS really is bootstrapping!"
-----------------------------------------
From my viewpoint, the bootstrapping arrangement consists in total of three elements - two voltage sources and one 2-terminal element (usually a resistor). We can look at it from two perspectives:
- In terms of ground, to bootstrap a floating 2-terminal element (resistor) means to apply voltage at its output terminal and to keep up it equal to the voltage of the input terminal.
- In terms of loop, to bootstrap an element means to insert a voltage source in series with the element and to keep up its voltage equal to the input voltage.
Thus formulated, the bootstrapping does not need any feedback - neither negative or positive - the two voltages can simply varying in the same way... then each of them will "see" the infinite resistance of the rest part of the circuit...
But a common practice is to make the output voltage follow the input one by means of the negative feedback (voltage follower). From this perspective, really the relatively large input resistance of the emitter follower is a result of a "bootstrapping".
The bootstrapping is not necessarily connected with positive feedback; it will appear only if the input voltage source has some internal resistance; in the case of a perfect voltage source, there is no positive feedback.
The role of the so-called "bootstrap capacitor", is to connect (AC) two points with different voltages (to translate the voltage variations).
And finally, it is important to clarify that the ohmic resistance of the bootstrapped element remains the same; only the total dynamic resistance of the whole network (resistor + second source) is artificially increased.
Glen,
Maybe your "pre-amp" circuit can act either as a voltage attenuator (0.005 - 1) or a voltage amplifier (1 - 100)?
Cyril - I am very sorry, but I cannot agree to your "definition" (and it is, indeed, a novel definition) because - according to my understanding - we have already a definition for the term "bootstrapping". And this commonly agreed definition is based on a specific positive feedback effect.
As a simple example, let`s take the classical emitter follower. Because of heavy negative feedback we have a gain of close to unity. And this effect - emitter follows the base voltage - is "bootstrapping" ??? Does this apply also to other positive unity gain amplifiers - for example, common base amplifiers?
If this is - in your view - already "bootstrapping" what then is the task of the capacitor between the emitter node and the base network which we call "bootstrap capacitor"?.
For my understanding, only such an additional element (capacitor) between two nodes with nearly the same signal voltage causes the bootstrap-effect because it connects (like a "bootstrap") both nodes.
Cyril,
It is both
depending on how the operator adjusts the gain during operations
in real on-going time,
from one moment through the next moment. .
If this were the single stage of a radio regenerative receiver
then
at low gain it would be a simple amplifier
and
at high gain it would be an oscillator ( which I would avoid ).
I simply see the inverting amplifier as an inverting amplifier.
The gain setting is not part of the definition.
However,
I will just learn the commonly agreed usage and move-on.
That is the method I use to hear what else is to be presented.
Some school classes were like that in my earlier life.
That reminds me of the "virtual" wild white rabbit
that passes thru my back yard.
Never touched, never measured,
and just as virtual as a dream.
Dear Lutz,
If you let me paraphrase your quote about the nature of religion and science, I would say that definitions are based on belief while reasoning - on doubt:) So, as we are here in the world of doubt, let's leave definitions and rely on our reasoning...
The differential input resistance of the op-amp in the circuit of a voltage follower (non-inverting amplifier) is connected between two voltages - input and following, so it is bootstrapped (artificially increased)... but the effect of this technique here is negligible because this resistance and so is high...
The base-emitter resistance of the transistor in the circuit of an emitter follower is also bootstrapped since it is connected between the input and following voltage... In the case of the AC follower below, the bias resistor R3 is AC connected (by means of the bootstrapping capacitor C2) in parallel to the base-emitter junction... and so it is also bootstrapped. This capacitor serves only as an AC coupling element like in the case of the AC amplifier.
In the common-base configuration, we cannot observe the bootstrapping idea since there is no "bridge element" connected between two equipotential nodes...
As for the presence of positive feedback, I still continue keeping my assertions above... but still reflecting on this... There is some mystery in this matter...
These were my doubts about the so-called "bootstrapping"...
My last thoughts are in the direction of the idea that there is a bootstrapping in all these circuits... but there is a positive feedback only in "true bootstraping circuits" (with an external biasing resistor)... Onlty, I am tempted to see a positive feedback also in the circuit of the ordinary voltage follower")...
The reasons for this assumption are that the internal resistance of the input voltage source and the equivalent resistance of the biasing circuit form a voltage divider... and the input voltage source is "helped" by the output voltage that increases the equivalent resistance of the lower branch of this divider...
Cyril - thank you for your reply. So - in your view - the shown circuit contains TWO bootstrapping effects, correct? Of course, I agree to all your explanations and descriptions - however, one should be aware that this contains (for my understanding) another novel definition for "bootstrapping". Of course, this is not a problem because everybody has the freedom to define certain terms. My only point was that - according to my knowledge - the simple follower property of a unity-gain feedback system is not classified in the literature as "bootstrapping".
Final question: As you know, the emitter follower gain is slightly below unity (the same applies to the opamp follower); what about a system with a gain of 0.7 or only 0.5?
Example: Emitter resistor RE=50 Ohms. For an emitter current of 1mA the gain is 50/(50+25)=0.666. Still "bootstrapping"? If "no" - where is the lower gain limit ?
The "bad" emitter follower will create a "bad" bootstrapping effect:)
But yet, what is the classic definition of "bootstrapping"?
Here is what Wikipedia says:
*Bootstrapping is a form of positive feedback in analog circuit design
* In the field of electronics, a bootstrap circuit is one where part of the output of an amplifier stage is applied to the input, so as to alter the input impedance of the amplifier.
This is a positive feedback - subcritical, which increases input resistance - ideally to infinite value.
Dear Lutz,
Your extracts from Wikipedia sound to me like texts from the Bible... to which I should believe unreservedly. But I can not do it... because they are quite unconvincing... and some of the texts inside the page are even wrong...
For example, only look at this sentence: "To minimize the size of the necessary capacitor, it is placed between the input and an output which swings in the opposite direction. This bootstrapping makes it act like a larger capacitor to ground."
If I had the desire from years ago to write in Wikipedia... and defend the written... I would add another "biblical text":
Bootstrapping in electronics is a technique for artifically increasing the resistance of a 2-terminal element by keeping up a constant voltage across it with the help of additional following voltage.
Quote: Your extracts from Wikipedia sound to me like texts from the Bible... to which I should believe unreservedly. But I can not do it... because they are quite unconvincing.
Quote: As you can see, in my definition of bootstrapping,
Cyril - we shouldn`t forget that we are speaking about DEFINITIONS. Definitions are arbitraire and commonly agreed "specifications". Hence, for somebody they might appear as "unconvincing". But - to allow technical communication between people - I think, we have no other choice than to rely on it and to use it.
Of course, I agree that some definitions are unfortunate (this may have historical reasons) - for example: Transistors "current gain", which in the past has caused many confusions about the real working principle of a BJT - up to today!
However, changing/modifying definitions is a very complicated and time-consuming process.
... but it is very desirable for the purposes of our discussion, where we try to reach the real truth about circuit phenomena...
(I will continue to expand my previous comment above about the role of the feedback... or I will move it below... OK this is better...)
Lutz and Josef,
As you can see, in my definition of bootstrapping, I have focused my attention on the essential property of the bootstrapping - the artificial increase in resistance, and so I have not mentioned any feedback. But let's consider its role here, for example, in the Josef's picture above (I have attached a similar picture below extracted from a similar question in the Stack Exchange forum where I - Circuit fantasist, have commented it).
1. NEGATIVE FEEDBACK. First at all, we can see that, with the help of a 100% negative feedback, a voltage follower (an amplifier with a gain of 1) is implemented. This perfect unity gain is achieved at the cost of an infinite loop gain. As a result, the voltage of the output terminal (the op-amp inverting input) of the input differential resistance (the internal "resistor" connected between the two inputs) is kept equal to the voltage of its input terminal (the op-amp non-inverting input)... the voltage across and the current through it are zero... so the input voltage source sees an infinite resistance... According to my definition of "bootstrapping", this means that the input differential resistance is bootstrapped.
So, we can conclude that, regardless of whether we want or not want it, the internal input differential resistance of every amplifying circuit with series negative feedback is bootstrapped.
Then, encouraged by the possibility of increasing resistance in this artificial way, we decide to apply this "magic" on an external resistor (the upper one on the figure) connected, for some reason, with its input terminal to the input voltage source. Following the "bootstrapping recipe" above, we apply the following op-amp output voltage to the resistor's output terminal. For some reason, we do it in an AC manner (via a decoupling capacitor).
The result is the same as above - the voltage across and the current through the resistor are zero... so the input voltage source sees an infinite resistance (connected in parallel to the infinite op-amp differential input resistance:) So, according again to my definition of "bootstrapping", this means that the upper resistor is bootstrapped.
2. NO POSITIVE FEEDBACK. Now let's see if there is another kind of feedback in this new arrangement. As we have connected the op-amp output through the resistor to the non-inverting input, we can expect that a positive feedback will appear. But in my opinion, this depends on the internal resistance of the input voltage source. If it is a perfect voltage source, it determines the voltage of the non-inverting input (directly through the zero internal resistance). The op-amp output voltage also is trying to influence this point (indirectly through the relatively high resistor's resistance)... but fails...
The conclusion is that, in the case of a perfect voltage source, there is no positive feedback... but still the resistor is bootstrapped...
3. POSITIVE FEEDBACK. If the input voltage source is not so perfect... and has some internal resistance (as in the Josef's picture above), this gives a chance to the op-amp output voltage to influence the common point at the non-inverting input. Actually, the two voltages (input and output) are summed by the resistive summing network of two resistors in series (a voltage summer with weighted inputs).
In this case, there is a positive feedback... but it is quite slight since usually the source internal resistance is lower than the bootstrapped resistance... It is also "subcritical", as Josef said, since the loop gain is always less than 1... and the circuit is stable (only, an interesting situation will be if there is no input source connected).
The final conclusion is that, in the case of an imperfect voltage source, there is a positive feedback... and the resistor is bootstrapped...
http://electronics.stackexchange.com/questions/77483/bootstrapped-resistor-a-current-source-or-open-circuit
Hi Cyril, here are my comments:
1. I fully agree with your DESCRIPTION - however, I must repeat that you have introduced another DEFINITION for "bootstrapping". This is not a problem at all - as long as one knows that it is not in agreement with the commonly accepted definition (in my view).
Here is your definition: According to my definition of "bootstrapping", this means that the input differential resistance is bootstrapped.
(One could ask: The resistance is bootstrapped? For my opinion, "bootstrapping" means that two NODES within a working system are tied together using an additional two-pole).
Example: Applying your definition to the emitter follower would mean: The dynamic base resistance rbe is "bootstrapped". For my opinion, this sounds a bit "weird".
2. and 3. Speaking about "positive feedback" we must define the point where the feedback is effective. Because you mention the internal source resistance Rs it is clear that you are considering the non-inv. input as relevant - and, yes it is true that this node does not see any feedback signal (Rs=0).
However, what about the common node of both resistors in your last figure ?
If these resistors are existent they are part of the input circuitry (and such a resistive path is required if we need an input coupling capacitor). Without the feedback capacitor (see your figure) the voltage at this node would be 50% of the input signal voltage. However, this capacitor forces the voltage to be identical to the output voltage (identical to the input voltage). Hence, the output voltage is fed back to the input circuitry without signal inversion - and this effect is called "positive feedback". .
Lutz, thank you for your understanding of my lengthy exposure... Maybe we should ask a separate question like, "What is actually 'bootstrapping'?"... where to copy the last comments from this question....? Below I will answer successively to your questions and will put some new ones...
Of course, "resistance" is a quantiry, not an element... To say it more correctly, I have named it figuratively "internal differential resistor"... So, this "resistor" internally connects the two op-amp input nodes... it is what is bootstrapped by the help of the series negative feedback...
In your example of an emitter follower, what is internally connected between the external base and emitter pins, is externally bootstrapped...
Your last thoughts about the biasing resistive network are extremely interesting and important... and I will try to answer them in a worthily way...
Well, I will use a more unusual way to comment them - instead to answer, I will ask you another question...
As you have said, the two resistors in series provide a DC path to ground... needed both for the non-inverting input (to not saturate the op-amp) and the input capacitor (to charge). As though, we have split a single resistor into two parts and AC connected (via a capacitor) the common node between them with the op-amp output.
My question (suggestion) is, "Can we remove the 'redundant' lower resistor and the capacitor... and connect the single upper resistor directly to the op-amp output?"
It seems this arrangement should work since there is a DC path (through the op-amp output to ground)... and the voltage at the lower resistor terminal will be equal to the voltage at its upper terminal... so it will be bootstrapped...
Cyril - of course, we can connect the upper of the two resistors to the inv. input - but WHY? We will have no "bootstrap" effect. In contrary - the total input resistor will be lowered!
My comment to:
So, this "resistor" internally connects the two op-amp input nodes.
I think, this is not true. It is a kind of "logical conflict". The two input nodes are the base (gate) terminals of the first diff. amplifier. Hence, the internal diff. input resistance cannot be removed from these terminals. It belongs to the whole circuitry which is defined by these two terminals.
However, the precondition for "bootstrapping" is the existence of a working circuit having two nodes which - in a second step - may be connected using an additional "bootstrap" element which has the task to tie these nodes together. But these nodes must exist before they can be "bootstrapped". And this is NOT the case for a "bootstrapped" base-emitter resistance. Without both nodes the internal resistance does not exist - and vice versa.
(I hope I was able to express myself clear enough).
In this context, one should remember that we can have feedback (a) which influences gain, input and output impedances or (b) which influences the output impedance only (cascode) or (c) which influences the input impedance only (bootstrapping).
Lutz,
Regarding the simplified bootstrapping arrangement... I have suggested to connect the upper resistor not to the inverting input... but to the op-amp output:) To avoid the appearance that this is the same thing, insert a resistor with the same resistance between the output and the inverting input (to equalize the voltage drops due to the input bias current).
Now, it seems, the upper resistor will be bootstrapped since the voltage of its lower terminal will be equal to the voltage of its upper terminal. My question was, "Will this arrangement work properly?"
Now about the "differential input resistor". I mean the convential op-amp parameter as it is explained in data sheets (e.g., on Page 31 of the attached link below): "Differential input resistance, rid, is the small-signal resistance between two ungrounded input terminals."
I have not suggested to remove this equivalent "resistor"... let it stays there as "connected" between the two input terminals...
Let's consider in more detail the example of the "bootstrapped base-emitter junction". When you apply the input voltage to the base... and the emitter is grounded (common-emitter configuration)... the p-n junction has relatively low resistance.... since the emitter voltage is firmly fixed.
Now, let's apply the clever "emitter degeneration" trick by inserting a resistor between the emitter and ground. The emitter is released and its voltage begins following the input base voltage... that means the p-n junction is bootstrapped... As though, the transistor itself "moves" the "ground" of its emitter... this is a kind of a "movable"... "following ground"...
I do not see a "logical conflict" here. The base-emitter junction (or the op-amp differential input) is inherently bootstrapped... and we cannot (and do not want to) remove this bootstrapping. Then, if we want, in addition to the existing bootstrapping, we can bootstrap an additional external element...
http://www.ti.com/lit/an/sloa011/sloa011.pdf
And what about frequency dependent bootstrap :-)
For example:
https://www.researchgate.net/publication/283123227_The_fourth_journey_from_low_input_voltage_BP_to_High-input-voltage_BP_filters_-_Bootstrap_Band_Pass_filters
Research The fourth journey: from “low input voltage BP” to "High-inp...
Josef,
Interesting concept.
I have done some minor experimenting with the idea,
related to the effects of varying the input signal
and thus altering the "effective" "Q" of the BP filter
...
but the Boot-Strapping positive-feedback
method is so very well developed by you.
I appreciate your careful observation
of the "BandPass Boot-Strap" effect on the Buffer OPA ,
and your excellent math documentation of that observation.
I see your circuit as a Novel Method
which achieves results by Filtering a Positive Feedback,
and this is very much backwards to my thinking.
I would ( ONLY ) suggest the use of a
MultipleFeedback BP filter in the positive feedback loop,
which would allow a single resistor to adjust f(0) ,
possibly followed by a buffer ( with shunt limiter )
to maintain phase control in the positive-feedback loop.
Now, I must go back and read in greater detail,
as I am cautious about stability ( No Oscillations )
if signals rise too high.
Josef, thank you.
Stability issues are resolved in
https://www.researchgate.net/publication/281625430_Zapojeni_pasmove_propusti_s_vazbou_typu_bootstrap
Czech only :-(
Article Zapojení pásmové propusti s vazbou typu bootstrap
Dear Cyril - I was afraid that I couldn`t express myself clear enough and seemingly I was right.
Quote: "I have not suggested to remove this equivalent "resistor"... let it stays there as "connected" between the two input terminals..."
If I remember correctly, I never have said that it was your suggestion to "remove" this resistance. In contrary, the resistance between base and emitter cannot exist without both nodes and vice versa: The nodes are physically defined by the B-E path and the corresponding resistance in between.
But the precondition for "bootstrapping" is the existence of two nodes within a working linear system which then - in another step - can be tied together via a "bootstrap" element. This is my understanding of "bootstrapping".
Quote: "Let's consider in more detail the example of the "bootstrapped base-emitter junction". When you apply the input voltage to the base... and the emitter is grounded (common-emitter configuration)... the p-n junction has relatively low resistance.... since the emitter voltage is firmly fixed.
Now, let's apply the clever "emitter degeneration" trick by inserting a resistor between the emitter and ground. The emitter is released and its voltage begins following the input base voltage... that means the p-n junction is bootstrapped.."
Cyril, here you again have described how emitter degeneration (negative feedback) works - and finally, you simply claim: "...the p-n junction is bootstrapped".
Sorry, but I can only repeat: For my opinion, this definition is not in accordance with the commonly accepted definition of bootstrapping.
Dear Josef - I like to express my deep interest in you paper on "Bootstrap Bandpass Filters". Of course, the basic principle is known: Positive feedback reduces bandwidth. For example, this principle is used also for the well-known positive-feedback Sallen-Key structures.
I am involved in Filter Design since many years (and I feel - more or less - somewhat experienced in this area). However, I must admit that I have never seen the bandpass structures as described in your paper. Very interesting. At first look I have seen a certain relationship to the "Primary Resonator Block" principle, where we have also one or two second-order low-Q filter blocks connected in a closed loop with an amplifier. However, a somewhat deeper analysis has revealed some basic differences between both concepts.
Finally, are you the inventor of this principle or is it described already elsewhere?
Thank you
Lutz
Hi Lutz
In 1987, I thought it was my discovery.
So - maybe yes :-)
Josef
Josef - I think, you have forgotten to mention in your paper another interesting property:
Your circuit belongs to the very few bandpass topologies which allows single-element tuning of the bandwidth (Q factor) without influencing center frequency and mid-band gain.
More than that, for Q approaching infinity (RB=0) it is an oscillator.
Hi Lutz
I was browsing through my old notes, first experiments with this structure I made at the beginning of 1985. "I played" with all pass filters - using a method: "what will be if" :-))
One of the trials (from 18 trials) "gave me" "bootstrap BP" :-)
Josef
...using a method: "what will be if"
Hi Josef - In this context, I remember William Shockley who has used the expression “creative-failure methodology” to describe his „what happens if...“ work.
More than that, reading your paper, the term „Q-multiplication“ came into my mind - and I remembered that I had read something similar before. Yesterday I have found it: In „Sergio Franco: Design with Operational Amplifiers...“.
In this book (2nd Ed., Fig. P3.28, page 158) there is a circuit which contains an amplifier (inverting) and a classical one-opamp-bandpass (MFB, inverting) in a positive closed loop - a structure principle, which is similar to Fig.3 of your paper.
Hi Lutz
a similar problem was investigated in my paper in 1990:
https://www.researchgate.net/publication/281625255_Pasmova_propust_se_snadnou_zmenou_cinitele_jakosti
It is obvious that the origin of circuit "is" before 1978 - see reference [1] in the paper.
Josef
PS
Sorry, in Czech only
Article Pásmová propust se snadnou změnou činitele jakosti
Interestingly, in Sergio Franco`s book we have - in principle - the same topology as shown in your last reference. However, both blocks are interchanged and the input is at the opamp inverter.
Thanky you for the link.
Lutz
Lutz
not at all, but a design of this type filter is somewhat uncomfortable :-(
Josef
Josef,
I did a quick SPICE circuit of my suggestion,
It is just addendum to your proposed circuit
about Positive Feedback and Boot-Strap ideas.
Also, I added more notes to the SPICE diagrams.
The Bode plot shows that analog filters
of this simple approach
will have wide skirts, with gausian shape.
Much more complex circuits can achieve
near DSP quality.
These Bode plots show the ability to shift f(0)
via a Single Resistor.
The versatility of the Inverting MFB filter is evident .
Transient plot shows effect of Q=7
on a Noise Signal Input.
The 700 Hz portion of the Noise is evident,
although it is small.
.
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