Looks like I'll have myself to answer the question...

Hi Cyril,

I can't understand why you give up the concept of virtual ground... it is so simple... In a classic textbook they would explain it quite simply: the op-amp has extremely high gain A... its output voltage Vout is finite... so the voltage between the op-amp inputs is extremely low (Vout/A → 0)... and the voltage of the inverting input is almost equal to the voltage of the ground... therefore this is a virtual ground... and the rest is easy...

Regards,

Mechkov:-)

Hi Mechkov,

If you say "virtual ground", you have to have a real ground, right? But if there is no real ground, how do we say that it is a "virtual ground"? "Virtual" means a copy of an "original"... but what do we do if there is no such an original?

Regards,

Cyril:-)

Hi Cyril,

Hmmm... there is always a ground... we have only to define what is a (real) ground... There may be no "global ground", but we can assign a "local ground"...

Regards,

Mechkov:-)

Hi Mechkov,

IMO "virtual ground" is a quite abstract term... it symbolizes a circuit point (node)... It is not clear where current flows and what the physical device behind the virtual ground point is...

Regards,

Cyril:-)

Hi Cyril,

Yes, I see... you move your focus from a point (virtual ground) to a branch (the network consisting of two elements in series - the ammeter and the op-amp output)... Thus you can show the loop where the current flows (input source -> ammeter -> op-amp output -> ground -> source)...

Regards,

Mechkov:-)

Mechkov, this network is very interesting... Is there any voltage across the ammeter? Yes, it is as it is an imperfect current meter; it has some resistance... Is there any voltage across the op-amp output? Yes, it is... and what is more interesting this voltage is equal and opposite to the voltage drop across the ammmeter... It is interesting that the resistance can vary... so the voltage drop will vary as well... and the op-amp voltage will vary staying always equal to it... Cyril:-)

Cyril, but this (the op-amp output) is exactly a negative resistor... Am I right? Is the op-amp output a negative resistor? See for example what is written in Wikibooks (skip Wikipedia, there will not answer this question):

http://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Current-driven_negative_impedance_elements

Regards, Mechkov:-)

Hi Cyril,

I must confess, I didn`t have time to go in detail through all the sites and pages you were referring to. But I understood that you have explained the circuit ("almost ideal ammeter") without using the concept of virtual ground. More than that, I understand that the ammeter would be ideal in case of an ideal opamp (infinite gain).

However, during the process of reading another question just came into my mind:

We could try to compare the circuit under discussion with another well-known concept, which is based on high-resistive measuring the voltage produced at a test resistor Rt in the milli-Ohm range. This also is something like an "almost ideal ammeter".

Question: For which value Rt are both circuits equal in performance as far as the systematic error introduced by the measuring device is concerned?

Yes Mechkov, you are right. The op-amp output behaves like a "current-driven true negative resistor" and even more... If it was a humble negative resistor, it would compensate only steady ammeter resistance (the negative resistor is not interested in the final zero voltage result)... Here the op-amp observes the final result - the voltage between the op-amp inputs, and keeps it zero even if it varies. This is an example of a full voltage compensation:

http://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation

Regards, Cyril

Hi Lutz, thanks for the support! I was just about to ask if someone famous circuit designer will join the discussion, but alas... аs usual we, educators, will have to do this "thankless" job...

Yes, I have explained the circuit without any mention of "virtual ground". I would not even mention the operational amplifier... and I would be content with just one small varying (copying, following) battery as shown in the attached picture... Now about your question... really, very interesting question... although it belongs to the next question about the transimpedance amplifier...

In the op-amp implementation, you probably mean the small voltage difference (error) between the two op-amp inputs due to the finite gain. Fortunately, here it is extremely low (Vout/A) since the op-amp output voltage is low (equal to the voltage drop across the ammeter). IMO to keep the same volage drop (at the same current) across the resistor Rt in the passive version, we need extremely low resistance and next high-gain amplifier... We have compared the two versions in the question below:

https://www.researchgate.net/post/Is_it_possible_to_use_a_Current_to_Voltage_Converter_as_a_non-inverting_amplifier

Here I would like to generalize the philosophy behind the two techniques. In the passive arrangement, we extremely decrease the losses into the resistor (in our case, this means to choose an ammeter with extremely low internal resistance) while in the active op-amp version we don't care about the losses leaving them to be all they want large... but then we destroy them with injecting equivalent energy...Analogical situations from our life are when we save money (a "passive" strategy) or we spend money and simultaneously earn the same money (an "active" strategy)... or when we isolate better windows (in winter) or simply include another heater in the room... or when we blew the tire while driving and we do not repair the but continue driving pumping it...

OK, that's a good discussion when we are already three (Cyril, Mechkov and Lutz:-) If Wangenheim joins the discussion, we will become four, and it will be even better:-)

Cyril, I want to ask another question. You have said above that "you I tried to adjust without any success the sensitivity of the ammeter by connecting in series to the movement a variable resistor". I don't understand why "in series"; as far as know, you should connect the resistor in parallel to the ammeter. Am I right? Mechkov

Yes Mechkov, you're always right... but do not forget that at that time I was freshman and I was just studying the classical instrumentation... Or maybe I had already heard what "virtual ground" was... and I saw it at the inverting input... Cyril

By the way, I remember, increasing the resistance of the resistor connected in series to the movement, I noted that putting too much resistance at one point the current started to decrease instead of staying constant... and it puzzled me more...

Then I remembered to look at what is happening in the amplifier output and what to see - the voltage had become almost equal to the power supply! Then I realized that, unfortunately, the good things in this world do not last indefinitely:-( and always comes the moment when they end...

... But what really amazed me, I remember, was that the magnitude of the voltage at the inverting input was also begun to change with increasing the resistance... and the virtual ground was no longer a virtual ground...

At that time (and now) I was greatly interested in the principles of inventiveness (de Bono, Altshuller, etc...); so I asked myself, "Can I use this harmful effect on something useful?" What do you think about this possibility?

Cyril, I have a question - and I apologize if you have covered it already in the various pages you have created (I suppose that was a very time-consuming task, was it not?).

But - as mentioned already - I didn`t go through all the pages up to now.

My desire is as follows:

Can you please show a simple circuit containing a voltage source that drives two arbitrary elements (active or passive, e.g. resistors) and the ammeter, which measures the current BETWEEN both elements?

Thank you.

Well, let's help you answer... if everything is fine, there is a virtual ground; if not - there is no virtual ground...

But now a new question came to me - what would happen if one connect a (perfect) voltage source directly to the input of this electronic ammeter (i.e., across to the input of the op-amp)? This is quite likely to happen, and every technician has done it at least once in his life (I - more than once:-) Or, as one friend of mine said, "if you leave the collector of a transistor open, there will always be a fool to connect it to +Vcc (n-p-n) or ground (p-n-p)".

It is clear what will happen if it was a humble movement ammeter - either the ammeter or the source will depart from this life... but what about the electronic ammeter?

Let's see where the current will flow to answer the question: input voltage source -> movement -> op-amp output -> power supply -> input voltage source. There is no any resistance somewhere in this loop -> the current will be unlimited -> the weakest element (most likely, the op-amp) will pass away (usually op-amps are ...

What do you suggest we do to prevent this from happening? Usually, operational amplifiers are protected against short connection to ground but what will happen if we apply (high enough) voltage to their output?

"There is no any resistance somewhere in this loop -> the current will be unlimited"

Cyril, this was not my question. Sorry.

My question was: How to measure the current BETWEEN two elements (like resistors) driven by a voltage source (circuit diagram)

Hi Lutz, the problem is that I cannot realize what exactly your question is. How are the two resistors connected - in parallel (current divider) or in series (voltage divider) and what does " the current BETWEEN two elements" mean (the currents flowing through the two legs of the current divider or something else? I am sure your question is clever but for now I cannot grasp what the idea behind it is... Regards, Cyril (have a nice trip)

Due to the strong interest of specialists and teachers from the fields of basic instrumentation and analog electronics to the original way in which the circuit of the "ideal" ammeter is explained by building, here's an animated version of this 4-step construction.

1. NO PROBLEM. A current flows through a loop...

2. A PROBLEM. To measure the current flowing through the circuit, we break it and connect a bare movement... but an undesired voltage drop appears across the ammeter... and the current decreases...

3. THE SOLUTION. We connect a humble variable voltage source in series with the movement and adjust its voltage equal to the adverse voltage drop... the voltage drop and the ammeter resistance "disappear"... and we obtain an almost ideal ammeter...

4. THE IMPLEMENTATION. Finally, we just replace the manual "op-amp" by a real one and thus we obtain a perfect op-amp ammeter...

5. THE FINAL STEP. Once we figured out the truth about the circuit, comes time to hide and keep to yourself the secret:-) So let's remove all "unnecessary" elements of the circuit diagram and "lengthy" explanations so that only a miserable "skeleton" to remain. Now we could get this and hundreds more such circuit diagrams in a kind of a "cookbook" of electronics :-)

We can enlarge the powerful idea of the “ideal” op-amp ammeter discussed here by replacing the movement with other current-driven loads (2-terminal elements like LEDs, solenoids, motors, rechargable batteries, etc.) Regardless of the device, the op-amp will always do the same – it will make its output voltage equal to the voltage drop across the corresponding device. As a result, the device “disappears” and the input current source “sees” just a “piece of wire”. We can think of this “wire” as of a kind of artificial “active superconductor” with zero resistance. We can also look at this phenomenon (aka “virtual ground”) from another perspective - shorting the input imperfect current source, we actually have provided it with ideal working conditions and it has began behaving as a perfect current source... and this is one of the possible ways of making perfect current sources.

But all these op-circuits have no (electrical) output; their outputs are light (LED), displacement (solenoids, motors...) or none (rechargable batteries). More frequently, we need circuits (converters) with electrical inputs and outputs; maybe the transimpedance amplifier is the most typical representative of this class of op-amp circuits. So, the next question should be about the ubiquitous transimpedance amplifier

https://www.researchgate.net/post/Is_there_any_connection_between_the_humble_resistor_and_the_transimpedance_amplifier_What_does_the_op-amp_really_do_in_this_electronic_circuit

This story would be an excellent continuation of the present story about the simpler op-amp ammeter...

Hi Cyril,

I think in this kind of problems, as in any possible engineering or life problem (not necessary in all scientific issues) we must keep cool and try to find the simplest way (sancta simplicitas). See https://en.wikipedia.org/wiki/Ockham%27s_razor the Ockham’s approach. I think if students are trained to see an AO amplifier as a control system (regulator) that is always trying to annul the differential input tension u+ - u- , and they perfectly understand how the input current is perfectly copied by the loop current, then they will easily understand how the AO’s output is automatically removing the undesirable effect of the ammeter’s internal resistance. Especially if you will associate this circuit with others like the repeater amplifier or the precision rectifier. I think trying to overload explanations is not really helping the majority of the students.

Best regards,

Marius

Sorry Marius, I saw just now your comment ... It is very interesting, I support it fully... even the last sentence ("trying to overload explanations is not really helping the majority of the students"). I know I am not a good (ordinary) teacher since I trouble good (ordinary) students with all these "overloaded explanations":) I am "bad" (extraordinary) teacher willing to teach a "bad" (super curious) students... But where to find them?

Now about your thought "...if students are trained to see an AO amplifier as a control system (regulator)..."

IMO one of the most powerful concepts of the op-amp is to think of it as of a sort of a "servo". Long ago, I have realized the paradoxical fact that although the op-amp is really a proportional ("non-inertial") device, in order to understand how op-amp circuits work, we have to think of it as of an integrating, inertial device what it actually is. Figuratively speaking, we have to think of the op-amp as of a "being" that "observes" continuously its input voltage and changes its output voltage until it manages to zero the input voltage. If we think of the op-amp in terms of UOUT = A.VIN (i.e., VIN and UOUT change simultaneously), we fall into a vicious circle traveling the feedback loop and we can never understand circuit operation. We have to assume that the output voltage stays late regarding the input voltage. More philosophically speaking, we have to see the causality in the op-amp operation; to think of the input voltage as of a cause and of the output voltage as of a result:

https://en.wikibooks.org/wiki/Talk:Circuit_Idea/Slowing_Circuit_Processes

Regards, Cyril

Yes - I agree that the opamp must be seen as an "inertial" device if we try to understand how it is possible to find a stable dc bias point after switch-on at t=0.

And a similar way of thinking is necessary to understand why/how an opamp based oscillator is able to start oscillations.

In this context, it is interesting to note that there are circuits which behave totally different during circuit simulation - depending on the opamp model used (with or without delay).

Dear Lutz, I know that you do not care about Wikipedia... but perhaps you might be interested to meet the sad (for Wikipedia and its administrators) fact that the guy who has rewritten entirely the Wikipedia article about transimpedance amplifier does not understand this extremely simple circuit:

https://en.wikipedia.org/wiki/User_talk:Circuit_dreamer#sad_fact

Here are his "brilliant" thoughts

"As far as putting a meter in the feedback path of an opamp being an invention you should try it and see what happens (ie: a latched opamp) before calling it an invention. You can't replace Rf of a transimpedance amplifier with a mA meter and get a useful circuit. Zen-in (talk) 4:54 am, Today (UTC+2)"

... and here is my answer at the bottom of my Wikipedia user talk page:

"A latched op-amp? Can you even realize how ridiculous is that said with respect to this as simple as possible circuit with negative feedback? Zen-in, only this text is enough to show that you do not at all understand what а transimpedance amplifier is... but you are the guy who has rewritten entirely the Wikipedia article about it... a sad fact about Wikipedia and its administrators... Circuit dreamer (talk, contribs, email) 7:40 am, Today (UTC+2)"

Have fun, Cyril