We can enlarge the powerful idea of the “ideal” op-amp ammeter already discussed in the question below by replacing the movement with other current-driven loads (2-terminal elements like LEDs, solenoids, motors, rechargeable batteries, etc.)
https://www.researchgate.net/post/How_do_we_improve_the_real_ammeter_How_do_we_create_an_almost_ideal_ammeter_What_does_the_op-amp_really_do_in_the_circuit_of_an_op-amp_ammeter?
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
http://www.circuit-fantasia.com/circuit_stories/understanding_circuits/current_source/inv_op-amp_current_source/inv_op-amp_current_source.htm
But all these op-circuits have no (electrical) output; their outputs are light (LED), displacement (solenoids, motors...) or none (rechargeable 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 question is, “Where do we take an electrical output from?”
Let’s solve this problem by building the ubiquitous transimpedance amplifier. Although we can do it by adding only one additional step to the 4-step ammeter building scenario, let’s for completeness do it from the very beginning.
1. NO PROBLEM. Imagine a current I = V/R flows in the simple Ohm's circuit... and it really is I = V/R... but we want to see if this is true. There is only one way to do it - by measuring the current.
2. THE PROBLEM. Contemporary instruments measure voltages; so, if we want to measure a current, we have to convert it into a voltage. For this purpose, we break the loop, connect a resistor and measure the voltage drop across it... but a problem appears. The voltage drop is desired for us and we want to be as much as possible higher to obtain a maximum "gain" (a minimum discrete error of the next ADC); so we need a maximum resistance. But, at the same time, this voltage drop disturbs the input (imperfect) current source; so to obtain a minimum error, we have to keep minimum resistance. If we measured the current of a perfect current source, there was no problem since the source would compensate the "undesired" voltage drop by increasing its own internal voltage... but in our case the source is imperfect. What do we do then?
3. THE REMEDY. We already know the remedy (from the question about the ammeter) - we can remove the resistance by an equal “anti-resistance”... or, in other words, the voltage drop by an equal (electromotive) "anti-voltage"... So, we add a humble variable voltage source in series with the resistor and adjust its voltage equal to the adverse voltage drop. As a result, the voltage drop and respectively the resistance as though disappear... and the whole circuit looks as a “piece of wire” to the input source. The additional compensation source actually "helps" the input excitation voltage source, which creates the input current (note that the two voltage sources are connected in series, in one and the same direction "- +, - +" so that their voltages are added).
4. THE OUTPUT. But where do we take the output from? Actually, the voltage drop across the resistor exists but it is not a good idea to use it as a circuit output (first, it is “floating”; second, the next stage may disturb it). Then? The clever idea is to use the compensating (op-amp) voltage as a "mirror" output (just like in life when we estimate indirectly some positive quantity by an equivalent negative "antiquantity"). All the circuits with parallel negative feedback (i.e., all the inverting op-amp circuits) exploit this idea. The advantages: first, the load is grounded; second, it consumes energy from the compensating voltage source instead from the input source (so, the load can be low-resistive enough). A disadvantage may be the inverted output...
5. THE IMPLEMENTATION. Finally, we have just to replace the manual "op-amp" by a real one and will obtain the so desired “ideal” current-to-voltage converter (transimpedance amplifier) well known from the classic electronics books.
This story about the transimpedance amplifier was the continuation of the previous story about the simpler op-amp ammeter. What do you think about it? Have you ever seen such an explanation before? Is it reliable? Is it useful for understanding compared with the conventional virtual ground explanation? If you like it, visit the links below and enjoy.
https://www.researchgate.net/publication/255696892_Transimpedance_amplifier_(my_comment_to_Bob_Peases_EDN_article)?
https://www.researchgate.net/publication/254864969_How_I_Revealed_the_Secret_of_Parallel_Negative_Feedback_Circuits_(a_circuit_story)? (a fancy version of this story)
http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/transimpedance_amplifier/transimpedance_amplifier.htm
http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/active_i-to-v_converter/active_i-to-v_converter.htm
http://en.wikibooks.org/wiki/Circuit_Idea/Op-amp_Inverting_Current-to-Voltage_Converter
http://en.wikibooks.org/wiki/Circuit_Idea/Op-amp_Inverting_Current-to-Voltage_Converter_Visualized
http://en.wikipedia.org/wiki/Current-to-voltage_converter
http://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation
And yet, is there any connection between the humble resistor and the transimpedance amplifier? What does the op-amp really do in this electronic circuit? Can we say the op-amp is a "negative resistor" and the transimpedance amplifier is a "zero resistor"? Or, even more figuratively, the transimpedance amplifier is a "neutralized positive resistor" (R - R = 0)? If so, it would be a good prelude to the subject of negative resistor... and we can continue this story...
Data How I Revealed the Secret of Parallel Negative Feedback Circ...
To arouse an interest in the subject, I would ask a little "provocative" question at the very beginning:
By adding a voltage that is less and equal to the voltage drop across the resistor, we have obtained a virtual resistor with decreased and zero resistance. But what will happen if we add a voltage that is higher than the voltage drop across the resistor?
Regards, Cyril
Dear Sir,
If we are adding voltage that is higher than the voltage across the resistor, it should follow kirchoffs law and obviously the voltage drop across the resistance would decrease accordingly. And hence there would be some output voltage in this case. And hence no Zero resistance. It would give some current, corresponding to difference in the voltages
Hi Parikshit,
Thank you for the reply (it happens so rarely these days, so I have to thank God when he sent me a contributor to my question:-)
To clarify the question, I have attached a picture where the additional voltage (-2Vr) produced by the variable "battery" Bh is two times higher than the voltage drop Vr across the resistor R (both the elements are connected in series inside the blue "balloon"). There is also an input voltage source Vin and a load Rl.
Note that, travelling along the loop, you can see that the additional voltage and the input voltage are connected in the same direction so that they are summed; the current and respectively the voltage drops across Ri and R, increase as well... Can you guess now what the phenomenon is? If not, visit the page below; it may help you.
http://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance
Regards, Cyril
Dear sir, That would give you a negative resistance, for an exmaple Resonant tunneling diode(RTD), which exhibits an negative resistance, wherein increasing the voltage would decrease the resistance.
Here is another interesting question somehow related to the transimpedance amplifier:
https://www.researchgate.net/post/What_is_electronics_Wikipedia_inside_and_what_are_Wikipedians_inhabitting_this_area_Can_we_beleive_it_can_we_rely_on_them_and_to_what_extent
Two days ago, I (Circuit dreamer) began discussing how to present circuits, particularly the transimpedance amplifier, in Wikipedia arguing with my main opponent (Zen-in):
https://en.wikipedia.org/wiki/Talk:Negative_impedance_converter#Questions_on_this_circuit_description (the beginning)
https://en.wikipedia.org/wiki/User_talk:Circuit_dreamer#A_discussion_about_NIC_and_TIA (the continuation)
Perhaps this discussion will awaken your interest in the famous transimpedance amplifier?
There...
https://en.wikipedia.org/wiki/User_talk:Circuit_dreamer#What_transimpedance_amplifier_is
...I have suggested to my opponent to put himself in the place of someone who does not know what "transimpedance amplifier" is and, to understand, visits his new Wikipedia article:
https://en.wikipedia.org/wiki/Transimpedance_amplifier
Seeing the picture at the top of the article, he/she will ask him/herself: "What is this resistor? What is its function? Why is it needed? What is this operational amplifier? What does it do? Why the resistor is connected exactly in this way - between the op-amp output and the inverting input?" The author has not answered these questions. He has said in the beginning "a transimpedance amplifier, (TIA) is a current to voltage converter". But it is quite possible that the visitor already knows (from the Ohm's law - V = I.R) that the simple resistor is such (but passive) current-to-voltage converter when passing the input current through it and taking the voltage drop across it as an output... and he/she discerns this resistor in his circuit diagram.
But has he told him/her that this resistor is exactly such a passive current-to-voltage converter? And that the attached op-amp converts it into an active current-to-voltage converter? And that the op-amp does it in an extremely simple, clear and perspicuous way - by adding the same voltage V = I.R which is lost across the resistor? If he had done this, the reader would make the connection with the familiar simpler passive device and would grasp the meaning of this active circuit solution.
The author has not mention even a word about the virtual ground - the basic idea on which this circuit is based (even you have removed the link to this article). Instead, he has filled the whole introductory part with all kinds of photo applications and common phrases...
Then I have reassured him not to be so upset because there are no answers to these questions even in the introductory part of the famous Bob Pease's article:
http://electronicdesign.com/analog/whats-all-transimpedance-amplifier-stuff-anyhow-part-1
https://www.researchgate.net/publication/255961049_Transimpedance_amplifier_(my_comments_to_Bob_Peases_EDN_article)?ev=prf_pub
He has only said, "let's connect a feedback resistor across it, from the output to the -input". But why? What is the problem? How is it solved by adding the op-amp and connecting the resistor in this way? However, the article is great for readers who want to know in detail the problems with the bandwidth and stability... and this makes unnecessary much of the last section (about bandwith and stability)... Who cares will go in the Pease's article... the encyclopedia is not the place for such details...
Finally, I have asked him a question (although I know that he does not like this subject): "Do you see any connection between transimpedance amplifier and circuits with negative resistance, particularly the NIC?"
Data Transimpedance amplifier (my comments to Bob Pease’s EDN article)
Cyril - I think, one simple way to explain the role of RF in the circuit is based on the classical inverting amplifier.
For this purpose, we have nothing to do than to imagine that the current I(in) is caused by a series resistor R1 beween Vin and the inv. opamp input.
Thus (because of the virtual ground principle): I(in)=Vin/R1.
Now - if we set R1=RF we have Vout=-Vin=-I(in)/R1 and because of R1=RF
we arrive at
Vout=-I(in)*RF.
Yes Lutz, one of the possible (top down) way of TIA explanation is to think of it as of a component of an inverting amplifier; the other component is the input current source consisting of Vin and R1. I prefer the dual (down top) way - first to "invent" the TIA and then to build the inverting amplifier.
Now let's see if there is a connection between this transimpedance amplifier and circuits with negative resistance. IMO we can see at least two connections:
https://en.wikipedia.org/wiki/User_talk:Circuit_dreamer#The_connection_between_the_transimpedance_amplifier_and_circuits_with_negative_resistance
1. THE OP-AMP OF TIA IS A NEGATIVE "RESISTOR". The input current IIN flows through the resistor R and creates a voltage drop VR = IIN.R. The op-amp produces the same but negative voltage VOUT = - IIN.R that compensates the voltage drop VR and the result is the zero voltage (virtual ground) at the input. In other words, the circuit consists of two equal resistors connected in series: the first is positive with resistance VR/IIN; the second is negative with resistance -VR/IIN. They mutually neutralize and the result is zero resistance. Looking at the circuit from the side of the input source, it is just a point with zero voltage and resistance towards the ground; we can think of it as of a "neutralized positive resistor". So the first conclusion is:
"The op-amp of a transimpedance amplifier (and in all the op-amp inverting circuits with negative feedback) acts as a (current-controlled or S-shaped) negative resistor that neutralizes the positive resistor R; thus the whole circuit acts as a zero resistor."
2. THE VNIC IS A TIA WITH ADDITIONAL POSITIVE FEEDBACK. The transimpedance amplifier is a circuit with (only) a negative feedback. If we introduce additional positive feedback by connecting a voltage divider R1-R2 between the op-amp output and its non-inverting input (see the figure below), we actually increase the negative resistance of the op-amp; now it dominates over the positive resistance R one and the whole circuit acts as a (current-controlled) negative resistor (VNIC). So the second conclusion is:
"The negative impedance converter (VNIC) is a transimpedance amplifier with an additional positive feedback."
If we compare the two circuits (TIA and VNIC), we can make a final conclusion:
"TIA is a 'compensated resistor' while VNIC is an 'overcompensated resistor'; so, as a whole, TIA is a 'zero resistor' while VNIC is a 'negative resistor' ".
Do you agree with these findings?
@Cyril, of course, we do agree! Happy new Year and wish You the best as to all of our colleagues!