Dear Cyril,

I will try to give an answer to this fine question. As a first step i will form the bridge configuration by mapping the circuit in the hand with a classical bridge configuration.

The type of this bridge is an ac bridge, fed by a voltage source connected by the input and the ground terminal. The balance port is between the inverting and the non inverting op amp input. From the characteristics of the op amp the the difference, the input difference voltage is nearly zero signifying near balance condition.  So,one of the half bridge consists of the capacitor C and the resistor R. The other half will consist of RL and an equivalent unknown branch to the ground, say having impedance Zx.

As the bridge is nearly is balanced then one can apply the bridge balance condition for ac briges. That is z1 z4= Z2 Z3, z1= 1/jwc, Z2= R, Z3= RL and Z4= Zx,,

It follows that Zx=jwC R RL= jwLx, where Lx is the unknown inductance,

Then Lx= CRRl,

This result means the port between the output of the op amp and the ground equivalent to an inductance Lx= CRRL.,

I t follows that the equivalent circuit given by Cyril is for the bridge half consisting of the Rl and the equivalent inductance.

Looking at the circuit from the input port  and representing it as a whole , then the side of the C-R must be with higher impedance than the R-Lx side. In this case the circuit analysis made previously is very expressive.

Thank you Cyril. That is the first time that i attack this circuit as a bridge. It is really much easier to analyse.

Best wishes

(copied from the related question about the long-tailed pair)

I begin commenting your speculations on the previous and this question...

Josef, regarding your guess "that this circuit is a "bootstrapped RL" - frequency dependent via C and R and follower with the OPA" I think that it would be so if RL would be connected between the common point of C and R... and the op-amp output (as far as I know, "bootstrapping" means to keep the voltage of the one end of a 2-terminal element equal to the voltage of the other end).

Abdelhalim and Lutz, I see that you have sensed the subtle joke when asking whether your formulas answered the question, "Is a gyrator a bridge circuit?"... and yet Abdelhalim managed to do it here "quantitatively":)

In my opinion, it is more difficult to answer such a "qualitative question" by using a quantitative analysis. It is much easier to do this qualitatively if we visually compare the two topologies...

Dear CYRIL,

IN MY ANSWER ABOVE , I MIXED THE QUALITATIVE AND QUANTITATIVE ANALYSIS..I THOUGHT THAT YOU WILL CONSIDER IT A MODEL ANSWER TO YOUR QUESTION. 

Best wishes

Yes, Abdelhalim... I have noted it above...

I continue thinking on this odd "balanced bridge"... and I should admit that this is a very strange way of balancing a bridge....

What puzzles me is that this bridge can be balanced without applying a negative feedback...

The classic bridge is manually balanced by means of a negative feedback... it is a kind of а manually-controlled servo system:) In a NIC, an op-amp does this donkey work:) by means of a dominating negative feedback.

Here, a voltage follower more precisely does this job... But a voltage follower can be implemented without applying a negative feedback... as a humble amplifier having a gain of 1...

So, what is confusing for me, is that this bridge is somehow automatically (mechanically, instantly, forcibly) balanced by copying the voltage from the left side to the right side.

Dear Cyril ,

It is the op amp here also that undertakes the automatic bridge balancing. By definition the negative feedback factor for a non inverting op amp configuration with  a feedback potentiometric divider say Rf and RL is =RL/ (Rf+RL)= 1 with Rf is equal to zero as in the op amp voltage follower. The feed back is negative 1. So there is a negative heavy feed back. The voltage follower is characterized by the heaviest negative feedback.

Here the op amp voltage follower is a differential unity gain amplifier. It is a non inverting amplifier with a negative feed back factor equals one. This makes any change in the non inverting input voltage will be inherited to inverting input by the feedback. So, so long the amplifier is linear, it will react correctly to balance the bridge by making tracking between its two terminals the + and the -.

This may shed light  how the bridge is  automatically balanced.

Best wishes

Abdelhalim, maybe the confusing fact is that here the bridge is balanced by changing directly the output voltage of the right side. This voltage does not depend directly on the supply voltage... it depends on the output voltage of the left side... which, in its turn, depends on the supply voltage...

In a true bridge circuit, both output voltages are produced by a common supply voltage... they both depend on this voltage...and, as a result, the supply variations are suppressed...

So, I would ask you what happens if we remove RL?. Will be this configuration a bridge as before? Since, as before, it will be well balanced... but one of its branch (RL) will be absent...

From this viewpoint, another question can be asked, "Is the op-amp follower a balanced bridge?" Since, its output voltage follows the input voltage and there is a " bridge" (the op-amp differential input) between them...

All these questions belong to the question below about the general philosophy of the balanced bridge.

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

Dear Cyril,

You are a great interactive interviewer. You are a great question synthesizer.

The conceptual answer on your question depends on our ability to map the new active circuits on the classical bridge circuit.  So, we preserve the mapping as made before while considering the changes of the values of the elements of the branches as special cases.  As for the gyrator circuit with RL=0,the bridge will be still balanced and the input current supplied by Vi will be zero. Applying the balance conditions one gets a zero value for the simulated inductance. One important conclusion is that one can not produce between the input port and ground an inductor with self resistance RL=0.

As for the other question, the judgement would be based on the comparison with the conventional bridge. Such a bridge has four branches , one feeding source and voltage detector. May be one may add a balancing mechanism. Based on this concept, one can consider that the op amp voltage follower an combined voltage detector and a an automatic balancing mechanism. And this exactly its role in the gyrator circuit brought by you.

Best wishes

Thanks, Abdelhalim! Greetings from the laboratory on Semiconductor devices where I am preparing the lab about diodes:)

The profound comments of Abdelhalim have awakened my interest in this bizarre circuit... I again began to think about it... and finally, by the help of the "bridge viewpoint", I completely realized the brilliant idea of this "simulated inductor"...

Indeed, I would call it more successfully "virtual inductor"... and will explain why below... Let's look again at the conceptual circuit diagram where the op-amp is replaced by the following voltage source VS.

In my opinion, this idea is based on the remarkable property of the balanced bridge that the voltage drops across the opposite branches are equal (here, VRL = VC and VS = VR). This is achieved by copying the picture of the voltage distribution across the elements of the left bridge side to the right side.

More specifically, the lower right voltage (the voltage VS of an aditional ideal voltage source) is adjusted equal to the lower left voltage (the voltage drop VR across the resistor R)... i.e. VS = VR. As a consequence, their complements (to the supply, i.e. input voltage) are also equal... i.e. VRL = VC.

As a result, on the left side we have a real CR circuit... and on the right side - an inverse copy of this circuit, i.e. a virtual RL circuit... and the correspondence between the individual components is as follows:

• The "copy resistor" RL (acting as a voltage-to-current converter) is an inverse copy of the "original resistor R" (acting as a current-to-voltage converter)
• The ideal voltage source VS is an inverse copy of the "original capacitor"... i.e. an "ideal virtual inductor"

Thus, the whole combination of VS and RL acts as a "real virtual inductor"... i.e. an inductor with losses...

Dear Cyril,

Welcome,

Well, this what i waited for to add another way of  analyzing the circuit to become glass clear.

Excuse me to make some developments to your bridge in the above comment. According to our analysis, one has to add the fourth element which you termed virtual inductor as a fourth dotted passive element between the negative terminal of the op amp and the ground. As you said the current in this virtual inductor will be equal to the current in RL and the voltage across it will be be equal to the voltage across R, the series resistance with the capacitor in the other side of the bridge. All this is a consequence of the balanced bridge model.

Now let us get the the value the inductor by means of simple circuit analysis:

The voltage across the indutor VL= V across R= iC R= RC dVc/dt= RCd( iL RL)/dt

= RRLC d iL/dt= L diL/dt, then L= RRLC, which is the same result from the impedance relationship of the bridge.,

Here one used that the current in a capacitor= CdVc/dt, and VL= ldiL/dt.

This means looking at the circuit between the inverting input and ground it is equivalent to an inductance with a value =R. RL C. 

Best wishes

After Abdelhalim has convinced us, "by means of simple circuit analysis", that in the lower right branch an "inductor" is connected, I will continue reasoning what is this "inductor"... and how it is created...

In the first question below, I have revealed... and posed for a discussion... how we can create artificial, "virtual" elements, by replacing them with varying (dynamic) voltage sources:

• If the source is connected in the same direction with the input source, it will add power (voltage) to the circuit... and this "helping source" will act as an active, negative element (resistor, capacitor, inductor) - see the second question.
• But with the same success we can connect it in the opposite direction. Then this "opposing source" will subtract power (voltage) from the circuit... and  will act as a humble passive element (resistor, capacitor, inductor)... and this is our case...

So, in this circuit, we have emulated a real inductor by connecting an opposing voltage source and varying its voltage through time in the same manner as in the case of the real inductor...

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

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

I continue reasoning about the unique properties of this "virtual inductor"...

A real inductor always has some internal resistance RL... that is distributed inside the coil... and cannot be separated from the inductance L. So, we cannot investigate the "pure inductance"... it is always "mixed" with some resistance...

Here we have the unique opportunity to investigate the "pure virtual inductance"... since the internal resistance is separated as an individual element - the resistor RL. For this purpose, we can observe the voltage across the "virtual coil" (the voltage of the emulating voltage source VS)... and the current through the "virtual coil" (the voltage drop across the "internal resistor" RL).

We can observe both by a clever setup (simple but perfect) where the ground of a floating scope is connected to the middle point between RL and VS... and its probes - to the upper and bottom rails. I use this brilliant idea in the laboratory to investigate in an elegant way various semiconductor diodes (the picture below is from The Art of Electronics Student Manual by Tom Hayes).

Dear Cyril, 

most coil windings present, also, a distributed Cap. Cwindings (shown more clearly at higher Freq.).

Note on : A real inductor always has some internal resistance RL... that is distributed inside the coil... and cannot be separated from the inductance L. So, we cannot investigate the "pure inductance"... it is always "mixed" with some resistance...

what is the easiest current sensor for smart meter design and fabrication

Maybe you should ask a special question about it?

Well, the "easiest" current sensor can be a humble resistor acting as a passive current-to-voltage converter (Ohm's law, written in a form of V = I*R). All kinds of digital multimeters exploit this idea.

You can make it "smarter" if add an op-amp connected in a circuit of an active current-to-voltage converter (transimpedance amplifier). If you want to grasp the basic idea of this solution, look at the links below.

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

http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/transimpedance_amplifier/transimpedance_amplifier.htm