I imagine that the long-tailed pair was invented in several steps:

1) We want an amplfier that is able to amplify voltage differences. Of course, we think of a BJT (transistor T2) because the base-emitter path provides such a differential input;

2.) Disadvantage: A very low input resistance at the emitter node of T2 (if compared with the mid-range value at the base node);

3.) Solution: Use of an emitter follower (transistor T1) as a first stage - connected through a coupling capacitor C to the emitter of T2;

4.) Thus, we have to arms of a bridge - connected with a C as a bridge element.

5.) When both transistor stages are symmetrical, the DC potentials at both emitter nodes are equal (balanced bridge) - and we are allowed to directly connect both emitter nodes (drop the coupling capacitor). As a result - we have the classical long-tailed pair with a common emitter resistor.

A very well fabricated circuit story:)

I also tell it to my students... indeed, up to the third paragraph (without introducing a capacitor)... I hope you do not forget to connect the emitter of T2 with a resistor to ground?

Also, "the DC potentials at both emitter nodes are equal" only in the case of equal input (base) voltages. If they differ, the emitter voltages (transistors) also try to differ. If their emitters were connected by some "bridge element" (like the capacitor above or simply a resistor), they will success... if not, they will tug... and will not success... but we can use their collector reactions (strains) as outputs...

I hope you do not forget to connect the emitter of T2 with a resistor to ground?

Cyril - thank you. I have modified step 2) correspondingly. Of course, you are right - T2 has an emitter resitor for allowing a DC current. Hence, as seen from the emitter of T2, we have a common base operation.

It seems the long-tailed pair is a very unusal bridge... especially when the emitter resistor is replaced by a current source...

For my opinion, it WAS a bridge before we have connected both emitter nodes.

Yes Lutz... then it was more bridge than now...

There was a question about this configuration (with unconnected emitters)... and we have compared it with the long-tailed pair... I will try to find out it...

OK, I have found it...

Another example of such a  "differential circuit" can be the input FET pair that I used in my diploma thesis, 1981...

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

Welcome respected colleagues Cyril and Lutz,

Bringing resemblance between the  things saves many efforts to study the new object. Simply, one can apply some acquired knowledge on the new object. The other way is to analyse each object as a separate entity. I think the last method is already used to work out the performance parameters of both circuits.

Coming now to the answer of the Cyril question, from my point of view.  Here i will strictly stick to the known topology of the bridge.

So, after inspecting the two circuits and mapping the elements of the differential amplifier circuit drawn in the question, one can make the mapping like that:

The pull up collector resistors of both transistors say RC1 and RC2 cab be mammped on R1 and R3 of the original bridge.  Then the elements connected directly to these resistors and having  a common point with hem are the transistors Q1 and Q2 . So, Q1 and Q2 represent the remaining two arms of the  bridge, they correspond to R2 and R4 respectively in the original bridge. This is very logical as the transistor between the collector and the emitter can act as  variable controlled resistor. It can be controlled by the base voltage. 

The source feeding the circuit similar to the bridge is VCC-VEE through the emitter tail resistance REE. This source can be substituted also by a current source. So, the bridge in its new form will be supplied by a current source. So, the remaining Branch is bridge branch where the detector is connected. It is now defined by the two collector points of the transistors C1 and C2. 

This is the new bridge from my point of view.

In summary, our new bridge is composed of two resistors, two transistors controlled by the base voltages , a current source and a  detector that is connected to the collectors of the two transistors.

This bridge can be balanced with VC2-VC2=0, OR can be unbalanced. The balancing process is normally accomplished by the base voltages.

Best wishes

And what about this pretty circuit :-)

Josef - for my opinion, the description as given by Abdelhalim still holds. The ohmic (static) resistors have been replaced by dynamic (differential) resistances of a current mirror. That`s all. Of course, the gain has increased considerably. But we are still able to measure corresponding voltage changes at both collector nodes.

Dear colleague Abdelhalim,

I think your nice desription clearly shows that the longtailed pair is (a) a balanced bridge for all common-mode signals and (b) an intentionally unbalanced bridge for all differential signals. Am I right?

Dear colleague Lutz,

Yes your predictions are in order. But now applying the conditions of balance of the bridge on our circuit, and assuming that RC1=RC2, Then the twp transistors are identical,and the same temperature, the bridge will be always balanced for the common mode input signals.

To disrupt the balance, one have to affect a difference between  the two input voltges between the two base of the transistors..

One can say the current distribution between the two transistors will be affected by the difference in their base voltage.

This is the first useful result  when considering the similarity of this circuit with the bridge.

Lutz - I think that situation is more complicated now. A collector voltage (signal) of the left transistor is almost constant while a collector voltage of the right transistor may be (is) very large.

Yes, Josef - I agree, there will be some kind of asymmetry.

A veritable cluster of ideas! I regret that now I have no opportunity to comment them... but I will in the coming days...

Regards,

Cyril

I begin commenting your posts...

I share your observation that the long-tailed pair can be considered as a current-supplied bridge circuit (by a bare resistor curent source - Abdelhalim, or by the more sophisticated transistor current source - Josef)... where "the pull up collector resistors of both transistors say RC1 and RC2 cab be mapped on R1 and R3 of the original bridge"... Also, R2 and R4 correspond to Q1 and Q2.

But after I have deepened into the essence of this configuration, I came to the conclusion that there is a significant difference between the classic resistor bridge circuit and the long-tailed pair when current-supplied:

• The whole classic bridge behaves as an equivalent ohmic resistor. So, when varying the supplied current, the (supply) voltage across this "resistor" varies as well.
• In contrast, the long-tailed pair (at constant input base voltages) behaves as an equivalent voltage-stabilizing resistor... voltage stabilizer... or zener diode... that keeps a constant supply voltage across this "zener diode"... This is because we can think of it as of two emitter followers in parallel that keep a constant emitter voltage when the current varies. So, when varying the supply current, the (supply) voltage across this "zener diode" does not vary.

The next situation is when we simultaneously vary, in the same direction, both resistances R2 and R4 (in the classic bridge)... and accordingly, both input base voltages of Q1 and Q2 (in the long-tailed pair)...

Seemengly, there is no difference between these circuits - in both, the lower resistances vary in the same direction... and as the current is kept constant by the supply current source... the voltage drops across them simultaneously vary in the same direction... and the differential output voltage stays constant (zero)...

But the mechanism of this action is different in the two circuits - in the classic bridge there is no feedback while in the long-tailed pair there is a negative feedback...

The final situation is when we simultaneously vary, now in opposite directions, both resistances R2 and R4 (in the classic brdge)... and accordingly, both input base voltages of Q1 and Q2 (in the long-tailed pair)...

Like above, seemengly, there is no difference between these circuits - in both, the lower resistances vary in opposite directions... and as the current is kept constant by the supply current source... the voltage drops across them simultaneously vary in opposite directions... and the differential output voltage vigorously varies...

But again the mechanism of this action is different in the two circuits - in the classic bridge there is no feedback while in the long-tailed pair there is a negative feedback. As a result, the long-tailed pair will vigorously switch between two states...

As a conclusion, I would say the differences in the operation between these circuits are caused by the fact that in the classic bridge the lower resistances are controlled independently... they do not mutually interact... while in the long-tailed pair, they interact because of the joined emitters.Thus in the long-tailed pair, the lower "resistances" (transistors) are controlled both from the side of bases and emitters.

These thoughts can clarify the role of the common emitter resistor (current source) in the long-tailed pair - it allows this interaction between transistors. If the emitters were grounded (as in the case of the classic voltage-supplied bridge circuit), they will be firmly fixed and separated from each other. In other words, the voltage of the common emitter node is cnahgeable... "soft"... "movable"... loose... and can be easily influenced by two sides...

But now I have realized that we not asked a separate question about the legendary long-tailed pair... we have only fragmentary comments about it in various other questions... at least, I have not found such. Am I right?

(... a continuation from Page 2...)

Another interesting topic of discussion (comparison) can be what happens when the supply voltage changes (even when the bridge is current-supplied, there is a voltage source V)...

There is a significant difference between the two circuits:

• In the classic bridge, since the lower elements are ohmic resistors R2 and R4, the two voltage dividers are linear devices whose output voltages VR2 and VR4 of the common nodes change proportionally to the voltage supply variations. Also, if the bridge is current-supplied, the bottom rail (the lower ends of R2 and R4) will vary.
• In the long-tailed pair, since the lower elements (transistors Q1 and Q2) are current-stabilizing, the two voltage dividers act as voltage followers. So, the collector voltages VQ1 and VQ2 of the common nodes follows voltage supply variations. The bottom rail (the common emitter node) will not vary since the emitter followers keep its voltage (almost) constant.

So, it is interesting that we can see in total three current "sources" in the long-tailed pair (such as the shown by Josef on Page 1) - one common transistor current source in the common emitters and two partial transistor current sources at the bottom of the "bridge".

The common emitter current source supplies the partial current sources with a total current; the differential input voltage distributes (steers) the current between the partial sources... 

By the way, the classic "ohmic bridge" can be also, for some reasons, current supplied...

Here is a "circuit story on the whiteboard" dedicated to such a solution. It was used as an example of a SPICE simulation... Years ago, I presented it in a more attractive way - as an 8-step "inventing scenario"... and dedicated to its author Rick Faehnrich. 

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

http://www.ecircuitcenter.com/Circuits/sbridge_i_bias/sbridge_i_bias.htm

Dear Cyril,

All these differences between the differential amplifier and the original bridge counted by you in your different presentations are okay. But i ask myself what are their usefulness for the explanation of the differential amplifier. I think after mapping the differential amplifier on the bridge configuration, one has to model the transistors as two branches of the bridge as well as the tail and the power supply voltage.

May be the most suitable model for the tail and power supply is a current source as nearly all colleagues agree on this model. So at the end effect, one has a bridge that is fed by a constant current source say IEE.

Now let us model the two transistors, as they must work as an amplifier they must be operated in the active region. Therefore, the transistors are modeled by voltage controlled current sources, with the controlling voltage is the base to emitter voltage of the transistor.

According to this model, the current IEE division between the two sides of the bridge, will be dictated by the difference between the base to emitter voltage of the two transistors assuming identical transistors.

Accordingly, If VB1=VB2, the current will be divided equally between the two branches such that IC1=IC2. To balance the bridge logically RC1=RC2, which means the pull up resistances of the two branches must be equal.  And this is  the balance condition where VC2-VC2=0.

On the other side, this condition can be found easily by applying the symmetry condition of the two branches of the bridges.

The other operating conditions of the differential amplifier, where VB1 IS NOT EQUAL VB2, the currents in the bridge side will not be equal and the bridge will become unbalanced such that VC2-VC1= Avd( VB1-VB2), with Avd is the differential voltage gain of the amplifier.

 Best wishes

Dear Abdelhalim,

I will leave you alone to answer your question about the usefulness of my observations above on the long-tailed pair seen as a bridge... and I will enrich my collection of 99 questions with another to make a total of 100 questions... in addition to my 2700 comments. Thus I will try to put a kind of record in the RG forum:)

My question will be about the essence of another abstract and not well understood circuit - the so-called gyrator. So, my question will be, "Can we present the gyrator as a bridge circuit?"

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

That seems me rather far-fetched :-)

Why not:)?

Cyril, this is no answer to your question - nevertheless: The passive R-L combination as shown by you is not the correct equivalent to the active circuit.

For the active circuit: Zin=(RL+sR*RL*C)/(1+sRL*C)

Congratulations dear colleague Cyril. Wish you more achievements.

Why not, Lutz? Compare the two drawings... they are the same (VS is a following voltage source... it follows VR)...

Obviously, the Zin as given in my former reply is not identical to the passive Zin=(R+sL).

Furthermore, it can be seen (visual inspection) that for very large frequencies the input resistance of the active circuit approaches Zin=R (in contrast to the passive circuit).

But Lutz it is not passive... it contains a varying voltage source (voltage follower)...

The only difference between the two versions is that in the "passive circuit" a conceptual follower is used... while in the active circuit it is replaced by a specific op-amp follower...

Cyril - are we speaking about the same circuits?

I am referring to (a) the shown active circuit (one opamp, one C, two R) and (b) the passice R-L circuit which should show the equivalence between both realizations (stating L=RL*R*C)). But they are not equivalent.

Dear coleagues,

If there is different visions for the same thing a third party openion can help. After analyzing  the op amp circuit  that is composed of unity gain amplifier and  an input connected via resistor RL connected to the inverting input and C-R divider to the noninverting input of the op amp unity gain amplifier, and found that the input impedance expression by Lutz is right. 

However i tried to resolve the issue further by splitting the the expression of the input impedance into a real pat and imaginary pert.

So, i found after some algebraic manipulations that;

Rin= Req+ jXeq,

with Req= RL( 1+w^2c^2RRL)/ (1+W^2C^2RL^2),

Xeq= wC(R-RL)RL/(1+W^2C^2RL^2),

It is clear that the reactance will be inductive if R>RL,and capacitive for R

OK Lutz, I have understood you... you were talking about the real LR circuit from the Wikipedia picture...

Abdelhalim, thanks for the competent analysis...

Quote Abdelhalim: "Since RL will be the inductor resistance, then it must be made much smaller than R and the expression for the Leq reduces to: Leq = wCRRL"

I think, an exact condition for the applicability of the simple R-L eqivalence can be derived directly from the correct expression:

Zin=(RL+sR*RL*C)/(1+sRL*C).

When RL*wC

Dear  Lutz and Abdelhalim,

I have not understood whether your formulas answered the question, "Is a gyrator a bridge circuit?"

Dear Cyril,

The formulas proved the equivalence between the the op amp circuit and the attached equivalent circuit in your question. It worked out the conditions of the validity of this equivalent circuit. That is when the the op amp circuit works as a gyrator. There was some doubt raised by Lutz about the equivalence of the two circuits in he question.

I think one can move to answer the resemblance of the gyrator circuit in your question with the bridge circuit.

I would propose that one has to give his answer in the new link of the question

Best wishes

I think that this circuit is a "bootstrapped RL" - frequency dependent via C and R and follower with the OPA :-))

I need time to realize this wisdom, Josef...

Quote Cyril: I have not understood whether your formulas answered the question, "Is a gyrator a bridge circuit?"

Cyril - of course, you are right. My contribution (formulas) couldn`t serve as answer - it was only a comment (correction) on the equivalence between two circuits shown in your text.

Here is my answer: Because each opamp with feedback (operating in the linear region) has a differential input voltage close to zero, we could formally compare such active circuits often with a (nearly) tuned bridge.

Most logical example: Opamp with positive and negative feedback (in particular, some oscillator circuits). But I am not sure if for all opamp-based circuits such a  view would be helpful.

I begin commenting your speculations...

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?":)

We know it is difficult to answer such a "quality question" by using quantitative analysis. It is much easier to do this qualitatively if we visually compare the two topologies...