Dear Lutz,

welcome,

Hope you are well!The input resistance of the differential transistor amplifier seen from the base of one transistor to the ground in the common mode can be expressed by

Ri = rbe + 2REE ( hfe +1), where rbe is the base to emitter small signal resistance, hfe is the small signal common emitter current gain of the transistor and REE is the common emitter resistance.

It is very large resistance.

The derivation is simple because of symmetry in the common mode one can split REE into two separate emitter resistances with every resistance equals 2REE. Then one has an input resistance looking from the base of each transistor as expressed previously.

If in the differential mode the input resistance from a base of the transistor to the ground is rbe and from base to base is 2rbe as the common emitter point will be at virtual ground.

Best wishes

I assume Q1 and Q2 are bipolar transistors ?.

In which case,

Rin = Beta * (Vt/Itot) * 4, with

Beta: The current amplification of the transistors.

Vt = kT/q , about 26 mV at room

Itot = The total emittor current of the transistors. (So, the sum for the 2 transistors.)

Mr. Henry Cloetens - thank you for being the first to give an answer. Yes - in my question I have mentioned the "base" of Q1 and both transistors are BJT`s.

However, I must point to the fact that I did not specify any particular operational mode (symm. or unsymm. differential or any thing between). That means: I expect a general answer...perhaps depending on the mode of operation.

Your question is too general.

If your circuit is, as described, an emitter-coupled pair with a straight resistor at the junction of the two emitters (commercial op-amps will use a current source) then you need to specify the bias configuration, which will probably dominate the input resistance. However, assuming pure voltage source drive, you will have high input resistance in common mode (since the transistors look like emitter-followers) and low impedance in differential mode since, thanks to the virtual ground at the emitters, the transistors now look like common emitter amplifiers.

Hi Mark - thank you.

Yes, of course the input resistance will be different for both extreme cases. With other words: The input resistance at one input will depend on the voltage at the other input node.

And this is the background of my question: I want a formula (if such a relation does exist) which shows how the input resistance at Q1 (Q2) depends on the voltage at Q2(Q1).

I think, this question is not too general. Or - perhaps it must be answerd in two steps?

Quick'n'dirty answer:

Input resistance of an emitter follower (ignoring bias circuits) is approximately hFE*Re,

that of a common emitter amplifier (ignoring bias circuits, and assuming a 'stiff' current source in the emitter) is approximately 1/gm = 1/(40*Ic)

Mark - thanks, but I know the input resistances for the three basic configurations (with and without feedbak). By the way - I think, your last sentence contains a misunderstanding (common emitter with an emitter current source?).

An interesting and "philosophical" question... I share the first Mark's answer above and will enrich it with some additional considerations...

In the common mode, we can think of this configuration as of two emitter followers working "in cooperation" at a common load. Figuratively speaking, they "help" each other in the common task of changing the voltage across the load by increasing/decreasing the partial emitter current in the same direction. So, "looking" at the inputs of the differential amplifier, input sources "see" higher resistance than if they were "looking" at the input of a single emitter follower. Of course, this matters if a simple resistor is included in the emitters; in the case of a current source, the difference would not be noticeable.

In the differential mode, we can think of this configuration again as of two emitter followers working at a common load but now interfering with each other. Figuratively speaking, they "oppose" each other in the common task of changing the voltage across the load by increasing/decreasing the partial emitter current in the opposite direction. So, "looking" at the inputs of the differential amplifier, input sources "see" extremely low resistance as though the transistor emitters are grounded... and really they are virtually grounded. Thus both emitter followers act not as common-collector but rather as common-emitter stages; so Lutz, I think there is a sense in the last Mark's sentence ("common emitter with an emitter current source").

Now about the Mark's remark "you need to specify the bias configuration"...

Usually, differential amplifying stages are biased from the side of emitters by sinking/sourcing constant emitter current requiring corresponding base bias currents. They flow through input sources (hence they have to be galvanic).

So, in this case, there is no need of any external biasing circuits that would decrease the input resistance.

I guess one should clarify first that we are talking about differential resistance dI/dv (not I/V or something). Also, the most common case of differential amplifiers are opamps and they usually operate in deep negative feedback loop.

That dramatically changes their effective input impedance as seen by the signal source.

Then, one can say that for MOSFET input pair the input resistance is usually enormously high, for JFETs is not that high especially at high teperatures and for differential BJT can be estimated as R = 26mv /Ibase

Here 26mv is kT/e, Ibase = Ibias = Itransistor/beta

That estimate comes as derivative of the IV of the BJT base diode.

Still, it might be affected by base and emitter resistance, if significant.

Yes, we are talking about differential resistance... and we can think of the common emitter "source" as of a dynamic resistor with finite static resistance V/I and infinite differential resistance dv/di...

Cyril and Nikolay - thank you both for your reply.

Of course, I share your opinion and the differential input resistance, of course, depends on the particular operational mode.

THIS IS THE REAL PROBLEM BEHIND MY QUESTION.

Each textbook contains the input resistance at Q1 for three particular cases - which are simple to compute: (a) common mode, (b) unsymm. diff. mode and (c) symm. diff. mode.

Of, course, I have the formulas/expressions for these (simple) cases.

But in many cases, the diff. amplifier is operated in none of these three specific modes.

The question is (again): What is the diff. input resistance at the base of Q1 - in dependence on the voltage at the base of Q2. And the three mentioned cases (a)...(c) are only special cases of the general expression, which I am looking for.

* Cyril , I don`t think that this is a "philosophical" problem since in some applications it is really important to know the value for the input resistance at the input node (base of the BJT). We have input resistance formulas for all amplifier configurations - discrete ot integrated..... why shouldn`t we be interested in these values for diff. amplifiers in case of arbritrary input voltages (of course, small enough to be considered as "small-signal" operation)?

EDIT: I was able to simulate the problme. I have produced a graph that shows the input impedance at the base of Q1 as a function of the input voltage at the base of Q2. Starting with Vb2=-Vb1 (symm. diff. mode) over Vb2=0 (unsymm. diff mode) until Vb2=Vb1 (common mode) this graph demonstrates how the input resistance increases approaching very large values (common mode).

Lutz, would you specify what resistance you mean when asking, "How do we calculate the input resistance at the base of Q1?"... and later, "What is the diff. input resistance at the base of Q1 - in dependence on the voltage at the base of Q2?"

Cyril - yes, it seems necessary to specify something more.

Lets assume

(a) a simple long-tailed pair with a finite dynamic resistance re in the common emitter leg (no further resistors between the emitters and the common point) and

(b) a small-signal low-frequency input signal at the base of Q1 (e.g. 1kHz, 10mV).

(1) If the same signal - with 180 deg phase inversion - would be connected to the base of Q2, the input resistance at Q1 will be identical to rin= rbe=h11=hfe. This is the so-called "symmetric diff. mode" which is used primarily for computational purposes only (no practical relevance).

(2) If the base of Q2 is grounded (unsymm. diff. mode) the input resistance at the base will be rin=2*rbe=2*h11=2*hfe.

(3) If the same signal - without phase inversion - would be connected to the base of Q2 (common mode operation), the input resistance at Q1 will be very large:

rin= h11+ h21*2re.

________________________

These three special cases are easy to compute - but the question arises:

What is the value of rin (at the base of Q1) for other signal amplitudes at the base of Q2?

For simplicity, we should consider only the following cases:

Other signal amplitudes (smaller, larger) with and without phase inversion.

Great presentation of the problem, Lutz!

You know I am not a fan of formulas, especially those that are not clear how they are derived. I can only guess that formulas about smaller signal amplitudes with phase inversion will stay between (1) and (2), and formulas about smaller signal amplitudes without phase inversion - between (2) and (3).

But the cases where there are larger signal amplitudes with and without phase inversion, are quite unclear. Maybe we will observe negative resistance in these cases?

Cyril you do not like formulas? But - what else can we do to analyze a circuit and to describe in detail the circuits behaviour? More than that, I think a formula is the only chance to see if a specific term may be neglected or not (against another term that is much larger).

What I do not like are eqivalent circuit diagrams because - rather often - they give a wrong picture and understanding of the circuit or the device.

Examples:

* A BJT equivalent small-signal diagram containing a current-controlled current source (Ic=beta*Ib). Although it works for calculation purposes, it is wrong from the physical point of view - because the BJT is a voltage controlled device.

* Even more "dangerous": An equivalent BJT small-signal diagram showing the inverse transconductance as a resistive element re=1/gm. This is - physically spoken - wrong . A so-called transresistance is not a "resistor" !!

Method Supplement to "The second journey: from the linear OPA model...

My English is not very good. Therefore, "formulas" :-)) Details can be found - see above.

Lutz von Wangenheim : "... because the BJT is a voltage controlled device ..."

I think you're thiinking about a FET. A BJT is current-controlled, and is totally characterised by the relationships between Ib, Ie and Ic - Vbe is just an unwilling passanger.

Hi Mark

The collector current depends on the base current, and the value of Vbe is a function of the current flowing through the junction. This is why the data sheet shows Ic against different steps of base current.

A possible answer is here: https://www.researchgate.net/post/Why_do_we_use_Ib_instead_Vbe_as_a_parameter_when_measuring_common-emitter_output_characteristics

Vbe = Ib *rb + Ut *ln(Ib/Ibo) + beta*Ib*re

Here

rb - base ohmic resistance

re - emitter ohmic resistance

Ut = kT/e = 26mV at 300K

Ibo - characteristic current for the base-emitter junction, usually very small

Quote: ".... A BJT is current-controlled, and is totally characterised by the relationships between Ib, Ie and Ic - Vbe is just an unwilling passanger. "

Mark - are you really sure? How should - let`s say - two charged additional carriers in the base be capable to release 400 additional charged carriers in the emitter region (assuming beta=200) ? Can you imagine?

No - thats wrong. That`s a misinterpretation of the relation Ic=beta*Ib. This relation must not be interpreted as effect=x*cause. It comes from Ib=Ic/beta because Ib is a certain (fixed) percentage of Ic.

Hence, Ib is an "unwanted passenger".

Of course, the BJT is voltage controlled - according to the famous Shockley equation. There are many effects which can be explained with voltage control only. Do you need examples?

I know that in many textbooks the BJT is described as a current-controlled device. It is a funny and surprising fact that many decades after invention of the BJT still two explanations can be found in the literature - one is correct and the other one is simply wrong!

Nikolay - thank you, but I must admit that I do not know how to interpret your answer. We are speaking about a long-tailed pair with a resistance in the common emitter lag - and the influence of a second voltage at the other transistor!!

What kind of circuit did you have in mind?

The derivation above is for common emitter. However I guess if we have symmetrical differential pair, the above diff impedance applies with x2.

Note that common mode opamp impedance usually huge but diff impedance of the opamp is rather small (not as small as usual common emitter RF amplifier because low current high beta transistor are used).

As far as the opamp stays in operating negative feedback, we don't really feel this rather low (maybe tens-hundreds Kohm) diff impedance (it is effectively multiplied by gain).

Nikolay - I agree to everything you wrote.. (except the last sentence), but this does not answer my question...sorry to say. Thank you.

Why yiu disgree with last sentense .. at least in case of simple follower..

Niklolay - why do speak about opamps? And - if you refer to opamps, which configuration (inverting/nin-inverting) ?

And if you say "multiplied by gain" - which "gain" do you mean? We have three options: (a) open-loop gain, (b) closed-loop gain and (c) loop gain.

You see, I cannot agree to a sentence that I do not understand...

The simplest case - noninverting 1:1 follower - open loop gain.

But I was asking for a diff. amplifier and the input resistances for different operational conditions with a second voltage at the second input.

Why do you speak about opamp followers?

I guess opamp followers is one of the most common cases..D I guess diff amplifier w/o strong feedback are used mostly in high dynamic range RF, or limiting (telecom )

Dear Lutz,

Perhaps Miller theorem (applied to Q1's rbe resistance) can help you to derive the so desired formula - https://en.wikipedia.org/wiki/Miller_theorem ?

Dear Lutz,

welcome,

Hope you are well!The input resistance of the differential transistor amplifier seen from the base of one transistor to the ground in the common mode can be expressed by

Ri = rbe + 2REE ( hfe +1), where rbe is the base to emitter small signal resistance, hfe is the small signal common emitter current gain of the transistor and REE is the common emitter resistance.

It is very large resistance.

The derivation is simple because of symmetry in the common mode one can split REE into two separate emitter resistances with every resistance equals 2REE. Then one has an input resistance looking from the base of each transistor as expressed previously.

If in the differential mode the input resistance from a base of the transistor to the ground is rbe and from base to base is 2rbe as the common emitter point will be at virtual ground.

Best wishes

Please look at the book

Microelectronics

By

Grabel

From

Northeastern University

Dear collegue Abdelhalim,

in your last contribution you have summarized the tree typical special cases, which I have mentioned already in my answer dated March 24.

But my original question was related to ALL OTHER conditions (and NOT only to these three typival cases)....which means: I am lookig for a formula for the diff. input resistance at one port - in dependence on the voltage at the other port!

Here is a preliminary result (for input 1) assuming an infinite common mode input resistance:

rdiff,1=2*hie(1+Vin2/Vd) with Vd=Vin1-Vin2.

Hence, the diff. input diff. resistance at prt 1 depends on the voltage at port 2.

For two special cases, the formula confirms the well known results:

rdiff,1=hie for Vin1=-Vin2, and

rdiff,1=2*hie for Vin2=0

For common mode operation (Vd=0) , we would get an infinite value which is not correct. Hence, the formula must be modified......supplemented.