Dear Cyril,

The differential amplifier is normally a DC coupled amplifier. Naturally, any input DC voltage will change the DC operating currents and voltages in the amplifier.If the two base resistances are connected in series with the base terminal of the transistors and the floating input voltage is applied the other terminals of the resistors, then there will be no path of the base bias currents to the ground. In this case floating voltage source will interrupt the the base bias currents leading to off the transistors.

When the base biasing resistors are connected as in your circuit and as i suggested in my previous comment , the bias base currents will always flow whether the input voltage is floating or grounded.

NOW assume complete symmetry of differential amplifier, the input base potentials will be equal and consequently, the potential difference between the two input terminals will be zero.

Therefore , one can apply a floating voltage source as in your figure and the circuit will operate well provided that:

1- the DC voltage drop on the base resistor is small compared to that dropped on RE. That is to provide a path for the base current without affecting its value as raised Cyril.

2-Rb must be at the same time >> the small signal resistance rbe of the transistor to avoid loading of the input voltage source.

That is B*RE >> Rb>> rbe

The gain of the amplifier will be as given in my previous answer.

My best complements to you.

Cyril, hello again.

I am afraid, such a floating voltage source will be not able to provide the necessary dc bias conditions.

Can we somehow solve this problem?

You'd have to introduce a common (shared) point (junction) where the current flow would produce alternating potential that you'd then have to ground (with some sort of impedance - in case of HF currents - or resistance for LF) to cause the excess change in the voltage to flow away to the ground and in that way to balance the potentials between two inputs.

However, this is highly discouraged, because it is hard to synchronise the frequency of the varying voltage source between two inputs to meat the signals on both ends.

Could you explain in more detail the nature of the two signals you are passing to these two amps? Just as Mr Lutz had mentioned in his answer, even if you manage to adequately achieve successful "mediation" between inputs in terms of alternating current, this will still fail to provide you with the DC bias conditions, because it is not guaranteed that the difference in potentials of two corresponding input points will always be positive, or always negative or even to always be null (unless grounded - which is not what you need here).

An interesting way of thinking... There is only one op-amp and one signal, Milan. I like especially your introduction about the common junction connected to ground. Do you know where the input bias currents flow (the current loops)?

Regards, Cyril.

Usually, I've seen internal input bias current cancellation being done with one resistor.

For inverting op-amps the + input is grounded over the bias current cancellation resistor, and for non-inverting the + input is tied to the positive potential point of the circuit source (through a resistor). This is a textbook way of wiring the circuit.

Could you attach at least the input part of the schematic so that I can take a look at where the signal is coming from, where you are leading it through and where you wish to add the floating voltage source.

I have an idea that might be a good solution to you. You might try to use an independent varying power source and in case the frequency of the change is high enough, you could create a phase shifted branch and connect one branch to the - input, and the other 180 deg shifted branch to the +. This way you can guarantee synchronism between currents that raise and lower the input signal amplitudes and always have the difference (input potential subtraction) constant (maybe not zero, but still constant).

Please add a schematic. I believe that I (or someone else) could help more if there was a schematic to take a look at and see where you are at right now with this problem.

Yes, connect a resistor between each base and the ground... and the bias currents will flow... but from where? It is interesting that they "spring" from the ground. I have explained this situation in the Wikipedia page about the long-tailed pair:

http://en.wikipedia.org/wiki/Differential_amplifier#Biasing

"...The biasing base currents needed to evoke the quiescent collector currents usually come from the ground, pass through the input sources and enter the bases. So, the sources have to be galvanic (DC) to ensure paths for the biasing currents and low resistive enough to not create significant voltage drops across them. Otherwise, additional DC elements should be connected between the bases and the ground..."

Now you have only to see how the floating differential voltage is splitted into two single-ended input voltages...

The picture is not exactly what you want but it gives some notion; try to imagine where bias currents flow...

Regards, Cyril.

Dear Cyril,

Please let me share in answering your question.

In order not to disturb the operating DC conditions of the differential amplifier indicated in your last comment , you can connect two relatively high value resistors, about 10 kohms between the two inputs and ground. These are the base biasing resistors in which the DC base biasing current flows. To apply the floating constant voltage source between the amplifier inputs without disturbing the biasing we have to connect a DC blocking capacitor in series with it. The reactance of the capacitor must be much smaller than the input differential resistance of the amplifier at the lowest frequency of the input voltage source such that we can neglect the its voltage drop. As said before, this is a common practice in radio frequency amplifiers and other AC amplifiers. Assuming that the input voltage Vi is sufficiently small then the DC bias Vbe of the transistors,then the transistors remains active and the input signal will be amplified. and the input voltage will be divided more or less equally among the emmiter to base junctions of the two transistors.This will affect collector currents Ic= gm*Vi/2. Ic1 flows down ward and Ic2 upwards.Then the differential output voltage Vout= - gm*2*RC*Vi/2. It is assumed that the collector resistors are equal. Then the voltage gain of the amplifier becomes gm*RC.Which is the small signal gain of such an amplifier.

WHEN the input voltage is increased to about four times the thermal voltage Vt =25 mV, the amplifier becomes nonlinear and the output voltage waveform will be clipped.

Increasing Vi further, the input sine wave will be converted to nearly a square wave.

I hope that my answer is satisfactory.

Thank you for your challenging questions!

Zekry

Hi, Zekry! Very interesting discussions...

Yes, I have meant exactly this arrangement (even with the favorite 10 k resistors-:) Only I think the input voltage source connected between the two amp inputs will not disturb the biasing currents since they are set by the emitter current source.This is a very clever biasing trick where we set the biasing current directly in the transistor's output (the emitter) but not in its input (base-emitter junction)... and this situation is an excellent illustration of the benefits of such an approach.

This "magic" (to reverse a unidirectional device) is possible only by means of negative feedback. I have showed another example in the discussion about the inverting versus noninverting configuration - the "transdiode", where we "push" a current into the collector as an input and take its base-emitter voltage as an output. For this purpose, the transistor (more precisely, the op-amp) adjusts its base-emitter voltage so that to pass this collector current). Now, we draw a current from the emitters and the transistors adjust their base-emitter voltages (base currents) so that to pass these emitter currents. So, the input voltage source will not prevent the biasing...

Regards, Cyril.

Cyril,

Your statement: "This is a very clever biasing trick where we set the biasing current directly in the transistor's output (the emitter) but not in its input (base-emitter junction)..."

I am not quite sure if this is really the case.

The common emitter path (the "long tail") contains a current source (very large resistor), which determines the sum of both currents through the transistors.

And, of course it is the base-emitter voltage VBE of each transistor that determines how this current is distributed between both transistors. I think, the working principle of a diff. amplifier is based solely on this effect.

More than that, the negative feedback effect (for differential as well as common mode signals) can be explained only based on the VBE role.

Well, I will explain it... This is a quite strange arrangement of two voltage sources (emitter followers) connected in parallel and a current source (in the emitters) connected in series to the voltage sources. All they are systems with negative feedback interacting each other - the first two keep up the voltage; the last - the current.

First set constant input voltages at the two bases (just ground them) and begin varying with the biasing current source in the emitters. This is a disturbance for the emitter followers and their transistors will react to this intervention by adjusting their Vbe so that the desired (by the current source) current to flow through the emitters.

This is the only way to pass the desired biasing current directly through the emitters - by making the transistors to do it themselves (by "moving" the emitters); we cannot directly "push" the biasing current into the emitters if the bases are noncontrollable (e.g., supplied by a constant current).

Setting the biasing currents from the side of the emitters has the advantage that the inputs (bases) are free from all kinds of biasing circuits. The only requirement is to connect them galvanically to the ground (to fix the base voltages).

The same idea is used in cascode circuits... This was my "system approach"...

Cyril,

yes - now I can follow this description and I agree (key word: automatic VBE adjustment).

I can see that you have managed to accomplish what you needed with resistor and capacitor in a serial link from bases for biasing, as Mr. Zekry has suggested.

This was an interesting discussion. Thank you for asking an interesting question :)

I will draw a picture to show where currents flow...

Here is the picture:

Dear Cyril,

The differential amplifier is normally a DC coupled amplifier. Naturally, any input DC voltage will change the DC operating currents and voltages in the amplifier.If the two base resistances are connected in series with the base terminal of the transistors and the floating input voltage is applied the other terminals of the resistors, then there will be no path of the base bias currents to the ground. In this case floating voltage source will interrupt the the base bias currents leading to off the transistors.

When the base biasing resistors are connected as in your circuit and as i suggested in my previous comment , the bias base currents will always flow whether the input voltage is floating or grounded.

NOW assume complete symmetry of differential amplifier, the input base potentials will be equal and consequently, the potential difference between the two input terminals will be zero.

Therefore , one can apply a floating voltage source as in your figure and the circuit will operate well provided that:

1- the DC voltage drop on the base resistor is small compared to that dropped on RE. That is to provide a path for the base current without affecting its value as raised Cyril.

2-Rb must be at the same time >> the small signal resistance rbe of the transistor to avoid loading of the input voltage source.

That is B*RE >> Rb>> rbe

The gain of the amplifier will be as given in my previous answer.

My best complements to you.

Hi, Zekry! You are precise and profound, as usual... thank you for the contribution... I am deeply impressed by and agree with your reasoning. I would like only to add that I have placed an emitter resistor instead a constant current source to simplify the circuit and to make it less "mystic"-:); people are usually frightened by current sources and do not understand their role-:)

Here is another picture of a differential amplifier (here, the emphasis is upon the dynamic load - another great idea deserving our attention).

Best regards, Cyril

I didn't suppose there are internally biased op-amps (with internal connections between the bases and the ground).

About the amplification of the common voltage - IMO there is no common voltage particularly in this example...

Hi, Adam!

The internal structure of the op-amp input stages and especially the eccentric biasing "from the side of the emitter" techniques used there are extremely interesting. A few years ago, I managed to reveal the secret of the sophisticated Widlar's (and to some extent - Fullagar's) ideas in this area. I exposed them in the Wikipedia page about the op-amp. The section about the 741 internal structure is almost completely written by me (user name Circuit dreamer, see the history):

http://en.wikipedia.org/wiki/Operational_amplifier#Biasing_circuit

To my great surprise it was not removed by the ortodox wikipedians although actually it was a pure original research, the worst thing in Wikipedia (these were only my speculations that were not approved by someone circuit genius-:) Only Widlar could say if they were what he thought in 60's when designing the legendary LM301... but he is not alive...

Best regards, Cyril.

Dear Adam,

Thank you for your congratulations; I sooner expected to see some remark that I should be more unostentatious-:) I would like to say that, although I have some op-amp practice, am not a circuit designer and a professional; I am just a circuit thinker, "fantasist" and" dreamer"-:). My pursuit is to unveil the mystery of clever, odd, paradoxical and even extravagant circuit ideas... and then to show the simple truth behind them to my students and to web readers...

I suggest to discuss all these clever biasing ideas and op-amp circuit solutions in separate questions. Here I will outline some thoughts about the biasing idea in the classic differential amplifier and in this clever op-amp circuit solution.

It is a charming idea to bias a bipolar transistor from the side of the emitter since the base remains completely free for the input source. The problem is that, when biasing a transistor from the side of the emitter, we lose the control from the side of the base-:) The reason is that, when we change the input base voltage, the emitter voltage changes in the same manner (follows) but the current does not change and, as a result, we do nothing - the output collector voltage does not change at all. This is the situation with the classic differential pair at common-mode input signals:

http://en.wikipedia.org/wiki/Differential_amplifier#Common_mode

To restore the base control, we have somehow to fix the emitter voltage and this is the situation in the differential pair at differential-mode input signals:

http://en.wikipedia.org/wiki/Differential_amplifier#Differential_mode

Widlar had to solve the same problem but with the Q3/Q4 bases since he decided to control the transistors from the side of the emitters through the input emitter followers (a "common-emitter - common-base" cascode connection) and all became "head downwards". In this "reversed" solution, the Q3/Q4 common base voltage freely follows the Q1/Q2 base common-mode voltages so the Q3/Q4 collector currents do not change and, as a result, we do nothing again - the Q4/Q6 common collector output voltage does not change at all. At differential input voltages, the Q3/Q4 common base voltage "freezes"; the Q3/Q4 collector currents and output voltage vigorously change.

I have attached the same picture as above but with a higher resolution.

Best regards, Cyril.

... Six years later, I go back to summarize the written above...

As a result of our discussion, it turned out we can connect a floating voltage source directly (without a decoupling capacitor) between the two inputs of a differential amplifier if we only provide (by two resistors) paths for the input bias currents. And what is more interesting, this voltage source does not affect the input bias currents. How is it possible?

To reveal the mystery of this paradox, I will develop the Abdelhalim's assumption in his second comment: "NOW assume complete symmetry of differential amplifier, the input in parallel potentials will be equal and consequently, the potential difference between the two input terminals will be zero."

In ordinary single-ended transistor amplifiers, the initial bias current flows through the input and determines corresponding bias voltage across it. So, if we connect the input voltage source directly to the input, it will divert the bias current and completely determine the input voltage.

We can think of this arrangement as of two sources (bias current and input voltage) in parallel. And to further dramatize the situation, let's imagine that the input voltage is zero. It is as if we have shortened the input with a piece of wire…

In the case of a differential amplifier, there are two equal bias currents causing two equal bias voltages across two inputs. And, in our case, we do not connect the input voltage source across an input; we connect it between the two inputs. Assuming again the input voltage is zero, we can think of the input source as of a piece of wire that connects the two bases. As the two base voltages are equal, no current will flow through this "wire"...

We can think of this arrangement as of a balanced bridge. First open its output and measure the currents flowing through its legs. Then short its output and measure the currents again - they will be the same since no current flows through the "wire"... it does not affect the currents...

Now replace the wire with the floating voltage source and vary its voltage - the two single-ended input voltages will vary as well...

Cyril...your circuit diagram overwhelms me...I am sorry.

On the other hand, I have a very simple explanation - perhaps too simple?

My explanation: When there is an IDEAL floating voltage source Vdiff between both base nodes, this voltage remains unchanged and is independent on the connected loads: Base input resistances, ohmic bias resistors from base to ground.

Lutz... I see... But if you look only at the drawn in black, you will see the "sterile" circuit diagram of the classic differential pair with bias resistors between bases and ground, floating input voltage source and split power supply.

Then, if you look at the drawn in green, you will see where currents flow in this circuit (a current map).

Finally, if you look at the drawn in red, you will see the voltages of sources and voltage drops across components (a voltage map).

Regarding your simple explanation: The question is not if the voltage of the input floating source will be changed but if the bias currents will be affected by the input voltage...

Only, if I have to be honest, a little while ago I found a small error in the Iin path. Now I think it passes not through Rb2 but through the base-emitter junction of T2... and futher through Re.

Under no circumstances it is not possible .

Why would anyone want to do this?

Without external gain defining components, the thing would have whatever gain the chip's production tolerances permitted, (like +/- 400%) which would be too unpredictable for any kind of serious product application.

As a result of running with this full (and random) gain, and a fixed gain/bandwidth product, the frequency response would probably roll off at about 10Hz or less.

Also, since the smallest potential available from any kind of battery is about 1 volt, just connecting it across the inputs would result in smoke and/or flames.... :-)

Ivailo, would you clarify what is not possible according to you - Iin passes through Rb2 or through base-emitter junction of T2?

Maybe I should give some further clarification...

I have long found that one of the most important things to understand circuits is to have a clear idea of ​​where currents flow; or if you do not know where currents flow, you do not really understand circuits. I always present currents in circuit diagrams by loops since, according to Mechkov's rule:), currents always return to where they started from.

In the differential pair, the bias currents are represented by two identical loops starting from the positive terminal of V-. Then bias currents pass through the base resistors R1,2... enter the base-emitter junctions of T1,2… pass through the emitter resistor Re... and finally return to the negative terminal of V-. The combination of the negative voltage source V- and the emitter resistor Re can be thought as of a current sink that draws equal bias currents from the emitters.

These currents have an average value and serve as initial input voltage sources. Then, the true input source has to add input current to one and to subtract equal input current to the other bias current.

If the input voltage source is connected as shown in the attached figure, it will subtract its input current Iin from the bias current Ib1 flowing through Rb1… and will add its current Iin to the bias current Ib2 flowing through Rb2. This means that it will divert Iin current from the base-emitter junction of T1 and will redirect Iin current to the base-emitter junction of T2.

So, the input current Iin will start from the positive terminal of Vin... turns down and enters the T2 base-emitter junction… flows through the emitter resistor Re... then passes through V-... and finally returns to the negative terminal of Vin.

Thus through the base-emitter junction of T1 flows the difference and through the base-emitter junction of T2 - the sum of the two kinds of input currents (Ib and Iin).

Conversely, if the input voltage source is reversed, it will subtract its input current Iin from the bias current Ib2 flowing through Rb2… and will add its current Iin to the bias current Ib1 flowing through Rb1. This means that it will divert Iin current from the base-emitter junction of T2 and will redirect Iin current to the base-emitter junction of T1.

So, the input current Iin will start again from the positive terminal of Vin... turns down and enters the T1 base-emitter junction… flows through the emitter resistor Re... then passes through V-... and finally returns to the negative terminal of Vin.

Thus through the base-emitter junction of T1 flows the sum and through the base-emitter junction of T2 - the difference of the two kinds of input currents (Ib and Iin).

Maybe I should finally comment on my mistake in the drawing above...

As shown there, two currents (Iin and Ib2) go in opposite directions through the same resistor (R2)... which is impossible. A resistor is a kind of "one way street":) When more than one current try to pass through it, only one total current passes...

In our case, since the input current Iin is smaller than the bias current Ib2, the latter determines the current direction.

Mark, I admire your courage to engage in this discussion. There is a wise Bulgarian proverb that says something like "While the wise are thinking, the crazy are acting". So, let's begin acting:)

With respect to your fear "if the smallest potential available from any kind of battery is about 1 volt, just connecting it across the inputs would result in smoke and/or flames.... :-)", I think you shouldn't worry so much.

Indeed, the voltage of the floating input voltage source is applied across a network of two base-emitter junctions connected contrary in series. So, one of them is reverse-biased and will break down at relatively low voltage… but not about 1 V...

I continue to think where the currents flow in this circuit... and I continue to wonder why this mental exercise is so interesting to me and not so to others…

In my opinion, there is still a situation where the input current Iin flows through the base resistor R2… and this situation happens even in the case of two grounded input sources. Of course, if there was any visible interest in what I am writing, I would gladly illustrate it with current and voltage maps. But since there is obviously no interest, I will be content only with verbal descriptions… and when a curious, benevolent and imaginative participant appears, I will draw them.

Here is what I want to add to my explanations above:

When the differential input voltage is small enough… or zero... or there is no input voltage source connected between the bases, the two input bias currents go out of the ground, pass through the base resistors and create small negative voltage drops across them. As described above, the input current (if there is such) leaves the positive terminal of the input source, turns down and enters the T2 base-emitter junction.

But if the input voltage is high enough, it will “lift” the left end of R2 “above” the ground and the T2 base voltage will become positive. Then the input current will enter R2 from the left, pass through it and go to ground. But what will happen with Ib2 bias current? Where will it flow?

Ib2 cannot continue flowing through R2… from “-” to ”+” (just as water cannot flow from lower to higher). So it will begin “looking for” a way to go… and will find it through the input voltage source… then down through the T2 base-emitter junction.

The situation is similar if we connect not one floating but two grounded input sources. Then the input bias currents will redirect from the resistors to the positive input sources.

Having nothing better to do, I thought I might provide a bit of help with this.

Attached is a circuit diagram, which I hope is exactly the same as yours, but with current sensors everywhere. Those currents missed can be inferred with the aid of Mr Kirchhoff.

I also enclose the SPICE netlist, which you can modify at will, and the text results of sweeping the input voltage from -0.6v to +0.6v.

The graphical results were too boring to include, and numbers speak louder than pictures.

(If you don't have a copy of SPICE, you can get one here: www.designsim.com.au/dnlds)

Enjoy!

You did a great job. It could be helpful to someone who thinks with numbers...

Oh, well, if you insist, but I told you it was boring...