Recently I came across with the shown circuit.
It has some interesting properties. Is there anybody who has used such a circuit already? Both opamps are assumed to be ideal.
Another odd circuit:)
At first glance... it resembles a fully differential amplifier (with a differential output)...
Well... let's first name the input resistors - R1 is the upper, R2 - the lower... and next - the output resistors of the first op-amp - R3 is the upper and R4 - the lower.
The input source V1 creates two equivalent voltages across the input resistors - VR1 = VR2 = V1/2 (because of the zero voltage across the op-amp input).
Then, VR4 = VR2... so v1 = -V1/2 (inverting amplifier).
This voltage is amplified two times by the second op-amp "non-inverting" amplifier... and the result is v2 = V1/2...
If this is true, the circuit really is a fully differential amplifier with a floating input source...
The two op-amps are comprised by an overall negative feddback... so the second op-amp amplifier acts as an inverting "non-inverting amplifier" (with a gain of two... but negative)...
The output amplfier has positive feedback, not negative (Maybe it's a Schmitt trigger?)
Sooner there are two feedbacks - a positive feedback (between the output and the non-inverting input), and a negative feedback (between the output and the inverting input of the first op-amp)... As if it begins to smack of a negative impedance circuit (NIC)...
Forgot about the working principle: though being ideal and whatever, the whole circuit is laid out to represent real resistance to load, at the same time acting as an inverting (V1) resp. non-inverting 2-fold amplifier.
Never seen such a thing before. But might come handy some time :)
Thanks a lot!
Aha... the output voltage is still equal to the input voltage but it is assymetrical towards the ground...
I asked my 'friend' once again:
NO - all voltages are perfectly symmetrical with respect to ground. (Symmetry cause by the single resistor to ground in the 2nd stage.)
The only 'odd' thing: If you split the voltage source into two, you will find the 'center point' traveling with the output frequency - with a reasonable amplitude :)
Real fun!
Hello All,
The circuit combines 100% negative feedback (via first stage), and 50% positive feedback (around second stage), thus it will be properly dc biased and functional. Its operation is as follows:
- the first stage can be assumed as four resistors one opamp diff amp with unity gain. The stage translates the input voltage VINdif to be appeared at V1 towards V2:
VINdiff = V1-V2
- the second stage, due to positive feedback, amplifies the voltage difference (V2-V1) by two, so:
V2 = 2*(V2-V1) = -2VINdiff
V2 = 2V1
Finally:
V1 = -VINdiff
V2 = -2VINdiff
Cheers,
Dobri
Hi Dobromir,
I got the same result as You.
With the ideal amplifiers, it is simple. And with the ideal amplifiers we could do the same to less exotic ways.
It would rather be interesting with real amplifiers????
Quote Dobri:
Finally:
V1 = -VINdiff
V2 = -2VINdiff
Hence: V1-V2=VINdiff.
Gentlemen - thank you for the various responses.
Yes - it is a tricky circuit I have discovered just some days ago.
And I agree - the difference between both opamp output voltages is identical to the floating input voltage Vin=V1-V2.
More than that - my analyses surprisingly have shown that the amount of positive feedback of the second opamp does NOT influence the result Vin=V1-V2. However, this applies to ideal opam models only. Analyses for realopamps are pending.
That means: We have a circuit which is able transform a floating voltage into a ground referenced differential voltage (with adjustable gain values).
Hi Josef,
Yes, why not. But, to preserve stability equal caps should be added in the first stage from V1 to opamp (-), and from V2 to opamp (+), as in usual diff amp. Because of extra gain of two, the bandwidth should be limited more than four times, let say ten times or more.
Dobri
Correction:
I have edited my recent contribution. (No change of common mode voltages).
Here is the result of simulations based on real opamp model (LM741):
Results are as expected (V1-V2=Vin,diff) for a maximum positive feedback factor k=1/3 (All other resistors are 1k). However, the gain of each stage depends on k.
For k>1/3 the circuit oscillates.
To analyze this, also power voltage for second OA should be considered (let's say +/- 10V). The second OA is not in linear mode, it is connected as inverting large-hysteresis discriminator. Its output flips just between two levels +10V and -10V. The thresholds are half of the power voltage, i.e. +5V and -5V. The first OA works in linear mode (with active virtual zero at input) converts input differential voltage to unipolar voltage (referred to ground), however its resistor on non-inverting input is not connected to ground (as in standard diff amplifier), instead, it is biased with output voltage +10V or -10V. So the whole circuit works as large-hysteresis comparator with differential (floating) input voltage, acting at input differential voltage +5V and -5V. In praxis, the power voltage for first OA should be larger than for second one. Let's say +/- 15V.
Dr. Balaz - with all respect, circuit simulations cannot confirm your observations.
In contrary, the second opamp is in the linear mode - in spite of (a certain amount of) pos. feedback which, however, must be evaluated together with the overall negative feedback path.
Dobri - I have some problems with your calculation.
At first, in your calculation: What is V1 and V2? Opamp output voltages or the voltages at the top and bottom of Vdiff?
Secondly: You assume a gain of two for the second opamp which has POSITIVE feedback. I don`t think thisd is allowed. This is a typical case of an unstable gain stage which is stabilized through an additional negative feedback path. And, therefore, we must perform another calculation (Note: The result is correct, however).
My approaches (Vo1 and Vo2: Opamp output voltages):
A) Simply using the superposition principle (for three voltage sources Vdiff, Vo1 and Vo2) ) we arrive after some manipulstions at the correct result (for all R being equal):
(Vdiff=Vo1-Vo2).
B) We can apply Blacks feedback formula for the closed-loop gain Acl,2 (opamp2). For this we (temporarily) define a new input Vin at the grounded resistor R1.
Note: We must start with THIS transistor because it must not be part of the feedback (for T1) due to its internal instability.
Acl,2=Vo2/Vin=Hf*Aol/(1-Aol*Hr) with open-loop gain Aol.
Note that this general formula is not tailored specifically to negative feedback . The feedback function is Hr=(Hr+) + (Hr-). We have two feedback branches with different signs: Hr+ = +R1/(R1+R2) (R1and R2=Pos. feedback resistors) and Hr- = -(1+Vdiff/Vo2).
After inserting these expressions into the closed-loop gain formula for Acl,2 assuming Aol infinite and setting Vin=0 we arrive at
Vo2=-Vdiff(1+R1/R2) and Vo1=Vdiff+Vo2=-Vdiff*(R1/R2).
For equal values R1=R2 we arrive, of course, again at Vdiff=Vo1-Vo2.
C) Loop gain (Stability):
It is possible to show that the second opamp (with a pos. feedback path) always has more than unity negative feedback (ideal opamps without additional phase shift). Hence the net feedback always is negative for any ratio R1/R2 - and therefore always stable!
However, as soon as the opamps are real with finite gain and additional phase shifts the amount of positive feedback R1/(R1+R2 )must be reduced to ensure stability.
Hi Lutz,
With V1 and V2 I denoted the voltages at nodes v1 and v2, VINdiff is the signal source V1.
I will explain again my simple (and intuitive) approach for easy understanding the operation of this circuit .
I think, the first stage operation is clear.
But again, it is a simple diff. amp. It translates input diff. voltage into a floating voltage between nodes v1 and v2. It is a floating signal source translating (shifting) VINdiff into a signal V(v1, v2).
Next, the second stage has positive feedback.
But, it can not operate in linear mode without deeper negative feedback!
Turning around the negative loop path from v2 to v1, via stage one, it is clear that V2 is divided by two, then is amplified by two, and so V1 is equal to V2, thus the second stage opamp will have gain of 1 in the loop from its output to its inverting input, that is why I say that the second stage has 100% negative feedback. Its output v2 is virtually shorted to the opamp inverting input, and only in such case the second stage can operate in linear mode with additional amount of positive feedback.
Next, the signal source for the second stage is the floating voltage source V(v1, v2).
When V(v1, v2) is zero, the second stage output v2 is also zero.
If V(v1, v2)=1V, v2=2V, because V(v1, v2) is converted to current via resistor in opamp feedback Rfb (from v2 to noninv. input). This current flows via resistor to ground Rgnd, and so V(v1,v2) is amplified by a positive fb gain (Rfb+Rgnd)/Rfb.
In our case both resistors are equal, and the positive fb gain is 2.
Pros:
Only one. Due to shared feedback, lower delay between v1 and v2
Cons:
Positive feedback. Stability problems.
Can be stabilized by two caps in the first stage (see my post above). These caps are mandatory for each diff. amp.
Especially here, a third cap can be added from the second opamp noninv. input to ground, in order to reduce positive feedback for high frequencies. The values of all caps should be optimized for a needed settling time and stable response.
Dobri
Hi Dobri - thank you for your additional explanations. But I still cannot fully follow your calculations (quoting your text in italic):
With V1 and V2 I denoted the voltages at nodes v1 and v2, VINdiff is the signal source V1.
Unfortunately, the floating input source was named as V1. Hence, it was good to rename it with V1=Vindiff.
I will explain again my simple (and intuitive) approach for easy understanding the operation of this circuit .I think, the first stage operation is clear.
Yes. Because of v1=Vindiff+v2 we have Vindiff = v1-v2.
But again, it is a simple diff. amp. It translates input diff. voltage into a floating voltage between nodes v1 and v2. It is a floating signal source translating (shifting) VINdiff into a signal V(v1, v2).
At which point can I measure this „signal“ V(v1, v2)?
Is it the difference (v1-v2) ? With other words: Vindiff=V(v1,v2) ?
The only remaining problem is to find the expressions for v1 and v2 because - up to now - we only know that the difference is (v1-v2=Vindiff.).
Next, the second stage has positive feedback.
But, it can not operate in linear mode without deeper negative feedback!
Turning around the negative loop path from v2 to v1, via stage one, it is clear that V2 is divided by two, then is amplified by two, and so V1 is equal to V2, thus the second stage op amp will have gain of 1 in the loop from its output to its inverting input, that is why I say that the second stage has 100% negative feedback.
Its output v2 is virtually shorted to the opmp inverting input, and only in such case the second stage can operate in linear mode with additional amount of positive feedback.
Yes, this is in accordance with my calculation.
Next, the signal source for the second stage is the floating voltage source V(v1, v2).
Why ???
To me, the signal source for the 2nd opamp is the first opamps output voltage v1.
When V(v1, v2) is zero, the second stage output v2 is also zero.
If V(v1, v2)=1V, v2=2V, because V(v1, v2) is converted to current via resistor in opamp feedback Rfb (from v2 to noninv. input). This current flows via resistor to ground Rgnd, and so V(v1,v2) is amplified by a positive fb gain (Rfb+Rgnd)/Rfb.
In our case both resistors are equal, and the positive fb gain is 2.
I must admit that I do not understand how V(v1,v2) is amplified by a „positive fb gain“ (1+Rgnd/Rfb).
Therefore, again my question: Where is the signal V(v1, v2) ?
Thank you
Lutz
Dobri,
Dobri wrote :
"""But, to preserve stability equal caps should be added in the first stage from V1 to opamp (-), and from V2 to opamp (+), as in usual diff amp."""
Are you referring to v1 and v2 ?
So, v1 cap would increase high freq negative feedback,
and
v2 cap would increase high freq positive feedback.
My first thought is that OA #1 would 'tend' to be damped.
My first thought is that OA #2 would 'tend' to oscillate.
Given an arbitrary 10nS transition time through each opamp,
then V1 is delayed by two different amounts of time,
compared to v1 and v2.
And v1 is delayed compared to v2.
Your added caps would alter these these delay times.
Questons :
(1) are you suggesting that the tendency to oscillate is an instability ?
(2) are you suggesting that the high freq negative feedback being applied from v2 onto OpAmp #2 (-) pin will tend to reduce oscillation ?
(3) are you suggesting that the high freq positive feedback being applied from v1 onto OpAmp #1 (+) pin will tend to cause oscillation ?
(4) are you suggesting that the added caps
will alter these time differences ?
Lutz wrote :
"""We have a circuit which is able transform a floating voltage into a ground referenced differential voltage (with adjustable gain values)."""
Question Lutz,
would this circuit have a higher than normal
Common Mode Rejection Ratio (CMRR) ?
Jan wrote:
"""inverting large-hysteresis discriminator""" ( ILHD )
and this is all becoming interesting discussion.
If this ILHD is balanced, then the whole circuit will not oscillate.
Jan, a Novel idea.
My Comment:
This all appears to be a balancing act,
and very interesting to visualize.
Lutz,
In spice language V(a, b)=V(a)-V(b), thus V(v1, v2)=V(v1)-V(v2)=V1-V2
For the first stage operation, see datasheet of e.g. INA105.
http://www.ti.com/lit/ds/symlink/ina105.pdf
The input source for the second stage is the voltage V(v1, v2)
Hi Glen,
I correct a bit my comment above about stability. So, these two caps in diff. amp is needed for reducing the bandwidth. They must be equal for HF common mode balance. These two caps should be enough for stability, but a third cap from the second stage noninv. input to gnd could be added for reducing positive fb in HF.
Dobri
"In spice language V(a, b)=V(a)-V(b), thus V(v1, v2)=V(v1)-V(v2)=V1-V2"
Dobri, yes - that`s what I have expected.
"For the first stage operation, see datasheet of e.g. INA105.
http://www.ti.com/lit/ds/symlink/ina105.pdf"
I have no problem with the first stage (see my calculation).
"The input source for the second stage is the voltage V(v1, v2)"
Looking at the circuit diagram, I see that the input for the 2nd stage is connected solely to the output of the first stage which is labelled as V1.
So - why do you consider V1-V2 as input signal?
@Glen Ellis:
"would this circuit have a higher than normal Common Mode Rejection Ratio (CMRR) "?
I think the CMRR solely depends on the resistor tolerances. For ideal conditions (opamp and parts tolerances) the common mode gain is zero.
Lutz,
The first stage is a diff amp similar to INA105. Please see page 1 of INA105 data sheet from the link above. The resistors are connected to pins Ref and Sense on the right side. The pin Sense usually is connected to the Output.
The pin Ref defines the potential where the output voltage is referred.
V(Inp, Inn)=Gain*V(Output, Ref).
Because Gain=1
V(Inp, Inn)=V(Output, Ref).
In our case V(Output, Ref)=V(v1, v2)
Do you agree?
Now the answer of your question:
So - why do you consider V1-V2 as input signal?
Because V1 is referred to V2, not to GND!
Dobri
Hi Lutz,
you are right (I was wrong - sorry for my mistake), both OA are in linear mode. I connected this on testing pinboard with LM358 - it works linearly and stable, the gain is G = 2. I have used a static floating input source (4V LiOn battery with potentiometer), also floating 50Hz source (6V transformer with potentiometer), it works normaly from zero to output clipping. There is of course low input impedance and poor common mode rejection - so I still do not know what is advantage of this circuit.
Dobri - sorry, but I cannot agree.
May I suggest not to use INA105 as a reference because (a) I know how such an amplifier works and (b) it is confusing to invent new terms and symbols.
I am afraid, we have a problem with the symbols and - in order to avoid misunderstandings - i recommend to stick to the symbols as given in the figure.
That means: The output voltage of the first opamp is called V1 and this is a ground referenced voltage. Of course - this voltage is identical to the superposition of two other voltages (V1=Vindiff+V2). But it is THIS voltage (V1) which is the input for the second opamp and not (as you wrote) the difference V1-V2.
In the figure, we have defined two ground refenced ouput voltages (V1 and V2) and I do not understand why you think that V1 would be not referenced to GND. Please, can you clarify?
Lutz
Lutz,
I do not understand why you think that V1 would be not referenced to GND
Because the first stage action is to generate V1 towards V2!
Because the first stage equivalent scheme is a Voltage Controlled Voltage Source (VCVS) controlled by Vindiff, and connected between nodes V1 and V2.
Is it clear now?
Dobri
No - sorry.
I don`t know the source of misunderstanding. To clarify this, would you please answer the simple question:
Is it correct that the output voltage of the first amplifier (called V1) acts as an input voltage for the 2nd amplifier ? Yes or No?
But I can not see any symmetry in this circuit... In a conventional fully differential amplifier, both single-ended output voltages (here, v1 and v2) are symmetrical to ground (in the sense that they are equal but with opposite signs), while here they have different values and same signs... they stand on the same side with respect to the zero voltage line...
Yes!
Something more, you wrote:
V1=Vindiff+V2
Lets consider the case when Vindiff=0.
V1=V2.
V1 is the first opamp output and is input for the second stage
if V2 trying to move, V1 will also move to make V1=V2
So, the 2nd opamp will see low impedance connections from its output to its inv. input.
now, lets Vindiff=1V
V1=1+V2
if V2 trying to move, V1 will also move with 1V offset to make V1=1+V2
so, there is a voltage source between V1 and V2 with value Vindiff.
Assuming the first stage as a voltage source between V1 and V2 makes easy to be analyzed the second stage (moreover, it can not operate linearly alone without this assumption!).
So, in simple hand calculations, the circuit should be analyzed in such way.
Of course, I agree that in extensive analysis all potentials should be referenced to GND.
Dobri
Lutz and Dobri, I can not understand at all what you are arguing here... There are two single-ended voltages (v1 and v2) in a proportion 1:2 that, of course, are referenced to ground... but we use their floating difference as an output voltage...
This is a typical "servo system" that changes two single-ended voltages until makes its floating difference equal to the floating input voltage... An example of a similar technique can be a bridge circuit with varying supply (input) voltage and three output voltages - two single-ended and one differential...
Negative impedance converters (NIC) exploit the same idea... but there the differential output voltage is kept equal to zero (balanced bridge).
Cyril,
I'm not arguing...I'm teaching you in easy understanding of electronic circuits :)
Anyway, attached is a schematic for hand calculations, illustrating the operating principle of this circuit.
Dobri
Dobri,
Thanks for the "lecture":) I fully share this approach. You can assure if you look at my story about NIC:
https://www.researchgate.net/post/What_is_the_basic_idea_behind_the_negative_impedance_converter_How_is_it_implemented_How_does_it_operate_What_does_the_op-amp_do_in_this_circuit
If you read it carefully, you will notice an amazing similarity with your presentation of the current circuit. In both circuits, an op-amp "copies" the input voltage across the upper resistor R2 (not as usual, across the lower R1)... and this voltage is floating. To do it, the op-amp creates two times higher output voltage referenced to ground. So, the feedback voltage divider is turned upside down here (R2 serves as R1)... the op-amp compares the input voltage with the voltage drop across R2 (not R1)... and keeps the difference between them equal to zero (see the last picture above)...
Let's now continue to teach this approach:)
In the VNIC above, an input current source passes its own current through a resistor and so it becomes an input voltage of the op amp. By analogy with this, we can assume that the entire left side (the first op-amp and the four 1k resistors) of the original Lutz's circuit is a voltage-controlled current source... and even I think this is a kind of the famous Howland current pump...
"Lutz and Dobri, I can not understand at all what you are arguing here..."
Hello Cyril - just to explain to you the point of disagreement :
At the beginning of this thread I have presented two different methods for calculating both opamp output voltages. In his contribution, Dobri has proposed a third method which, however, I could not follow - in particular the sentence: "The input source for the second stage is the voltage V(v1, v2)" .
I am sure, you will understand that I am interested to clarify such points of technical disagreements.
Cyril - regarding you last contribution ("There are two single-ended voltages (v1 and v2) in a proportion 1:2" ), I am sure you are aware that the core of the discussion with Dobri was how to find this relation of 1:2 and - more than that - to calculate the exact values of both voltages (not only the ratio). Do you no understand ?
If the circuit under discussion can be explained with the help of the NIC principle is something I have to think about.
Regards Lutz
No, Cyril.
It is not VCCS!
The first op-amp and the four 1k resistors represent VCVS!
Hi Dobri,
You wrote: "Lets consider the case when Vindiff=0. V1=V2."
Yes - of course, BUT: V1=V2=0 (because there is no signal source in the whole circuit)
...there is a voltage source between V1 and V2 with value Vindiff.
Yes, no doubt about it. Therefore, the negative feedback factor kn between the output of opamp 2 (V2) and its inverting input contains Vindiff - and is, therefore, even larger than unity. It is important to note that this factor is NOT constant but depends on V2.
Negative feedback factor for opamp 2: kn=1+Vindiff/V2.
This ensures good stability properties in spite of pos. feedback.
Yes Dobri, the "building formula" is as follows:)
VCCS (the op-amp + the tree 1k resistors) + CVC (the fourth 1k resistor) = VCVS (the op-amp + the four 1k resistors)
(CVC means "current-to-voltage converter")
Dear Lutz,
I hope that I have successfully clarified the Dobri's "odd" idea that the input voltage of the second stage is applied between the op-amp input and its output instead between its input and ground. For this purpose, I have shown that NICs exploit the same odd trick (the input loop includes the op-amp differential input and the output resistors)...
Hi Lutz,
Assuming ideal opamps, the negative feedback is 100%, kn=1, see my equivalent circuit attached above. Do you agree with it?
Why are you including Vindiff in the formula for kn?
Dobri
Now about the proportion 1:2, Lutz...
I prefer to you to draw where currents flow in this circuit (the current loops). Then you will realize my idea above that the second op-amp and its two 1k resistors form a little odd "non-inverting amplifier" with a "turned upside down" voltage divider. If we suppose that the upper resistor is R2 and the lower is R1 (as in the Dobri's picture), its gain is not 1 + R2/R1 but 1 + R1/R2. But since here the two resistors are equal, the proportion is the same (1:2) as in the conventional non-inverting amplifier.
The problem is that if we think of this circuit as of a conventional non-inverting amplifier (with a grounded input voltage), it seems impossible... since it seems as a positive feedback circuit (Schmitt trigger). And really, how it would be possible the voltage of the non-inverting input to be equal to the voltage of the inverting input... as we applied the input voltage to an inverting input?
"Assuming ideal opamps, the negative feedback is 100%, kn=1, see my equivalent circuit attached above. Do you agree with it?
Why are you including Vindiff in the formula for kn?"
Dobri - the answer is simple: Because I have calculated the feedback factor from the original circuit. And - these calculations lead to the correct result for V2 and V1. (see calculation in my former post under (B). More than that, circuit simulations confirm the result kn>1.
To me, it is clear - looking into the circuit - that the voltage Vindiff must contribute to the feedback signal. Why do you think it would have no influence on feedback for the second opamp ?
In linear circuit analysis, voltage sources are replaced with short circuit
If we suppose that the upper resistor is R2 and the lower is R1 (as in the Dobri's picture), its gain is not (1 + R2/R1) but (1 + R1/R2.)
Cyril - yes, this is in accordance with my findings (as given in one of my former posts):
Output voltage (opamp 2): V2=-Vindiff*(1+R1/R2)
However, this is the expression for the output voltage (with Vindiff as "fictive" input). In reality we have V1 as input for this opamp. For this reason, we cannot say that (1+R1/R2) would be the gain of the second stage.
For my opinion, the gain of the second stage cannot be defined because this stage cannot work in a "stand-alone" mode because it needs for linear operation the negative feedback loop via opamp 1.
And exactly this was the reason I have used THIS amplifier as a starting point for my calculation by "temporarily" defining a new (fictive) input voltage into the resistor R1 (which, at the end of the computing process, was set to zero again). I have described this "tricky" procedure in one of my earlier contributions under (B).
"In linear circuit analysis, voltage sources are replaced with short circuit"
Does this apply here? I think, this is true for AC analyses and DC supply voltages only! But not in our case where a source produces a SIGNAL to be amplified! More than that - the voltage Vindiff can be, of course, any ac voltage.
"In reality we have V1 as input for this opamp"...
Dear Lutz,
First at all, the input voltage of the second op-amp is the difference between the voltages at its inputs... so we can talk about the input voltage of the second op-amp circuit. But If we assume that v1 is the input of this circuit, we should assume that this circuit is a Schmitt trigger... since v1 is applied to the inverting input, and the feedback voltage divider is connected between the output and the non-inverting input...
So, we should assume that the input voltage source can be floating and connected between the op-amp output and input as in the Sheingold's paper below (Fig. 4, page 7):
http://www.philbrickarchive.org/1964-1_v12_no1_the_lightning_empiricist.htm
Ey Cyril - interesting paper. Thank you so much!.
In principle, I agree with you. Yes - it is somewhat "odd" to say that V1 would be the input to the 2nd opamp - because the whole block (including R1, R2) is no linear amplifier at all. On the other hand, the opamp alone works in its linear range because of the dominating negative feedback.
Therefore, my description of the whole arrangement is as follows:
* We have an opamp (opamp 2) with positive resistive feedback kp=R1/(R1+R2) and an active (opamp 1) negative feedback factor kn which dominates as long as kn>kp. Therefore, opamp 2 works in linear mode.
* Anywhere within this active negative feedback path (opamp 1) is a signal source Vindiff , which is responsible for the two opamp output voltages V1 and V2.
* This case can be treated as a "disturbance transfer effect" known from the general control theory. This is because such "disturbances" may be effective anywhere within the loop. Hence, we can define two "disturbance" transfer functions T1=V1/Vindiff and T2=V2/Vindiff. And the results are:
T1= (- R1/R2) and T2= -(1+R1/R2)
What do you think about this view?
Interesting thoughts, Lutz... With your enthusiasm you made me start thinking seriously about this "devilish" circuit:) ... but it takes time...
Dear Lutz,
I needed some time to give meaning to what you said. Now I completely second your idea about the input signal acting as a disturbance. Do you remember my question about the nature of the input signal in a negative feedback system? It can help this chain of thoughts...
https://www.researchgate.net/post/What_are_actually_the_input_and_output_of_a_negative_feedback_system_Can_we_consider_the_input_quantity_as_a_disturbance_and_the_output_quantity-as_a_reaction
Lutz,
Voltage sources are shorted always when superposition is applied.
I wrote this only to show you that there is a short connection from V1 to V2, in case of ideal opamps.
Next, the circuit is perfectly linear. There are no nonlinear elements like multipliers, diodes, switches, saturated transistors, etc. Please, stop discussing nonlinearity, because this is another story.
"disturbance transfer effect" - what is this animal?
Signal Gain, Noise Gain, Loop Gain, Closed Loop Gain or what?
I'm guessing that this may be a noise gain, but you wrote two ideal signal gains!
So, T1 and T2 are two ideal signal gains:
T1=V1/Vindiff
T2=V2/Vindiff
Dobri
Helo Dobri:
"Voltage sources are shorted always when superposition is applied."
Why do you think, I have used superposition? It is obvious (and I have mentioned it) that I have applied rules of the LINEAR control theory.
"Next, the circuit is perfectly linear. There are no nonlinear elements like multipliers, diodes, switches, saturated transistors, etc. Please, stop discussing nonlinearity, because this is another story."
Sorry, but I did not use the term "non-linearity" at all. As mentioned, I have used the linear control theory with forward amplifier and feedback blocks. And this reqires that the amplifier is operated in the LINEAR region - otherwise I am not allowed to use the term "gain" at all. And for this purpose, the negative feedback for opamp2 must dominate. I am afraid, you didn`t understand the calculation principle at all (I will give again a short summary at the end).
"disturbance transfer effect" - what is this animal?"
This is no "animal" - instead, it is an element of the classical control theory. Let me explain to you:
A control loop has two tasks: (1) The system output must follow the "guidance" or "reference" variable and (2) it must attenuate all "disturbances" which may be coupled into the loop anywhere. For this purpose, two fundamental transfer functions are defined: (1) "Reference" transfer function and (2) "Disturbance" transfer function.
Do you understand?
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Background of my approach: Of course, we can analyze the whole circuit based on (A) superposition rule with three voltage sources. I did this - and the result is correct.
However, in order to understand the working principle of the circuit it is beneficial to treat the system as a control loop and apply (B) Black`s classical feedback formula. This requires to define a forward amplifier and a feedback path. Here, we have no choice because opamp 2 must not be seen as a feedback block because it operates not in the linear amplifying region (because of positive feedback). In contrary, opamp 1 works as an amplifier and, therefore, is considered as feedback device. Hence, opamp 2 is our forward amplifier - equipped with a positive and a negative feedback loop. Now we are able to apply Black`s famous formula for the closed loop gain.
That`s what I have done based on the definition of two feedback factors (positive kp and negative kn, see my earlier post). Assuming ideal opamps with infinite open-loop gain, the closed-loop gain depends on the feedback elements only.
It is important to note that the negative feedback factor kn contains the voltage ratio Vindiff/V2 (V2: ouput voltage of opamp2).
However, a small "trick" was necessary: For defining opamp 2 as a forward amplifier in a closed loop I need an external input signal. For this reason, I have lifted the resistor R1 from ground and connected to a signal source (Vtest). Now - after applying Black`s formula I could set Vtest=0 again - and solve for the only remaining external voltage in the equation: Vindiff. Of course, the ratio Vindiff/V2 appears in the formula because it is part of the feedback path (Vindiff acts as a "disturbance" variable).
The result is identical to the result obtained by superposition (and is confirmed by circuit simulation).
Lutz,
...the negative feedback factor kn contains the voltage ratio Vindiff/V2...
wrong! there are no multipliers in the circuit!
... a small "trick" was necessary...
no need of any tricks!
..in order to understand the working principle of the circuit...
the working principle I already described above!
If you are writing an article, I propose you to start it with this simple explanation.
Next you should define the open and closed loop gains and evaluate them.
You can start your analysis from the attached picture.
kp=R1/(R1+R2)
A2ol is the the open loop gain of opamp 2.
Dobri
Lutz,
I updated the picture in the previous post. V1 should be after the first summing junction.
With this scheme easily can be defined signal gains V2/Vdiff and V1/Vdiff.
Next, to add influence of the first stage, two transfer functions should be defined:
T1fb(s) = V1(s)/V2(s), when Vindiff=0
T1sig(s)=V1(s)/Vindiff(s), when V2=0
T1fb(s) should be added in the feedback from V2 to the first summing junction.
T1sig(s) should be added between signal source Vindiff and the first summing junction.
Next, breaking the loop between V2 and T1fb(s), we can evaluate the loop gain starting from T1fb(s) and finishing to V2 for phase margin optimization in AC analysis.
Vindiff=0, and T1sig(s) does not take a part in this loop.
T1sig(s) influences only on signal gain V2/Vindiff and V1/Vindiff
Dobri
"With this scheme easily can be defined signal gains V2/Vdiff and V1/Vdiff."
Dobri,
your block diagram looks nice and simple - you certainly have used it for calculation of the ratio V2/Vdiff (as announced in the quoted sentence at the top). May I kindly ask you to show us how you "easily" could arrive at the correct result?
(More than that, I must admit not to understand the background of the second part of your contribution; what do you want to proof or to show?)
Lutz,
(-Vindiff-V2+Kp*V2)*A2ol=V2
V2/Vindiff=-1/(1/A2ol+1-kp) ~ -1/(1-kp)
Attached two pictures could help.
Used a first order transfer function for the first stage signal gain and feedback gain.
Hope you will understand, see fig 2.
Gcm and Gdiff are common mode and differential mode gains of the first stage.
note, that V2/V1 depends on the common mode gain of stage 1.
worst case Gcm=4*delta, delta is the tolerance of the used resistors.
Fig. 3 is the loop gain.
Will discuss later, have to go.
Dobri
Hi Dobri,
Finally I can agree - your block diagram reflects the given circuit and gives the correct results.
Some posts earlier I have described two other methods directly based on the given circuit diagram.
Thus, we see that there are always different ways to analyze such feedback systems.
If you are interested - here is another (very simple) approach:
For ideal opamps (Aol infinity) we always set the opamps input differential voltage Vd=Vp-Vn=0.
Hence, we have Vp=Vn.
From the circuit:
Vp=V2*R1/(R1+R2) and Vn=V1=V2+Vdiff (function of the adder).
Hence:
Vp=Vn
V2*R1/(R1+R2)=V2+Vdiff
V2[R1/(R1+R2)-1]=Vdiff
V2= -Vdiff*(1+R1/R2)
V1=V2+Vdiff=- Vdiff*R1/R2.
I think, the circuit under discussion has produced many interesting analyses and contributions.
Thank you to all participants.
Hi Lutz,
It was pleasure to discuss this interesting circuit.
Although, its application is limited to education.
Regards,
Dobri
Dobri,
Having spoken that it is limited to "education",
what possible applications come to mind ?
Is Lutz' circuit better at something ?
Where in the world 'might' this circuit find application ?
Glen,
My first thoughts were about pros and cons of this circuit, because it is nothing more then diff. amp + next opamp stage, combined in a composite amplifier.
The main difference is that the input voltage for the second stage V1 is referred to its output V2.
I do not see applications of exactly this configuration, but rearranging a bit with ac coupling etc., it could have application in analog filtering.
Imagine that the second stage e.g. is a band-pass filter.
Combined with the positive feedback, this could have advantage in analog filtering for implementing high-Q functions.
Dobri
As another possiblke application: There are sensors which produce a floating voltage between two nodes only. This sensor voltage is transferred into a ground referenced voltage difference which then can be further processed.
Right to the point! just think of inputs to commonplace cheap digital signal acquisition boards; the really simple ones don't necessarily have floating inputs but surely must have ground referenced ones (think most direct- to-ADC micro controllers) ..then such a circuit might help but as presented the loading on the signal source is considerable; perhaps higher value input resistors and some decoupling capacitors could indeed benefit this layout. Thank you all for an interesting circuit and analysis.
Perhaps an interesting supplementary remark:
The circuit under discussion works also (with the same results) if the input terminals of the opamps are interchanged. In this case, the second opamp works as a classical non-inverting amplifier. The first (summing) amplifier has positive feedback - but is stabilized due to a dominating negative feedback loop (it overrides the pos. feedback because of the gain 1+R2/R1).
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