Werner said:
Thanks for that great non-answer.
Now, what is the common mode signal again in this discussion? Did I hear "input signal"?
Error signal?
SY's Second Law Time: Here's one for you (Millwood) to try. Breadboard up a simple opamp inverter with some moderate gain (say, 20 dB). Feed it some sine waves and square waves and, using a 10x probe, look at the waveform, magnitude, appearance, and phase wrt to the input. Do the same for a noninverting amp of the same gain. Post your results and then everyone can stop arguing and do something productive.
unfortunately I don't have a probe.
I had hoped that some wiser folks can explain in plain english why one particular topology is superior than the other. I guess that's not going to happen.
I had hoped that some wiser folks can explain in plain english why one particular topology is superior than the other. I guess that's not going to happen.
I'm not an expert on this topic, but I'll try:
The difference between the two circuits is that the substraction of signals is carried out inside the opamp for the non-inv. circuit and outside the opamp for the inverted circuit.
In the case of the inverted circuit, the +input is at zero (or near zero in case of a resistor there). The current of the two signals (input and feedback signal) are subtracted at the -input. This is kirchhoff (i1=i2+i3) and theoretically ideal.
For the non-inv. circuit, the signals are subtracted in the opamp. So first the signals go through some stages that have non-linearities before they are substracted. This means that the substraction is not ideal. So the distortion at the output increases compared to the inverted circuit!
Another advantage is that the inverted circuit works near 0V on both inputs (instead of the signal levels of the non-inv. circuit). This will very likely also help the linearity...
Fedde
The difference between the two circuits is that the substraction of signals is carried out inside the opamp for the non-inv. circuit and outside the opamp for the inverted circuit.
In the case of the inverted circuit, the +input is at zero (or near zero in case of a resistor there). The current of the two signals (input and feedback signal) are subtracted at the -input. This is kirchhoff (i1=i2+i3) and theoretically ideal.
For the non-inv. circuit, the signals are subtracted in the opamp. So first the signals go through some stages that have non-linearities before they are substracted. This means that the substraction is not ideal. So the distortion at the output increases compared to the inverted circuit!
Another advantage is that the inverted circuit works near 0V on both inputs (instead of the signal levels of the non-inv. circuit). This will very likely also help the linearity...
Fedde
Friendly advice from someone who also takes a reductionist view of audio electronics: get out of the armchair. Spend the paltry bucks it takes to equip yourself in a basic manner (I managed on a salary considerably below that of my own dumb banker). Learn how to use it by actually using it. Then you have the possibility of being able to speak with some authority.
Konnichiwa,
Actually, Douglas Self has already done this and posted the results on his website.... And for virtually all Op-Amp's his measurements show very material differences in THD between the different topologies.
Other than that I must agree with Dieckmann San, if you rely on D. Selfs views you are in the wrong camp, however, his view on audibility aside, he has analysed the common structure (Lin) type amplifervery extensively. And my point is a little more suble, more like "EVEN D. SELF HAS NOTICED THE DIFFERENCE" (he is usually the last one).
Sayonara
SY said:
Actually, Douglas Self has already done this and posted the results on his website.... And for virtually all Op-Amp's his measurements show very material differences in THD between the different topologies.
Other than that I must agree with Dieckmann San, if you rely on D. Selfs views you are in the wrong camp, however, his view on audibility aside, he has analysed the common structure (Lin) type amplifervery extensively. And my point is a little more suble, more like "EVEN D. SELF HAS NOTICED THE DIFFERENCE" (he is usually the last one).
Sayonara
SY said:
no matter how they tell you, they don't pay bankers that well. In between the taxes, mortgages, partnership puchases and capital calls, there aren't much left, no matter how dumb the banker is.
Kuei Yang Wang said:
take a shot at someone who is unable to defend him/herself isn't the most honorable thing to do.
You know that, I know that, but it's often helpful and educational for people to go do it themselves.
Not everything that everyone puts on their websites is true, I'm shocked to say. I do like Self's, too, and his data have been, IME, reliable and well-interpreted.
After having asked a couple of questions, and having received several intelligent, factual, insightfull and helpfull answers ranging from "go back to 101" via " nobody knows anything about nothing anymore these days" to "man, are you stupid", let me float the following:
From Operational Amplifiers (2nd edition), (Jiri Dostal, Butterworth-Heineman, 1993), chapter 5 para 2.2. we find the operational equation for a non-inverting amplifier as:
Vo{1+1/A[1+R2/R1//Zd//Zcm-+Zo/Rl(1+(R2+Rl)/R1//Zd//Zcm-)]}
=Vs[(R2/R1+1)(1+1/CMRR)+R2/Zcm-+Zo/A(1/Zd+1/CMRR(Zd//Zcm-))]
+Er(R2/R1+1+Zo/A(Zd//Zcm-))+Ir-.R2(1-Zo/A.R2) -----------(1)
Where
Vo=output voltage
Vs =input signal at non-inv input
A=open loop differential gain
R2=feedb res from output to inv input
R1=res from inv input to gnd
Zd=differential mode input impedance
Zcm=common mode input resistance
Zcm-=inv input common mode input impedance
Zo=open loop output impedance
CMRR=common mode rejection ratio
Er=input error voltage
Ir-=inv input error input current
We make the following assumptions:
Zcm=¥; Zd=¥; Zo=0; Er=0; Ir-=0
We then get:
Vo[1+1/A(1+R2/R1)]= Vs[(R2/R1+1)(1+1/CMRR)] -----------(2)
Reworking to the familiar V0/Vs for the closed loop gain, knowing that R2/R1+1=1/b, we get:
Vo/Vs=[(1/b)(1+1/CMRR)]/[1+1/A(1/b)] ------------(3)
Which can be reworked to:
Vo/Vs=[1+1/CMRR]/ [b+1/A] -------------(4)
Now let’s do a sanity check to see we didn’t lose it on the way down: assume that A (open loop gain) and CMRR both are infinite. We then get the familiar Vo/Vs=1/b, which we know to be correct, so equation (4) appears to be correct.
Let us do one more substition. Because A = differential open loop gain, we like to rework the CMRR to the common mode open loop gain. So, we say C= common mode open loop gain=1/CMRR.
We then get finally Vo/Vs=(1+C)/(b+1/A) ----------------(5)
Now to the 64k$ question: what does it mean? I don’t know what it means in detail, but we can draw some conclusions. (It is easier if you write it out in the traditional way with a horizontal divide line). It appears from (5) that the CMRR has a direct impact on the closed loop gain, and that this impact is of the same order as of the open loop differential gain A. Furthermore, it appears that this similarity doesn’t depend on the feedback factor.
One unknown (for me) is the degree of non-linearity of the CMRR or rather C. We know that A is non-linear, which in fact gives rise to the distortion. If C is of the same order of non-linearity as the differential gain A, we can conclude that CMRR DOES matter for the closed loop gain, in the same way as A matters. That again means that the inverting gain configuration has an advantage here over the non-inverting gain configuration.
To quote from the same reference, chapter 5 para 4.3:
“ The common mode rejection ratio is the most serious interfering factor in [non-inverting] operational circuits[..]. In many cases, it would be possible to compensate the [static] factor (1+C) by adjusting the feedback impedances [R1, R2]. However, troubles arise from the non-linearity of [C], i.e. the variation of [C] with signal. The resulting non-linearity of the [closed loop gain] is independent of the loop gain A.b (independent of the closed loop gain), and can only be removed by selecting a better and more expensive operational amplifier. This gives the [inverting amplifier configuration] a possible advantage in those applications where the relatively low input impedance does not matter.”
And, please if you see 'b' instead of the greek letter beta, you can substitute one for the other in the formulas. The 'yen' symbol should read infinity.
Anybody willing to take this further? I would love to see how we can put some numbers on this.
Jan Didden
From Operational Amplifiers (2nd edition), (Jiri Dostal, Butterworth-Heineman, 1993), chapter 5 para 2.2. we find the operational equation for a non-inverting amplifier as:
Vo{1+1/A[1+R2/R1//Zd//Zcm-+Zo/Rl(1+(R2+Rl)/R1//Zd//Zcm-)]}
=Vs[(R2/R1+1)(1+1/CMRR)+R2/Zcm-+Zo/A(1/Zd+1/CMRR(Zd//Zcm-))]
+Er(R2/R1+1+Zo/A(Zd//Zcm-))+Ir-.R2(1-Zo/A.R2) -----------(1)
Where
Vo=output voltage
Vs =input signal at non-inv input
A=open loop differential gain
R2=feedb res from output to inv input
R1=res from inv input to gnd
Zd=differential mode input impedance
Zcm=common mode input resistance
Zcm-=inv input common mode input impedance
Zo=open loop output impedance
CMRR=common mode rejection ratio
Er=input error voltage
Ir-=inv input error input current
We make the following assumptions:
Zcm=¥; Zd=¥; Zo=0; Er=0; Ir-=0
We then get:
Vo[1+1/A(1+R2/R1)]= Vs[(R2/R1+1)(1+1/CMRR)] -----------(2)
Reworking to the familiar V0/Vs for the closed loop gain, knowing that R2/R1+1=1/b, we get:
Vo/Vs=[(1/b)(1+1/CMRR)]/[1+1/A(1/b)] ------------(3)
Which can be reworked to:
Vo/Vs=[1+1/CMRR]/ [b+1/A] -------------(4)
Now let’s do a sanity check to see we didn’t lose it on the way down: assume that A (open loop gain) and CMRR both are infinite. We then get the familiar Vo/Vs=1/b, which we know to be correct, so equation (4) appears to be correct.
Let us do one more substition. Because A = differential open loop gain, we like to rework the CMRR to the common mode open loop gain. So, we say C= common mode open loop gain=1/CMRR.
We then get finally Vo/Vs=(1+C)/(b+1/A) ----------------(5)
Now to the 64k$ question: what does it mean? I don’t know what it means in detail, but we can draw some conclusions. (It is easier if you write it out in the traditional way with a horizontal divide line). It appears from (5) that the CMRR has a direct impact on the closed loop gain, and that this impact is of the same order as of the open loop differential gain A. Furthermore, it appears that this similarity doesn’t depend on the feedback factor.
One unknown (for me) is the degree of non-linearity of the CMRR or rather C. We know that A is non-linear, which in fact gives rise to the distortion. If C is of the same order of non-linearity as the differential gain A, we can conclude that CMRR DOES matter for the closed loop gain, in the same way as A matters. That again means that the inverting gain configuration has an advantage here over the non-inverting gain configuration.
To quote from the same reference, chapter 5 para 4.3:
“ The common mode rejection ratio is the most serious interfering factor in [non-inverting] operational circuits[..]. In many cases, it would be possible to compensate the [static] factor (1+C) by adjusting the feedback impedances [R1, R2]. However, troubles arise from the non-linearity of [C], i.e. the variation of [C] with signal. The resulting non-linearity of the [closed loop gain] is independent of the loop gain A.b (independent of the closed loop gain), and can only be removed by selecting a better and more expensive operational amplifier. This gives the [inverting amplifier configuration] a possible advantage in those applications where the relatively low input impedance does not matter.”
And, please if you see 'b' instead of the greek letter beta, you can substitute one for the other in the formulas. The 'yen' symbol should read infinity.
Anybody willing to take this further? I would love to see how we can put some numbers on this.
Jan Didden
janneman said:
I think we are one step closer to getting the answer as this deals with distortion not some hypothetical signal on both inputs discussed before.
I like to see some reality check on this cmrr-induced distortion.
most amps have good cmrr. the lm3875 from 120db (dc) down to 40db (@50khz), and opa541 from 110db to 65db (@50khz), and the lowly tl082 from 110db to 80db (@1mhz).
1+1/cmrr range from 1.01 (@50khz) to 1.000001 for the lm3875 and even smaller for others.
I don't know about you guys and my rubber ears certainly aren't good enough to hear that kind of difference.
Konnichiwa,
Well, leaving aside all the math and theoretical evaluation, how about some hard numbers?
Look here:
http://www.dself.dsl.pipex.com/ampins/webbop/2134.htm
This shows the difference between non-inverting (series feedback) and inverting (shunt feedback) operation of the OPA2134 (generally reckoned to be a good Audio Op-Amp). Some others also are shown via this page:
http://www.dself.dsl.pipex.com/ampins/webbop/opamp.htm
I would expect the LM3875 to do notably worse BTW....
Sayonara
janneman said:
Well, leaving aside all the math and theoretical evaluation, how about some hard numbers?
Look here:
http://www.dself.dsl.pipex.com/ampins/webbop/2134.htm
This shows the difference between non-inverting (series feedback) and inverting (shunt feedback) operation of the OPA2134 (generally reckoned to be a good Audio Op-Amp). Some others also are shown via this page:
http://www.dself.dsl.pipex.com/ampins/webbop/opamp.htm
I would expect the LM3875 to do notably worse BTW....
Sayonara
That depends on the nature of its LTP. Self has examples of bad-rep opamps that do not exhibit CM problems.
By the way, common mode distortion is as a rule problematic with JFET opamps, something identified and remedied by Jung a long time ago.
http://www.analog.com/UploadedFiles/Application_Notes/742022599AN232.pdf
Wow, I didn't know that people are concerned about THD 0.01% vs. 0.02% or even less for the 627 opamp.
millwood said:
This reminds me....
The plain and simple has become mysterious because the world of art as a whole has been so full of unrest, din, excitement, and delirium for so long... The drunkard finds it hard to be content with spring water, the harlot with morning prayers, the gambler with playing forfeits. Yet they were all unspoilt at birth.
But they have forgotten it, and it is hard to get back to the simple and primitive.
Carl August Nielsen, in Living Music, 1925
Joe R.
Werner said:
Werner, that ref you give from Walt is about non-linear input capacitance, which is not the same as CM distortion. CM distortion also exist when the input cap is perfectly voltage independent.
To answer another post above, apparently the measurements of Self clearly identify CM distortion in well regarded opamps, quote:
" The two THD plots show the device working at a gain of 3x in both shunt and series feedback modes. It is obvious that a problem emerges in the series plot, where the THD is higher by about three times at 5 Vrms and 10 kHz. This distortion increases with level, which immediately suggests common-mode distortion in the input stage. This is ironical since this input stage is supposed to be a selling point. "
Looking at the LM3875 data sheet which I happen to have here, I see a CMRR of 60dB at 6kHz. Static, that means a gain of 1.001 x the expected gain, or a deviation of 0.09dB. At higher freqs, that will increase, but indeed I doubt that anybody hears such a LINEAR distortion.
But Selfs results show that the CMRR is not constant during the signal cycle, much as the open loop gain is not constant during a signal cycle. (If it were, there would be no non-linear distortion). Especially around the zero crossing. open loop gain can vary by as much as a factor two. IF CMMM would also exhibit a similar non-linearity (which I am not sure, but Selfs results are convincing), that 0.09dB would vary during the signal cycle, which if it were symmetrical would give rise to 3rd order harmonics.
So however you cook it, CM distortion seems to be real, at least in some devices.
Jan Didden
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