| JoeBob |
What would the gain of the regular Thorsten/Peter Daniels gainclone be? 220k/10k, so a gain of 22, right?
Also, how high a gain could this design be pushed to? 40, 80, 100? |
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| needtubes |
| Actually, with 10k and 220k, the gain is 23... I don't know howhigh I would push the gain, though. (personally, for me, g=23 was higher than I like... but it worked out ok) |
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| moamps |
| quote: | Originally posted by needtubes
Actually, with 10k and 220k, the gain is 23... I don't know howhigh I would push the gain, though. (personally, for me, g=23 was higher than I like... but it worked out ok) |
Hi,
Actually, gain is 22. For inverting topology gain is R2/R1. Maximum gain is limited with bandwith you need.
Regards |
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| needtubes |
| Oops... I was thinking non-inverting... forgot that their circuits were inverting... :o Sorry. |
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| BrianGT |
| quote: | Originally posted by moamps
Hi,
Actually, gain is 22. For inverting topology gain is R2/R1. Maximum gain is limited with bandwith you need.
Regards |
If you account for the potentiometer in the gainclone, which is in series with R1, then it is not always 22, and changes with the movement of the wiper on the potentiometer.
I am wondering if the actual perfect gain for the gainclone is quite a bit lower then 22, since the average listening level isn't full volume. It might be interesting to figure out what exactly the gain is at the setting that you use on the gainclone in inverting configuration, and setting the same volume level and same gain in non-inverting configuration to compare.
Assuming a 100k potentiometer (thorsten's schematic), and the desired listening level is in the middle (linearly in the middle), then the gain would be 220k/(10k+50k) = a gain of 3.667.
--
Brian |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by BrianGT
Assuming a 100k potentiometer (thorsten's schematic), and the desired listening level is in the middle (linearly in the middle), then the gain would be 220k/(10k+50k) = a gain of 3.667.
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Your calculation is based on assumption (middle of travel) and incomplete math. The reality looks a lot more complex due to a number of interactions.
Sayonara |
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| travis |
| my sound card kept making my amp clip so i changed feedback resistors. i have 20k and 1.3k or 1.6k, i don't recall. doesn't oscillate though. |
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| Peter Daniel |
| I'm using 250k for feedback, 10k series resistor and 50K log. pot. I don't know about other people, but my source outputs only so much signal that I usually have the pot set at around 28th pos. on 31 detented potentiometer. It sounds pretty good this way and the gain is close to 30. |
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| Joe Rasmussen |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Your calculation is based on assumption (middle of travel) and incomplete math. The reality looks a lot more complex due to a number of interactions.
Sayonara |
Incomplete maths indeed - at mid position of pot... a quick check on the ol' sliderule... trumpets blaring... 6.286 .
The 100K pot must be regarded as 25K as the assumption must be that the source impedance, say from your CD Player, is low. So:
(220)/(10+25) = 6.286 .
When wiper is at the top (hold on to your ears) the gain = 22. So likewise when at the bottom, except gain of 22 times zero (no signal to amplify).
Now the math are complete! :nod:
Seriously though, anyone got a diff result?
Isn't it interesting that the calculation is different from shunt feedback, where gain is (R1+R2)/R2, whereas series (inverted) feedback is R1/R2.
Joe R.
PS: Oh yes, my preferred gain for LM3875 is +33dB. |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by Joe Rasmussen
Incomplete maths indeed - at mid position of pot... a quick check on the ol' sliderule... trumpets blaring... 6.286 .
Seriously though, anyone got a diff result?
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Well, if you calculate the actual gain from the input of the Pot with the pot in the center....
Then you get 50K : 50K//10k = 1:7 which in turn is amplified by 22, giving a net of 3.1428, yet the noisegain is as you have pointed out 6.286.
As said, it is a complex sort of circuit (simplicity is deceptive) but it works okay, especially if the feedback resistor is increased to 330K, which should avoid the sound degradation observed with a (22k) resistor from negative input to ground.
There are a great many ways to skin this particular Dog (I never skin cat's, I love cat's - can't stand dogs though).
Sayonara |
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| UrSv |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Well, if you calculate the actual gain from the input of the Pot with the pot in the center....
Then you get 50K : 50K//10k = 1:7 which in turn is amplified by 22, giving a net of 3.1428, yet the noisegain is as you have pointed out 6.286.
As said, it is a complex sort of circuit (simplicity is deceptive) but it works okay, especially if the feedback resistor is increased to 330K, which should avoid the sound degradation observed with a (22k) resistor from negative input to ground.
There are a great many ways to skin this particular Dog (I never skin cat's, I love cat's - can't stand dogs though).
Sayonara |
Interesting. What is the reason for the sound degradation as noted by some and how can the increase to 330 K avoid it? And what about the 18 K from pos input to ground? |
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| carlosfm |
| quote: | Originally posted by Kuei Yang Wang
As said, it is a complex sort of circuit (simplicity is deceptive) but it works okay, especially if the feedback resistor is increased to 330K, which should avoid the sound degradation observed with a (22k) resistor from negative input to ground.
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My GC is a power amp.
I have the 22k resistor to ground and I don't notice any degradation.
But I don't have any resistor to ground before de input cap.
And no pot.
It seams to me that if people use the 22k resistor to ground together with a resistor to ground before the cap they may be messing things up.:dodgy: |
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| millwood |
my way of determining amp gain is to consider the following factors:
1) most pre-amps output 1VRMS (or 1.4Vpp).
2) if you use 40v rail, the maximum output is probably 35v rail (assuming some losses at transformer and output devices).
3) so you want to amplify the 1.4Vpp signal to 35Vpp output and the gain is 25x.
of course, you may want to factor in output power, etc. as well in a real calculation but the above will set you to the ballpark.
My favorite is actually 28x as it gets a very "scientific" 100w out of a 8ohm load. |
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| carlosfm |
| quote: | Originally posted by millwood
My favorite is actually 28x as it gets a very "scientific" 100w out of a 8ohm load. |
:confused: :eek: |
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| moamps |
| quote: | Originally posted by millwood
[B]my way of determining amp gain is to consider the following factors:
1) most pre-amps output 1VRMS (or 1.4Vpp).
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1VRMS is ca 2.8Vpp or 1.4Vp ;) |
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| millwood |
| quote: | Originally posted by carlosfm
:confused: :eek: |
with a 1Vrms signal, and 28x gain, the output signal swings 28Vrms, which on 8ohm speakers generates (28v)^2/8=100w.
| quote: | Originally posted by moamps
1VRMS is ca 2.8Vpp or 1.4Vp ;) |
my fault, :) |
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| carlosfm |
| quote: | Originally posted by millwood
with a 1Vrms signal, and 28x gain, the output signal swings 28Vrms, which on 8ohm speakers generates (28v)^2/8=100w.
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Even using the maximum allowed PSU voltage, you can configure an LM3875 with 28x gain (or more) and you won't get 100w (I'm talking RMS).
It would be nice... |
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| millwood |
| quote: | Originally posted by carlosfm
Even using the maximum allowed PSU voltage, you can configure an LM3875 with 28x gain (or more) and you won't get 100w (I'm talking RMS).
It would be nice... |
I was thinking about a normal amp, not the gainclone specifically.
Yeah, it would have been nice, :) |
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| Joe Rasmussen |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Well, if you calculate the actual gain from the input of the Pot with the pot in the center....
Then you get 50K : 50K//10k = 1:7 which in turn is amplified by 22, giving a net of 3.1428, yet the noisegain is as you have pointed out 6.286.
As said, it is a complex sort of circuit (simplicity is deceptive) but it works okay, especially if the feedback resistor is increased to 330K, which should avoid the sound degradation observed with a (22k) resistor from negative input to ground.
There are a great many ways to skin this particular Dog (I never skin cat's, I love cat's - can't stand dogs though).
Sayonara |
I concur, in fact I loaded up CircuitMaker and constructed the circuit, put 1V into it and got 3.14V out. Interesting thing is that varying and even deleting the 22K made no difference, still around 3.14V out. It varies the noisegain but not the voltage gain. This is still assuming mid position of 100K pot. It occurs to me that it doesn't change the actual voltage gain because the 22K is across the virtual earth (- input) and real earth, so no change as there is no signal across it - at least in theory there should be a complete nul.
But changing the 'noisegain' is an improvement in two ways that I can see, 1) better stability (enhanced further by increasing 220K to 300K) and 2) controlling DC off-set.
It struck me from early on with your first schematic, that the middle position of the 100K pot was its Achilles heal - the 20dB gain required by National to ensure stability - but that has now been corrected.
Am I reading your thinking correctly on this?
Joe R. |
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| Joe Rasmussen |
| quote: | Originally posted by UrSv
Interesting. What is the reason for the sound degradation as noted by some and how can the increase to 330 K avoid it? And what about the 18 K from pos input to ground? |
I'd like to know that too... sound degradation. Ideally the 18K should be by-passed, does the fact it isn't create a sonic degradation?
The 18K, if I may answer this, is close in value to 220K in parallel with the 22K. This keeps the DC Off-set on the output low. The bipolar inputs draws current and if you don't balance the circuit it shows up on the output and puts DC into your speakers.
Joe R. |
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| UrSv |
| quote: | Originally posted by Joe Rasmussen
I'd like to know that too... sound degradation. Ideally the 18K should be by-passed, does the fact it isn't create a sonic degradation?
The 18K, if I may answer this, is close in value to 220K in parallel with the 22K. This keeps the DC Off-set on the output low. The bipolar inputs draws current and if you don't balance the circuit it shows up on the output and puts DC into your speakers.
Joe R. |
Thanks Joe, I was afraid the answer was already posted somewhere in the thread and I simply had missed it. I knew about the 18 K value being the value of 220 K in parallell with 22 K so my comment should rather have read:
Does the 18 K resistor also degrade the sound quality and if so how do I remedy that? |
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| Joe Rasmussen |
| quote: | Originally posted by UrSv
Thanks Joe, I was afraid the answer was already posted somewhere in the thread and I simply had missed it. I knew about the 18 K value being the value of 220 K in parallell with 22 K so my comment should rather have read:
Does the 18 K resistor also degrade the sound quality and if so how do I remedy that? |
By-passing it with a cap? It would need to be a large value, say 10uF, and not electrolytic, stating the bleeding obvious. But Thorsten reckons increasing 220K to 300K remedies it. Like to know what the thinking is behind that? Or have I missed something already posted?
Joe R. |
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| UrSv |
| Joe, actually I think KYW states that increasing the feedback resistor remedies the sound degradation of the 22 K from the negative input. I would also like to know why exactly that is but also as in my previous reply to you the question is how, and if, the 18 K from the POSITIVE input degrades the sound and how to remedy that. |
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| Peter Daniel |
| quote: | Originally posted by UrSv
... and if, the 18 K from the POSITIVE input degrades the sound and how to remedy that. |
How about shorting it?;) |
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| UrSv |
| Oh, it is short enough for my liking. About 10 mm.. ;) |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by UrSv
Joe, actually I think KYW states that increasing the feedback resistor remedies the sound degradation of the 22 K from the negative input. |
Nope, if the feedback resistor is increased to 330k the (noise)gain is always > 10, so 22k and 18k can be removed, Pot is 100k linear. Maximum gain is slightly > 30db.
Sayonara |
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| UrSv |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Nope, if the feedback resistor is increased to 330k the (noise)gain is always > 10, so 22k and 18k can be removed, Pot is 100k linear. Maximum gain is slightly > 30db.
Sayonara |
OK, clear. |
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| Peter Daniel |
| I did some listening comparisons, and the sound is different with 250K and 300K (what I used). You might try it yourself and decide which you prefer. I'm also using log pot and the control range is very good, both with average efficiency speakers and horns (100db efficient). So you might also consider using that type of pot instead of linear. |
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| carlosfm |
| quote: | Originally posted by Peter Daniel
How about shorting it?;) |
Simple answer, but with the manufacturing drift these chips have you may have from as little as 7~8mv DC Offset to around 50mv.
And I love my speakers.:bawling:
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Nope, if the feedback resistor is increased to 330k the (noise)gain is always > 10, so 22k and 18k can be removed, Pot is 100k linear. Maximum gain is slightly > 30db.
Sayonara |
I did a board with LM1875s and 330k as feedback resistors.
I used the 22k resistor as always, but for initial tests I ommited the 18k (direct wire).
I had almost 40mv DC.
So, I removed the direct wire and put 50k multi-turn pots.
Final resistance of the pots at 0.0mv: 18.9k.:devily:
But the 22k resistor remained in place, it's not as you're saying, Kuei.
Maby removing the 22k resistor...
Ahhh... nevermind, the board is finished and looks loovely with the 50k multi-turn pots.:devily: |
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| Peter Daniel |
| quote: | Originally posted by carlosfm
Simple answer, but with the manufacturing drift these chips have you may have from as little as 7~8mv DC Offset to around 50mv.
And I love my speakers.:bawling:
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I believe I love my speakers even more, but even 100mV won't make any damage to them. |
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| fedde |
No, and it keeps them comfortably warm... :)
Fedde |
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| Joe Rasmussen |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Nope, if the feedback resistor is increased to 330k the (noise)gain is always > 10, so 22k and 18k can be removed, Pot is 100k linear. Maximum gain is slightly > 30db.
Sayonara |
OK, let's get a handle on this, we seem to be discussing moving goalposts.
On the basis of above re 22K & 18K, it should look like this now: (?)
Won't this now give us 300mV DC on the output? The circuit (my apologies for amending it) seems to be missing something?
IF we left the 22K from (-) to ground, then we'd get 22mV DC Off-set, that is... not great, but not that bad.
| quote: | Originally posted by Peter Daniel
I did some listening comparisons, and the sound is different with 250K and 300K (what I used). You might try it yourself and decide which you prefer. I'm also using log pot and the control range is very good, both with average efficiency speakers and horns (100db efficient). So you might also consider using that type of pot instead of linear. |
Sure does change the sound... seems maybe there is a consensus that +30dB is good? If so it's come to the same one I made about 10 months ago.
IF I had 100dB sensitivity speakers, I'd choose a Log pot too, but with normal sensitivity - circa 90dB or a little less - I'd stick with Lin.
Joe R. |
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| Peter Daniel |
| quote: | Originally posted by Joe Rasmussen
Won't this now give us 300mV DC on the output? The circuit (my apologies for amending it) seems to be missing something?
Sure does change the sound... seems maybe there is a consensus that +30dB is good? If so it's come to the same one I made about 10 months ago.
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I built the circuit with 300k and the offset didn't change much, being at 28mV average. And I prefer 250k in feedback. |
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| millwood |
one thing I had already wondered about.
On the schematic Joe had posted. The wiper is essentially grounded, for A/C signal, by a combo of the input resister (10K) and feedback resistor (220K/300K) in parallel (the small speaker impedance not withstanding). so the wiper has a A/C load of less than 10K.
so even if the pot is log, is the output really log given that the wiper is loaded by a <10K resistor network? |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by Joe Rasmussen
OK, let's get a handle on this, we seem to be discussing moving goalposts.
On the basis of above re 22K & 18K, it should look like this now: (?)
Won't this now give us 300mV DC on the output? The circuit (my apologies for amending it) seems to be missing something?
IF we left the 22K from (-) to ground, then we'd get 22mV DC Off-set, that is... not great, but not that bad.
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Well, enough folks build the same circuit with a 220K feedback resistor and have low offset. Going up from 220k to 330k (NOT 300k - I'm not sure why keep insisting on that value) will increase the offset at the worst by 1/3rd.
| quote: | Originally posted by millwood
so even if the pot is log, is the output really log given that the wiper is loaded by a <10K resistor network? |
No, it is "hyper log", if you so want. Also, the fairly low AC load will cause all of sorts of interesting effects if conductive plastic (YUCK) pot's are used, as the wiper impedance on these is dependent on the current through the wiper....
Sayonara |
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| Fred Dieckmann |
| Walt Jung reported on the wiper contact nonlinearity for carbon pots over 25 years ago in The Audio Amateur PAT-5/WJ-lA preamp mod article as I remember, and recommended several Meg as a minimum load. I changed my 100K Alps Black Beauty pot for law faking cermets and will never go back, The Alps is very colored in tonal balance but does complement slightly hard or bright electronics. the loss in resolution is what really made me give up on them. There are some cheaper better sound conductive plastic pots than the Alps pot, but I take cermet pots over them as well. |
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| Pedja |
Hello Joe,
I think that bypassing supposed 22k from – input to ground would be very risky step. :att'n: I thought it wouldn’t be bad to tell this before someone tries that.
Pedja :goodbad: |
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| millwood |
| quote: | Originally posted by Joe Rasmussen
Won't this now give us 300mV DC on the output? |
in the particular schematic, the inverting input is essentially DC grounded via the feedback resistor (the speaker DC resistance is too little vis-a-vis the feedback resistor). National says that the typical input bias current for LM3875 is 0.2ua. and since the non-inverting end is grounded, so the output offset is 0.2ua*300K=60mv.
it would then reason that for lower DC offset, one shouldn't be using the coupling cap on the input so that the input resistor would help lower the resistance from the input to ground, thus lowering DC offset. |
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| Fred Dieckmann |
| Eliminate the coupling cap on the input and you then have a DC gain equal to the AC gain instead of unity gain as exist in the present topology. There is also a contribution from offset voltage between the inverting and the non-inverting inputs which is not a function of input bias current through the feedback resistors. Remove the cap and you also have the contribution of the preamp offset voltage times the closed loop gain. This will vary with pot setting. Offset due to bias current through the pot will change for different volume settings as well. This is all very basic op amp theory which is in most op amp tutorials. You will very likely be better off with an input capacitor for low and consistent offset voltage, which also can be nulled further by a DC voltage on the non-inverting input. This is very basic stuff that one should really make an effort to understand if your going to build even very simple Chip amplifiers. |
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| Joe Rasmussen |
| quote: | Originally posted by Peter Daniel
I built the circuit with 300k and the offset didn't change much, being at 28mV average. And I prefer 250k in feedback. |
As they said at Pearl Harbour... too many Zeros.
I posted that 300mV figure just as I was about to hop into the car and ten minutes later... :bulb: . My JLTi & The DIY Hybrid IGC uses 1M feedback resistors and gives 100mV DC Off-set if left uncorrected - so naturally 300K should give 30mV and not 300mV!
In fact the input currrent is 0.1uA, so:
0.0000001A x 300000R = 0.03V or 30mV, so this tallies nicely with your 28mV (250K, right).
While I don't think that is great, it is not a major downer. In fact, even if we had 4 Ohm speakers with 3 Ohm DCR, it is only 0.3mW dissipation in the voice coil (and a very slight pull of centre which will reduce effective Xmax), and it won't affect weeney tweeter voice coils due to high-pass C/O.
One thing I've noticed, the DC parameters of the 3875 are rather good, in fact especially that this isn't just an opamp but a power opamp!
Joe R. |
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| Joe Rasmussen |
| quote: | Originally posted by Fred Dieckmann
Eliminate the coupling cap on the input and you then have a DC gain equal to the AC gain instead of unity gain as exist in the present topology. There is also a contribution from offset voltage between the inverting and the non-inverting inputs which is not a function of input bias current through the feedback resistors. Remove the cap and you also have the contribution of the preamp offset voltage times the closed loop gain. This will vary with pot setting. Offset due to bias current through the pot will change for different volume settings as well. This is all very basic op amp theory which is in most op amp tutorials. You will very likely be better off with an input capacitor for low and consistent offset voltage, which also can be nulled further by a DC voltage on the non-inverting input. This is very basic stuff that one should really make an effort to understand if your going to build even very simple Chip amplifiers. |
I don't anyone would be game to eliminate the input cap.
Joe R. |
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| carlosfm |
| quote: | Originally posted by Peter Daniel
I believe I love my speakers even more, but even 100mV won't make any damage to them. |
I love my speakers more!:D |
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| carlosfm |
| quote: | Originally posted by Joe Rasmussen
I don't anyone would be game to eliminate the input cap.
Joe R. |
I wouldn't do that. |
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| janneman |
Guys,
I may have missed the arguments, but why not run the input on a 10k pot to the +input and scale the feedback network back to 1 K and 22K?
All those large values only generate unnecessary noise as well.
Every standard text will tell you that this setup gives unnecessary offset, which may not damage the speaker but puts the voice coil eccentric.
Why this peculiar ranges of values?
Jan Didden |
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| carlosfm |
| quote: | Originally posted by janneman
Guys,
I may have missed the arguments, but why not run the input on a 10k pot to the +input and scale the feedback network back to 1 K and 22K?
All those large values only generate unnecessary noise as well.
Every standard text will tell you that this setup gives unnecessary offset, which may not damage the speaker but puts the voice coil eccentric.
Why this peculiar ranges of values?
Jan Didden |
Jan, are you talking non-inverted topology?
Yes but it seams that this chip works better in inverted.
And you can't lower the input resistor on the inverted mode or the input impedance will be too loow. |
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| millwood |
| quote: | Originally posted by carlosfm
And you can't lower the input resistor on the inverted mode or the input impedance will be too loow. |
wouldn't an ideal opamp have infinite input impedance on both inverting and non-inverting ends? regardless of what input resistors you use. |
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| Joe Rasmussen |
| quote: | Originally posted by millwood
in the particular schematic, the inverting input is essentially DC grounded via the feedback resistor (the speaker DC resistance is too little vis-a-vis the feedback resistor). National says that the typical input bias current for LM3875 is 0.2ua. and since the non-inverting end is grounded, so the output offset is 0.2ua*300K=60mv.
it would then reason that for lower DC offset, one shouldn't be using the coupling cap on the input so that the input resistor would help lower the resistance from the input to ground, thus lowering DC offset. |
You are right re 60mV, but...
When I designed the JLTi I actually measured it @ 0.1uA...
So this circuit with 300K is the circuit I understand Thorsten now recommends, it should give 30mV DC. This tallies well with what Peter is getting, he got 28mV using 250K.
The 0.2uA is mentioned in the application note as typical, but several dozen samples I've used it is nearer to 0.1uA - maybe that is because I got them of the same batch? If so, then so has Peter?
But then we could expect that the new 'approved' circuit, which I gather looks like above (with 300K) will give DC Offsets in the range of 30-60mV.
So I think we are in heated agreement.
Joe R.
PS: Can KYW confirm the 'approved' part of my statement?
PPS: Eliminating input cap will amplify DC, so maybe not a good idea. |
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| carlosfm |
| quote: | Originally posted by Joe Rasmussen
But then we could expect that the new 'approved' circuit, which I gather looks like above (with 300K) will give DC Offsets in the range of 30-60mV.
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I woudn't sleep well.
I would test a multi-turn pot on the non-inv. input, measure it and then keep the pot or put a resistor.
That's how I do it.
These chips have a huge manufacturing drift. |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by janneman
I may have missed the arguments, but why not run the input on a 10k pot to the +input and scale the feedback network back to 1 K and 22K?
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Well, if you actually measure inverting vs. noninverting opertion (or listen) you will find that inverting operation sounds and measures notably better (see under Early Effect - common mode distortion).
| quote: | Originally posted by janneman
All those large values only generate unnecessary noise as well.
|
Do they now? How much Ein is generated under "worst case" conditions for the Inverting Amp? The 100k Pot is at middle travel and thus the effective impedance to ground becomes 25k from the Wiper, in series with 10k. The feedback resistor is 330k and thus we have 31k impedance to ground from the inverting input.
This equals around 25nV |/ Hz Resistor Johnson noise. The circuits noisegain at this worst case condition is a little above 10, so for a 20KHz bandwidth the output becomes 250nV * 141 = 35uV or 0.000035V, which is -98db below 2.83V / 1 Watt or -112db below nominal output of 25W/8R.
Okay, it is possible to have less noise, but not really with the LM3875 wich is pecified as having an Ein of > 2uV A WEIGHTED, which once linearised no lower if not higher than the resistor noise.
| quote: | Originally posted by janneman
Every standard text will tell you that this setup gives unnecessary offset, which may not damage the speaker but puts the voice coil eccentric.
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!!!??? This setup gives several quite neccesary effects (lowered distortion, better transient perfomance) compared to the conventional circuit, the rest (primarily offset) is minimal. Typhical offset reported with this circuit is < 40mV, the same is often found on many a commercial "High End" or "HiFi" Amplifier (examples - measured in stereophile Perraux R200 Amp at 9/13mV and Musical Fidelity A3.2 21/26mV). Using a law-faked linear pot as volume control also has many advantages, most of them sonically.
What you illustrate to me is what I feel is wrong with both the AES/JAES and the "objectivist" side of audio. You see something unusual and instead of actually trying to understand you simply bash as "bad design" and suggest that same self (inherently BAD) design that is being avoided for perfectly good reasons.
The fact is that those who choose unusual design approaches usually do so for VERY GOOD reasons. Another fact is that those choosing unusual approaches CAN very well design using the usual ones, but have for any number of reasons choosen to not do so. What would behoove those seeing such a design well is to try to understand the why (which may take some research) instead to simply say "you should have just done the same as everyone else"....
Sayonara |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by Joe Rasmussen
PS: Can KYW confirm the 'approved' part of my statement?
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I do not understand the part "approved"?
The IGC is a public domain circuit which has seen a few variations, all concerend with keeping the circuit stable, offset low and so on. There is not, nor can there be an "approved" circuit.
The 330k simply come from comments that people who added the 22k from the negative input to ground felt it sounded "worse". So, increasing the feedback resistor from 220k to 330k will make sure that the circuit is stable without added resistors.
Yes, in theory a 330k resistor in parallel with a 10k/4u7 series circuit should be added on the positive input to ensure identical offset and similar AC impedances, yet again people have found that doing so makes the sound worse, so if you leave this out and simply put a wire bridge in you will have 60mV worst case and probably < 30mV normal DC offset, which is normally considered unproblematic.
What has "approved" to with anything?
Sayonara |
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| janneman |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
Well, if you actually measure inverting vs. noninverting opertion (or listen) you will find that inverting operation sounds and measures notably better (see under Early Effect - common mode distortion).
Do they now? How much Ein is generated under "worst case" conditions for the Inverting Amp? The 100k Pot is at middle travel and thus the effective impedance to ground becomes 25k from the Wiper, in series with 10k. The feedback resistor is 330k and thus we have 31k impedance to ground from the inverting input.
This equals around 25nV |/ Hz Resistor Johnson noise. The circuits noisegain at this worst case condition is a little above 10, so for a 20KHz bandwidth the output becomes 250nV * 141 = 35uV or 0.000035V, which is -98db below 2.83V / 1 Watt or -112db below nominal output of 25W/8R.
Okay, it is possible to have less noise, but not really with the LM3875 wich is pecified as having an Ein of > 2uV A WEIGHTED, which once linearised no lower if not higher than the resistor noise.
!!!??? This setup gives several quite neccesary effects (lowered distortion, better transient perfomance) compared to the conventional circuit, the rest (primarily offset) is minimal. Typhical offset reported with this circuit is < 40mV, the same is often found on many a commercial "High End" or "HiFi" Amplifier (examples - measured in stereophile Perraux R200 Amp at 9/13mV and Musical Fidelity A3.2 21/26mV). Using a law-faked linear pot as volume control also has many advantages, most of them sonically.
What you illustrate to me is what I feel is wrong with both the AES/JAES and the "objectivist" side of audio. You see something unusual and instead of actually trying to understand you simply bash as "bad design" and suggest that same self (inherently BAD) design that is being avoided for perfectly good reasons.
The fact is that those who choose unusual design approaches usually do so for VERY GOOD reasons. Another fact is that those choosing unusual approaches CAN very well design using the usual ones, but have for any number of reasons choosen to not do so. What would behoove those seeing such a design well is to try to understand the why (which may take some research) instead to simply say "you should have just done the same as everyone else"....
Sayonara |
Well, yes, I understand all that, but I also see pages and pages of posts on that offset. I didn't say "you should do like everyone else", I just pointed out a way to get rid of the offset.
And your shot at the engineering world is totally misplaced. Do you really think that those engineers that designed your beloved GC chip did this by listening? Its the guys and girls from the AES and JASA that gave you those wonderfull chips and drivers and DACs that makes this forum possible at the first place. You wouldn't take such cheap shots if you had only a minimum of historical perspective.
And I DID say: "I may have missed the arguments", so you COULD have answered: "yes Jan, indeed this is because you didn't read all those previous posts, we've been there" or to that effect.
It seems you have an uncanny ability to read about 120dB more from my post than I put into. It really scares me. I say one thing and presto: out comes a complete analysis of the way I think, work, what have you.
Why is it that so many people on this forum, in particular you and a few others, are so focussed on scoring? You don't strike me as very insecure, so I am really at a loss here.
Maybe I am too naive, maybe this is all about scoring and not about audio at all?
Jan Didden |
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| millwood |
| quote: | Originally posted by millwood
wouldn't an ideal opamp have infinite input impedance on both inverting and non-inverting ends? regardless of what input resistors you use. |
I thought about it a little on the way to work and I think I was wrong. in an inverting design where the non-inverting input is grounded, the inverting input then acts like ground so the input impedance should be the input resistor in parallel with the feedback resistor. Since the feedback resistor is much bigger, the input impedance is mostly that of the input resistor. |
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| millwood |
| quote: | Originally posted by Joe Rasmussen
PPS: Eliminating input cap will amplify DC, so maybe not a good idea. |
two different sources of DC offset. The one you are talking about in this particular sentence is from a pre-amp, and the DC offset we had been talking about is one generated by the amp. |
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| janneman |
| quote: | Originally posted by millwood
two different sources of DC offset. The one you are talking about in this particular sentence is from a pre-amp, and the DC offset we had been talking about is one generated by the amp. |
I think he is talking about the GC. If you eliminate the input cap, the CG gain goes all the way to DC, meaning that the DC offset at the input is also amplified. It will also make the offset dependent on the volume setting.
Jan Didden |
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| millwood |
| quote: | Originally posted by janneman
I think he is talking about the GC. If you eliminate the input cap, the CG gain goes all the way to DC, meaning that the DC offset at the input is also amplified. |
you are right, having the input cap essentially makes the amp look like a follower (1x gain) for DC offset thus reducing DC offset. However, the source of the DC offset I was talking about here is input biasing. and it is only dependent on the resistance from the inverting input to ground. so it stands to reason that lower resistance there (for example, lower feedback resistor, or elimination of the input cap) would help lower (that kind of) DC offset.
I would agree with you 100% if the DC offset were caused by something else.
while I agree that eliminating the input cap helps from the perspective of reducing input bias induced DC offset - got to highlight it, it has many down sides as had been pointed out by many, including the one below.
| quote: | Originally posted by janneman
It will also make the offset dependent on the volume setting.
Jan Didden |
one more thing. If you lower the voltage on the non-inverting end, you may be able to offset the DC offset caused on the inverting end. Take the 60mv DC offset for example. The closed loop gain of the amp is 30x, right? so a 60mv/30x=2mv negative voltage on the non-inverting end will do it. since the input bias on the non-inverting end is also 0.2ua, you can ground a resistor of 2mv/0.2ua=10K (?) to eliminate DC offset caused by input-bias.
And if you work through the math, I am sure that that resistor takes always the value of the input resistor, an assertion that had been made a few times before. |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by janneman
And your shot at the engineering world is totally misplaced.
|
Is it?
| quote: | Originally posted by janneman
Do you really think that those engineers that designed your beloved GC chip did this by listening?
|
Well, there is a telling little note at the end of the BP200 Application note for the Overture chips....
Also, certain measurements have shown behavious that suggest that the results of a number of studies on hearing physiology/psychology where taken into account with the LM Overture series, but NOT it virtually ANY other series of chips.... I find that interesting. I guess someone should talk to the Head Dude at NS when they developed the "Overture" core.... I do have sneaky suspicions.
| quote: | Originally posted by janneman
Its the guys and girls from the AES and JASA that gave you those wonderfull chips and drivers and DACs that makes this forum possible at the first place. You wouldn't take such cheap shots if you had only a minimum of historical perspective.
|
I am not complaining so much about history (other than fairly recent one), but present.
| quote: | Originally posted by janneman
And I DID say: "I may have missed the arguments", so you COULD have answered: "yes Jan, indeed this is because you didn't read all those previous posts, we've been there" or to that effect.
|
Okay Jan, I THOUGHT the argument for inverting mode use of Op-Amp's for removal or reduction of several distortion mechanisms was one that is well published all across the board, including the writings of Hawkesford, Duncan, Self and others, so that it was "obvious" to those skilled in the art (of electronics).
| quote: | Originally posted by janneman
It seems you have an uncanny ability to read about 120dB more from my post than I put into. It really scares me. I say one thing and presto: out comes a complete analysis of the way I think, work, what have you.
|
Actually, perhaps I am wrong and mistaken as to my assumptions what constitues "general knowledge" among other EE's? Given that I no longer actively, professionally work in the field and in fact note Solid State only in passing I make the assumption that if I have heard about it and am aware of it, that is (should) constitue(es) general knowledge.
So if it is generally known (and it really should be) that inverting operation leads to more inherent linearity in Op-Amp's having differential input structures (discrete or monolithic makes ZIP difference) then a criticism that says "you shouldd really use non-inverting mode" says to me:
"Look buster, just toe the party line or we run out of the club!" (incidentally, I'm no longer in the club for exactly those reasons)
Now I have mistaken this myself and what you where saying to me was "I don't understand". In which case I will still ask you why did not cary out some research of your own on the topic?
| quote: | Originally posted by janneman
Why is it that so many people on this forum, in particular you and a few others, are so focussed on scoring?
|
I am not focused on scoring. I do however as vigerously attack complacency and mediocracy as well as mindless ratteling of the orthodox position as your average "objectivist" takes any opportunity to attack subjects from Resistor/Capacitor/Cable "sound" to Jah know what. I do this partially as "excercise" and secondly because I feel that too rarely a well founded active defense is mounted in public groups against this extreme orthodox objectivist position.
In addition I do have an extremely low tolerance to what I percieve as Trolls and also people who I percieve to behave deliberatly Idiotic. My perception may be very flawed, true.
But let me ask you honestly, when you see an unorthodox circuit, is your first reaction "I wonder why they did this and what improvement they gain?" or is it "They really should have done A/B/C/X like everyone else does!". The feeling I get from your postings is the latter, a simple defense and re-guritation of orthodoxy.
It is something that I have problems with in ANY subject area. By definition the orthodox position is ultimatly 99% wrong, though often only slightly wrong, but enough to make it rather unusable.
If I mistake you, I am sincerely sorry. However in this case I would perhaps suggest to avoid to suggest as "fix" to any unorthodox circuit you see a reversion to orthodoxy. Maybe just ask "why" or look it up?
Sayonara |
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| millwood |
| is there really a detectable difference between inverting and non-inverting? and if so why? |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by millwood
is there really a detectable difference between inverting and non-inverting? |
Yes.
| quote: | Originally posted by millwood
and if so why? |
Isn't it obvious? Long tailed pair inputs have a very limited Common Mode rejection (regardless of device, be it Valve, BJT or FET) and a non-inverting Amplifier has a common mode signal equal to the signal, an inverting amplifier equal to ZERO.
The CMRR generally falls off according to 2nd order function and is pretty low to start with, causing added distortion (usually around 4 - 6db over inverting mode), which rises above a designdependent point with a 2nd order slope. If you actually make a resonably linear amplifier (in other stages) you may find this nonlinearity at higher frequencies to determine the behaviour of the (looped feedback) Amplifier!!!!
The implications for non-harmonic distortion components (especially all sorts of noise modulation and PIM) should also be obvious.
Again for starters check Self on the subject. He very well describes the mechanisms and some possible partial remedies (apart from my own - "Invert Da Suckah!!!!" or "Death to the Long Tail Pair - CFB RULEZ").
If you are asking - is there an audible difference?
My experience is "Yes", but I also hear differences between different Capacitors, Resistors and Cables, so WTFDIK?
Sayonara |
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| millwood |
| quote: | Originally posted by Kuei Yang Wang
Isn't it obvious? Long tailed pair inputs have a very limited Common Mode rejection (regardless of device, be it Valve, BJT or FET) and a non-inverting Amplifier has a common mode signal equal to the signal, an inverting amplifier equal to ZERO.
|
so do I get you right that having a common mode signal on both inputs is better than not having a common mode signal on both inputs?
If so, why don't we all inject identical random noise on our amplifier's input and feedback ends and our $10 boombox will sound as nice as our Krell / or whatever of your favorite brands, :)
Seriously, I don't understand your reasoning. CMRR is what it is, no matter what signal you have on either end. |
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| Joe Rasmussen |
| quote: | Originally posted by Kuei Yang Wang
Konnichiwa,
What has "approved" to with anything?
Sayonara |
I know what you mean, but I think you misunderstood me.
I suppose I felt apologetic for modifying/amending/appending a circuit diagram/schematic that I attributed to you as the source (is it?). Like violating 'copyright' but hopefully for the right motive.
In a sense if someone posts a schematic or anything graphic, that doesn't entirely make it PD, does it?
Also, I wanted to elicit from you a nod (in that sense 'approval') that the circuit shown is now what you had in mind. Nothing more than that. I think we got it, the nod I mean. And I agree with the change, for whatever that is worth.
At least we are not discussing 'moving goalposts' as we were earlier on, the schematic is now there for us to reference to.
Joe R. |
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| Kuei Yang Wang |
Konniciwa,
| quote: | Originally posted by millwood
so do I get you right that having a common mode signal on both inputs is better than not having a common mode signal on both inputs?
|
No, you got it wrong way around.
| quote: | Originally posted by millwood
If so, why don't we all inject identical random noise on our amplifier's input and feedback ends and our $10 boombox will sound as nice as our Krell / or whatever of your favorite brands, :)
|
Sorry? I seem to miss your point. What has random noise to do with this?
| quote: | Originally posted by millwood
Seriously, I don't understand your reasoning. CMRR is what it is, no matter what signal you have on either end.
|
Do me a favour, PLEASE. Read Self's bloody book and articles.
It saves me miles of explaining and his Book really should be a standard on everyones shelfs (if only to understand why making amplifiers the usual way is usually a bad idea). You can have my one if you put cover price plus shipping to a childrens charity (I learned everything to learn from it ages ago).
Sayonara |
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| millwood |
| quote: | Originally posted by Kuei Yang Wang
Konniciwa,
No, you got it wrong way around.
|
here is what you wrote:| quote: | | a non-inverting Amplifier has a common mode signal equal to the signal, an inverting amplifier equal to ZERO. |
so what exactly is a "common mode signal"? |
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| Kuei Yang Wang |
Konnnichiwa,
| quote: | Originally posted by millwood
here is what you wrote:
|
Thanks for reminding me. I could have otherwise forgotten (sarkasm alert for non europeans).
| quote: | Originally posted by millwood
so what exactly is a "common mode signal"?
|
!!!???? You NEED TO ASK????
It is a signal (actually AC or DC, but we are talking AC here) that is different from the "reference" (usually the negative rail) and common to both inputs. Any such signal will have any number of effects if applied to a LTP (Long tailed Pair) stage.
These have been documented for ages and ages since around the 1950 or was that the 1590 - ahhm, no, no electronics then, so 1950's it is (I do have problems with ancient history - too much of it to keep current - sorry).
Of course, my "mistake" is not to treat the "Op-Amp" (monolithic or dicrete) as black box, but as a circuit usually composed usually out of LTP, VAS & CAS with megatons of loop feedback....
Sayonara
BTW, for the uninitated, LTP = Long Tailed Pair, VAS = Voltage Amplification Stage, CAS = Current Amplification Stage |
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| millwood |
| quote: | Originally posted by Kuei Yang Wang
It is a signal (actually AC or DC, but we are talking AC here) that is different from the "reference" (usually the negative rail) and common to both inputs. Any such signal will have any number of effects if applied to a LTP (Long tailed Pair) stage.
|
Thanks for that.
So a common mode signal is one that is present on both inputs of an opamp.
And for a non-inverting opamp, both inputs have the same signal (which is the input signal). to which I said earlier| quote: | | having a common mode signal on both inputs |
.
And for an inverting opamp, there is no common mode signal (both ends are AC grounded - unless you consider the ground as a signal source). to which I said earlier| quote: | | having a common mode signal on both inputs |
.
and you had stated earlier:
| quote: | | Long tailed pair inputs have a very limited Common Mode rejection (regardless of device, be it Valve, BJT or FET) and a non-inverting Amplifier has a common mode signal equal to the signal, an inverting amplifier equal to ZERO. |
questions:
1) why was my understanding not consistent with your statements?
2) why is having common signals on both inputs better than NOT having common signals on both inputs?
I fully expect that you can explain this to us all without resorting having us read a book, :) |
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| Kuei Yang Wang |
Konnichiwa,
Bludie quotes.... Lets do it olde style....
> So a common mode signal is one that is present on both
> inputs of an opamp.
AND is different from the "reference" primarily in AC terms.
> And for a non-inverting opamp, both inputs have the same
> signal (which is the input signal). to which I said earlier quote:
> having a common mode signal on both inputs
Yes.
> And for an inverting opamp, there is no common mode
> signal (both ends are AC grounded - unless you consider
> the ground as a signal source).
Not quite - the positive input is (AC) ground and becomes the "reference" input for the feedback system. The inverting input will be FORCED by the feedback to follow the non-inverting input, whch has no AC voltage.
> to which I said earlier quote: having a common mode signal on
> both inputs
And I said - NO. You MAY have forgotten to insert the key word "NO", or you may have the "common mode" definition wrong (a Common Mode signal BY DEFINITION exists on both inputs).
Yet you claim an (AC) common mode signal exists for the two inputs of the Op-Amp's in inverting mode, based on your writing. And I cannot agree with that, sorry. It is wrong.
> 1) why was my understanding not consistent with your
> statements?
See above.
> 2) why is having common signals on both inputs better
> than NOT having common signals on both inputs?
You make no sense. Common Mode Signals are COMMON to both inputs ALL THE TIME.
As to why is having NO common mode signal (my claim - please re-read what I wrote) is better than having one - isn't it BLINDINGLY OBVIOUS? The CM signal will bleed heavily distorted into the rest of the amplifier circuit.
> I fully expect that you can explain this to us all without
> resorting having us read a book,
I fully expect that you AT THE VERY LEAST exhaust the generically availabe sources on a topic before bothering me.
I guess that leaves us at an impasse? I expect you to inform yourself if you wish to talk as "INFORMED" participant (you may participate as admitted ignoramus as much as you like BTW).
You expect me to teach you for FREE what you omited to study and which is widely and readily available.
I THINK NOT. I am happy to comment on "what works", but I do not run on-line EE101 and 201 courses, sorry. This well's winch creeks on that stuff.
Sayonara |
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| millwood |
| quote: | Originally posted by Kuei Yang Wang
You expect me to teach you for FREE what you omited to study and which is widely and readily available.
Sayonara |
No, I wasn't expecting you being such a bad teacher, ;)
I mistakenly thought that you said an non-inverting design is better. My bad. and appologies.
I figured out what went wrong in the above exchange. Your argument is essentially that as long as CMRR isn't infinite, any common mode signal (presenting on both inputs) will be amplified by the amp and be presented at the output.
an inverting design doesn't have a common mode signal (unless you consider ground being a singal source);
a non-inverting design has a common mode signal (the input signal itself).
and it beats me on two things:
1) why is having a common mode signal which is identical to your input signal not desirable?
2) maybe you are the exception but how many of us can hear a -120db signal?
I cannot.
BTW, have you thought about taken the audio challenge posted recent at the other form? the prize is $10K, too little for you I suppose? |
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| Kuei Yang Wang |
Konnichiwa,
| quote: | Originally posted by millwood
I mistakenly thought that you said an non-inverting design is better. My bad. and appologies.
|
No, my point was that inverting designs are "better" (in the way Winnie the Poo considers "better") and in certain (most) applications more apropriate.
| quote: | Originally posted by millwood
1) why is having a common mode signal which is identical to your input signal not desirable?
|
If you still don't got it - RTFB (Read The *******g Books).
Come on, this EE101 stuff.....
| quote: | Originally posted by millwood
2) maybe you are the exception but how many of us can hear a -120db signal?
|
Such a blanket statement (first, where does you "120db" figure originate, second, once you have adjusted it to reality - have you cionsidered "nonharmonic" distortion mechanisms?
And have you ever considered the CMRR of Op-Amp's past DC? It makes for interesting Graphs. The closed loop CMRR for the rather "wideband" and "audio grade" OPA604 drops below 90db @ 10KHz. Now this matters not much, unless you have a lot of noise (from delta signa converters?) creeping in above say 100KHz where the CMRR is only a little above 60db.
And I notice the NE 5532 is not even charaterised for CMRR vs. Frequency. You may perpetuate the myth of the dstortion free op-amp, except under noiseload conditions things still look way dire.... Of course, that is not "THD & N", it's just ALL nonlinearities lumped - no more "cherry picking....
| quote: | Originally posted by millwood
BTW, have you thought about taken the audio challenge posted recent at the other form? the prize is $10K, too little for you I suppose?
|
No, not particulary, primarily because all this challenge business is statistically so rank, it makes 2 Week old rotten fish smell sweet, but secondly because I was not aware of such.
Now, if I can define the audition process and the statistics such as to even out the likelyhood of type 1 and type 2 errors for small audible differences, I'll have no need to even be involved in first place anyway, as my own contribution (even with the hearing of a Vampire Bat) would be infinitisimal, the first and foremost point of entrance for statistics anyway.
So please, do yourself a favour, try studying the subject AT THE VERY LEAST to the limits documented by "objectivist" contibutors and then also study physio/psychoacoustocs to a level of reasonable competence. Combining both fully "objectivist" branches of science gives interesting inshighst, add o this a little understanding of dynamic systems (electronic or not) and you may go far, assuming you wish to.
It all boils down back to doing your homework, instead of substituting assumptions of others for your own knowledge.... Hence I usually try to avoid teaching my own prejudices and views to other and instead expect them to go and find their own darn answers, if they really want answeres (as opposed to re-inforcement of a position already held)....
Sayonara |
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| millwood |
| quote: | Originally posted by Kuei Yang Wang
Such a blanket statement (first, where does you "120db" figure originate, second, |
Come on, Mr. Sayonara. It is so EE101 stuff and you still ask questions about it?
No more free education of you on this forum OK. Go read the *******g book!
:)
This is turning into such a useless discussion.
BTW, I laughed so hard when you had to rely on alpha and beta to get yourself out of a challenge. that's so classic Mr. Sayonara. |
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| Bricolo |
Why do you all call Kuei Yang Wang Mr Sayonara?
If I'm right, Konnichiwa means "Hi" and Sayonara means "goodbye"
I don't thinkt that Sayonara means the same thing as when other posters wite "Joe", "Alex", "Peter" at the end of theyr posts |
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| Joe Rasmussen |
| quote: | Originally posted by millwood
Come on, Mr. Sayonara. It is so EE101 stuff and you still ask questions about it?
No more free education of you on this forum OK. Go read the *******g book!
:)
This is turning into such a useless discussion.
BTW, I laughed so hard when you had to rely on alpha and beta to get yourself out of a challenge. that's so classic Mr. Sayonara. |
Hi millwood
Check out Douglas Self's web site and then bookmark it!!!!!!! :nod:
http://www.dself.dsl.pipex.com/ampins/ampins.htm
Perhaps more specifically the page: Distortion In Power Amplifiers
http://www.dself.dsl.pipex.com/ampins/dipa/dipa.htm
Joe R. |
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| millwood |
thanks for the link.
here is what it says about common mode distortion:
| quote: | 3.2 Input-Stage Common-Mode Distortion.
This does not appear to exist at detectable levels in normal amplifier circuitry. |
the discussion about thermal distortion is also quite interesting. He essentially said that for discrete devices, it is unlikely we will see thermal distortion above 10hz due to large thermal inertia. I remember reading from another thread where another author said that thermal distortion can be upto 10KHz. |
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| janneman |
I looked at the data sheet for the LM3857 yesterday, IIRW the CMRR is about 40dB at higher audio freqs. That means that a common mode signal of say 100mV generates up to 1mV of CM error that is amplified and appears at the output. Wouldn't that be significant? I think I am with KYW here.
Jan Didden |
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| millwood |
| quote: | Originally posted by janneman
I looked at the data sheet for the LM3857 yesterday, IIRW the CMRR is about 40dB at higher audio freqs. That means that a common mode signal of say 100mV generates up to 1mV of CM error that is amplified and appears at the output. Wouldn't that be significant? I think I am with KYW here.
Jan Didden |
and that 1mv is amplified to a 20mv signal on the speaker (of 8ohm), or how many watts is that?
plus, what is that common mode signal again? it happens to be the input signal. |
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| jh6you |
I do not understand very well.
But, CMRR = differential gain/ common mode gain.
If it is 40dB, then differential gain is 100*common mode noise gain. So what?
:yinyang: |
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| Werner |
| quote: | Originally posted by millwood
plus, what is that common mode signal again? it happens to be the input signal. |
Put that man in room 101 with Naim-driven Bose Acoustimess speakers at full blast.
the schoolbook transfer of an opamp is
Vo = Aol (Vp - Vn)
with Aol likely being some ugly non-linear coeffient.
if you stay in school a day longer than that, the equation becomes
Vo = Aol (Vp - Vn) + Acm (Vp + Vn)
with Acm another nasty non-linear thingy.
I sort of admire T's patience in this ... |
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| jh6you |
| quote: | Originally posted by millwood
common mode signal again?
|
Am I wright? Two signal voltages of the same phase, frequency and amplitude applied to two inputs. Ideally, they are cancelled at the output, to zero.
:yinyang: |
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| Fred Dieckmann |
Using inverting mode for op amp circuits is recommended by very good op amp applications guys that know what they are talking about. Relying of Mr. Self opinions on amplifiers is usually a pretty good indication you're on the wrong side in my experience.
John Curl designed with an inverting topology for the JC-2 about a quarter of century ago. Nelson Pass's Aleph X designs use an inverting topology. The reason most amps do not use an inverting topology is the input impedance must be lower the 100K ohms that everyone seems to use. It is a better topology and for reasons including some independent of signal level.
It appears to me that difference between the DIY God of Thunder's stance and that of the non inverting crowd is, as Mark Twain put it, "the difference between the lightning and the lightning bug.
http://www.geocities.com/Hollywood/...96/optimist.wav
You're out of the woods,
you're out of the dark,
you're out of the night.
Step into the sun,
step into the light,
then open your eyes |
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| Werner |
Cancelled at the output???
No. Read what I wrote.
(
Kuei, with a concept like CM/CMRR apparently not belonging to common electronics knowledge anymore (*) I smell an opportunity for a new amplifier brand and matching Marketing Campaign.
(*) I wonder when they'll start forgetting about Miller.
) |
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| jh6you |
| quote: | Originally posted by Werner
Cancelled at the output???
|
I said, "ideally."
If CMRR is 40dB,
when differential gain (Adf) is 100, common mode gain (Acm) is 1,
when Adf is 90, Acm is 0.9,
when 80, 0.8,
when 70, 0.7,
when 60, 0.6 ...
according to books.
:yinyang: |
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| millwood |
| quote: | Originally posted by jh6you
Am I wright? Two signal voltages of the same phase, frequency and amplitude applied to two inputs.
yes.
Ideally, they are cancelled at the output, to zero.
:yinyang: |
yes you are. But none of us live in that ideal world, :)
take numbers for example. Suppose you have a less than ideal opamp, and its differential gain is -20x (the inverting gainclone is in the ballpark), and CMRR of 40db (at 50Khz I think). and you apply two voltages, 10v (at the inverting end) and -11v at the non-inverting end).
the common mode signal (10v) will be amplified by the model at -40db of its differential gain. that is 20x/100x (40db is 100x)=0.2x.
so the end result is 1v (the differential signal)x20x+10v (the common mode signal)x0.2x=22v, vs. 20v of an ideal opamp.
now, which one of us has built a gain clone with 10v 50khz common mode signal applied to both ends? |
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| millwood |
| quote: | Originally posted by Werner
blahblahblah deleted |
Thanks for that great non-answer.
Now, what is the common mode signal again in this discussion? Did I hear "input signal"? |
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| jh6you |
millwood
Pls don't do overaction if you understand what I mean.
:yinyang: |
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| SY |
| quote: | | 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. |
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| millwood |
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. |
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| Peter Daniel |
| quote: | Originally posted by millwood
unfortunately I don't have a probe.
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I could send you one, if it helps;) |
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| SY |
| Well, even a 1x probe should work. You do have an oscilloscope, don't you? |
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| millwood |
| quote: | Originally posted by SY
You do have an oscilloscope, don't you? |
no. maybe I can ask my employer for one, :) |
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| fedde |
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 |
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| SY |
| quote: | | no. maybe I can ask my employer for one, |
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. |
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