People who design commercial audio equipment know that audio journalists usually have sufficient technical skills to take the lid off something and have a quick look inside; similarly their readers think they know what a cap does so if a cap is good then a big cap must be better? The rows of big caps tell a good story, but they are irrelevant (or maybe counter-productive) to sound quality.ginetto61 said:
Actually, when you see something like this (large electrolytics around OPamp packages) there are two scenarios:
1) What DF96 above said, or
2) You might be missing something - possibly two small resistors per OPamp in series with the power supply feed.
In the latter case what is going on is an attempt to isolate all currents relevant to the particular OPamp except input and output related, to a small 'island' of ground plane or a local star ground around the OPamp. It prevents the power supply distribution from conducting unwanted current components around the complete system. In this scenario, each OPamp is fed from tha common power supply via a small resistor (sometimes a lossy inductor - with relatively high series resistance) and has a local decoupling network made out of a large and small cap from it's pin to ground. Sometimes there might be other components such as a zobel networt to compensate for some misbehavior of the large cap ESL. What this really is, is a (not so) 'poor man's' shunt regulator local to the OPAmp. It does indeed help in some cases.
That being said, you will see caps in the 100s uF range but rarely in 1000s of uF, unless the manufacturer has a huge surplus on stock and is trying to get rid of them, while attempting to look clever and mysterious 😛
1) What DF96 above said, or
2) You might be missing something - possibly two small resistors per OPamp in series with the power supply feed.
In the latter case what is going on is an attempt to isolate all currents relevant to the particular OPamp except input and output related, to a small 'island' of ground plane or a local star ground around the OPamp. It prevents the power supply distribution from conducting unwanted current components around the complete system. In this scenario, each OPamp is fed from tha common power supply via a small resistor (sometimes a lossy inductor - with relatively high series resistance) and has a local decoupling network made out of a large and small cap from it's pin to ground. Sometimes there might be other components such as a zobel networt to compensate for some misbehavior of the large cap ESL. What this really is, is a (not so) 'poor man's' shunt regulator local to the OPAmp. It does indeed help in some cases.
That being said, you will see caps in the 100s uF range but rarely in 1000s of uF, unless the manufacturer has a huge surplus on stock and is trying to get rid of them, while attempting to look clever and mysterious 😛
Dear self-thinkers!
You have forgotten both mister Ohm and Kirchoff. Big cap near the poewr pins in a power source in the rightest place: as near to the power consumer as it can be. Stop thinkig about impedances etc, you don't understand the basics.
Try a 4700 uF Panasonic FC at every opamp power pin and use fast amp, like LM 6171, and you won't believe it's opamp playing.
You have forgotten both mister Ohm and Kirchoff. Big cap near the poewr pins in a power source in the rightest place: as near to the power consumer as it can be. Stop thinkig about impedances etc, you don't understand the basics.
Try a 4700 uF Panasonic FC at every opamp power pin and use fast amp, like LM 6171, and you won't believe it's opamp playing.
On the contrary, it is those who place big caps where smaller caps should be who forget their circuit theory. Thinking about impedances is the only way to go; those who don't understand the basics may have to think about something else, such as 'bigger=better'.Ola said:
It is particularly important that fast opamps have good HF power pin decoupling; this means small caps. Get this wrong, and instead of sounding like an opamp playing (i.e. good sound reproduction) you may get distortion from parasitic oscillation which some may confuse with extra 'detail' in the music.
Hi and thanks again. I understand much better now that it could be a case of "marketing driven perverse engineering".
But still i have not completely realized why it could be even detrimental for performance. Maybe irrilevant but even counter-productive. Why ?
and as the caps look like average Rubycon a re-capping with something higher grade could be advisable ? and which caps ? Maybe Nichicon gold 105C ?
Thanks again, gino
Hi and thanks a lot for the very interesting advice
You mean a small resistor (how much ?) on the + pin of the capacitors acting like an RC filter ?
unfortunately i do not have the schematic
However i was also thinking another thing
If the max current from an opamp to the load is let's say 1 mA in the absolutely worst scenario, how much would be the current draw of the op-amp in that conditions ?
i think extremely low ... i am wrong ?
But the idea of a RC filter close to the op-amp seems very sane
Thanks again, gino
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Physically big caps introduce inductance. This degrades HF decoupling, yet it is HF decoupling which is most needed.
Hi and thanks a lot again.
Strangely i do not see inductance mentioned in the data sheets
http://industrial.panasonic.com/lecs/www-data/pdf/ABA0000/ABA0000CE22.pdf
I see only mentioned:
Ripple Current (100 kHz) (+105 °C)
Impedance (100 kHz) (+20 °C)
thanks again, gino
You won't see inductance in most cap datasheets. This is because in most cases it is dominated by lead length. To estimate, replace the cap by a piece of wire. Then find the inductance of the resulting circuit loop from the chip back to the chip. Bigger the cap, bigger the loop so more inductance.
Hi thanks a lot again and i promise the last question really 😱
If a line stage will be required to put out let's say max 1 mA (to be generous) to the load i guess that in those conditions the current drawn by the opamp will be something like 0.1 mA ?
So there will be really no need to go over 50uF x voltage rail (again to be generous). Am i right ?
I was clearly overrating the actual power consumption of the opa even in the worst scenario
Thanks a lot again, gino
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Where does the load current come from? Draw a loop; current always flows in loops.ginetto61 said:
To go from current to capacitance you have to travel via frequency, and then arrive at voltage.
The opamp is driving the (1mA) current into the load? In which case the opamp can't draw less than its putting out unless its got some clever buck switching stage internally.
If you're concerned about SQ then you'll need to know the PSRR of your opamp to arrive at a suitable power supply impedance. Hint - PSRR plots in opamp datasheets typically aren't measurements, they're from simulations and those sims aren't normally done with any output signal from the opamp, just the test stimulus on one or other supply.
Hi and thanks for the kind reply but i do not understand.
You are talking with an ignorant really ... 😱
Let me rephrase the question.
When the opamp is outputting 1mA its consumption is equal to the quiescent current ? less ? more ?
you mean that i should measure it to be sure of the value ?
That is difficult. My only chance is to start from a kit an maybe carry out marginal mods to it. Measurements and even more design are out of my reach completely.
Thanks again, gino
The quiescent current passes from +ve supply to -ve supply.
There is no current to power ground and no current to the load.
If you connect a load and pass current to that load then the +ve and -ve supply currents change.
If the quiescent current for the whole opamp is 2mA, half of which is the output pair bias current, then when output current = zero then the TWO +ve supply currents are 1mA & 1mA. The bias currents through the front end of the opamp hardly change for changes in output current.
The TWO -ve supply currents are 1mA & 1mA. The total add up to the 2mA passing from +ve to -ve.
The output current at any instant in time is the DIFFERENCE between the +ve supply bias current and the -ve supply bias current
Take 1mA as the output current.
The DIFFERENCE in the output pair bias currents has to be 1mA. They started out as 1mA each and CHANGE to 1.5mA and 0.5mA.
The total +ve supply current is 2.5mA and the total -ve supply current is 1.5mA and the output current is +1mA flowing from output pin to the load and from the load back to the supply.
If the output current became -1.6mA then the +ve supply current would be 2mA + (-1.6/2) = 1.2mA and the -ve supply current would be -2mA + (-1.6/2) = -2.8mA ( the minus indicating flowing into the supply ).
Important note:
All the above has applied when teh opamp remains in ClassA for the whole output current changes.
This also applies to Power Amplifiers.
The supply currents CHANGE when ClassA output current flows !!!!!!!!!!!!!!
The supply rail currents are NOT CONSTANT when the opamp, or the Power Amp are in ClassA.
There is no current to power ground and no current to the load.
If you connect a load and pass current to that load then the +ve and -ve supply currents change.
If the quiescent current for the whole opamp is 2mA, half of which is the output pair bias current, then when output current = zero then the TWO +ve supply currents are 1mA & 1mA. The bias currents through the front end of the opamp hardly change for changes in output current.
The TWO -ve supply currents are 1mA & 1mA. The total add up to the 2mA passing from +ve to -ve.
The output current at any instant in time is the DIFFERENCE between the +ve supply bias current and the -ve supply bias current
Take 1mA as the output current.
The DIFFERENCE in the output pair bias currents has to be 1mA. They started out as 1mA each and CHANGE to 1.5mA and 0.5mA.
The total +ve supply current is 2.5mA and the total -ve supply current is 1.5mA and the output current is +1mA flowing from output pin to the load and from the load back to the supply.
If the output current became -1.6mA then the +ve supply current would be 2mA + (-1.6/2) = 1.2mA and the -ve supply current would be -2mA + (-1.6/2) = -2.8mA ( the minus indicating flowing into the supply ).
Important note:
All the above has applied when teh opamp remains in ClassA for the whole output current changes.
This also applies to Power Amplifiers.
The supply currents CHANGE when ClassA output current flows !!!!!!!!!!!!!!
The supply rail currents are NOT CONSTANT when the opamp, or the Power Amp are in ClassA.
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Assuming the opamp's feeding a load connected to 0V, one rail will take more (almost 1mA more) the other slightly less. Which rail takes more depends on the output voltage polarity from the opamp.
It would be the only way to be certain yes. Do you need absolute certainty? I agree the measurement would be difficult - so one alternative is to adjust the required amount of decoupling by ear.
Thanks a lot indeed. I was completely wrong. About both the current delivery and the power consumption. The values are pretty ridiculous ... no problem at all for any power supply.
I see. Unfortunately i do not have a schematic to understand if the caps are connected in some original way.
Electronics design is really an art and some details can give an impressive boost to performances indeed. There lies the difference between good design and excellent design i guess. I should try to see how this caps are connected.
Thanks a lot again for the very valuable advice.
Kind regards, gino
Hi !
Some years ago I've tried to fined out the current ability of the AD843 output stage, while it remained in class A. I used the following simple, but unfortunately, not precise procedure.
I added two small resistors (~ 1 om) in the supply circuits in order to control by the oscilloscope the distortions of the chip supply current, when the sine input signal is applied. Increasing it gradually, I caught the moment, when the distortions started. It indicated, that the output stage leaved class A operation. For instance, if you watch the current on the positive pin, the negative part of sine must begin distorts.
Distortions began at output voltage ~ 4v with 1 kom OpAmp load.
It gives for AD843 4 ma maximum output current in class A, that corresponds to 2 ma quiescent current of it's output stage.
So, there is a big difference between the quiescent currents of the chip (~ 12 ma) and it's output stag (~ 2 ma)!
Hi and thanks a lot for the very interesting information
Could you elaborate this because i am trying to learn something.
I looked at the DS but found nothing except THD at a low 1.5V out.
Thanks again. gino
your method makes sense to me and your result seems reasonable. This probably indicates that your method is good.
Now look at 2mA opamps and then try 0.5mA opamps.
Their output bias must be tiny. As a consequence their ClassA capability must also be tiny.
Could you adopt the forced ClassA (half the output stage turned on as single ended and the other half effectively turned off) and see if your method works for that as well.
Hi Gino, Hi Andrew! Thank you for replay.
As far as I understand, the situation you described has the following reason.
The maim goal of the ordinary OpAmp development is to obtain the maximum chip speed by the coast of minimum quiescent chip current. The result is that the main part of chip quiescent current is devoted to achieve the high speed, rather then the depth of output stage class A operation.
That is, the high output stage bias current is not an agenda mainly, when the OpAmp is developed.
Yes Andrew, really, the method of forced Class A seems to be the only one to achieve that class of chip operation, when modern OpAmp's with their tiny quiescent current are applied in audio. At the same time they are immensely high speed, but what's the good of it for audio
.
It seems to me, forced Class A approach is efficient enough and there's no need to measure their tiny class A capability.
Returning to the thread theme, I'd like to advise two links of practical OpAmp usage in audio:
Notes on Audio Op-Amps
Op-Amp review page
With best wishes, your Igor Shukov
As far as I understand, the situation you described has the following reason.
The maim goal of the ordinary OpAmp development is to obtain the maximum chip speed by the coast of minimum quiescent chip current. The result is that the main part of chip quiescent current is devoted to achieve the high speed, rather then the depth of output stage class A operation.
That is, the high output stage bias current is not an agenda mainly, when the OpAmp is developed.
Yes Andrew, really, the method of forced Class A seems to be the only one to achieve that class of chip operation, when modern OpAmp's with their tiny quiescent current are applied in audio. At the same time they are immensely high speed, but what's the good of it for audio
It seems to me, forced Class A approach is efficient enough and there's no need to measure their tiny class A capability.
Returning to the thread theme, I'd like to advise two links of practical OpAmp usage in audio:
Notes on Audio Op-Amps
Op-Amp review page
With best wishes, your Igor Shukov
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