In many thread in this post I see some people asking how to control the volume/voltage of a line level signal. In many cases, people recommend using a 10k potentiometer (optimally it should logarithmic but let's not take this into account right now) as a voltage divider connecting the left pin of the pot to GND, the middle pin will be the line level with its volume decreased and the right pin would be the line level.
Some people says it works just fine, but I am wondering here: the line leve spec recommends using a 100 to 600 ohm resistance in the line out level. Using the 10k pot as above, would result in a much much higher resistance.
Also, many line inputs have already 10k ohm (or even 100k) resistance so using a 10k ohm at the line output shouldnt be a good idea (I dont know why, but I think it shouldnt be correct).
So I ask you guys: why most people recommend using a 10k ohm potentiometer at the line level out? Wouldnt it be much better to use a 1k ohm potentiometer since you would be much closer to the spec (600ohm) and much further away from the 10k-100k resistance of most line input?
Some people says it works just fine, but I am wondering here: the line leve spec recommends using a 100 to 600 ohm resistance in the line out level. Using the 10k pot as above, would result in a much much higher resistance.
Also, many line inputs have already 10k ohm (or even 100k) resistance so using a 10k ohm at the line output shouldnt be a good idea (I dont know why, but I think it shouldnt be correct).
So I ask you guys: why most people recommend using a 10k ohm potentiometer at the line level out? Wouldnt it be much better to use a 1k ohm potentiometer since you would be much closer to the spec (600ohm) and much further away from the 10k-100k resistance of most line input?
1) 1k pots are hard to find
2) some circuits will not drive 1k to HIGH level without a rise of distortion. (Your 100-600r is not a loading specification; hi-fi boxes are not simple resistive sources.)
If the pot directly drives a cable, the Rout should be low to preserve HF, but this can strain the previous stage.
You often will need a buffer before the pot to drive < 2k volume controls.
Like everything, there are compromises. Usually a 10k pot is suitable, especially with a stage following it.
You often will need a buffer before the pot to drive < 2k volume controls.
Like everything, there are compromises. Usually a 10k pot is suitable, especially with a stage following it.
I tried googling next stage load but came up with nothing. You would mind explaining to me what I am mistaking?
I am becoming to understand how stupid I am in this audio stuff... I really didnt understand anything you said. I just googled again and it says that line level impedance should be between 100 and 600 ohm. So how 10k is suitable?
I see on many experts on this forum and others too saying that the output impedance should be at least 10x lower than the input impedance of line level. That's why I am having a pretty hard time trying to understand how a 10k ooutput impedance looks fine when most input impedance are also 10k. The 10:1 ratio is not being respected..
In this post https://www.swamp.net.au/swamp-content/guides/line-level-signals-explained.html#:~:text=Line outputs have signal impedance,used to drive the signal. it says:
""Line outputs have signal impedance ranging typically from 100ohm to 600ohm, while a line input has an impedance of 10,000ohm (10k ohm). The difference in impedance between the IN and OUT is what is used to drive the signal.""
Also, this highly voted answer at StackExchange shows how important is to keep impedance of line level at most 600 ohm.
https://electronics.stackexchange.c...t-is-impedance-matching-in-audio-applications
There are many many places on the internet talking about the 10:1 rati and 100-600 ohm. So I ask you: if I dont respect this advice, and use, for example a 10k ohm potentiometer at the line level to attenuate the output... will the resulted audio be perceptible bad?
In this post https://www.swamp.net.au/swamp-content/guides/line-level-signals-explained.html#:~:text=Line outputs have signal impedance,used to drive the signal. it says:
""Line outputs have signal impedance ranging typically from 100ohm to 600ohm, while a line input has an impedance of 10,000ohm (10k ohm). The difference in impedance between the IN and OUT is what is used to drive the signal.""
Also, this highly voted answer at StackExchange shows how important is to keep impedance of line level at most 600 ohm.
https://electronics.stackexchange.c...t-is-impedance-matching-in-audio-applications
There are many many places on the internet talking about the 10:1 rati and 100-600 ohm. So I ask you: if I dont respect this advice, and use, for example a 10k ohm potentiometer at the line level to attenuate the output... will the resulted audio be perceptible bad?
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Yes, lets say you have signal source like cd player. But any other line level signal like fm radio, phono output, reel to reel or digital tape player applies. They all have line level output with anywhere between 200mV and 1V output, meant to drive next stage, which is preamp or integrated amp. This line level signal is best tranfered from the source to next stage, when its not loaded too much. As a matter of fact, it should not even know its driving some load. If you load it with low impedance, signal will drop and it will roll of too soon. And it will sound like am radio.
Please do not mistake 600 microphone load, or 32 ohm mc load, that is totaly different thing.
In short, even if the line level output has some output impedance, its meant to be loaded with many times of that value.
Ps: try to plug your 600 ohm sennheiser headphones directly to cd player line level output (terrible idea, low damping factor) and than plug it into dedicated headphone amplifier (which has output imedance few ohms, very high damping factor)
Ps2: amp with output impedance of 0.02 will never drive 0.02 ohm, that would be terrible damping factor, but will happily drive 4 or 8 ohms
You keep repeating the same non sense. Output impedance of signal may be 100 to 600, but even google is clear that next stage should be 10k minimum.
This is first google hit:
https://www.google.com/search?q=lin...-verizon-sscr&sourceid=chrome-mobile&ie=UTF-8
This is first google hit:
https://www.google.com/search?q=lin...-verizon-sscr&sourceid=chrome-mobile&ie=UTF-8
So when you say "next stage level" is actually "input line level". BUT I am talking about the OUTPUT line level. When I talk about 100 - 600 ohm, I am talking about the output line level.
https://www.google.com/search?q=out...i30j0i390l4.5357j0j7&sourceid=chrome&ie=UTF-8
Google says clearly it shoudl be 100-600 ohm. So why are people recomending using a 10k potentiometer at the output line level to control volume?
Please see the image above which I extracted from the reference you posted. You see, the link also says 100 to 600 ohm. So how do people use 10k potentiometer to attenuate line output level signal?
In good old time of tubes, with grid impedance of megaohms, it was normal to have 100k to 500k volume pot.
As a matter of fact, i have goldmund line stage which has first stage pre with 100k input impedance (otherise it would be megaohms since it has jfets on input) followed by 25k volume pot followed by the buffer. Sounds good to me.
As a matter of fact, i have goldmund line stage which has first stage pre with 100k input impedance (otherise it would be megaohms since it has jfets on input) followed by 25k volume pot followed by the buffer. Sounds good to me.
Again, you are mixing two things. Line level ouput, like cd player, or fm tuner, has no volume pot on output.
Next stage has volume control.
I am out.
The 600 ohms is just a standard impedance level for pro equipment, originally related to line transformers,
which are very seldom currently used in home audio equipment.
The Rout actually can be anything that does not excessively attenuate the highs or the overall amplitude,
so it can be anywhere from 0 ohms to about 5k ohms, depending on the requirements.
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A potentiometer is a voltage divider, which is a series resistance, followed by a shunt resistance.
Vout = Vin x Rshunt / ( Rshunt + Rseries )...........where ( Rshunt + Rseries ) = the value of the volume control (10k, etc.).
The Rout of the previous stage is not intended to be involved with the control of volume, though it may have a slight effect.
The position of the wiper in the volume control is what determines the amount of attenuation of the input voltage,
as shown in the above equation. Google may give some information, but it will often lack context.
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10X minimizes the drop of level when a load is connected.
But here we have volume control on a simple system. The user WILL adjust level as desired. So strict 10:1 ratio (1dB) is not important.
We have three units. Source is usually <500r. Load is usually >10K. And a pot which is made as say 10K but is used as a potentiometer which means the worst-case impedance is 2.5k. (Try it. Short both ends of a 10k pot with a 500r resistor and measure the wiper to either end.) And it is 2.5k at "mid-point" so any loading is masked by user setting.