Post20 is nearly useless.
A 1/3rd value resistor will attenuate the siganl by an extra 2.5dB. Hardly noticeable.
attenuation = 20(log{1/1.333}) = 20*log0.75 = 20*-0.1248 = -2.4966dB
post18 makes much more sense.
2times vol pot for the added resistor gives an extra attenuation of -9.5dB
4times gives -14dB
And fit a switch across the added resistor.
That allows full volume for parties and switch to muted volume for lone listening.
A 1/3rd value resistor will attenuate the siganl by an extra 2.5dB. Hardly noticeable.
attenuation = 20(log{1/1.333}) = 20*log0.75 = 20*-0.1248 = -2.4966dB
post18 makes much more sense.
2times vol pot for the added resistor gives an extra attenuation of -9.5dB
4times gives -14dB
And fit a switch across the added resistor.
That allows full volume for parties and switch to muted volume for lone listening.
.
<< Post20 is nearly useless...A 1/3rd value resistor will attenuate the siganl by an extra 2.5dB. Hardly noticeable... >>
Unfortunately, comparing resistance to decibels is meaningless. They don't convert back and forth.
But I'm glad you said that because it reminds me to mention:
All through this thread there's a tendency to speak of gain and decibels as if they were interchangeable. But thinking this way will lead to confusion because electronics is one thing, and audio is another thing. You can relate them, yes, but they are not the same.
Op amps (chip amps) don't work in decibels, they work in gain. This distinction is very important because, for instance, doubling gain does not double decibels, and vice versa. This can get confusing, but only if you let it. It's not complicated, just don't mix apples and oranges.
In the present case we add in pears--percentages--which can really mess things up if you don't keep your thinking straight.
Sticking with the present case we divide the audio input signal into thirds. Why? Why not. The actual input level in volts (bananas?) is unknown, so we apply the SWAG principle and come up with thirds.
The only known quantity is the value of the input potentiometer, so having decided on thirds, we assign two of them to the input pot. We want the volume potentiometer to drop (control) 66 2/3% of the input signal, and we want to burn the rest (which is already being burned anyway, but not in a way that benefits us).
DO NOT at this point start thinking about decibels, gain, volts, or anything else except getting rid of 1/3 of the input signal.
Obviously this calls for an additional resistor. Obviously this new resistor must be 1/3 the value of the existing resistor, which is the volume pot. One third plus two thirds equals one, so all quantities are accounted for.
Now that a decision has been made there's a reference point, which is what we needed in the first place. Now we can convert back and forth into decibels, volts, gain, or next week's lottery numbers if you like, knock yourself out.
Assuming an existing input level of 2 volts (as good a guess as any), we now have a new maximum input level to the amp of about 1.3 volts, which is not an unreasonable level.
Is this assumed input level going to be the real one? How would anybody know, since nobody knows the existing input level. Maybe the new dropping resistor will actually have to be 1/2, 2/3, or who-knows the value of the existing volume pot.
The important thing here is to keep your apples and oranges straight. You're working with electronic circuits, so you have to think in terms of electronics. The important thing here is to keep your apples and oranges straight. You're working with electronic circuits, so you have to think in terms of electronics. Doing otherwise can have unhappy results.
.
<< Post20 is nearly useless...A 1/3rd value resistor will attenuate the siganl by an extra 2.5dB. Hardly noticeable... >>
Unfortunately, comparing resistance to decibels is meaningless. They don't convert back and forth.
But I'm glad you said that because it reminds me to mention:
All through this thread there's a tendency to speak of gain and decibels as if they were interchangeable. But thinking this way will lead to confusion because electronics is one thing, and audio is another thing. You can relate them, yes, but they are not the same.
Op amps (chip amps) don't work in decibels, they work in gain. This distinction is very important because, for instance, doubling gain does not double decibels, and vice versa. This can get confusing, but only if you let it. It's not complicated, just don't mix apples and oranges.
In the present case we add in pears--percentages--which can really mess things up if you don't keep your thinking straight.
Sticking with the present case we divide the audio input signal into thirds. Why? Why not. The actual input level in volts (bananas?) is unknown, so we apply the SWAG principle and come up with thirds.
The only known quantity is the value of the input potentiometer, so having decided on thirds, we assign two of them to the input pot. We want the volume potentiometer to drop (control) 66 2/3% of the input signal, and we want to burn the rest (which is already being burned anyway, but not in a way that benefits us).
DO NOT at this point start thinking about decibels, gain, volts, or anything else except getting rid of 1/3 of the input signal.
Obviously this calls for an additional resistor. Obviously this new resistor must be 1/3 the value of the existing resistor, which is the volume pot. One third plus two thirds equals one, so all quantities are accounted for.
Now that a decision has been made there's a reference point, which is what we needed in the first place. Now we can convert back and forth into decibels, volts, gain, or next week's lottery numbers if you like, knock yourself out.
Assuming an existing input level of 2 volts (as good a guess as any), we now have a new maximum input level to the amp of about 1.3 volts, which is not an unreasonable level.
Is this assumed input level going to be the real one? How would anybody know, since nobody knows the existing input level. Maybe the new dropping resistor will actually have to be 1/2, 2/3, or who-knows the value of the existing volume pot.
The important thing here is to keep your apples and oranges straight. You're working with electronic circuits, so you have to think in terms of electronics. The important thing here is to keep your apples and oranges straight. You're working with electronic circuits, so you have to think in terms of electronics. Doing otherwise can have unhappy results.
.
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😉 An old and imo worthwhile 'trick' is to add a resistor inline to the RCA plug of a purpose dedicated interconnect. Even into the tubular centre prong has been managed
Make up a couple of cables in varying R values, to zero in the 'loudness' deemed appropriate to needs if you don't have a precise value yet nailed down.
Subsequently Swap out Interconnects for party uses... If there is need 🙂
Make up a couple of cables in varying R values, to zero in the 'loudness' deemed appropriate to needs if you don't have a precise value yet nailed down.
Subsequently Swap out Interconnects for party uses... If there is need 🙂
ya, but he is absolutely right...
does not mean he can act like that freely,
but... what the heck, if you look around Andrew has allways been like that.
in fact i kind-of enjoy the style.
the borg come into my mind 😀
resistance is futile.
All right I'll be more honest:
Post20 is useless. He has not got a clue.
Compare to Post18, that Member has much more knowledge and sense.
I'm not sure a single series resistor will do.
I remember seeing a set of DIY cables with L-pad attenuators shoehorned into the RCA connectors at one end. I could be wrong, but I think the values were;
R1 = 36R
R2=75R
And I say, just point 2. of post#18 is all that's needed.
What KieranC describes in post#1 and post#3 is screaming: "linear pot" (or defective pot)
There is NO way a log pot will overload the amp after the first 10% of it's range, even using a 10V source. Resistance to ground will be almost always less then 1K at this setting (probably ~70 to 200ohms) for a 50K log pot.
So, KieranC, are you sure that input pot is not a linear one?
What KieranC describes in post#1 and post#3 is screaming: "linear pot" (or defective pot)
There is NO way a log pot will overload the amp after the first 10% of it's range, even using a 10V source. Resistance to ground will be almost always less then 1K at this setting (probably ~70 to 200ohms) for a 50K log pot.
So, KieranC, are you sure that input pot is not a linear one?
Why?
Actually there is 2 resistors, second one is volume pot.
Or we can see 3 resistors here - input resistor, part of volume pot from top to slider and part of volume pot from slider to ground.
And thees 3 resistances together make variable 2 resistor divider 🙂 In this 2 resistor divider first "resistor" is input resistor + top half of pot, second "resistor" is bottom half of the pot.
Bare, clever idea, I like it, even no need to open amp for changing gain.
bentsnake, your previous post is very useful, especially schematic drawing. Except resistor value. If too much gain and hard volume regulation become a real problem, x2 pot value is to start from, probably more is needed. This is from my personal experience with headphones amps.
your second post probably tell true, however make things more complicate to understand 🙂
I like to see things simple - calculate total gain of amp (with amp in this case, I mean everything between input and output jacks, not only amp chip)
Example:
20K input resistor, 10k volume pot and LM chip, configured to gain 21.
Input resistor together with volume pot make 1/3 divider.
Total amp's gain from input to output is 21/3=7
your second post probably tell true, however make things more complicate to understand 🙂
I like to see things simple - calculate total gain of amp (with amp in this case, I mean everything between input and output jacks, not only amp chip)
Example:
20K input resistor, 10k volume pot and LM chip, configured to gain 21.
Input resistor together with volume pot make 1/3 divider.
Total amp's gain from input to output is 21/3=7
If a lower gain is needed, why configure the LM3886 for a gain of 21? How about a gain of 11 (20.8 dB)? Rf1 = 10k, Ri = 1k would give you that.
Then drop the gain further by adding an attenuator up front. Either a resistive divider or a resistor in series with the volume pot. I suspect a 20 dB attenuation would work just fine. This means the resistor in series with the pot will need to be about 9x the resistance of the pot. Recall:
Attn(dB) = -20*log(Rpot/(Rpot+Rin))
Rin = 9*Rpot ==> Attn(dB) = -20*log(1/10) = 20 dB.
Rin = 10*Rpot would give you 20.8 dB of attenuation for a total system gain of 0 dB (1 V/V) at the highest volume setting.
~Tom
Then drop the gain further by adding an attenuator up front. Either a resistive divider or a resistor in series with the volume pot. I suspect a 20 dB attenuation would work just fine. This means the resistor in series with the pot will need to be about 9x the resistance of the pot. Recall:
Attn(dB) = -20*log(Rpot/(Rpot+Rin))
Rin = 9*Rpot ==> Attn(dB) = -20*log(1/10) = 20 dB.
Rin = 10*Rpot would give you 20.8 dB of attenuation for a total system gain of 0 dB (1 V/V) at the highest volume setting.
~Tom
<< your previous post is very useful...Except resistor value. >>
.
Your point is valid, but we keep running up against the fact that nobody knows what the input level really is, so one guess is as good as another--although your experience with headphone amps certainly should be given weight.
All of which seems to not matter because original poster KieranC apparently has given up on the idea of getting any help and left. And who can blame him.
.
.
Your point is valid, but we keep running up against the fact that nobody knows what the input level really is, so one guess is as good as another--although your experience with headphone amps certainly should be given weight.
All of which seems to not matter because original poster KieranC apparently has given up on the idea of getting any help and left. And who can blame him.
.
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be sure to understand that for example -3 dBL stays -3 dBL.
if the input is a gigavolts, then it will be half of that.
same goes for 1 V input, or any input You can figure out.
Then the amplifier applys a gain of whatever dBL.
the exact value of the input does not have anything to do with that.
thing is, if at whatever input its too loud, then an L-pad giving -3 DBL will make the output half.
Just.. be sure to get this straight.
then the world will make a lot more sense.
if the input is a gigavolts, then it will be half of that.
same goes for 1 V input, or any input You can figure out.
Then the amplifier applys a gain of whatever dBL.
the exact value of the input does not have anything to do with that.
thing is, if at whatever input its too loud, then an L-pad giving -3 DBL will make the output half.
Just.. be sure to get this straight.
then the world will make a lot more sense.
3 dB is a doubling or halving of power.
6 dB is a doubling or halving of voltage or current.
10 dB is a change of 10x in power but only a 3.2x change in voltage/current.
20 dB is a change of 100x in power but a 10x change in voltage/current.
Power follows: dB = 10*log(ratio).
Voltage and current follow: dB = 20*log(ratio).
~Tom
6 dB is a doubling or halving of voltage or current.
10 dB is a change of 10x in power but only a 3.2x change in voltage/current.
20 dB is a change of 100x in power but a 10x change in voltage/current.
Power follows: dB = 10*log(ratio).
Voltage and current follow: dB = 20*log(ratio).
~Tom
Ah yes, sorry. I was considering the attenuated cable as a stand alone solution. I missed the other resistors in the chain.
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Hi,
I have same problem with LM3886 from Chipamp..
My dac give about 2 V, so I build a passive preamp with two pot (10 k) in serial wiring.
Output of the first is connected to the input of the second, and output of the second to LM.
Ground pot's are connected to input ground (same ground).
First is main level, and second adjust level for LM.
It work's well, and in my case pot are near 12 h for reasonable level.
I can also use it for bi amping, output of th first is connected to bass amp, in parallel with the second pot.
Phil.
I have same problem with LM3886 from Chipamp..
My dac give about 2 V, so I build a passive preamp with two pot (10 k) in serial wiring.
Output of the first is connected to the input of the second, and output of the second to LM.
Ground pot's are connected to input ground (same ground).
First is main level, and second adjust level for LM.
It work's well, and in my case pot are near 12 h for reasonable level.
I can also use it for bi amping, output of th first is connected to bass amp, in parallel with the second pot.
An externally hosted image should be here but it was not working when we last tested it.
Phil.
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