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Post by JFrankParnell on Feb 22, 2012 0:14:16 GMT -5
Ok, so i got tired of fiddling with the volume thingy on my puter, so I went to to RS and got 2 1/8th in stereo jacks and a 100K dual pot and a little plastic box. I went out to the garage, fired up the iron and commenced to soldering... I just hooked up the grounds straight accross between the jacks. Then a hot wire to the pot (left side) and from the pot (middle) for each channel. Great... Went back in and hooked it up, the results were less than perfect. It never turns all the way down. And the taper is like 0-7=very quiet, 7-11=loud with a very sharp knee in between. My first thought was that I needed more resistor, 100k wasnt enough to quiet my signal down to nothing. Went back to the shack, but that 100k was the only dual pot they had. And the guy working wasnt the brainiac one. So, after all this, I started thinking...  Wait, this isnt how we wire guitar volumes... In guitars, we shunt some of the signal to ground with the pot...but, but, but... this is stereo... there's only one ground...if I shunt some signal from each channel to ground, wont that mono-fy my stereoishness? Fine, I'll google it. 1st hit: beavishifi.com/articles/Volume_Control/ Which does look earily similar to our guitar volume wiring.  I'm not sure why it looks like the signals get crossed in his diagram, but I dont really care if its Angus in the right and Malcolm on the left or vice-versa. So, do you guys vouch for this diagram?  |
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Post by ashcatlt on Feb 22, 2012 11:10:08 GMT -5
The 1/8" stereo jack only has one "ground" connection coming in and going out. Think of a mixing board. You've got all kinds of separate jacks there, all with their sleeves connected to the same chassis ground. I guess I'm not qualified to explain why it doesn't matter, but it doesn't.
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Post by JFrankParnell on Feb 27, 2012 0:11:28 GMT -5
ok, i wired it up per diagram above, works great. Turns down all the way, turns up all the way. The taper is still kinda lame, though. From about 2 till 7 or 8 there is very little volume change. Any suggestions for this? Linear taper pots? More or less ohms?
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Post by reTrEaD on Feb 27, 2012 13:35:09 GMT -5
Under normal circumstances, a linear pot is not so good as a volume control. During the first part of the rotation from counterclockwise, you'll have a large increase in volume. From the middle to clockwise, not much more happens.
If you use audio taper pots, the change in volume is even across the entire rotation.
But if the resistance of the audio taper pot is high and the load has a relatively low resistance, the change is all pushed up toward the clockwise end of the rotation. This can be solved by using an audio taper pot of lower resistance, or by using a linear pot.
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Post by JFrankParnell on Feb 27, 2012 15:24:47 GMT -5
Hmm, ok. The source is the headphone jack of my puter (as opposed to the 'line out' in back. I forget why, but it might just be the convenience of the jack in front.). The load is a pretty nice Altec system, a sub and two tweeters. The pots are 100k dual gang audio taper. I will typically be using it on the lower end of the dial, tunes playing low while I work.
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Post by ashcatlt on Feb 27, 2012 19:46:17 GMT -5
I think those powered computer speakers expect to be fed from a headphone out anyway. You could try to google for specs and see if you can find input impedance to help find an appropriate pot value. You could also try sticking resistors across the outside lugs of the pots to bring the total effective value down. That will mess with the taper too, though.
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Post by reTrEaD on Feb 28, 2012 9:20:36 GMT -5
I think those powered computer speakers expect to be fed from a headphone out anyway. You could try to google for specs and see if you can find input impedance to help find an appropriate pot value. This is probably the best approach. You could also try sticking resistors across the outside lugs of the pots to bring the total effective value down. That will mess with the taper too, though. If the input impedance of the powered speakers is lower than the resistance value of the pots, that's already messing with the taper. Putting resistors between the CW lug and wiper of the pots will mess with it, in the opposite direction. This might help make it useable, but the better solution would be to use pots that are lower than the input impedance of the powered speakers but higher than the output impedance of the headphone output. |
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Post by JFrankParnell on Feb 28, 2012 19:32:41 GMT -5
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Post by reTrEaD on Feb 28, 2012 20:06:16 GMT -5
I'd suggest 1k~10K.
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Post by sumgai on Feb 28, 2012 23:12:54 GMT -5
JFP, You should be using an L-pad (Google it, at your convenience). The 1/8" stereo jack only has one "ground" connection coming in and going out. Think of a mixing board. You've got all kinds of separate jacks there, all with their sleeves connected to the same chassis ground. I guess I'm not qualified to explain why it doesn't matter, but it doesn't. I am. Qualified, that is. But even so, I'll take the easy way out, and quote ChrisK - - "Normally when we say ground, we mean a connection common to several circuit elements that is nothing more than a reference point - a measured quantity (volts, whatever) is shown as X quantity compared to ground. However, when it carries part of the signal, you should think of it as "signal return", the other leg through which the signal must return to the source, to complete the circuit." I'll add only that as a common reference point, ground (chassis, a buss wire, what have you) "plays nice" with all signals and/or DC voltages, no problems at all. There are occasions where separate reference points are called for, but for most ordinary electronic gadgets, a single reference point is pretty much the norm. HTH sumgai |
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Post by reTrEaD on Feb 28, 2012 23:34:36 GMT -5
Why an L-pad? They cost more and I don't see and advantage for an L-pad over a simple pot in this particular application.
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Post by cynical1 on Feb 28, 2012 23:45:38 GMT -5
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Post by sumgai on Feb 29, 2012 2:12:42 GMT -5
c1,
Good catch, should be nearly required reading. +1
reTrEaD,
You hit it on the head the first time - impedance matching. An L-pad is usually a fancy name for an attentuator, but most often the ordinary usage is for that class of attentuator meant for power amp output levels (i.e. titi volts, beaucoup amps, in turn meaning high power).
Headphone outputs qualify in this regard because the great majority of them are meant for either 16 or 32Ω. Here, the amount of power available isn't the concern, but we still need to match the impedance closely, or suffer the consequences.
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Post by reTrEaD on Feb 29, 2012 11:59:40 GMT -5
What consequences would we suffer?
If the load was highly reactive (like a speaker), damping factor would be an important issue. The shunting across the speaker's terminals mitigates decrease in damping factor that occurs when increasing series resistance.
But the load isn't a (passive) speaker. It's a powered speaker.
In a tube amplifier, keeping the load the output transformer sees reasonably close to the specified value is important. An unloaded or lightly loaded output transformer can result in arcing on the primary side.
But this isn't a tube amp. It's a modern solid state design.
As long as the pot resistance is equal to or greater than the minimum impedance the headphone amp can drive, there will be no danger of damage.
As long as the pot resistance is equal to or less than the impedance of the load, the change in the pot's taper (due to loading) won't become a significant issue.
For this application, I don't see any benefit to using an L-pad instead of a simple pot. But using an L-pad won't cause any harm, other than to the wallet.
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Post by sumgai on Feb 29, 2012 12:54:59 GMT -5
reT, For the best transfer of voltage/current/power across a connection, it's important to match the impedance, regardless of the source (active device or a matching tranformer) and regardless of the destination - it's called 'best practice' for a variety of reasons. However, there are times when we do violate that best practice, I agree. But not when estimating the abilities of a device, be it on the input or the output side. I don't know for certain, but I'd rather like to think that the 'follower' amp, such as the active speaker set mentioned by JFrank, is meant to be driven by a very low impedance source, on the order of 16 or 32Ω, possibly even lower. That argues against using attentuator values well in excess of two orders of magnitude greater than the expected value. Accordingly, as stated before, L-pads are usually associated with higher power levels, that's true, so there is, normally, an expense encountered over these parts - they're heavier duty, to be certain. However, they're attentuators first and foremost, meaning that they'are three-pole devices, just like our normal volume pots. From that, we see that a normal-sized potentiometer is what we want, but at a very low ohmage value. Such values are not often encountered in this physical-sized casing, sorry to say. Hence, we opt for spending a bit more moola to get the desired result. As usual, monkeying with external resistors to lower a pot's value has deleterious effects on both the taper and the overall performance of the pot, and such is to be used only when all other avenues are exhausted. The sole exception, and a big one for most non-professionals, would be the ratio of the cost to the perceived value - if it's more expensive than the rest of the whole bleepin' device, then of course it's OK to ignore the so-called 'best practice', and use what you've got on hand.  HTH sumgai |
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Post by reTrEaD on Feb 29, 2012 17:59:12 GMT -5
reT, For the best transfer of voltage/current/power across a connection, it's important to match the impedance, regardless of the source (active device or a matching tranformer) and regardless of the destination - it's called 'best practice' for a variety of reasons. If the intention was achieve impedance matching between a 30 ohm output and a 10k load, an L-pad won't get you there. You'd need a transformer. But in this application, impedance matching and best transfer of power is unimportant. We're interested in scaling the voltage down, as needed. However, there are times when we do violate that best practice, I agree. But not when estimating the abilities of a device, be it on the input or the output side. Let's examine the "abilities" of these devices: The headphone driver is capable of driving fairly low impedances. If we use a higher impedance load, we won't get the maximum power transfer into the load. Is this a problem? Absolutely not. As long as the amplitude (voltage) of the signal at the input of the amplified speakers is sufficient, all is well. What about the next device? I don't know for certain, but I'd rather like to think that the 'follower' amp, such as the active speaker set mentioned by JFrank, is meant to be driven by a very low impedance source, on the order of 16 or 32Ω, possibly even lower. The amplified speaker system has an input impedance rated at 10k. Driving a 10k load with a 10k source would result in a very modest decrease in signal amplitude (voltage). Driving a 10k load with a 1k source would result in a decrease of signal amplitude that's statistically insignificant. Having a source impedance that's even lower won't change matters at all. That argues against using attentuator values well in excess of two orders of magnitude greater than the expected value. This argument is severely flawed. Thinking that a 10k load needs to be driven by a 32 ohm or less output impedance is wrong thinking. From that, we see that a normal-sized potentiometer is what we want, but at a very low ohmage value. A pot whose resistance value is equal to or less than the load impedance is sufficiently low. One that's a tenth of the value of the load impedance will give marginally better performance. Anything less than that is insignificantly better. At full CW rotation, the output impedance of the headphone amplifier is driving the load directly. Worst case, when the pot is in the middle of its resistance, the "source" impedance (at the wiper) is approximately 1/4 of the pot value. For a 10k pot, that would mean about 2.5k. Would there be any adverse affects of a 2.5k source driving a 10k load? No. Well, maybe just a little. If you had a VERY long cable after the pot feeding the amplified speakers, this would be more susceptible to hum and noise than a 30 ohm source. But unless the cable was 50 ft or more, I would have no concerns about that. If so, a 1k pot (still a commonly available part) could be used, keeping the maximum "source" impedance down to just slightly over 250 ohms. Long story short, L-pad have their place. But in this app, they don't have an advantage over a simple inexpensive pot, that would justify the additional expense. |
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Post by sumgai on Mar 1, 2012 12:51:27 GMT -5
'TrEaD, All you say above is true, providing that the input impedance is indeed 10KΩ. I don't know that it is, I don't even know what kind of amplified speakers JFrank is using.... So the only "flaw" would be in my assumption as to the capabilities of the equipment under discussion, not in the logic flow itself. And for the record, an attentuator of any design is not meant to match impedance, it's meant to maintain the impedance match (or mismatch, if that was the design goal), or alter it as little as possible. That is all, over and out. sumgai |
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Post by ashcatlt on Mar 1, 2012 16:56:43 GMT -5
JFrank posted a link to the specs for his speakers. How are those speakers? I have an old set of Altecs that I love, and wish I had another 6 or 7 sets! I also have a newer set which is passable, but not quite as nice. On yours the satellites seem to have two speakers, one pointing forward and the other at an angle. WTF is up with that? Some sort of pseudo-surround thing, or...  |
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Post by reTrEaD on Mar 1, 2012 17:35:45 GMT -5
JFrank posted a link to the specs for his speakers. Apparently Sumgai missed that and we ended up in an apples and oranges debate. (sh)It happens. |
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Post by JFrankParnell on Mar 1, 2012 18:44:58 GMT -5
That speaker set is really nice. I think i looked up that they retail for a couple hundy. Haha, thrift store didnt know what they had, i got em for $20 i think. I believe you are right in that the angled tweets are sposed to be surroundish. I think that part only works if you hook it up with usb to the puter. And prolly have to have some driver and whatnot. I can still adjust the sub from the dial and buttons on top.
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Post by JohnH on Mar 2, 2012 15:19:48 GMT -5
The older Altec Lansing's are really good - and I think Ash and I have the same set (ACS45.1). Its my one piece of computer gear that remains from the 1990's and will be kept through this decade too I hope. JFPs look like this?  Similar vintage and the sub-woofer looks the same, and the satellites look similar with an extra bit stacked on top. I expect they sound great. So from the specs, we now know the input Z is >10k. That explains why a 100k pot was giving very low volume until turned up high. If you are running from an output which can also drive headphones, then it has a low output impedance, so there's a big comfortable factor between input and output impedance. You pot should be somewhere in the middle of this range. Id say any pot from 100 Ohms up to about 5k will be fine, 10k will also work (better than the 100k but with a change to the pot taper), and 1k would be ideal. John |
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Post by JFrankParnell on Apr 1, 2012 19:57:24 GMT -5
Thats it, John. Just put a dual 5k in there, works great, nice taper and all. Thanks for the advices, guys.  |
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Wait, this isnt how we wire guitar volumes... In guitars, we shunt some of the signal to ground with the pot...but, but, but... this is stereo... there's only one ground...if I shunt some signal from each channel to ground, wont that mono-fy my stereoishness?
