Volume control - can someone explain for me please? - diyAudio
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Old 10th July 2015, 07:49 AM   #1
Katch is offline Katch  United Kingdom
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Default Volume control - can someone explain for me please?

I'm pretty good at repairing amps and building amps by following "recipes" but I'd be the first to admit that actual maths and science of it all might as well be black magic (I've certainly let the magic smoke out on occasions...)

Which brings me to todays question born of ignorance;

I've just built a passive volume control to handle the level put out by my Quad 303s (using a Yaqin tube phone stage with no volume control and line level stuff)

The circuit is as simple as they come. 21 step 10k SMD DACT attenuator wired into 2 pairs of RCAs. That's it.

It works exactly as expected.

However, I don't understand it. I've measured the impedance at the inputs and read 10k ohm regardless of the attenuator's position.

At the outputs, I read 10k ohm at maximum volume and 0 at minimum. This seems backwards to my logic?

Can someone explain why 10k resistance at the input and the output is louder than 10k at the input and 0 at the output?
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Old 10th July 2015, 07:57 AM   #2
Mooly is online now Mooly  United Kingdom
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You are reading 10k across the attenuator. That's from input to ground. If it were a pot it would be across the two end terminals. The setting of the switch (wiper for a pot) does not alter that value.

At the output (wiper) its a different story and the result you get depends which terminal you measure to.

At minimum volume the wiper is down at the earthy end of the pot and so does read 0 ohms between those two points. At max volume the wiper connects to the input and so then reads 0 ohms to that point instead. And from ground to wiper would read 10k at max volume.
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Old 10th July 2015, 08:05 AM   #3
Katch is offline Katch  United Kingdom
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So if I read the resistance across the input and the output of one channel and move the attenuator - it should give readings that make more sense, i.e. decreasing resistance as the volume increases?
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Old 10th July 2015, 08:12 AM   #4
Mooly is online now Mooly  United Kingdom
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I'm not following you on that one

The readings on the attenuator or pot assume that the wiper is not connected to anything. Any external resistance across the wiper/ground connections will alter the absolute readings but not the principle.
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Old 10th July 2015, 08:17 AM   #5
Mooly is online now Mooly  United Kingdom
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Look at the diagram here, the top channel,
Passive preamp (passive volume control)

Across top and bottom terminals it will read 10k. From the bottom terminal to the middle moving contact will read from 0 ohms at minimum volume through to 10k at max volume. It increases in resistance because the contact is moving away from the lower one where we are measuring from.

From the top to middle contact it is reversed. At max volume we read 0 ohms (becase the wiper is contacting the input terminal) and at min volume we read 10k. Its 10k because that is the total resistance of the attenuator. The wiper is now contacting ground and so that is why the volume is minimum.
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Old 10th July 2015, 08:25 AM   #6
Katch is offline Katch  United Kingdom
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Ok, making sense now. Thank you.
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Old 10th July 2015, 09:08 AM   #7
DF96 is offline DF96  England
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Google 'potential divider'. A volume control does not work by simply putting resistance in the signal path, but it is amazing how many newbies think it does.
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Old 11th July 2015, 09:21 AM   #8
AndrewT is offline AndrewT  Scotland
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Go to ESP and read all about it.
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Old 11th July 2015, 10:15 AM   #9
Mooly is online now Mooly  United Kingdom
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Andrew means here,
Elliott Sound Products - The Audio Pages (Main Index)

And under 'Articles' you will find Beginners' Luck - The beginners' Guide to ... where there is a Potentiometers article.

(Unless you know the site, a general search for ESP will never bring it up)
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Old 14th July 2015, 07:08 PM   #10
benb is online now benb  United States
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Quote:
Originally Posted by DF96 View Post
Google 'potential divider'. A volume control does not work by simply putting resistance in the signal path, but it is amazing how many newbies think it does.
Or as I learned it, 'voltage divider.'
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