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Circuit for controlling speaker volume

Discussion in 'General Electronics Discussion' started by Breakshift, Mar 15, 2010.

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  1. Breakshift

    Breakshift

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    0
    Mar 15, 2010
    I have a set of 5.1 speakers which are all controlled via a single volume knob on a very basic amp. I want to build a circuit/s to allow me to adjust the volume of each speaker individually. Currently I'm looking into simply connecting a potentiometer between the amp and the speaker, but my problem is that I can't find a log potentiometer that has a power rating of over 7.5W. I require this rating as my speakers are 7.5W each and hence the potentiometer needs to be able to handle this in the case of turning the volume of the speaker right down.

    So I'm not sure how to proceed. I'd appreciate any help!

    Thanks.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,387
    2,772
    Jan 21, 2010
    If you're very lucky, there may be 5 audio amplifiers inside the amp, each connected to the decoder which generates the 5 channels (I'm presuming it's not got 5 separate inputs).

    If you can find these, then you could break the circuit here and take each one to their own log potentiometer and back via shielded cable. As a guess, I'd start with a 20K pot (or whatever's close that's available) and see if it works for one channel before doing the others.
     
  3. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    It's a very old system - it came with a computer years and years ago. And yes, it has separate inputs for each channel! It relied on the sound card to output 5 channels to it.

    Because of human hearing being logarithmic - I'd need the pot to also be log right?

    The speakers impedances are 4 Ohms - so I'd need a resistor in parallel with the pot in order to not drop all the voltage across the pot/resistor.

    But I can't find a pot with a high enough power rating at a decent price - is there a way around this or am I going to have to shell out some good money?

    By the way, I'm doing this mainly for interest, the practical result is just a bonus really!

    Thanks.
     
  4. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Cheap old car stereos sometimes used speaker pot's for balancing front/rear, but it's really a poor solution.
    With separate analog inputs available it's much better to put volume pot's there, say two stereo and one mono to cover all five channels. 10-20k should do the trick.
    The concept has a name; passive preamp or pad I think.
    Yes, all volume controls are logarithmic due to the way ears work.
     
  5. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    I see - so by putting a pot in before the signal reaches the amp, it doesn't require such a high power rating? That makes sense, thanks.

    Something I forgot to mention is that I'm not using it as an actual 5.1 surround sound system, I'm just connecting my TV to its auxiliary input which is a 3.5mm plug. So it's essentially just a stereo system.

    So how would I connect a 3.5mm jack to a pot? Would the easiest way be to cut the plastic coating and solder the two wires inside to the pot?

    Thanks for the help - I really appreciate it. I'm very new to this!
     
  6. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    Sorry for the double post - I can't see an option to edit a previous post (because I wasn't logged in at the time! My bad).

    Am I right in saying that the input impedance of an amp is extremely high? Ideally, infinite?

    I may be wrong, but the way I see this is as follows: in order to turn the volume of a speaker down, I want to cause the audio signal voltage to drop across some resistive element as well as the speaker itself. By putting a pot before the amp, I can avoid the necessarily high power rating required to place it directly before the speaker. However, if the input impedance of the amp is extremely high, then the signal voltage will be dropped almost entirely across the amp. I would need a resistance equalling the input impedance of the amp to even half the signal voltage it receives.

    I just tried measuring the input impedance of the amp, and my multimeter didn't give a reading which means it's either too great for it to measure, or it's an open circuit.

    Am I right?

    Thanks.
     
  7. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Yes, in most aspects, except a pot has three terminals (one ground) and so divides the signal as the center tap slides along the track.
    Amps usually have at least 33k input impedance and are "insulated" from DC (so you usually can't measure that with a multimeter).
    Volume pot's should have a resistance value considerably less than the input impedance of the following amp.
    Stereo jack (twin) cables will have a shield (ground) wire around the center (signal) wire, one pair for each channel. Grounds are connected together at the plug.
    You'll find plenty of images if you Google for potentiometer wiring.
     
  8. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    Oh I see. So by connecting the ground of the pot to the ground of the signal line, I can vary the voltage at the input to the amp from max to zero. So around 10k would be good?

    Thanks. Things are a lot clearer now :)

    Edit - I'm confused. Essentially I want to take a 3.5mm stereo wire, split it into the left and right channels, attach the pots to these lines and then feed it to the amp. However, the amp has 6 RCA inputs (left/right front, left/right rear, sub and centre), so what do I do for the sub and centre speaker? As they're central, it seems wrong to connect them to either the left or right channel. What exactly is fed to subs (just in commercial equipment) when the input is just a stereo signal?

    Thanks.
     
    Last edited: Mar 18, 2010
  9. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    Can't find a certain pot from UK distributor

    Hey, I can't find a 10k (or around there) log pot with a power rating of at least 0.5W from a UK dealer. I don't want to pay £5 for delivery from the US for a single pot. What are the leading UK electronics suppliers? I've searched Google for UK electronic suppliers and been through them all and can't find one. Any suggestions/help?

    Thanks! :)
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,387
    2,772
    Jan 21, 2010
    If connected to the input of the amplifier, the pot will be dissipating almost no power at all. Just get an ordinary 10k log pot from anywhere.

    Almost no power at all means about 3 mW (three thousandths of a watt).

    (And please don't start new threads for something like this or you just might get someone answering your question literally rather than being able to see what you really need).

    edit: if connected correctly, even to the output of an amplifier it wouldn't dissipate much power.

    Connect the potentiometer like this to the amplifier. The signal travels from left to right, so to the left is your signal source, and to the right is your amplifier.
     
    Last edited: Mar 19, 2010
  11. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Why didn't you say right from the start that you wanted a surround decoder rather than just individual volume controls?
    Googling for simple passive surround decoder will bring you loads of examples.
     
  12. Breakshift

    Breakshift

    7
    0
    Mar 15, 2010
    Steve - the output of my TV is just under 0.5W max, so surely the pot needs to have that power rating in the case of adjusting it such that all the voltage is dropped across it and none across the amp? I know this probably wouldn't happen as I'd just use the main volume control on the amp but it's a possibility.

    Resqueline - I do want just individual volume controls! One for each speaker, including the sub and the centre speaker.

    Thanks.
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,387
    2,772
    Jan 21, 2010
    No, the pot does not need to be 0.5W. The output is 0.5W into 8 ohms.

    It would be a whole 0.4 mW into 10k
     
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