Volume pedal for FRFR monitor?

andyg_prs

ICBM said:

andyg_prs said:

You couldn't direct me to, or sketch out the wiring could you?... Please

Very simple - the incoming balanced signal goes to the two outer connections on the pot, and the outgoing is from the centre and the anticlockwise (the one that would normally be grounded). ie, the same as a normal volume except that the anticlockwise terminal is not grounded. The grounds should still be connected to the casing of the pedal to prevent noise.

In theory, it should be done with a dual-gang pot so the two halves of the signal are reduced symmetrically, but since the centre point is floating it actually makes no real difference.

Thanks - if I go ahead with this, I'll sketch this up and check I'm correct.....but actually, it occurs to me, if I'm making something custom, why not do it with XLR sockets......and cut out all the XLR to TRS adaptors......

Again though, how would that be wired?

Thanks,
Andy

andyg_prs

Is this anywhere near correct?

ICBM

andyg_prs said:

Is this anywhere near correct?

Yes .

andyg_prs

ICBM said:

andyg_prs said:

Is this anywhere near correct?

Yes .

What type / value pot would you use?

ICBM

andyg_prs said:

What type / value pot would you use?

It probably doesn’t matter much - if you’re re-purposing an existing volume pedal then whatever it has, although lower resistance will be better than high, and log better than linear.

If you’re building one or going to replace the pot anyway then I would probably do it properly with a dual-gang pot, something like 10K log. In that case connect the input pins 2 and 3 to the clockwise terminals, the output pins 2 and 3 to the middle terminals, and the anticlockwise terminals together but still not to ground.

andyg_prs

ICBM said:

andyg_prs said:

What type / value pot would you use?

It probably doesn’t matter much - if you’re re-purposing an existing volume pedal then whatever it has, although lower resistance will be better than high, and log better than linear.

If you’re building one or going to replace the pot anyway then I would probably do it properly with a dual-gang pot, something like 10K log. In that case connect the input pins 2 and 3 to the clockwise terminals, the output pins 2 and 3 to the middle terminals, and the anticlockwise terminals together but still not to ground.

I love the opportunity to learn so....with the single pot, one side of the stereo signal seems to be controlled by the pot.  So from the source audio, pin 3 goes to the outer pin (one end of the resistive strip), and then for the output the connection to pin 3 is from the wiper.

The source audio from Pin 2 goes to the other end of the resisistive strip, and the output goes there too - e.g. wired to the same lug.  So I'm thinking that path of least resistance, literally, is that the signal on pin 2 passes straight through.

But I'm guessing there must be a reason that it is connecting to the other end of the resistive strip rather than being hardwired between the two XLRs and missing out the pot altogether.  Would you be able to explain to this idiot what is happening there?  Something to do with this 'floating point'?

With the above arrangment, I was going to use a molded box.....as it's just far easier to make the quite large holes for the XLR sockets.  In that case would I just wire pin 1 of the input and output xlr together and then to the case of the pot?

For the dual gang version - which is what I will do as I am going to make something from scratch.....I can see that the source audio from pins 2 and 3 are each respectively going to go to the clockwise terminal of their own pot.....the output is the corresponding middle wiper that makes sense.  The anticlockwise terminals are connected together.  Again, why is this......and again I guess it's to do with the inter-play of the two sides of the stereo signal and noise cancellation?

Do I do the same as above for the dual gang version for the earthing arrangement?

For both arrrangements, single pot / dual pot......would it make and difference if you swapped how you wired the outer lugs of the pot......or would swapping it make the pot all the way anti-clockwise full on rather than full off?

Finally, how do I logically look at a pot and understand which lug is anticlockwise vs clockwise?

Thanks,
Andy

robinbowes

Why not put the volume control before the captorX?

andyg_prs

robinbowes said:

Why not put the volume control before the captorX?

The speaker output of the amp goes to the CaptorX. I assume that cable is expensive in comparison to other cables. 

I also assumed that it was supposed to be kept as short as possible. 

robinbowes

andyg_prs said:

robinbowes said:

Why not put the volume control before the captorX?

The speaker output of the amp goes to the CaptorX. I assume that cable is expensive in comparison to other cables. 

I also assumed that it was supposed to be kept as short as possible. 

Ah yes, of course. It’s a load box, not just an IR loader.

ICBM

andyg_prs said:

I love the opportunity to learn so....with the single pot, one side of the stereo signal seems to be controlled by the pot.  So from the source audio, pin 3 goes to the outer pin (one end of the resistive strip), and then for the output the connection to pin 3 is from the wiper.

The source audio from Pin 2 goes to the other end of the resisistive strip, and the output goes there too - e.g. wired to the same lug.  So I'm thinking that path of least resistance, literally, is that the signal on pin 2 passes straight through.

But I'm guessing there must be a reason that it is connecting to the other end of the resistive strip rather than being hardwired between the two XLRs and missing out the pot altogether.  Would you be able to explain to this idiot what is happening there?  Something to do with this 'floating point'?

With the above arrangment, I was going to use a molded box.....as it's just far easier to make the quite large holes for the XLR sockets.  In that case would I just wire pin 1 of the input and output xlr together and then to the case of the pot?

For the dual gang version - which is what I will do as I am going to make something from scratch.....I can see that the source audio from pins 2 and 3 are each respectively going to go to the clockwise terminal of their own pot.....the output is the corresponding middle wiper that makes sense.  The anticlockwise terminals are connected together.  Again, why is this......and again I guess it's to do with the inter-play of the two sides of the stereo signal and noise cancellation?

Yes. You've got most of it . It's not stereo though - it's balanced. In a balanced signal, there are equal and opposite AC voltages on the two conductors - the idea being that noise is induced on both equally, and hence then cancelled when the two are recombined either in a differential amplifier or a transformer - so simply reducing the amplitude of the difference between them will work, thus you can do it with one pot. However, that does add series resistance on one side and not the other as you've spotted, so it could potentially cause noise since there may no longer be complete cancellation of noise picked up at the pot itself.

If you use a dual-gang pot, even with the centre point floating and not grounded, then this problem is eliminated because the whole thing is now symmetrical, so noise will be picked up equally by both sides. You don't want to ground the floating point as this prevents the guarantee of the two signals being equal and opposite as they must be in a true balanced cable - and also, it can cause trouble if there's phantom power present, since it's a voltage applied between both the conductors and ground.

If you're using a non-conductive enclosure, simply connect pin 1 of the input to pin 1 of the output.

andyg_prs said:

For both arrrangements, single pot / dual pot......would it make and difference if you swapped how you wired the outer lugs of the pot......or would swapping it make the pot all the way anti-clockwise full on rather than full off?

Yes, it will reverse the operation.

To know which end is which, just hold the pot and turn the shaft, looking at which way the internal rotor will turn. Or, if you're holding the pot with the terminals towards you and the shaft facing down - as you would see it if you were looking into the back of the enclosure with the pot in place - then the clockwise terminal is on the left and the anticlockwise on the right.

andyg_prs

To know which end is which, just hold the pot and turn the shaft, looking at which way the internal rotor will turn. Or, if you're holding the pot with the terminals towards you and the shaft facing down - as you would see it if you were looking into the back of the enclosure with the pot in place - then the clockwise terminal is on the left and the anticlockwise on the right.

Hmm, in my single gang pot diagram above, I drew it as though the shaft was facing down.....so I think I had it wrong as you can see that the output was coming from the centre and the left (clockwise) terminal rather than the centre and anti-clockwise terminal as you stated?



I have ordered the parts and it is an ABS case, so is the diagram below correct please?  For ease of drawing, I've got the input XLR on the left now....



Thanks,
Andy

ICBM

Yes, that's it . As long as by 'down' you mean the shaft is pointing down and 'towards' the viewpoint. (I know it's hard to draw a dual-gang pot clearly.)

andyg_prs

The nearest real photo would be this image:

So is the terminal that is connected to the white sheathed wire and the back of the case the clockwise terminal?   that's the orientation I'm trying to show in my child's drawing

ICBM

andyg_prs said:

So is the terminal that is connected to the white sheathed wire and the back of the case the clockwise terminal?   that's the orientation I'm trying to show in my child's drawing

No, that's the anticlockwise terminal.

Look at it from the point of view of turning the knob and it should make sense .

andyg_prs

Got it

andyg_prs

Hmm, I found a 10k log dual gang pot at CPC - with holes in the lugs which helps my rubbish soldering.....but they cancelled my order....they no longer stock them.  I can't find any others like it!  CPC do a linear version, and on Amazon I can find ones that are 47k log but £17 and very slow delivery.......any ideas of where to source something?

ICBM

I've got one of these - https://uk.rs-online.com/web/p/potentiometers/0168320 - old stock but unused. PM me your address and I'll post it

andyg_prs

That's very kind.....I will pm you, but also will need your paypal or other details so I can reimburse you

andyg_prs

Hmm, my XLRs appear to have an earth tag as well as pin 1 (I wish all the terminals had holes in )

Isn't earth, earth?  In other words, do I also need to interconnect pin 1 on both XLRs and the tags on the XLR and the rear casing of the pot?

Does my enclosure being plastic rather than metal make any difference to this?

ICBM

Pin 1 is audio ground, and the casing is the chassis ground - not necessarily the same thing. They can be connected together, and often are, but it's not a certainty that they always should be... it depends on various other considerations.