Help Understanding 'Current Control' Configuration using Pot + Fixed Resistor

[al_m]

Hi all,

I have been working to fix a broken dual gang pot on a Boss guitar FX unit i recently bought. Luckily there is a detailed service manual with nice clear schematic available. Unfortunately the datasheet and pin configuration for the pot (ALPS RK09K series) is very difficult to find and i'm still looking. The only datasheet i've found doesn't have the dual gang pot i need even listed on it. In addition, it's a bit of an oddity, as there are only 5 pins. It seems to be a ganged ground connection. Yet on the schematic, Boss list six pins even though the board layout shows five 

The pot controls output level but is not a gain control, it's a variable 'pad' of sorts, to set the level between -10dBV and +4dBu standards. The unit functions even with the pot completely removed from circuit, as i discovered when i went to take it out and found that the force which damaged it had broken all the pins free from the solder joints. What it appears to be doing is tapping off part of the current from the output, and sending it to ground. The output is very hot with nothing in circuit.

So far, i replaced the pot with two 20k 1% matched resistors, and all seems fine.

What i found interesting was the configuration with a fixed resistor in series, and the wiper and one side of the pot sharing a connection to the series resistor. I found a page on the ALPS website that lists this as a 'current control' configuration, which makes sense, but i was wondering how exactly it works. Searching online has brought very little mention of this usage for a potentiometer and i was wondering if anybody here would be kind enough to explain to me a bit about the general theory behind it.

And am i right in taking it that when the wiper it at the top of its travel (as layed out in the schematic) it's effectively out of circuit, and when it's at the bottom it's like it's shorting the 20k resitance of the pot out of the circuit? The bit i'm interested in in particular is what's happening in between, is it acting like two parallel resistances, one the 20k resitive element of the pot, and the other the value of the wiper to ground, or is it acting as a divider? A bit confused about this.

Thanks.

Here's the schematic for the section i am working on:

[TimFox]

R102 and the pot form a voltage divider with R104.
With the wiper at the top, the full resistance of the pot is in series with R102, and with the wiper at the bottom only R102 is in circuit.
Therefore, the output is lowest with the wiper at the bottom and highest with the wiper at the top (maximum total resistance).
With the pot removed entirely, there is negligible load on R102 so the output is even higher.

[al_m]

Thanks for the reply. Yes, if you read my post towards the end, that's what i surmised from looking at the schematic. But what happens when the wiper is halfway between the top and bottom? And why is it a 'current control'?

[fourfathom]

In that circuit, the pot is essentially a rheostat (two terminal variable resistor).  You can ignore the top terminal, since the wiper shorts out the pot between wiper and top.  In your mind, disconnect the top connection and you will see that the resistance between the wiper and the bottom of the pot (the grounded terminal) is what matters.

So the circuit is a resistive divider, where the resistor to ground can vary from 470 and 20470 Ohms.  The attenuation is set by the ratio of the 2.2K input series resistor and the pot + 470 Ohms.  When calculating the attenuation range you also need to include the (unknown) output impedance of what is driving that 2.2K resistor, the 100K resistor in parallel with the transistor, and the input impedance of whatever gets plugged into the output jack.

"Current control" ?  I have no idea.

[TimFox]

Assuming a linear taper pot, the resistance of that leg will be R102 plus (half the pot resistance) and the output voltage is a simple calculation.
"Current control" is a misnomer.

[al_m]

Thanks guys for the answers. Seems my initial supposition that i was looking at a voltage divider with a variable leg was correct then (which i forgot to mention in my first post). What threw me was looking at the PCB and finding that the wiper and top terminals share a single trace leading back to the series resistor. I suppose for layout's sake it might be easier, but is there any other reason not to have the top terminal just left floating?

[TimFox]

The designer had a choice between using the pot itself as a voltage divider (the original meaning of "potentiometer") or using it as a variable resistor ("rheostat") as part of a voltage divider.
When choosing the latter, it is normal to connect the wiper to one end of the pot.  If nothing else, this means that the two-terminal device never goes "open" when hitting a bad surface on the wiper.  Also, if substantial current is flowing, the maximum current through the wiper is less than that through the element.  Other than that, it is a common practice that does no harm:  it's probably better than leaving a connection literally floating.