Resistor between Pins 1 and 3 of Pot on Treble Booster

[nickbungus]

Hi

A friend has asked me to repair his Covington Treble Booster (which seems to be mostly the Greg Fryer treble booster albeit a few changes).

It was an easy fix, a wire had come away at the pot, but the Gain pot (10k) has a resistor between pins 1 and 3.  Can anyone please tell me what the purpose of this is? 

[mcknib]

It's usually there to lower the overall pot resistance i.e if you've got a 100K and you need 50K you'd place a 100K across pins 1 and 3.

I think Alchemy's done a pot taper calculator somewhere on here that I can't find at the moment

[Gargaman]

Nice!
So you paralell the resistor with the total value of the pot.
If I want to make a 1k pot into 900ohm pot, would 10k resistor do the job?

Are you reffering to this, Billy?
http://www.diystompboxes.com/analogalchemy/emh/emh.html

[antonis]

So you paralell the resistor with the total value of the pot.
If I want to make a 1k pot into 900ohm pot, would 10k resistor do the job?

Actually, you make pot's total resistance 909Ω ..

Basically, this method is used for altering sweep responce (i.e. from linear to hill or banana..)

[Gargaman]

I see, Antonis.
Maybe a 9k, according to the formula Req=(R1xR2)/R1+R2?

[mcknib]

Found it here's the calculator by Alchemy

http://www.diystompboxes.com/analogalchemy/emh/emh.html

Edit: yes that's it Gabriel should have read your post before searching woulda saved me some time

[nickbungus]

Brilliant.  That all makes sense now.  There's a 10K pot with a 22k resistor across it to give 6.875K.  Which also explains why there is a 6K8 marking on the board which I couldn't fathom what component it was related to.

[anotherjim]

I wonder if this isn't without consequences? Consider the pot at 50%, then the output impedance now includes the pot resistance (half of it?) as an additional path to the  collector resistor and this changes the high pass frequency of the output capacitor. This is not necessarily bad, I haven't done any calculations. It might even make the circuit more interesting to use if the tone changes usefully as you vary the output level.

[Gus]

This makes me wonder if they know what they are doing

Two correct ways to do this

Use a 10KA and adjust resistors two easy places to do this 2.2K emitter resistor is the one I would adjust

Use a 5KA and a 1.8K fixed in series 5KA to collector 1.8K to +9VDC you will always have a min volume.  It is not a gain control it is a volume control.  A volume control is a voltage divider a gain control changes gain they are different.

Why two threads about the same effect?

[induction]

Basically, this method is used for altering sweep responce (i.e. from linear to hill or banana..)

Assuming the pot is set up as a voltage divider, using bridging resistors between lugs 1 and 2 or between lugs 2 and 3 will change both the taper and the effective value of the pot (the new value will depend on the pot setting).

But I don't understand how using a bridging resistor between lugs 1 and 3 would change the taper. The effective value of the pot is constant in this case, it doesn't depend on the pot setting. For a given pair of voltages at lugs 1 and 3 (and assuming the input impedance of whatever comes next is comparatively high), the voltage divider math shouldn't be affected by a resistor in parallel with the whole pot. As anotherjim says, the output impedance will be affected, but why would the taper change?

Maybe my explanation is oversimplified and I've left out some crucial detail. Anyone care to set me straight?

[nickbungus]

Sorry about the two threads.  One as meant to be about the effect the other was just a question on pot wiring

[davent]

Basically, this method is used for altering sweep responce (i.e. from linear to hill or banana..)

Assuming the pot is set up as a voltage divider, using bridging resistors between lugs 1 and 2 or between lugs 2 and 3 will change both the taper and the effective value of the pot (the new value will depend on the pot setting).

But I don't understand how using a bridging resistor between lugs 1 and 3 would change the taper. The effective value of the pot is constant in this case, it doesn't depend on the pot setting. For a given pair of voltages at lugs 1 and 3 (and assuming the input impedance of whatever comes next is comparatively high), the voltage divider math shouldn't be affected by a resistor in parallel with the whole pot. As anotherjim says, the output impedance will be affected, but why would the taper change?

Maybe my explanation is oversimplified and I've left out some crucial detail. Anyone care to set me straight?

A single resistor from the wiper to one of the outside lugs or 2 resistors,each a different values from the wiper to each outside lug will change the taper.

http://sound.westhost.com/pots.htm#chg-law
http://www.geofex.com/article_folders/potsecrets/potscret.htm

For parallel resistor calculation i find a reverse calculator far more useful. Tell it what you want the results to be and it gives you all the pairs to achieve it.

http://www.sengpielaudio.com/calculator-parallel.htm


dave

[induction]

A single resistor from the wiper to one of the outside lugs or 2 resistors,each a different values from the wiper to each outside lug will change the taper.

Agreed. But I've seen several claims that a resistor between lugs 1 and 3 will change both the pot value and the taper. I don't understand why the taper would be affected in this case, and I wonder if I'm missing something.

[slacker]

Agreed. But I've seen several claims that a resistor between lugs 1 and 3 will change both the pot value and the taper.

It will do if the pot is used as a variable resistor, if you're using lugs 1 and 2 as a variable resistor and put a resistor in parallel with lugs 1 and 3, then the resistor and the portion of the pot between lugs 2 and 3 are in parallel with the resistance between lugs 1 and 2.
If it's being used a voltage divider then yeah it won't affect the taper, providing the impedance of the thing connected to the wiper is high compared to the value of the pot.

[Gus]

Draw it out and then do some math

Remember current divides up in a parallel circuit and is the same in a series circuit
SO if you have 10K in parallel(outside lugs) with 22K about 1/3 of the current is in the 22k section 22k is from this thread http://www.diystompboxes.com/smfforum/index.php?topic=112665.msg1040555#msg1040555
If you use the wiper to drive another stage you will not have as much current to drive might be an issue or might not depends on what is after it

Now the way this TB is wired is different it seems if the trace is correct the 10K is a variable resistor in the collector leg in parallel with a 22k?  They could have just used a 10k IMO then the bias gain change would be even greater. 0 to 6.8k vs 0 to 10k

Using a fixed resistor across the outer lugs make no sense to me, adjust the circuit to the potentiometer value you are using.

[anotherjim]

I've finally read the other thread.
It turns out the pot IS a variable R. 1&2 are common so the parallel resistor is both correcting the total R and altering the taper. For all I know it's a linear pot too.
What is weird is that the pot as wired allows 0 collector resistance in a common-emitter amp  - but go to the other thread for that.