Help with upgrading pot for headphone amp

I've built a small pocket headphone amp. For my volume pot I chose a Bourns 50k pot. A great pot, however I find that the volume goes from silent to very loud quickly. Can I switch to a 100k pot to get something a little smoother? Bourns or Alps is ok with me. I am a green amateur so any suggestions would be greatly appreciated. Thanks to all.
 
There have been some very good studies of this -- it's just math after all . .

Usually about 5:1 or so makes a half-decent approximation of a log-taper pot -- your 50k linear pot with a 10k resistor, wiper to Ground, will come pretty close to what you can expect from a much-more-expensive, 2-matched-gang logarithmic pot.

Cheers
 
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...your 50k linear pot with a 10k resistor, wiper to ground, will come pretty close to what you can expect from a much-more-expensive, 2-matched-gang logarithmic pot.
The added fixed resistor is sometimes called a "taper resistor", and Rick's statement is absolutely true if the pot is:
  1. Fed from a voltage source, i.e., very low source impedance, and
  2. Is very lightly loaded, i.e. is connected to a resistance much greater than 10k.
In a properly designed circuit, the second requirement is always satisfied. The output of a pot should never be loaded down heavily, so if the circuit designer knew their job, it won't be.

The first requirement is also most often satisfied in modern solid-state circuits, where the pot is likely being fed from the output of an op-amp, which is very low output impedance.

But this isn't always true. The first requirement may not be met if the pot is, for example, fed from a valve (tube) stage. In that case, the added taper resistor will mess up the behaviour of the circuit, and it won't work as intended.

The reason for this is simple: the added taper resistor has two effects on the circuit. One of them is desirable (it turns a linear pot into a log pot).

But the other effect is undesirable: it causes the end-to-end resistance of the potentiometer to vary with the position of the pot wiper.

In the example being considered, the 10k taper resistor will appear in parallel with the 50k pot resistance when the pot is turned up to maximum. This has the effect of dropping the end-to-end resistance, from 50k, all the way down to 8.33k.

More generally, if we use a taper resistor that's one-fifth of the pot resistance, then the end-to-end resistance of the combination drops to one-sixth of the pot resistance, when the pot is turned up to maximum.

This is not a problem if the preceding stage has a low enough output impedance, and sufficient output current capability, to comfortably drive the reduced load (Rpot/6). In our example, just about any op-amp will comfortably drive an 8.33k load.

However, in tube amplifiers using, say, a 12AX7 triode, the output impedance of the tube gain stage may be as high as 40 kilo ohms. This may be perfectly happy driving a 250k pot - but if we add a 50k taper resistor to the pot, the tube is then loaded down heavily with a 41.67 k load when the pot is turned up to max. This will most definitely adversely affect the gain (and maybe frequency response) of the preceding tube stage.

The problem isn't restricted entirely to tube amplifiers. It is perfectly possible to make a transistor gain stage with a high output impedance; for example, a BC148 silicon transistor in common-emitter mode, with a 220k collector load resistor. It makes an excellent gain stage, so long as you don't load down the output...and a pot with a taper resistor might be disastrous in this case.

Basically, Fahey is right. The added taper resistor is a kludge. In some circuits, it is an excellent kludge. But in other circuits, it is a very bad kludge, one that will produce very undesirable results.

Caveat emptor - use a taper resistor only if you understand your circuit well enough to be sure that the kludge will work as intended.

-Gnobuddy