Detrmining what value pot to use

[jeff]

I think I understand how a pot works. It's a variable voltage divider, right. But here's the question: if you have a 1M pot and a 1K pot both set to half way the voltage out is half the voltage in so what determines the value of the pot? I suppose you don't want to change the source's load(much) but how do you figure the value of the pot to use?

[HotBluePlates]

The intended function of the pot, and the circuit it is placed in.

I'd love to be more specific, but will need a specific application to more fully answer your question.

[jeff]

Specifically I had in mind the champ as a preamp thread/schematic. The output knob is a 1meg. I was wondering why. To get a voltage for the NFB of a regular champ you use a voltage divider made up of 820 ohm and 47 ohm(seems close in value to a 1K pot turned down to me). The 1K wouldn't really load down the 4 ohm resistor/speaker much so I was wondering why for a line out they used a 1Meg pot instead of a 1K and how to calculate what size pot to use in general.

With a 1K at half or a 1Meg at half do you still get half the voltage out? Won't you have more series resistance with a 1 Meg though? Does it have to do with the resistance of what you may plug it into?

[jeff]

Am I correct in assuming that if you put a voltage across the lugs of a pot that, set to half, from wiper to lug will be half the voltage?

And are there simple guidelines in picking a value to the pots resistance so that it's:
Big enough too.... but
Small enough so...or
X times greater than...
etc.

[HotBluePlates]

Specifically I had in mind the champ as a preamp thread/schematic. The output knob is a 1meg. I was wondering why. To get a voltage for the NFB of a regular champ you use a voltage divider made up of 820 ohm and 47 ohm(seems close in value to a 1K pot turned down to me). ...

With a 1K at half or a 1Meg at half do you still get half the voltage out? Won't you have more series resistance with a 1 Meg though? Does it have to do with the resistance of what you may plug it into?

The NFB voltage divider values are partly determined by the reuiqred resistance of the shunt resistor (47 ohm in your example). Where/how the NFB loop is created and injected dictates that value, and the desired amount of feedback determines the series resistor value (820 ohm in your example).

How familiar are you with Ohm's Law and series/parallel circuits?

If you have an existing circuit (like the 4 or 6 ohm resistive load on the Herzog), and slap another pot/resistor in parallel, you add another path for current to flow. You also lower the overall resistance of the circuit, pulling more current and placing a heavier load on the source. Heavier load => more current drawn from the source => more voltage dropped across the source's internal impedance => less output voltage from the source.

If the resistance of the newly-added branch is very much higher than the existing branch, total current might not increase a lot and vary the voltage across the circuit. But these things are all relative.

The source impedance of the Herzog's output is not exactly low. But there are another several considerations. If the input of the following amp has a 1M impedance, and we create a 1M output impedance from the Herzog, there will be maximum power transfer between the two, but the voltage will drop by half. If we provide a source with less than 500k output impedance, the following amp's input will more accurately track the output voltage. Don't ask why right now, but a 1M pot used as a volume will look like at most something like a 250k source impedance; it will look like less or a LOT less depending which way you twist the knob and how far.

Regardless, thanks to the relatively high input impedance of an amp's input (the intended load of the 1M pot), low source impedance is not a requirement. The Herzog is not intended to drive long cable runs, also allowing a higher source impedance for this circuit.

The other big consideration is the 4-6 ohm resistor that the pot is paralleling is dissipating significant power. We'd like to use a typical 1/2w (or smaller) pot; they're cheaper! By the nature of parallel branches, the voltage across each branch is the same. If we make the pot's resistance very high, there will be very much less current through the pot, and therefore very much less power dissipated in it compared to the 4-6 ohm resistor.

The Herzog (and a Champ) might make 4w full-tilt. This implies ~4.9vac across a 6 ohm load. A 1M resistor across 6 ohms only drops the total resistance to 5.99996 ohms. So we can guess at a glance that almost no power is dissipated in the 1M pot. 4.9v across 1M results in 4.9 microamperes of current. The pot will dissipate 4.9uA * 4.9v = ~24 microwatts of power.

Therefore any 1M pot will be satisfactory.

[PRR]

> if you put a voltage across the lugs of a pot that, set to half, from wiper to lug will be half the voltage?

Only if the pot wiper is UN-loaded.

For "expected action": generally the input to the pot must be lower impedance than the pot||load, the load on the pot must be greater than the pot's worst-case wiper impedance (roughly Rp/4).

Eggample. Champ Volume control. The source is a 12AX7 plate which we know is near 39K impedance. The pot load is either open-grid (near infinite) or a 1Meg grid resistor. We want pot >39K but <4Meg. 250K, 500K, 1Meg are all suitable here.

> To get a voltage for the NFB of a regular champ you use a voltage divider made up of 820 ohm and 47 ohm...

I think the AA-Champ uses 2,700 against 47? 820+47 sounds like a bigger amp.

The "wiper" of this voltage divider drives a cathode. Cathode impedance is close to 1K. A 1K load on a 1Meg pot would load it very seriously. Anything below 99% rotation would be about zero. Even 1K impedance in the divider cuts tube gain to half. Therefore the divider was scaled for <<1K tap impedance. 100 ohms would be suitable. We could also note that speaker-side impedance is under 10 ohms and use say 100 ohm total divider, perhaps 100:1.7. I assume Fender had an excess of 47r parts, hence 2700:47.