series R vs V divider question

[squidsquad]

I have a one transistor boost & wanna put an input pot on it...I see 2 ways:
Joe G's Easy Face uses a 250k pot as a divider to ground after the in cap.
Fuller's FF (as seen at GEO) has a 50k as series resistance before the cap.
Is one way "better" or are both the same?

[R.G.]

There is a hidden resistor in both of them - the resistance of the transistor's base input.

Here's a bit of knowledge, perhaps more than you want. If I have some voltage, say 9.0V, and I want a different voltage. Let's make it easy, 3.0V. I know that I can get 3.0V by using a resistor divider. But how is that different from a 3.0V battery?

Obviously, a battery can put out a lot of current, and a resistive divider can't. But how do we determine how big the differences are?

With about a page and a half of math I can derive this (or at least I could, back when it was on the circuits course final exam), but I'll just state it. A 9V battery producing 3.0V by a resistive divider looks exactly like a 3.0V battery with a single resistor in series with it. The value of that series resistor is equal to the parallel combination of the two resistors making up the divider.

Example: 9.0V battery, 3.0V out. R1 (connected to battery +) = 2.0K, and R2 (connected to ground in series with R1) =1.0K. The voltage produced at the junction of R1 and R2 is 9.0V*R2/(R1+R2) = 9*1k/(1k+2k)=9*(1/3) = 3.0v.  The 3.0V looks to anything it's connected to as though it's connected through a series resistor of R = (R1*R2)/(R1+R2) =  (1K*2K)/(1K+2K) = 2k/3=0.666k = 666 ohms.

Notice: there is no electronic test which you can do on the 3.0V output that can distinguish it from a perfect 3.0V battery with a 666ohm resistor in series with it. They are identical electronically.

So back to the question - series resistor versus voltage divider. If the signal is the same level in both cases (that is, both of them are your guitar), then the differences are entirely in the divider pot versus the series resistor. The divider pot causes a voltage equal to the divider ratio to appear at the input, but the series resistance that appears to be in series with that divided signal is different for every divider level. At the top of the pot and at the bottom the source impedance appears to be zero. It is a maximum in the center of the pot resistance of 1/4 of the pot value. So at half travel on the pot. the signal is 1/2 through a resistance of 250K/4 = 62k. For the series 50K resistor, the signal level is always unity, and the series resistor is always 50K. If you make the 50K a variable series resistor, the signal level will be unity through a resistance of 0 to 50K.

Then there's that input resistance. The output of either one drives the circuit input at the cap. Exactly what the resistance on the boost is depends on the boost circuit.

For best flexibility, the divider offers a widely different signal level. But the exact effect depends on the circuit input impedance.

They are certainly not the same. I would use a divider pot in front of a high impedance booster, and a series resistor in front of a low impedance booster.

[QSQCaito]

Glad to learn this R.G. I didn't got all the part, but this part is very useful:

"Obviously, a battery can put out a lot of current, and a resistive divider can't. But how do we determine how big the differences are?

With about a page and a half of math I can derive this (or at least I could, back when it was on the circuits course final exam), but I'll just state it. A 9V battery producing 3.0V by a resistive divider looks exactly like a 3.0V battery with a single resistor in series with it. The value of that series resistor is equal to the parallel combination of the two resistors making up the divider."