hi and low inputs. what for?

[martin manning]

Of course it does Martin. I don't know what theoretical realm you're in, but the series resistance between the guitar signal and the grid definitely attenuates the signal level at the grid. Try increasing the 68k's to 1 meg and see what happens to your guitar signal level at the grid.

I think my theory is correct-

For input #1, you have the 1M to ground, and after that, 34K in series to the grid. There is virtually no current flow through the 34K, so whatever voltage is at the top of the 1M will show up at the grid. The series resistance after the 1M doesn't matter.

For input #2, you have a voltage divider made up of two 68K resistors (the 1M is shorted). That cuts the signal at the grid in half, but any two equal-valued resistors would do the same, as all the current is flowing to ground. The impedance seen by the guitar is the two 68K's in series, or 136K.

MPM

[bnwitt]

Whatever Martin. You believe what you want to believe but I suggest you do some hands on testing to verify your theory.

[cminor9]

Cminor9, think of the two resistors as pipes....

Ok, now that you explain it that way, it makes sense. Thank you!

[bnwitt]

Ok Martin, I slept on it and I think I understand where you are coming from on this. I get the resistance to ground for signal loss point. I'm not sure the statement that any two same value resistors would perform the same however. I'll have to slap in some 1 meg resistors in place of the standard 68k values and test that theory out. Somehow I think I'll find a huge reduction in signal on the grid but maybe you're correct.

[GibsonLover]

Back to the original question of "what for?" remember that in the good ol' days PA systems were slow to develop. Most small combo groups did not have PA's, and what was available looked like another guitar amp anyway. Your guitar amp was your PA. Mic in one and guitar in the other. The hi-low option is for the user to to suit himself based on his equipment and the guidance of information already posted.

[tubeswell]

or for mic'-ing a harp?

[martin manning]

Ok Martin, I slept on it and I think I understand where you are coming from on this. I get the resistance to ground for signal loss point. I'm not sure the statement that any two same value resistors would perform the same however. I'll have to slap in some 1 meg resistors in place of the standard 68k values and test that theory out. Somehow I think I'll find a huge reduction in signal on the grid but maybe you're correct.

Okay, cool, I'm interested to hear what you find out, since I've never tried it myself. For the input #2 case, the two 68K resistors will cut signal to the grid in half, and show input imedance of 136K. If you put in two 1M, then you still have half the signal at the grid as you have at the tip of the cable, but now you have 2M input impedance. You won't load down the source as much, and that in itself will affect the signal level at the grid. In that respect the 1M's won't perform exactly the same as the 68K's. Also, the high frequency break point will move down to 1.5 or 3 kHz. Going from 68K to 1M is a huge change, far from the 2x difference between 34K and 68K as seen in the original #1 vs #2 inputs.

MPM

[tubeswell]

Well the way I've had it explained to me by the pros, a voltage divider works on the ratio of the two resistors. The actual size of the resistors affects the amount of current getting through (and there is virtually no current on a grid anyway - unless its forward grid current - but that's another story), but it is the resistors' ratio that determines the voltage, e.g.;

The 1M is the bottom leg of the divider, so the voltage across it will be its percent of the whole resistance times the voltage. If you used a 100k ground leg resistor and 3k4 series resistor, the same voltage division would occur, except now there would be only 103k4 in total and therefore more of a load on the input device (because more of the already insignificant current would be getting to ground, so maybe you would opt for a 470k ground leg resistor and 16k series resistor?) As long as the overall resistance is high enough to prevent loading, use any resistor values you want as long as the ratio between them yields the voltage you want.

JM2CW