Posted by Damir [ 213.202.85.140 ] on May 04, 2005 at 13:44:22:
In Reply to: Re: Channel jumping inputs fender to Matchless posted by walters on May 03, 2005 at 14:30:27:
If we count in 68k resistors (and we should), then the situations are like that:
a) 2 amps, both input 1 with Rin=1Meg with "Y" cord - guitar "see" the two 1Meg loads in parallel, or 1Meg/2=500k. Three amps - 1Meg/3=333k.
b) 2 amps, "daisy chaining" - guitar in input 1 (Rin=1Meg), then out through input 2 to the second amp and its input 1 again (Rin=1Meg). BUT, we now have two 68k resistors in series between inputs 1&2 on the first amp, or in another words, guitar "see" 1Meg in parallel, then 2*68k=136k in series, then 1Meg in parallel. The resultant resistance is now (1000k+136k)//1000k = 531,8 kOhms. Not much of the difference, but if we connect the third amp on the same way (another 2*68k in series, and 1Meg in parallel) then resultant resistance is (531,8+136)//1000k = 400k, larger then "triple" Y-cord result of 333k. But, first amp get full signal voltage, and other two little attenuated signal through 136k/1Meg combinations.
Normally, this is correct if the inputs are wired at the same way, you can try it with ohmmeter, amps switched off, cables installed.
-All in all, you can try both solutions, but personally I don`t like either of them, as I said, our tiny voltage/high impedance source (guitar PUs) probably needs better solution, see splitter, etc, theory.
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