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  #1  
Old 09-23-2008, 05:06 AM
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selectable input gain

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Hey Guys,

I'm building a new pedalboard(for the fourth time or so....) and I have an extra hammond box lying around, and a couple of input jacks, so it got me wondering of building a box where I could leave all the cables coming out of the board to be left plugged in, so I only have to plug in my bass cable and the cables to my amp...

slight problem: I have passive and active basses, and I want 2 input jacks, one for passive and one for active, would it be sufficient to just solder a transistor(/resistor?) between them so that I have a normal, unaffected input and one that has less gain?

if so: which transistor(/resistor?)?

if not: how would I do this?

Thanx!
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Last edited by bass_fish : 09-23-2008 at 05:18 AM.
  #2  
Old 09-23-2008, 09:22 AM
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A resistor is what you want, and yes it would work. The value of it would depend on your signal level. What I'd do is get a pot, say a 100K linear (though other values of pot would work), put it in series where the resistor would go, and adjust the pot until the level is where you want it. Then use a multimeter to measure the value of the pot, and select a resistor of that value. If you don't feel like buying a pot and a DMM, then you could just buy a few different values of resistor and experiment to see which ones get you close. Resistors are cheap.

Note: if you just connect the jacks together with no switch, then you cannot leave two instruments plugged in at the same time. Doing so would make the passive bass sound like crap, and might affect the active one as well, due to loading.
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  #3  
Old 09-23-2008, 09:38 AM
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I don't plan on leaving the basses plugged in at the same time, so I wouldn't need a switch right(do have them laying around the house... so it wouldn't be a problem to put one in anyway)

but I also could just use the (on/on)switch to select the resistor, right? so one side would be no resistor and the other side would be a resistor...

I'll just buy some resistors and try some stuff out, just as easy....

thanx
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  #4  
Old 09-23-2008, 09:40 AM
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if I happen to have some 100K linear pots laying around, I wouldn't even need a resistor right?
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  #5  
Old 09-23-2008, 11:11 AM
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A pot is a variable resistor.
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