I'm going to be wiring up my Samson wireless into my Washburn active 5 string. I've done this on another (passive) bass and it works great! But before I left the transmitter in a position where I could access the battery and on/off switch without having to pull the wireless or open the cavity completely, and I did not need to worry about draining an Active circuit's 9 volt battery.
Now because this bass is Active, anytime a plug is inserted into the bass' jack it will drain the battery.
I plan on wiring the transmitter's jack leads directly to the bass' jack. I will be wiring up the transmitter's battery terminals to a battery compartment (single AAA battery) and having an on/off switch to kill a battery lead to the wireless (because the Samson's transmitter will be inside the cavity and I will have no access to it's on/off switch) thus turning off the wireless inside.
My question is, if I solder the transmitter jack leads to the bass' jack, will killing the battery to the transmitter be enough to also kill the bass' jack connection so it doesn't drain my
Active's 9 volt battery, or will I need another switch to kill that connection also?
I know that if left plugged in (even without a power source) the Active battery will drain, and I'm assuming that soldered leads from the transmitter to the bass jack would probably be the same as plugging a cable in the jack, would killing the transmitter be enough? Correct me if wiring to the jack leads is the same as leaving it plugged in. I'm not sure, but I think it is.
Here's the transmitter circuit board outside it's case...
Link Removed
Now because this bass is Active, anytime a plug is inserted into the bass' jack it will drain the battery.
I plan on wiring the transmitter's jack leads directly to the bass' jack. I will be wiring up the transmitter's battery terminals to a battery compartment (single AAA battery) and having an on/off switch to kill a battery lead to the wireless (because the Samson's transmitter will be inside the cavity and I will have no access to it's on/off switch) thus turning off the wireless inside.
My question is, if I solder the transmitter jack leads to the bass' jack, will killing the battery to the transmitter be enough to also kill the bass' jack connection so it doesn't drain my
Active's 9 volt battery, or will I need another switch to kill that connection also?
I know that if left plugged in (even without a power source) the Active battery will drain, and I'm assuming that soldered leads from the transmitter to the bass jack would probably be the same as plugging a cable in the jack, would killing the transmitter be enough? Correct me if wiring to the jack leads is the same as leaving it plugged in. I'm not sure, but I think it is.
Here's the transmitter circuit board outside it's case...
Link Removed