Active/Passive Electronics Question

[Jd4003]

Hello,

I have a fair amount of wiring experience, but not with active electronics and before I get started I was hoping someone could help clear up my design.

For my next project I am planning on throwing in a 3pdt switch with two magnetic pickups and a piezo bridge. It uses a Bartolini MPB2-918 dual buffer fed into two volume pots (1 mag, 1 piezo) when active. When passive the neck and bridge pickups will (in theory) feed into those same two pots.

My question is: with the power off to the buffer, will its output leads be open, short to ground, or something else?

Look at the pot in the attached diagram to see what I mean. The green lead is the output from the buffer and the black lead to that same terminal will be the bridge pickup in passive mode.

Sorry for the long post! Any help will be highly appreciated. Thanks!

[line6man]

It's not a good idea to switch the battery with the signal path routing. You may end up with loud pops when you switch. Plus, there really is no reason for it, unless you forget to unplug your bass frequently.

Several connections are unspecified, so I cannot tell how the rest of the wiring goes.

[Jd4003]

Originally Posted by line6man

It's not a good idea to switch the battery with the signal path routing. You may end up with loud pops when you switch. Plus, there really is no reason for it, unless you forget to unplug your bass frequently.

Several connections are unspecified, so I cannot tell how the rest of the wiring goes.

Yeah, I apologize for the incomplete diagram.. I tried to capture the section with the part in question as best I could.

Thanks for the tip. I can certainly remove the battery switch.

If the battery switch were removed though, the original question still stands. Is it a problem to have both the buffer output and the unbuffered pickup running to the same lug on the volume pot as shown? I know it won't be a problem when the buffer is active as the wire running back the the 3pdt switch will just be an open circuit, but I have no idea how it will behave in passive mode.

To clarify: hot output from the pickup leaves the switch and enters the pot shown via the black line, what effect if any does the green wire running back to the buffer have?

Thanks!

[line6man]

Originally Posted by Jd4003

Yeah, I apologize for the incomplete diagram.. I tried to capture the section with the part in question as best I could.

Thanks for the tip. I can certainly remove the battery switch.

If the battery switch were removed though, the original question still stands. Is it a problem to have both the buffer output and the unbuffered pickup running to the same lug on the volume pot as shown? I know it won't be a problem when the buffer is active as the wire running back the the 3pdt switch will just be an open circuit, but I have no idea how it will behave in passive mode.

To clarify: hot output from the pickup leaves the switch and enters the pot shown via the black line, what effect if any does the green wire running back to the buffer have?

Thanks!

You can't run passive magnetic pickups parallel to a preamp output*, or parallel to piezo elements. Doing either would leave you with an impedance mismatch that would place a relatively direct load from the lower impedance circuit on the higher impedance signal, which would kill a lot of output. (*Unless the preamp output has an unusually high output impedance that is comparable to that of the magnetic pickups.)

Why are you trying to bypass a the magnetic pickups' buffer while continuing to buffer the piezo signal? There is no desirable reason to do this, and it would be difficult to pull off, anyway.

[Jd4003]

Thanks for the quick replies Line6, I really am in the dark when it comes to active electronics..

I haven't done a good job of explaining it. The idea is not to run the piezo buffered and the magnetic pickups unbuffered simultaneously. It's simply to bypass the buffer altogether when the piezo is not in use. The reason behind using the same pots for both situations has to do with the physical design of the instrument.

In this case, does the impedance mismatch still apply and cause problems if there is no output from the preamp when the pickups are bypassing it?

[line6man]

Originally Posted by Jd4003

Thanks for the quick replies Line6, I really am in the dark when it comes to active electronics..

I haven't done a good job of explaining it. The idea is not to run the piezo buffered and the magnetic pickups unbuffered simultaneously. It's simply to bypass the buffer altogether when the piezo is not in use. The reason behind using the same pots for both situations has to do with the physical design of the instrument.

In this case, does the impedance mismatch still apply and cause problems if there is no output from the preamp when the pickups are bypassing it?

All you have to do is put in a standard active/passive switch. The output from the magnetic pickups and their controls will go to the switch, along with a connection to the output jack. In passive mode, the pickups will go to the jack, without anything else in the signal path. The magnetic pickup input and output of the buffer will then be wired to be switched in in active mode, so that that wiring will stay as it was before.

[Jd4003]

Well I guess that simplifies things. I have a better, working schematic now that shouldn't have any issues with different impedances in parallel. Thanks for all the input.