Impedance issue between DBX and GK bass amp

Mattattaker

New member
I’m trying to figure out how I can use a DBX 1066 compressor with my Gallien-Krueger bass amp. I have both active and passive basses. There seems to be a huge impedance issue with the IN/OUTS, so I’m wondering how to do it. Someone advise me to use an in line preamp and a reamp box, but I lack details as how to wire everything. Suggestions/solutions please?
 

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bouldersoundguy

Well-known member
It's entirely normal, actually preferred, to go from a low impedance output to a high impedance input. That's why they're arranged as they are in both those units. The opposite condition, high impedance output to a low impedance input (e.g. piezoelectric pickup into a mic input) is to be avoided.
 

Mattattaker

New member
It's entirely normal, actually preferred, to go from a low impedance output to a high impedance input. That's why they're arranged as they are in both those units. The opposite condition, high impedance output to a low impedance input (e.g. piezoelectric pickup into a mic input) is to be avoided.
Fair enough, but how do I wire this together?
I read that if you plug a low z output into a high z input, you get noise…
 

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bouldersoundguy

Well-known member
Impedance is not level. While sending low level (voltage) signal into something meant for a higher level could set you up for noise, it's entirely normal and desirable to connect a low impedance (Z) output to a high impedance input.

The person writing some of what's in those pics is confusing level and impedance.
 

bouldersoundguy

Well-known member
Assuming the Out and In connectors are an effects loop on the amp, simply connect the Out to the input of one channel of the 1066 and connect the output of that channel to the In using standard guitar cables.
 

ashcat_lt

Well-known member
Yeah you can plug the dbx straight into the amp without issue. The idea to use a reamp box for that is asinine.

You will have issues connecting a passive bass directly to the end input, though. Youll lose a bunch of treble unless you put a high impedance buffer or preamp between. The active bass should drive it fine.
 

rob aylestone

Well-known member
It's so easy to misunderstand how AC circuits work. Even for those who did physics at school, it usually boils down to Ohms Law - and we worry about voltages, resistance and currents and making circuits. They deliberately don't go into AC circuits because then resistance becomes impedance and there's a big step up in the things that impact on AC - so inductance, capacitance and resonance suddenly creep in. Teachers of the subject have to take a deep breath before leaping in, because misunderstanding is so easy, and when non-teachers who do understand it try to explain it they fail, and those that haven't quite got it (like in those examples above) fail miserably. In fact, when you get it right, you can squirt audio down extremely long lengths of cable with decent results. Mic wise - we all kind of take 150-600Ohms or so as 'normal' for mics - but back in the 60's/70's we had high impedance versions or switchable ones) so you could plug a mic into a jack on a guitar amp. Many also forget that at this period we also had some mics in very low impedance versions - 25-50 Ohms.

The reality of connecting devices is to get the matching as equal as you can, with the source impedance always being lower than the impedance of the destination. Optimum level matching without unwanted loss of HF. Cable has capacitance, especially in longer lengths, and this is what generates the old rule about high impedance guitar pickups and shorter cables. Look at a typical guitar tone control. What exactly is it? A single capacitor, wired through a variable resistance across the pickup output. What does it do? It reduces the level of the higher frequencies. If you change the capacitor value, it just alters where the cut begins and how sharp that cut is. If you wanted to, you could use two knobs and a little inductor and have a treble and a bass roll off. Anyway - if a little capacitor can make the bright pickup become mellow, then a long length of cable that has capacitance does exactly the same thing. If you look up the specification of cables it tells you the capacitance per foot or metre.

If you get the source and destination impedance around the wrong way - as in connecting a piezo pickup to a mic input, the effect is of putting a virtual short-circuit on the very high impedance - so level drops quite a bit, but then you restore it with the gain control and that is where the noise comes from. This near short on a wirewound pickup also deadens the strings. You've played with magnets and how they can push against each other - and a coil, partially shorted, tries to limit the forward and backward voltage the string is inducing in it - kind of putting the brakes on. If you twang a guitar with good sustain with the guitar unplugged, you can time how long it takes to decay. If you short a jack plug and stick it in, the sustain time goes down. Not much, but at college we measured it and proved the science.With modern DAWs, you can edit the twang start and end to a pretty accurate level - say -3dB and trim the clip start. Then find a very low point you can measure and trim again. If you do the short and trim that one - to the same levels, you'll find it's not the same length - it decays quicker.
 

ashcat_lt

Well-known member
If you change the capacitor value, it just alters where the cut begins and how sharp that cut is.
Course most of the action of a passive guitar’s T pot comes from the inductance of the pickup. Lowering the resistance of the load on that inductor lowers the cutoff of the RL lowpass that is formed. In a lot of ways the capacitor actually limits how low that can go so it doesn’t end up going all the way to silence

This, BTW, is why you lose treble when you plug a passive guitar (or bass) into a lowZ input. “Everything useful is a voltage divider”. The pickup has a largish resistance at low frequencies and gets very large at higher frequencies, and is the “top resistor” in the divider. The input is probably capacitor coupled so is very large at some probably subsonic frequency and then will be most just static’s resistance for the rest of the spectrum. That’ll be your “bottom resistor”. The simple divider equation will show that at the point in the frequency range where the ”top resistor” is large compared to the “bottom”, you get significant attenuation. The bigger the input Z is, the higher frequency that point will be. Well, until the point where cable capacitance makes that load start to look small again...
 

Mattattaker

New member
Thanks to everobody for helping me out. The correct setup was to use a preamp (the FMR really nice preamp that was advised to me by Austin Ledyard from Radial was indeed transparent and didn’t color the sound of my bass) to plug the passive bass, then go to channel one of the DBX, then use a reamp box to the input of the amp. After, I used Send/Return of the amp directly to channel two of the DBX.
It sounds amazing🤘🏼!
 

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