Adjustable crossover

[rankot]

I have built an adjustable crossover according to this schematic:


But I modified it to dual supply like this:


I experience a large drop at the output on both bands, and I have no idea why. Do I miss something obvious? There is not a ground connected at the input jack in my drawing, but it's done on the prototype, so that is not the case.

[antonis]

Try to make R7 12k (as in original ciruit..)

[Rob Strand]

Input impedance isn't guitar friendly - too low.   Will need to add an input buffer.

[rankot]

Try to make R7 12k (as in original ciruit..)

It doesn't help, it just makes different slope.

[rankot]

Input impedance isn't guitar friendly - too low.   Will need to add an input buffer.

OK, I will. Thanks!

Those impedance topics are something I really can't comprehend. Too much beyond my architectural mind

[rankot]

Added a NPN buffer from AMZ site and it does the job. No more sucking! Thanks to everyone!

[iainpunk]

so here i was, exited about a cool crossover distortion scheme, but it was a fancy filter 

Added a NPN buffer from AMZ site and it does the job. No more sucking! Thanks to everyone!

we also recommend NOT adding a buffer to girlfriends/wives for the same reason.

cheers, Iain

[antonis]

Those impedance topics are something I really can't comprehend.

Just imagine them as (variable/frequency dependent) resistive voltage dividers..

By considering C13 almost short (very low impedance / capacitive reactance) and R2 open (very high resistance compared with R1)(*), circuit input impedance simply is 5k6(R1) 'cause U4.1 inverting input is considered as "virtual ground" (summing point of input and feedback resistors)..
So, you can consider that point "wired" to signal source negative terminal (- pole) due to zero voltage difference between them..
This makes R1 the only resistive element across ideal signal source +/- poles hence the whole signal current is set by signal voltage divided with R1 (Ohm's law)
For real world voltage sources, "internal" resistance (Zs) should be set in series with any external circuit resistance so we result into Zs + R1 total resistance across signal source terminals..
(Zs can be guitar or preceding effect output impedance)
The greater the R1 value the higher the voltage drop across it hence the higher the voltage level on Zs/R1 junction..
(more signal enters into the circuit..)

(*) R2 & R1 are set in parallel when facing signal input 'cause both end to 0V (GND), consequently both are set across signal source +/- terminals..

[rankot]

Those impedance topics are something I really can't comprehend.

Just imagine them as (variable/frequency dependent) resistive voltage dividers..

By considering C13 almost short (very low impedance / capacitive reactance) and R2 open (very high resistance compared with R1)(*), circuit input impedance simply is 5k6(R1) 'cause U4.1 inverting input is considered as "virtual ground" (summing point of input and feedback resistors)..
So, you can consider that point "wired" to signal source negative terminal (- pole) due to zero voltage difference between them..
This makes R1 the only resistive element across ideal signal source +/- poles hence the whole signal current is set by signal voltage divided with R1 (Ohm's law)
For real world voltage sources, "internal" resistance (Zs) should be set in series with any external circuit resistance so we result into Zs + R1 total resistance across signal source terminals..
(Zs can be guitar or preceding effect output impedance)
The greater the R1 value the higher the voltage drop across it hence the higher the voltage level on Zs/R1 junction..
(more signal enters into the circuit..)

(*) R2 & R1 are set in parallel when facing signal input 'cause both end to 0V (GND), consequently both are set across signal source +/- terminals..

Thanks Antonis, I get it now! Nice explanation!