Buffered splitter...really buffering?

[lukatosh]

Hello!

Many years ago i've found this schematic on an article made by codda effects... it's a buffered splitter circuit. The article is no longer online.

The circuit is powered by 9v and va is tied to 4.5v

how can i know the input and output impedances and look if it's really buffering the splitted outputs signals? (or at least adapting their impedances)




i've build it many times and it works, it splits the signal... but i would really want to know the values behind it.


Thanks so much!

[EBK]

Looks like your input impedance would be roughly 500k ohms (the two 1M resistors in parallel to AC ground, and taking the cap to be roughly zero ohms and the op amp inputs to be roughly infinite ohms).

Someone wiser than me will be along in a minute to chime in further. 

[Fancy Lime]

As Eric said, the input impedance for JFET-input opamps in this configuration is effectively determined by whatever goes from signal to ground or any other DC reference value before the opamp. So in this case 1M parallel with 1M = 500k. The output impedance is the internal output impedance of the opamp + the series resistance behind it. So in this case a few hundred Ohm from the opamp (you can look that up in the datasheet) + 1k.

The caps could play into this if their values were much smaller but as it is, they don't really matter in the audio band.

Andy

[ElectricDruid]

The caps could play into this if their values were much smaller but as it is, they don't really matter in the audio band.

+1 agree wot they sed.

Just to put values on that, the RC combinations both give a highpass filter with cutoff at about 3Hz or so. Since that's significantly lower than you can hear, and *way* lower than the bottom end of a guitar (even a bass) there's no problem from the caps. Here's the numbers for the output, for example:

https://electricdruid.net/rc-filter-calc/?f=&r=100k&r_series=3&r_errors=1&c=470n&c_series=1&c_error=10