shift register sequencer - buffer/voltage divider question

[artilect99]

Hi all,
I am building a CGS gated comparator-style shift register sequencer using CD4094. The GC uses a standard NPN transistor buffer output arrangement with a voltage divider to reduce it to ~5v gates, along with a resistor driving an LED. The standard KLEE-type mod is to then run the gates through passive attenuators, which then get summed by an op-amp to a CV out.

However I noticed that when using the individual gate outs, the summed CV out would sag. So for my version I wanted to switch to op-amp buffers instead of transistors, and put the voltage dividers before the buffer instead of after so they're isolated from the load.

I also want to retain the R2R ladder DAC output of the original, which on Ken's circuit takes its inputs straight off the shift register.

So my question is, would the schematic shown below accomplish this? Will it cause problems to divide down the shift register outputs BEFORE going to the R2R? (I imagine the DAC would be more precise with the full 12v, but then the output is 0-12v and will have to get scaled down quite a bit eg for melodic sequences...)

Will the resistor values chosen for the voltage dividers interact with the R2R in some unexpected way? (I also noticed Ken has his R2R grounded at both ends... I wonder why?)

I just want to settle on the optimum arrangement before I start doing the layout.

Any help/insight on this would be appreciated, as well as any general comments about the circuit. Thanks!

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gated comparator.png
Description:	original buffers
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current output arrangement.png
Description:	arrangement I'm considering
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[mskala]

Seems to me it would be better to attach the R-2R network to the outputs of the buffers. I realize you're trying to "isolate" the gate output jacks from the R-2R network, but the voltage divider's 4.3kΩ output impedance is going to interact with the R-2R network, and although that may not be a big deal if it's consistent, you could remove doubt by connecting the op amp's very low output impedance instead. Because the op amp has low output impedance, it provides isolation even between two things that are both connected to its output. Changing loads on the op amp output will change its current without changing its voltage and so shouldn't affect other loads that are watching the voltage, up to the limits of the op amp's current-driving capability; and the 1k resistor is providing some isolation, too.

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Screenshot_2022-09-23_16-42-37.png
Description:
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