[sdiy] Buchla 266 Quantized Random Voltage
Pete Hartman
pete.hartman at gmail.com
Sun Dec 15 07:49:34 CET 2013
Many thanks to Dave Brown for helping me off list to grasp the "how"
that I was asking about yesterday.
I found the problem. The output of the XOR gate was not feeding back
into the shift register, because the 100K resistor on its output
didn't get soldered on one end. Thankfully, not my build ;-). There
was another resistor in a similar state, and a couple others which
were were questionable because I could see light through them (though
they did appear to have some solder). Touching all that up, and it
appears to be all good now.
Thanks
Pete
On Sat, Dec 14, 2013 at 12:17 AM, Pete Hartman <pete.hartman at gmail.com> wrote:
> I'm troubleshooting one of Roman's 266 clones, and while I understand
> what the QRV is supposed to DO, I don't yet grasp entirely HOW it is
> supposed to do it.
>
> The behavior I see is that on the start of my clock, I get quantized
> voltages, correctly varying among a number of steps based on the input
> pot, for 20 clocks. After that I get nothing, for as long as I've
> been able to observe.
>
> I understand that we're looking at a shift register. I've read Dave
> Brown's discussion of the startup voltage to start things off with all
> 1's....what I suspect is happening is that after those ones are
> shifted out, I'm only getting zeros in. But I'm not grasping (at
> midnight anyway ;-) ) how I might test that hypothesis, nor how it is
> that it should be getting things that aren't zeroes shifted back in.
>
> I've googled a couple of things trying to find something discussing it
> in step by step detail, but found nothing. I scanned Bernie's
> Electronotes article about random voltages from issue #64, but nothing
> jumped out at me as matching the topology (though I do see he talks
> about shift registers, so when I'm more awake I will try to read it
> more closely).
>
> But if anyone knows of a resource where this is discussed in more
> detail, I would very much like to read it.
>
> Thanks!
>
> Pete
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