Hi all, I removed the CD401068CM and capacitors and made up an identical startup circuit on a breadboard with a pair of LEDs and it worked fine, 3 second delay for ‘reset’ (yellow LED) and another 3 second delay for ‘pwrup’ (red LED). Just before reassembly I noticed the holes on the board for the PWRUP cap were dirty so I cleaned that and the surrounding area off with IPA and soldered everything back in.
It works! It only reports 16K RAM though so there’s a fault in bank 1, I’ll deal with that later. Thanks for the pointers :) — Adrian/Witchy Binary Dinosaurs - Celebrating Computing History from 1972 onwards > On 28 Sep 2017, at 03:08, Jerry Weiss <j...@ieee.org> wrote: > >> On Sep 27, 2017, at 8:14 PM, Charles Dickman via cctalk >> <cctalk@classiccmp.org> wrote: >> >> On Wed, Sep 27, 2017 at 8:23 PM, Adrian Graham via cctalk >> <cctalk@classiccmp.org> wrote: >> >>> Schematic for the circuits is here - >>> http://www.binarydinosaurs.co.uk/newbrainPowerupCircuits.png >>> <http://www.binarydinosaurs.co.uk/newbrainPowerupCircuits.png> - top >>> circuit is PWRUP and the bottom one is RESET that goes straight to the Z80. >>> >> >> CD4000 logic can't handle inputs pulled above the power pin or below >> the ground pin without the possibility of malfunctioning or even >> complete failure. Google for "CD4000 parasitic SCR" or "CMOS >> latch-up". When power is cycled rapidly, the caps will still have >> charge which could cause the latch-up. I did some industrial control >> designs with CD4000 and used similar timing circuits, but always >> included diodes to prevent inputs from being driven too high or too >> low. I would have put diodes across both of the resistors. >> >> Adding your probes may change the currents inside the chip that change >> the latch-up behavior. Of course it also only makes sense if the power >> is cycled quickly for some value of quickly. Others may have an >> explanation, but that was my first thought from looking at the >> schematic. >> >> -chuck > > The capacitors may have become leaky, and start to as resistors in the 100K > ohms range. > > If so, the circuit now contains a voltage divider and and not meet the > threshold needed for > the logic gate to switch. The probes have their own resistive component and > thus > shift the other side of the divider, which puts the signal back into a > functional range for > the Schmitt Trigger. > > Jerry > > >
