Strange grounding problem

Adrian Graham binarydinosaurs at
Thu Sep 28 18:51:54 CDT 2017

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 :)

Binary Dinosaurs - Celebrating Computing History from 1972 onwards

> On 28 Sep 2017, at 03:08, Jerry Weiss <jsw at> wrote:
>> On Sep 27, 2017, at 8:14 PM, Charles Dickman via cctalk <cctalk at> wrote:
>> On Wed, Sep 27, 2017 at 8:23 PM, Adrian Graham via cctalk
>> <cctalk at> wrote:
>>> Schematic for the circuits is here - <> - 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

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