Hi Witchy,
Your not the first to be scratching their heads over the reset circuit, there were quite a few attempts over the years to deal with the very basic design.
If you have a look at the last revision of the SRAM Coreboards MB8328-6, you will see the same basic design, but some significant changes over time.
the 8328 is flipped from the drawing on your 1732 boards D1 (on the 1732) splits the Reset into two sections.
All the left of D1 is the power on Reset, with that in the middle is the Reset circuit driven from the Reset Key on the Keyboard, finally the three inverters (starting at pin3) through to D3 are buffer / driver for pulling the reset line low in multiple areas of the computer.
The Power on Reset looks at the 12V supply via the resistor divider R12 / R11, as this has changed over the years to reduce power dissipation in the onboard regulators so has the need to change the value of the resistors in the divider. The final design incorporated a Zener diode to reduce the effect of a range of power sources.
R10 is to limit the input current & D2 is to ensure that voltage doesn't exceed 5v quite important when you consider the design has the inverters running on the Vbb power source (The backup battery).
Now that is why I asked about the battery, if a battery is fitted or even just removed the best the power to the inverter will be your normal 5V - 0.6 V so at best 4.4V this will give you a threshold as an absolute minimum of around 3V required on the input pin 13 to take the output pin 10 out of the Power on reset state to allow the system to run. changing the Power voltage into the Microbee to lower values (some operate down to 8.5V) you may not be able to get to the threshold on the input of Pin13 and the system remains in perpetual reset. So if your power in is significantly below 12 V you may need to address the resistor divider values. However as you have one board operational on you supply and if the resistor values are the same across the three Coreboards then you should be able to stay with them as they are.
First up confirm your D1 is actually behaving as a diode and not leaky in the reverse direction before reconnecting it back into circuit.
As the voltage at pin 3 is getting to around 2.8 volts when D1 is connected I would say that the output from pin 10 is not all the way to a Low output state, otherwise your voltage on pin 3 should be pulled to about 0.6v. Based on that, my suspicion would be the 4584 CMOS Schmitt inverter, if you have trouble getting one of these you can use a 74HC14 Pin for Pin compatible but only suited to 5 Volt logic, but as we are running on 5 Volt logic, will be suitable in this location.
The final caveat is "No Board Damage" from historic battery leakage, which could be causing partial shorts etc.
Ernest
Your not the first to be scratching their heads over the reset circuit, there were quite a few attempts over the years to deal with the very basic design.
If you have a look at the last revision of the SRAM Coreboards MB8328-6, you will see the same basic design, but some significant changes over time.
the 8328 is flipped from the drawing on your 1732 boards D1 (on the 1732) splits the Reset into two sections.
All the left of D1 is the power on Reset, with that in the middle is the Reset circuit driven from the Reset Key on the Keyboard, finally the three inverters (starting at pin3) through to D3 are buffer / driver for pulling the reset line low in multiple areas of the computer.
The Power on Reset looks at the 12V supply via the resistor divider R12 / R11, as this has changed over the years to reduce power dissipation in the onboard regulators so has the need to change the value of the resistors in the divider. The final design incorporated a Zener diode to reduce the effect of a range of power sources.
R10 is to limit the input current & D2 is to ensure that voltage doesn't exceed 5v quite important when you consider the design has the inverters running on the Vbb power source (The backup battery).
Now that is why I asked about the battery, if a battery is fitted or even just removed the best the power to the inverter will be your normal 5V - 0.6 V so at best 4.4V this will give you a threshold as an absolute minimum of around 3V required on the input pin 13 to take the output pin 10 out of the Power on reset state to allow the system to run. changing the Power voltage into the Microbee to lower values (some operate down to 8.5V) you may not be able to get to the threshold on the input of Pin13 and the system remains in perpetual reset. So if your power in is significantly below 12 V you may need to address the resistor divider values. However as you have one board operational on you supply and if the resistor values are the same across the three Coreboards then you should be able to stay with them as they are.
First up confirm your D1 is actually behaving as a diode and not leaky in the reverse direction before reconnecting it back into circuit.
As the voltage at pin 3 is getting to around 2.8 volts when D1 is connected I would say that the output from pin 10 is not all the way to a Low output state, otherwise your voltage on pin 3 should be pulled to about 0.6v. Based on that, my suspicion would be the 4584 CMOS Schmitt inverter, if you have trouble getting one of these you can use a 74HC14 Pin for Pin compatible but only suited to 5 Volt logic, but as we are running on 5 Volt logic, will be suitable in this location.
The final caveat is "No Board Damage" from historic battery leakage, which could be causing partial shorts etc.
Ernest